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The White House has been briefed on the Mueller report and, while it will not invoke executive privilege, it has "significant concern" about what is going to be released, according to ABC News' Jon Karl.
"There is significant concern on the president's team about what will be in this report and will be unredacted," Karl told ABC News' "This Week" on Sunday. "The good news is already out there."
Karl was referring to Attorney General William Barr's summary of the Mueller report, which found no evidence of President Donald Trump's campaign conspiring with Russia to meddle in the 2016 presidential election.
Now, the White House will have to take some potential political hits from aspects of special counsel Robert Mueller's investigation, especially the unknown of what former White House legal counsel Don McGahn told investigators during what is reportedly "significantly more than 30 hours with" Mueller's team, according to Karl.
"There is significant concerns about what will be in here – new information on the obstruction of justice question, on what the president was doing regarding some of the big questions," Karl told host George Stephanopoulos.
". . . What worries them most is what Don McGahn told the special counsel. Former White House counsel Don McGahn has visibility on all of this."
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You don’t even have to modify your existing Win XP computer to take Linux for a spin. How is that possible? Many Linux distributions allow downloading of a file which can be burned to a DVD or run from a USB drive. Boot your PC from the DVD or USB drive to test drive Linux.
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Free Fall Scavenger Hunt Printable
This post may contain affiliate links. Your cost is the same, but it helps support the site and keep our freebies FREE! Thanks for your support!
Fall is the perfect time of year for a Fall Scavenger Hunt and I created this cute printable just for you guys!! We enjoy going for walks in the fall and feeling the cool brisk air blowing our hair around while we are bundled up in our jacket’s and hoodies! What better way to enjoy a walk with your little one than by going on a super cool and fun Fall Scavenger Hunt.
Don’t forget to hop over and check out these 50+ Thanksgiving Activities for Kids that we have on the blog! There is enough there to keep you busy for the next month!! Interested in more Free Printables?? We have plenty for you to check out now and plenty more to come. Make sure you sign up for our email updates at the top of the page!
There are many other things to learn while on your fall scavenger hunt. It’s the perfect time to talk to your child about the changing of the seasons and the many signs of fall.
Signs of Fall to Look for…
Leaves will begin to change
Leaves will begin to fall off the trees
The weather begins to get colder
Animals will begin preparing for winter
The farmers are in the fields
Dad starts watching football
Those are just a few off the top of my head. You will find many teachable moments while taking your child on a fall scavenger hunt walk if you just take the time to look around. Watch what they are interested in and just go with it!
[…] is nothing better in the Autumn Season than taking a walk and doing a fun fall scavenger hunt with the kids! our favorite thing to collect in the fall is pinecones!! Then we take tham back home […]
[…] is definitely here and we took off the other day on a nature hunt using our fall scavenger hunt printable to find some fantastic fall nature items!! I had already made some of our Acorn Cookie snacks to […]
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package fssh
import (
"encoding/json"
"errors"
"github.com/libragen/felix/util"
"github.com/spf13/viper"
"golang.org/x/crypto/ssh"
"time"
)
type fsshToken struct {
Uid uint `json:"uid"`
Mid uint `json:"mid"`
Ex time.Time `json:"ex"`
}
func TokenGenerate(userId, machineId uint, ex time.Duration) (secret string, err error) {
t := fsshToken{
Uid: userId,
Mid: machineId,
Ex: time.Now().Add(ex),
}
bs, err := json.Marshal(t)
if err != nil {
return
}
key := viper.GetString("app.secret")
return util.AesEncrypt(bs, key)
}
func TokenToSession(token string) (c *ssh.Client, err error) {
key := viper.GetString("app.secret")
bs, err := util.AesDecrypt(token, key)
if err != nil {
return
}
t := fsshToken{}
err = json.Unmarshal(bs, &t)
if err != nil {
return
}
if t.Ex.Before(time.Now()) {
return nil, errors.New("token is expired")
}
sshConf, err := util.NewSshClientConfig("pi", "000", "000", "", "")
if err != nil {
return nil, err
}
// Connect to ssh server
return ssh.Dial("tcp", "home.mojotv.cn:22", sshConf)
}
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External ear
The external ear comprises the auricle (or pinna), the external auditory meatus, and the tympanum (eardrum). The pinna concentrates and amplifies sound waves and funnels them through the outer acoustic pore into the external auditory meatus, which carries them to the tympanic membrane.
Gross anatomy
Auricle (pinna)
The auricle is the part of the ear that projects laterally from the head. It is composed of an irregular concave plate of elastic cartilage and dense connective tissue, covered by skin which contains short hairs (tragi), sebaceous glands, and ceruminous glands.
Structure
The auricle has a complex shape that is composed of several ridges, notches, and grooves (see Figure 1):
helix: posterior free margin of the auricle
anthelix: a ridge parallel to the helix
crura anthelicis: a pair of limbs located above the external acoustic pore
fossa triangularis:tiny depression between thecrura
scaphoid fossa: the depression between the helix and anthelix
tragus: prominence in front of the external acoustic pore
can be manually pushed back over the pore, to mitigate noise
antitragus: situated in the lower part of the anthelix and faces the tragus
intertragic incisure: a notch separating the tragus from the antitragus
cavum conchae: the deepest depression in the auricle, inferior to the crus of the helix
cymba conchae: depression surrounded by the crus of the helix below and the inferior crus of anthelix above
ear lobe: the lowest part of the ear and the only part that does not contain cartilage, situated below the intertragic incisure
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outerbanks - camping Rig - Photo 9731883
We saw some unusual rigs on the beach, including this Ford which was designed for long stays on the South Core Banks. It belongs to Terry and Carolyn McLamb from Goldsboro, North Carolina, who stretched and stuffed it with everything needed for extended beach fishing expeditions, including lots of fresh water and an onboard generator
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Big Thoughts for Little Thinkers: The Trinity eBook (EPUB, PDF)
$3.99
The Trinity promotes the Christian doctrine that there is one God who exists as three persons the Father, Son, and Holy Spirit. This teaching is important for children to relate rightly to each member of the godhead. Our ideas about God affect every area of our lives.
Description
The most foundational teachings of the Christian faith are presented in the Big Thoughts for Little Thinkers series at a level preschool and elementary children can understand. In simple and precise language, God centered theology is promoted, giving children a firm foundation in God’s timeless truth.
The Trinity promotes the Christian doctrine that there is one God who exists as three persons the Father, Son, and Holy Spirit. This teaching is important for children to relate rightly to each member of the godhead. Our ideas about God affect every area of our lives.
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I'd have put Halo on the list for a truly excellent co-op experience, or Rainbow Six for the one-shot death. The "storyline interaction crap" for HL doesn't cut it though; Duke3d did it just as well (both were Sci-Fi, one was campy and the other was "legit").
AvP should have been higher on the list; I also would have put an honerable mention to TimeShift (IIRC). And WHAT about Jedi Knight??? Christ, that storyline was WAY better than HL1...
* Purge is now grumpy
Logged
"If it weren't for Philo T. Farnsworth, inventor of television, we'd still be eating frozen radio dinners." - Johnny Carson
Halo only made a difference for the console and even for a console, it was pretty bland. Poor level design. What halo did for the console was, made FPS games workable using a joystick, and offered great multiplayer. It did nothing significant for the pc crowd. The pc version was bland and offered nothing new.
While FPS games were popular before Half Life, Half Life was the one that merged everything together, solid graphics, awesome sound, great AI, and a compelling story. Before that, you would get pieces of that in different FPS games, never found in one game.
There's comedy, there's high comedy, and then there's guys trying their best to ignore the 800 lb gorilla.
I remember when major companies had to specifically enact rules against playing any of those top 5 games during company time...Oh, wait, that never happened.
I'm sorry but I have absolutely no idea what you're trying to say. Are you saying that since the top 5 games weren't played during company time they didn't advance the genre very much? I'll try my best not to ignore the 800 lb gorilla but at this point I think he just flew right over my head.
No Duke 3D? Wasn't it the next big game after Doom? It should have at least gotten an honorable mention.
they mentioned it in the opening.
As to Co-op, Halo didn't punish people for dying, and their waypoint system kicked ass. I agree with LE (I was trying to remember the nagging title I couldn't recall). UU was freakin' awesome, and since you could shoot a bow, it was a first person shooter.
ROTT was a standout for gore, Serious Sam for the 10000000 monsters rushing you, and Hexen / Heretic for the goth stuff.
I'd also add Unreal / Wheel of Time for storyline, and Q3 Arena and UT series for multiplayer.
Logged
"If it weren't for Philo T. Farnsworth, inventor of television, we'd still be eating frozen radio dinners." - Johnny Carson
Pssh...we all know Duke Nukem Forever is #1... .... ......bhahahhaha I can't say it with a straight face!
Personally, I'd bump Tribes up a notch or two. With a good team that was one hell of a great game. I'd also deflate the hypemachine that is Goldeneye. I've played it. It isn't all that and a ham sammich. I'll totally agree with the number 1 pick though.
You know, Chronicles of Riddick: Escape from Butcher Bay, while not strictly a shooter in the traditional sense did bring normal mapping to the table first. That has gotta count for something, give its widespread use now.
It's kind of silly for Halo not to be on the list. Goldeneye may have brought FPS games to consoles, but Halo opened the floodgates (no pun intended ). If there had been only Goldeneye and no Halo, I doubt we'd see anywhere near the number of FPS games on consoles that we do now. Regardless of what anyone thinks of the game, it definitely had a large impact on the FPS genre.
« Last Edit: September 10, 2006, 06:17:45 AM by EddieA »
Logged
"Why did the chicken cross the Mobius strip? To get to the same side." - The Big Bang Theory
I'm sorry but I have absolutely no idea what you're trying to say. Are you saying that since the top 5 games weren't played during company time they didn't advance the genre very much? I'll try my best not to ignore the 800 lb gorilla but at this point I think he just flew right over my head.
Wonderpug, the 800lb gorilla I'm talking about is the one game on that list that had a cultural impact on NON-gamers. Doom transcended gaming, and brought attention to gaming like no other title on this list. No, it was not the first, no it wasn't that best, no, it's not event the best selling (in total numbers sold) FPS. What it is is the first game to broaden the market IMMENSLY. All of the other titles here were great/important/etc to Gamers, and Doom went well beyond gamers, which is why I had the point that corporations started having rules of using network connectivity for Doom. All of the geeks can whine all they want, Doom is the most important game on this list, by a mile.
As an aside, what are the chances we will ever see another Descent game? I loved those. That, and how do you mention Marathon (I've always included it on my list of favorite games ever) and not mention the physics editor? The Phorte editor still blows almost any non-map editor I've seen out of the water. Hell, most games don't even have such things yet.
I agree that Doom should absolutely be high on the list if not number one. As Big Jake said it brought gaming to a huge audience and had a level of immersion that surpassed everything before and after for years. It also introduced elements that would mold the industry such as it's multiplayer options and open architecture that allowed player-made maps and mods.
It's really hard to think of any recent Multiplayer FPS that hasn't borrowed multiple features from Tribes...
1] What, T and S aren't good enough? (TeamSay and Say) Available in almost any iD game after quake2. (1997, Tribes 1 is 1998)
2] Team Fortress (1) mod had classes before Tribes 1.
3] CounterStrike again.
4] is a mishmash of examples, multiple game types for FPSs dates back before Tribes.
1) Was referring to QuickChat, very much similar to BF2's "CommoRose" or whatever you call it. And a lot of games picked up on it, Counter-Strike was the first to somewhat automate it, though the "Fire in the hole!"s get kind of irritating.
2) I never played the original Team Fortress, but I was under the impression you couldn't customize your class beyond just a basic few selections... no custom load outs, ect.
3) I'm going to assume you meant Team Fortress, and I never played the Quake mod. I know they were in the HL mod... but again I'm not too sure what was in the original Team Fortress.
4) I don't actually remember all of them... it shipped with like 7 different game types out of the box, though, if I remember correctly. I don't think too many where that unique, the point I was trying to get at more was how they were laid our with turrets, inventories, power generators, ect - enhanced gametypes.
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About how turnovers cost them the game, Mack said:
“You can’t win when you turn the ball over six times. San Jose was an excellent football team and I really saw some improvement from our football team. I thought the defense played their tails off.” (UH)
“You can’t win a ballgame when you turn the ball over six times.” (HSB)
“I thought every phase on our team improved, except we threw six turnovers. That’s one that got away from us and I’m very disappointed in that.” (HSB)
“Defensively we kept answering the challenge. The 13 points we gave up in the second half were all off turnovers. I thought we did a great job holding San Jose on quick-change possessions.” (UH, HSB)
“We got beat. We sort of gave it away. We sort of beat ourselves. You can’t win a ball game when you turn it over six times.” (SJMN)
Praising their OL, Mack said:
“Our offensive line played very well against a San Jose defensive front that is one of the best in conference. I thought we did some things well on offense, but again, you can’t turn the ball over six times.” (UH)
About how they won last season despite giving up 6 turnovers at Idaho and 5 turnovers at San Jose, Mack said:
“We did it last year, but we were getting turnovers, also. We’re going to lead the world in turnovers — that’s what we have to improve on.” (HSB)
.
About switching their offense to fit with Inoke’s strengths, Ron Lee said:
“(Graunke) was still struggling (in practices), and Funaki had a couple of good weeks of practices.” (HA)
HA Note: “The Warriors also went with quarterback Inoke Funaki, who is best when he is on the move. The past two weeks — UH had a bye last weekend — were spent honing a revised offense with Funaki in mind. The Warriors added bootlegs, rollouts and, when the defense cramped the tackle box, play-action passes.”
.
——————– Quotes from the UH players ————————–
About how they need to stick together, Sol and Adam Leonard told the team after the game:
“No pointing fingers. Unless you played a perfect game, don’t dare look at anyone else’s performance. We’ve all got to evaluate ourselves.” (HSB)
About how they need to focus on Fresno State now, Adam said:
“We talked about not pointing a finger at anyone. Nobody played a perfect game, we all win together and lose together. We are going to learn from this game, we can’t sit around and feel sorry for ourselves. We have to get prepared for Fresno State.” (UH)
.
About how they were excited at halftime with their 17-7 lead, Inoke said:
“(At halftime) we felt good. You know we were excited, we wanted to keep the intensity up and keep trying to move the ball. The second half I thought our offense did well. It’s just me.” (HSB)
About his INT on their first drive of the 2nd half, Inoke said:
“Yoda was there. I just underthrew it.” (HA)
About his final INT, where he thought that Salas was supposed to run a fade-and-stop route but Salas ran by CB Coye Francies, Inoke said;
“It was a bad communication. He was going over the top, and I was thinking he would come back for it.” (HA)
About how his turnovers put the defense into tough situations, Inoke said:
“For the defense it was bad, very, very bad situations. No matter who you play, you don’t give them any rest and you give the other team good field position, you’re not gonna win.” (HSB)
Taking responsibility for the loss, Inoke said:
“I take responsibility for it. The ball slipped out of my hand (on the fumble). The picks, so many. It doesn’t matter who you play, you turn the ball over that many times, and then you give your defense no rest … it was a tough loss.” (HA)
“I feel like a big part of it was me, all the turnovers. Our defense did all that, our defense came out firing. We turned the ball over, too many turnovers in that second half. … You turn the ball over that many times … you can’t win.” (HSB)
.
After the game, Tyler said:
“We lost as a team. It sucks.” (HA)
About how they didn’t take care of the ball, Tyler said:
“It’s nobody’s fault, it’s just that we had the ball in our hands and it’s up to us what we do with it. Gotta do the right thing and we haven’t been. Tonight, we did the same thing, we lost, and we didn’t take care of the ball, didn’t score.” (HSB)
About how he struggled in the game, Tyler said:
“I couldn’t get it done. I got put in a situation to get it done, to win the game, and I didn’t. My hand had nothing to do with it. I felt fine.” (HA)
“I had a chance to win it, a chance to lead the team to victory but I blew it. Or, we blew it as a team. It just doesn’t feel good, same feeling at Oregon, at Florida. Just doesn’t feel good to lose.” (HSB)
About his first drive, which ended in a missed 55-yard FG, Tyler said:
“I had to read every throw and made a couple bad ones, a couple good ones, but … we were in field-goal range, then I took a sack, and that kind of messed us up. I’ll take the fall for that because half the linemen heard the wrong call and that’s my fault for not speaking up in the huddle.” (HSB)
.
About how their 6 turnovers cost them the game, John Estes said:
“We can’t turn the ball over six times. That’s what it came down to. They didn’t beat us. We beat ourselves.” (HA)
About how their turnovers kept making things hard for their D, Estes said:
“We kept putting our defense in bad positions. Our defense played (its) ass off. They (the Spartans) had the ball at the 20-yard line, the 30-yard line, the 10-yard line.” (HA)
.
About how they improved on offense this game, Aaron Bain (Yoda) said:
“They capitalized on some big plays and they made some defensive adjustments in the second half. I think tonight we continued to show improvements on offense and we came together as a team. Coach told us and we know we can’t win with that amount of turnovers.” (UH)
.
About how he suffered what is thought to be a concussion while scoring his second TD, Pilares said:
“I got hit on the top of the head.” (HA)
.
About how he suffered a concussion during their first drive of the game, Malcolm Lane said:
“I was just out of it, I was dizzy and out of it. I feel OK right now.” (HSB)
.
About how their turnovers made it hard on their D, Sol said:
“We have no excuses. That’s our motto. We love the challenge.” (HA)
HA Note: “At one point the UH defense held the Spartans from scoring on nine consecutive series. But a Warrior offense that ranks 119th in turnover differential was unable to put much breathing room between it and the Spartans and it came back to haunt the Warriors.”
About how their play on defense carried over from last year, Sol said:
“We played together, we played with emotion and we played with love for each other. That’s what carried us all last year. In practice, we came together as brothers and took it on ourselves to understand that if they can’t score, they can’t win, no matter what kind of field position they have. Our defense has got to lead this team, and that’s what we’re going to do. Tonight, the team fed off of us. The crowd had our back the whole night. It’s fun playing with love for each other. This team grew as a team tonight. We might not have won in the stats, but we earned a lot of victories in other areas tonight.” (HA)
About how their defense played well in the game, Sol said:
“This is always sort of a rivalry game. The emotions were high throughout the game and games like this are always going to be tough. The emotions helped us tonight, I think that was one of our best defensive games.” (UH)
.
About how he hates to lose, Adam Leonard said:
“I don’t hate a lot of things in this life, but I hate losing. No matter how it comes, whether it’s a blowout or a close loss.” (HSB)
.
About how their defense played well, Keala Watson said:
“I think we finally woke up to the fact that we play a big role in this team’s success. You play how you practice, and we’ve been practicing with a lot of emotion and intensity these last couple of weeks. That’s what we brought out there on the field.” (HA)
“That’s what defined the defense tonight. We came out ready to play.” (HA)
.
About how the plays they were making on defense got their defense even more excited, Keao Monteilh said:
“We were just trying to defend our (Western Athletic Conference) championship. They were coming into our house and trying to take it from us, so we had to do what we had to do. Every time somebody made a big play, we got pumped.” (HA)
.
About how his left triceps was reinjured and he could possibly be lost for the season, Rocky Savaiigaea said:
“They said I tore it again. I won’t know anything until I get it looked at tomorrow, but I heard it pop.” (HA)
.
About his miss on the 55-yard attempt, Kelly said:
“It was a 100 percent shank. It was just a bad kick on my part. I over-thought the kick. I came in too fast. As soon as I kicked it, I knew it wasn’t going anywhere.” (HA)
.
——————– Quotes from the SJSU coaches ———————–
After the game, coach Dick Tomey said:
“The toughest thing, is to win in the fourth quarter on the road … and against a team that has had a bye week.” (HA)
HA Note: “So, for the first time in 10 games over three season, there was jubilation flowing in the visitors’ locker room in Halawa. For the first time in 16 WAC games over three seasons, somebody other than UH had something to celebrate. And did.”
About how this was a great win for their team, Tomey said:
“It was just a great win for us. We had lost two games in the fourth quarter on the road, and we were playing the defending champions – a team that has won 15-straight conference games. The team that won was going to be in first and the team that lost was going to be last. We just shut them out in the second half and made so many plays on defense. We didn’t have a take away and they had too many. I just credit our guys for battling in the fourth quarter on the road and getting the win.” (UH)
“We have a really good kicker and he kicked two amazing field goals. It was just a great win for us. We had lost two games in the fourth quarter on the road and we’re playing the defending champions who won 15 straight.” (SJMN)
Happy with their play on special teams, Tomey said:
“I just thought we needed a touchdown; we had good field position. If we didn’t make it the ball was going to be down there. So I think that was a good chance. We had a good chance to complete the ball; we had someone open. We have a good kicker. His kick offs are unbelievable and he kicked two amazing field goals late in the game – long field goals – to win. He’s one of the best around. Everything he does really helps us. But just a great team effort overall. The offense did not turn over the ball. The kicking game was tremendous. Kick off coverage, punt coverage, field goals, that’s the best job we’ve done on the kicking game. Hawaii’s going to have a really good team. They’re going to win a lot of games. And I think San Jose State is going to have a really good team too.” (UH)
About the pressure their DL was able to apply, Tomey said:
“We had a bunch of sacks. I don’t know how many, but we had a bunch. We had five? Six? We have some good ends. But they doubled (Carl) Ihenacho, who’s our best guy. It freed up some other guys. Guys that haven’t played much like Pablo Garcia, who came into play and did a great job. And Mo Marah came in. We had a lot of guys make plays that haven’t played before.” (UH)
About their halftime adjustments, Tomey said:
“We made a few adjustments (at halftime) to help us with the shovel pass because they gashed us with a couple things. Inoke’s ability to run was real difficult. But we have a good defense; guys with the ability to make plays. We have some secondary guys with good instincts. We have a real disciplined team; we did not turn the ball over in a tough environment. When Kyle Reed was in, he had a lot of pressure on him. He was able to manage the game enough to be a winner. So I’m really thrilled about that.” (UH)
Asked if he thought it would take 6 turnovers for SJSU to finally be UH after 7 losses, Tomey said:
“I don’t give a bleep. We won a close game in the fourth quarter.” (HA)
About the 6 turnovers, the most that SJSU has forced while he has been the head coach there, Tomey said:
“We have a good defense and we have some guys that can make plays. We have some secondary guys who have some good instincts and we have a real disciplined team.” (HA)
About how they avoided turnovers while UH had too many, Tomey said (I think that Tomey said “giveaway” but the reporter wrote “takeaway’):
“We shut them out in the second half and just made so many plays on defense. We didn’t have a takeaway, and they had too many.” (SJMN)
About how UH didn’t score in the second half, something that hadn’t happened at Aloha Stadium since 1998 (before JJ came to UH), Tomey said:
“We shut them out in the second half and just made so many plays on defense. I credit our guys for battling in the fourth quarter and getting the win on the road.” (HSB)
Praising Strubeck for the two long FGs, Tomey said:
“He’s one of the best around. He said (tonight) wasn’t nearly as hard as having all your teammates around you hitting you in the head. We gave him the game ball. He just rallied to the cause.” (HSB)
AP Note: “Early in the third, Spartans coach Dick Tomey chose to go for it on fourth-and-4 from the Warriors 16 instead of attempting a 33-yard field goal that could have trimmed the lead to a touchdown. They didn’t make the first down.”
About the 50-yard FG that Strubeck hit (his previous season-high was 41 yards and he had already missed a 47-yard FG in the game), Tomey said:
“If you stand back there and see how far that is, it’s a damn long way. On the road, with the crowd and the fact that he missed some this year. That really was huge for us.” (HSB)
“The [kick] previous to [the 50-yard] had plenty of leg. And at that time we thought we’d take a swing at it. But he has been a terrific kicker for us. And the way we practice is that we get the guys around the kicker and we throw things at him, and we have people jumping up and down. So that (50 yarder) is not nearly as hard as having your teammates throw things at you, hitting you in the head. We just gave him the game ball and he rallied to the cause. And if you stand back there at fifty yards, that’s a damn long one, on the road with the crowd. But he made two in a row.” (UH)
.
About how he encouraged the defense even though UH was moving the ball in the first half, defensive coordinator Keith Burns said:
“I told (our team) to just keep playing. We’ve got a code that we play by defensively and just had to stick to it and play hard. It was a great team win, a great team effort.” (HA)
“We just stuck to what we believe in. We played with a great deal of pride. We have this defensive motto called The Code and it showed up in the second half.” (HSB)
About how they have struggled against UH’s offense in the past, Burns said:
“You know, we’ve gotten lit up for five, six touchdowns in the past (against Hawai’i), but (our team) believed in themselves and that’s the most special thing about this win.” (HA)
.
——————– Quotes from the SJSU players ————————
About how they never gave up on Strubeck despite his struggles this season, QB Kyle Reed said:
“We never gave up on him. We knew he had it in him. He made big plays for us.” (HA)
.
About how he hit FGs to tie and win the game, Jared Strubeck said:
“It felt great.” (HA)
“I made it for the team. I had to do my part. I was happy to contribute.” (HA)
Happy that Coach Tomey didn’t lose faith in him despite how he had missed 5 of 8 FGs before this game, Strubeck said:
“It feels great knowing Coach still believes in me. It was such a dire situation and we needed a kick even though I’ve struggled.” (HSB)
HA Note “Strubeck tied it at 17 with a career-long field goal of 50 yards with 9:32 to play. It also was the Spartans’ farthest field goal in 13 years. He then won it with 1:49 remaining when his attempt from 47 yards was true.”
About how his teammates prepared him in practice so that the crowd would not bother him, Strubeck said:
“The guys on our team know what to do to get to me. All the noise here is just white noise. The guys on my team — they know exactly what to say.” (HSB)
About how he was 3-for-9 on FGs before his two long FGs to tie and win the game, Strubeck said:
“Was I really that bad?” (HSB)
.
About how their scouting of UH led to his INT of UH’s first offensive play of the second half, safety Kyle Flynn said:
“We went over that play this week and we knew that they would send one guy to the flat and the slotback to the corner. Our linebacker covered the flat, so I went to the corner and that’s where he threw it.” (HA)
Praising their front 7 on D, Flynn said:
“The front seven (on defense) played a whole new game — with a whole new attitude — in the second half, they really made the difference for us. They put pressure on them and (the secondary) was able to come up with some turnovers.” (HA)
.
About how it is hard to beat UH at home, backup OL Steve Lightsy (who graduated from Kahuku) said:
“It’s a tough game playing against Hawaii at home. Words can’t even say what it’s like to come here and win.” (HSB)
.
—————— Quotes from others ———————–
About Colt’s love for Hawaii, Terry Brennan said:
“He has a strong relationship with Hawai’i, and I think he wants to keep a part of himself here.” (HA)
Betsy Brennan said that not having Colt play in the UH game made the experience of watching it:
“a little less stressful. I still screamed my head off.” (HA)
HA Note: “The Brennans were happy to see so many of their son’s jerseys being worn by UH fans. The Brennans met with UH coaches and players during the team’s walk-through and had dinner with friends from the department.”
About how Colt hasn’t been playing much, Terry Brennan laughed and said:
“At least he’s got a great seat to all the games.” (HA)
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The South Australia health chronic disease self-management Internet trial.
To evaluate the effectiveness of an online chronic disease self-management program for South Australia residents. Data were collected online at baseline, 6 months, and 12 months. The intervention was an asynchronous 6-week chronic disease self-management program offered online. The authors measured eight health status measures, seven behaviors, and four utilization measures; self-efficacy; and health care satisfaction. Two hundred fifty-four South Australian adults with one or more chronic conditions completed baseline data. One hundred forty-four completed 6 months and 194 completed 1 year. Significant improvements (p < .05) were found at 6 months for four health status measures, six health behaviors, self-efficacy, and visits to emergency departments. At 12 months, five health status indicators, six health behaviors, self-efficacy, and visits to emergency departments remained significant. Satisfaction with health care trended toward significance. The peer-led online program was both acceptable and useful for this population. It appeared to decrease symptoms, improve health behaviors, self-efficacy, and reduce health care utilization up to 1 year. This intervention also has large potential implications for the use of a public health education model for reaching large numbers of people. It demonstrates that an Internet self-management program, which includes social media, can reach rural and underserved people as well as be effective and reduce health care costs. If this intervention can be brought to scale, it has the potential for improving the lives of large numbers of people with chronic illness. It represents a way the medical care and public health sectors can interact.
|
Q:
How to make a global property (setter and getter)?
I need to create an object for example NSString and let other classes get/set the value for it.
thx. for help :)
A:
The easiest way to do this is to attach the value as a property to a singleton instance of some class. One singleton instance that already exists in your application is the application delegate. So, just add an NSString property to your application delegate and you can access it from any class in your app (as long as you #import your application delegate).
In your application delegate:
@property(nonatomic, strong) NSString* someString;
In your other classes:
[self doSomethingWithAString:((YourAppDelegateClass*)[[UIApplication sharedApplication] delegate]).someString];
|
Q:
converting google api search results obtained in json format into json object
I am using google api to get search result for certain queries in JSON format. Now I want to convert it into JSON Object in java and access only particular values.
The JSON response format is :
{
"kind": "customsearch#search",
"url": {
"type": "application/json",
"template": "https://www.googleapis.com/customsearch/v1?q={searchTerms}&num={count?}&start={startIndex?}&lr={language?}&safe={safe?}&cx={cx?}&cref={cref?}&sort={sort?}&filter={filter?}&gl={gl?}&cr={cr?}&googlehost={googleHost?}&c2coff={disableCnTwTranslation?}&hq={hq?}&hl={hl?}&siteSearch={siteSearch?}&siteSearchFilter={siteSearchFilter?}&exactTerms={exactTerms?}&excludeTerms={excludeTerms?}&linkSite={linkSite?}&orTerms={orTerms?}&relatedSite={relatedSite?}&dateRestrict={dateRestrict?}&lowRange={lowRange?}&highRange={highRange?}&searchType={searchType}&fileType={fileType?}&rights={rights?}&imgSize={imgSize?}&imgType={imgType?}&imgColorType={imgColorType?}&imgDominantColor={imgDominantColor?}&alt=json"
},
"queries": {
"nextPage": [
{
"title": "Google Custom Search - apple",
"totalResults": "531000000",
"searchTerms": "apple",
"count": 10,
"startIndex": 11,
"inputEncoding": "utf8",
"outputEncoding": "utf8",
"safe": "off",
"cx": "013036536707430787589:_pqjad5hr1a"
}
],
"request": [
{
"title": "Google Custom Search - apple",
"totalResults": "531000000",
"searchTerms": "apple",
"count": 10,
"startIndex": 1,
"inputEncoding": "utf8",
"outputEncoding": "utf8",
"safe": "off",
"cx": "013036536707430787589:_pqjad5hr1a"
}
]
},
"context": {
"title": "Custom Search"
},
"searchInformation": {
"searchTime": 0.206589,
"formattedSearchTime": "0.21",
"totalResults": "531000000",
"formattedTotalResults": "531,000,000"
},
"items": [
{
"kind": "customsearch#result",
"title": "Apple",
"htmlTitle": "\u003cb\u003eApple\u003c/b\u003e",
"link": "http://www.apple.com/",
"displayLink": "www.apple.com",
"snippet": "Apple designs and creates iPod and iTunes, Mac laptop and desktop computers, the OS X operating system, and the revolutionary iPhone and iPad.",
"htmlSnippet": "\u003cb\u003eApple\u003c/b\u003e designs and creates iPod and iTunes, Mac laptop and desktop computers, \u003cbr\u003e the OS X operating system, and the revolutionary iPhone and iPad.",
"cacheId": "5iRmnZTn43cJ",
"formattedUrl": "www.apple.com/",
"htmlFormattedUrl": "www.\u003cb\u003eapple\u003c/b\u003e.com/",
"pagemap": {
"cse_image": [
{
"src": "http://images.apple.com/home/images/ipad_hero.jpg"
}
],
"cse_thumbnail": [
{
"width": "348",
"height": "145",
"src": "https://encrypted-tbn3.google.com/images?q=tbn:ANd9GcQRUCTcMJO12wSHtTA8iXXzdoaHo1ssBW8cyP5ZONgIdpFtr9gNxmRdruk"
}
],
"metatags": [
{
"author": "Apple Inc.",
"viewport": "width=1024",
"omni_page": "Apple - Index/Tab"
}
]
}
},
{
"kind": "customsearch#result",
"title": "Official Apple Store - Buy the new iPad, Apple TV,
.
.
Now I just want to access "items" array and my code is:
org.json.JSONObject json=null;
json = new JSONObject(jsonResponse);
org.json.JSONObject queryArray=json.getJSONObject("queries");
org.json.JSONArray itemsArray=queryArray.getJSONArray("items");
for(int i=0;i<itemsArray.length();i++)
{
org.json.JSONObject newJSONObj=itemsArray.getJSONObject(i);
System.out.println("Title ::"+newJSONObj.getString("title"));
System.out.println("Link ::"+newJSONObj.getString("link"));
}
This code is giving NoSuchElementException for "items".
Please help...
A:
If my brain is parsing this correctly, "queries" does not have "items" within it.
"queries": {
"nextPage": [
{
"title": "Google Custom Search - apple",
"totalResults": "531000000",
"searchTerms": "apple",
"count": 10,
"startIndex": 11,
"inputEncoding": "utf8",
"outputEncoding": "utf8",
"safe": "off",
"cx": "013036536707430787589:_pqjad5hr1a"
}
], // end nextpage
"request": [
{
"title": "Google Custom Search - apple",
"totalResults": "531000000",
"searchTerms": "apple",
"count": 10,
"startIndex": 1,
"inputEncoding": "utf8",
"outputEncoding": "utf8",
"safe": "off",
"cx": "013036536707430787589:_pqjad5hr1a"
}
] // end request
}, // end queries
You want
json.getJSONArray("items");
I think.
|
SKY – 04
Chapter 4
Cruntiq said, “Everyone, thank you
for coming here today on such short notice. As well as mobilizing and being so
cooperative with regards to the abrupt and no doubt seemingly strange orders we
gave you this morning.” He paused a moment and looks around the hall.
The conversations quickly died as
the audience shifted their attention to Cruntiq.
“I know everyone has questions as
to what’s been going on recently and why you’ve been going on these unexplained
patrols for the last two months.” He paused again, his words briefly echoing
throughout the spacious dome. “One reason we’ve called this conference tonight
is to discuss this with you all, but before we get to that, I’m sure all of you
know that today’s patrol was by no means standard. And I’m also sure most of
you are aware that one squad did not return from this mornings mission.”
“Unit 144 did not return this
morning… We cannot at this moment say that they are deceased, but having not
heard from them at all today, as well as them falling off the ship tracking
system, we must declare them MIA for the time being. News has already been
delivered to their families just before this gathering.”
Hleis looked over at his friends.
They looked back at him; all mirroring the same solemn look on their face, one
that they were all familiar with.
Since birth they were taught the kingdom
was in a state of war, a kingdom that had been in a state of war for millennia,
a kingdom that had been created for the purpose of war. The feeling of
unexpectedly losing friends, without saying a proper goodbye, was a familiar
one. Through repetition it transformed into am aching feeling of emptiness.
Cruntiq continued, “With that in
mind, for tomorrow we will be doing the same patrol procedure again. All units
are to keep their eyes and radars open for any traces or signs of unit 144,
either for rescue or recovery. We will also be assigning three units to travel
the same direction as 144 tomorrow.”
Before he could continue, someone
yelled from the mid rows of the conference hall, “Where the hell are the mages!
My brother died for this and yet they’re nowhere to be seen!? They don’t even
speak to the people they’re making die for their missions!?”
Murmurs and whispers echoed
throughout the hall.
A few people shouted at Cruntiq,
“Yeah! Where are the mages!?”
“Bring them out here! They owe us
an explanation!”
“They’re orders got four of ours
killed! Make them answer us!”
Cruntiq stood for a moment, before he
turned his eyes at Veindelio.
Veindelio did not return the look
and kept his eyes forward at the crowd the entire time, maintaining a stoic
expression on his face. Cruntiq turned his eyes away, takes a blink and a deep
sigh before stepping up to address the hall once again.
“People, do not forget that we are
a strong group. Our small town has been a pillar of the outer trade routes for
decades due to the strength of the individuals that we consider our neighbors.
We owe it to their strength and skill to not assume them to be dead.”
When Cruntiq spoke the people of
West-Hail listened. The trust they had in him translated to the respect they
showed him when he spoke.
“It’s true. The mages have been out
of sight for the majority of you all, but please remember that they only number
in the twenties while we number in the thousands. I have been in contact with
the captain of the mages, the man many of you saw this morning and the man standing
besides me now.”
From the seats, someone shouts
loudly, “Explain yourself mage!”
This was followed by an assorted
array or “Yeah!” and “C’mon!” All from frustrated soldiers seeking to provoke
an answer from the head of the mages.
Veindelio, unfazed by the provocations,
rather assertively stepped up to the microphone.
He was slightly taller than
Cruntiq, while being exceptionally well built despite being a magical
specialist. He had been on many battlefields from his demeanor. His stature
revealed that could hold his own in close combat with any of the people here.
This was apparent to nearly all the soldiers here just from his calmness alone.
He began speaking.
“Although we may come from
Center-Storm in the heart of our kingdom, do not think that we do not recognize
your strength. As an unfortunate occurrence happened today I owe it you all to
give some context as to why we’re here. The Order of Capital Mages is stretched
thin during these times, and as such deployment of our units is reserved for
critically important tasks. I apologize with all due respect, but we have been
assigned to complete a task in West-Hail. One that we must carry out in secrecy
and that has important implications for our entire kingdom.”
Veindelio looked around to gauge
the atmosphere in the room before continuing, “As Cruntiq pointed out earlier,
there are but a handful of mages here in West-Hail. Our numbers may seem small,
but that is no reason to be concerned. Leadership in the capital is aware of
the competence of West-Hail and decided to not to send any more reinforcements.
It is a testament to your town that official military forces was not deployed
here and that the capital trusts you all to be able to help us accomplish our
task.”
He paused yet again before making
his closing statement. “I thank you for your trust, and I promise that all will
be clear in time. The role that you play will be crucial for the success of our
kingdom.”
Murmurs yet again filled the room.
Cruntiq stepped up this time and addressed the group. “Everyone, Veindelio and
I will remain in this hall for a while longer. If you have questions we’ll try
to answer the ones we can. Everyone else is dismissed for the night. I’ll see
everyone tomorrow.”
Traer turned to Hleis and co. and
says, “I’ll stay for a bit, I have a few questions I want to ask. You guys head
on home. You need the rest since you’ll all be heading out tomorrow. I’ll stay
since I don’t patrol and update you guys tomorrow.”
“Only if we make it back,” Maltii
says with a slightly sarcastic disdain for the situation in general.
“Was that it? That’s all they had
to say? What was even the point of this meeting?” Vayling exclaimed.
“Good luck Traer,” Hleis said.
***
The next morning carried on
smoothly but with a more serious tone than the day before. Supplies were packed
onto the ships, the units gathered, and preparations were made in an almost
silent atmosphere.
The defense force of West-Hail now knew
there was real danger in the skies, confirmed by the existence of a missing
squad. They also were aware that they were alone and that whatever they found,
they would have to deal with it on their own.
For Hleis, Vayling and Maltii, this
was especially true as they presumably lost a good friend. Flask pulled their
ship out of port and begun the patrol with a solemn but serious mood permeating
the trip.
The patrol was uneventful that day.
The squad kept their eyes on their surroundings and the radar, but nothing out
of the ordinary was spotted.
They reached the end of their four
and a half hours going outwards away from West-Hail.
On their way back, Vayling said to
Hleis and Maltii, “You know, after the conference yesterday I was almost
expecting us to run into something today. Honestly kind of surprised that we
ended up fine especially since Tyrize did not…”
Hleis jumped in, “Vay, think about
it, of course we were going to be fine. There’s a reason they sent three squads
out in the same direction squad 144 went in. We should focus on returning
safely and preparing for whatever it is that those three groups found.”
Flask interjected, “Aye Hleis,
that’s why they sent out three groups after all. They didn’t send them all out
at once too I hear. Gave them the fastest ships and staggered them. Make sure
that at least one of them makes it back to town to report.”
“Despite what the fancy mage said
yesterday, we are still just a local militia. Gah, how worried do you think our
families and the other townsfolk are right now after all that’s been going on?”
Maltii exclaimed exasperated.
***
It was just before sunset when
Hleis’ squad returned from their patrol. They weren’t the first group to
return, nor would they be the last. Pulling into port, the tension in the air
was palpable as nearly every unit was on edge, waiting on word of the three
units sent out in 144’s direction. The last communication was received around
midday, and none of the three groups had yet returned.
Hleis and Vayling left port
together on their way to the town hall to meet with Traer and Vayling’s father.
Both were hoping to see if any information about Tyrize surfaced today. Outside
of the docks, there was an unfamiliar sense of disorder about. Civilians were
running about in a chaotic fashion, clearly distressed.
Vayling pulled a local shopkeeper
over and said to him, “Hey man, what’s going on? Why’s everyone freaking out?”
“You didn’t see anything out there?
You guys just came back from patrol right? You didn’t see anything?” he replied.
“No, what are you talking?” Hleis
asked him.
“We got a sudden order half an hour
ago from the mayor! We’re supposed to all be gathering supplies right now and
bring them to lowest level! Prepare to fortify! State of Emergency!” He
exclaimed before running off to continue his duties.
Vayling looked at Hleis. Wearing a
confused, but alarmed look on his face. “What the!? They found something!? None
of the guys at the port said anything about fortifying the town!”
Hleis, in a low voice, responded,
“Vay, this was after you left, but at the docks yesterday Traer said he
overheard the word ‘evacuation’. Sounds like they might have found something then,
not today. We have to find Traer.”
Vayling’s normally cheerful
demeanor disappeared. He gave Hleis a head nod and the two take off, running to
the center of town, heading towards the town hall where the defense force
strategic team meets.
To their surprise, three mages were
sitting outside the main entrance, helmets off. Their frustration was on
display by their body language.
Both Hleis and Vayling looked over at
them, but did not stop nor slow down as they entered the hall. Hleis looked
back and made eye contact with one of them, a rather young mage. By appearance,
she couldn’t have been much younger or older than he or Vayling, but wore an
ornate hairstyle and had vibrant golden eyes unlike the magenta eyes of the denizens
of West-Hail.
She and Hleis maintained eye
contact until the town hall broke their line of sight.
Although the exchange was a little
strange, Hleis thought nothing of it; he had no time to even begin thinking
about it, as the main hallway was littered with mages, West-Hail militia and
the scattered and panicked townsfolk. The main door at the end of the hall was
the Mayor’s office.
It was shut closed, but the entire
town hall could hear that something was going on inside.
All the other major members of town
leadership, aside from Cruntiq and the Mayor, were sitting in the main hall, waiting,
along with everyone else, for whatever was happening in there to conclude.
Traer noticed the guys before they
notice him. He waved them down, Hleis and Vayling go over to meet their friend.
“We’ve heard back from all three
teams sent out, no sign of 144, but they spotted some strange aerial
disturbances on their radars, I have no idea what it could be, probably
Mastigs, but Cruntiq does. He ordered a full lockdown of the town and for us to
make siege preparations,” Traer replied.
“144…damn…What about that thing you
mentioned yesterday?” Hleis asked,
“I think that may have been the
stirrings of what’s going on now. The mages were not aware of the lockdown
order. Cruntiq gave the order on his own, without consulting them. Right now he
and the mage captain are in the mayor’s office discussing things, but there’s
no doubt that he’s upset.”
Vayling remarked, “So that’s why
the mages are all scattered about here waiting for something.”
“I’m guessing if we’re fortifying
the town, we’re to make preparations also? Are there any orders for the defense
force? We didn’t hear anything at the dock,” Hleis said.
“Yeah, the vice captain left to
initiate a high level fortification procedure just now. The mage captain pulled
Cruntiq aside before further orders could be given though and that’s the reason
for the state of panic right now.” Traer looked around before continuing, “The
mages look like they’re waiting around as well. From the looks of things, for
our survival, I hope Cruntiq and the mages work it out all right. We’ll need to
cooperate otherwise we’ll be doomed.”
The three of them continued to
chat, keeping an eye out for the occasional messenger relaying information
about the fortification process, returning scout ships and overall supply
levels.
Suddenly the mayor’s door opened
and Cruntiq walked out with a serious look on his face. Veindelio followed
behind him.
“Veindelio, we’ll have to continue
this later, I need to oversee my town’s defenses.”
“…Fine Cruntiq, but if the Mastigs
don’t destroy us, the capital certainly may. You’ll have to deliver impossible results
against these odds,” Veindelio remarked.
Cruntiq replied while still walking
forward, “I’ve seen how this goes before. The capital wastes exceptional and
capable soldiers with their foolish tactics. They waste the resources that they
never even bother to come see with their own eyes while they spend all their
time hiding in Center-Storm.”
“Cruntiq, watch yourself.
Center-Storm is the heart of our nation. They must oversee the defense of the
entire kingdom.”
“A shrinking kingdom Veindelio…Ever
thought how strange it is that in the first five hundred years of CloudStock’s
history, we were basically undefeated against the Mastigs? When the Mastigs
were at their strongest.”
“We had the King of Legends then.
You know this.”
“A warrior king that could live
five hundred years? No, however strong he may have been individually, his real
power was creating a kingdom that could get a high level of usefulness from the
entire population. Not the wasteful recklessness that’s set in since his death
on the battlefield.”
By this point in the conversation,
the two of them had reached the main entrance. Just before exiting the
building, Cruntiq turned to face Veindelio and said, “I’ll take command of my
men, you do the same with yours, I’ll trust you to assign them where you see
fit.”
“Fine, we’ll reconvene an hour
prior to the Mastic arrival.” Veindelio replied as the two go their separate
ways.
The mages naturally flocked to
Veindelio, while everyone else gathered around Cruntiq.
He addressed the local leaders and
all those present outside the hall. “Okay, I’m sure by now everyone’s aware of
the lockdown. We’ve located a class three Mastig attack fleet. A slow moving
one, but they’re on the way to us now that they know our general direction. ETA
is around five hours. No reinforcements, we’re on our own here. We’re too far
away from any military garrison and can’t risk giving away the location of
another settlement. Help the civilians that you can, fortify and supply the
walls and take up defensive positions as directed by the strategic team. Let’s
get through this one alive alright?”
|
---
abstract: 'We consider the wave equation for sound in a moving fluid with a fourth-order anomalous dispersion relation. The velocity of the fluid is a linear function of position, giving two points in the flow where the fluid velocity matches the group velocity of low-frequency waves. We find the exact solution for wave propagation in the flow. The scattering shows amplification of classical waves, leading to spontaneous emission when the waves are quantized. In the dispersionless limit the system corresponds to a 1+1-dimensional black-hole or white-hole binary and there is a thermal spectrum of Hawking radiation from each horizon. Dispersion changes the scattering coefficients so that the quantum emission is no longer thermal. The scattering coefficients were previously obtained by Busch and Parentani in a study of dispersive fields in de Sitter space \[Phys. Rev. D [**86**]{}, 104033 (2012)\]. Our results give further details of the wave propagation in this exactly solvable case, where our focus is on laboratory systems.'
author:
- 'T. G. Philbin'
title: An exact solution for the Hawking effect in a dispersive fluid
---
Introduction
============
The scattering of waves by material inhomogeneities occurs in many guises and is most familiar through the reflection of light at sharp boundaries. The basic phenomenon of scattering may seem intuitive, but new types of scattering become possible when the inhomogeneity consists of a position-dependent velocity of the medium. The flow of the medium can allow additional propagating modes that are absent in the non-moving case. Scattering into the additional modes may be accompanied by wave amplification in which energy is transferred from the flow to the wave. When this amplification occurs in a region where the flow velocity exceeds the group velocity of the wave, the process is analogous to that underlying the Hawking effect at a black-hole horizon [@unr81; @bar05; @rob12]. Inhomogeneous flow profiles and black-hole horizons can both act as amplifiers for waves. The quantum Hawking effect [@haw74; @bro95] of black holes is due to the coupling of the horizon amplifier to the wave’s quantum ground state, the latter being “amplified" into real quanta or quantum noise [@Gardiner]. Both classical wave amplification and spontaneous quantum emission can occur in inhomogeneous flows and several theoretical proposals have now been investigated experimentally [@gar00; @sch02; @rou08; @phi08; @bel10; @lah10; @wei11; @ste14; @ngu15; @ste15; @euv15a].
The Hawking effect in moving media differs most significantly from the astrophysical case in the essential role played by dispersion. The lack of short-wavelength dispersion for waves in space-time renders the astrophysical Hawking effect somewhat singular, as its derivation features infinite wavelength shifts at the horizon [@Jacobson]. For waves in material media, dispersion necessarily limits the wavelength shifts and the Hawking effect has no unphysical features [@unr95; @cor96]. Dispersion has been shown to alter the spectrum of Hawking radiation from the original thermal result for black holes [@rob12; @mac09; @leo12; @fin12; @cou12; @mic14; @rob14; @euv15]. In general the spectrum of quantum Hawking emission depends on the dispersion and the shape of the flow profile. Moreover a flow velocity that exceeds the group velocity of the wave is not necessary for measurable wave amplification to occur [@wei11; @euv15a; @euv15]. Waves in moving media thus offer a rich theoretical and experimental arena where the quantitative dependence of the Hawking effect on dispersion and flow profile can be explored.
A less welcome effect of dispersion is to make wave scattering in the experimental systems very difficult to solve analytically. The most accessible experimental system to date uses surface waves on flowing water, where Hawking amplification of classical incident waves has been observed [@wei11; @euv15a]. The dispersion of water waves is complicated, with regions of normal dispersion ($d^2\omega/dk^2<0$ in the fluid frame) and anomalous dispersion ($d^2\omega/dk^2>0$ in the fluid frame) giving various types of horizon effects [@sch02; @rou10; @pel15]. The system most studied theoretically is sound waves in a flowing Bose-Einstein condensate (BEC), and experiments with BECs have detected quantum Hawking emission of phonons [@ste14; @ste15]. A fourth-order dispersion relation is widely used to model the BEC system [@bar05], and a polynomial dispersion relation can also be used in a simple model for water waves. The resulting wave equations have not been exactly solved for an inhomogeneous flow and theoretical predictions of the Hawking effect in these systems are based on approximate analytical techniques and numerical simulations [@unr95; @cor96; @bro95a; @cor98; @him00; @sai00; @unr05; @mac09; @mac09b; @leo12; @rob14; @fin12; @cou12; @mic14; @euv15; @rob16; @cou16; @mic16].
Here we give the exact solution for wave scattering in a flow whose velocity changes linearly with position, where we use a simple BEC model featuring a fourth-order anomalous dispersion relation [@bar05]. The one-dimensional linear flow profile has regions of positive and negative flow velocity, so there are two horizons, one for left-moving and one for right-moving waves. The flow is thus analogous to a black-hole or white-hole binary (depending on the sign of the velocity gradient). A linear flow profile is often assumed in the neighbourhood of a single horizon, and approximate treatments of this case using the same techniques employed here were given in [@unr05; @cou12]. But the exact treatment of the linear profile necessarily involves two horizons. On the other hand, the wave scattering does not depend on the profile being linear at large distances, so that the scattering coefficients obtained here are also valid for profiles that slowly change from linear to flat far from the horizons.
The linear flow profile in the dispersionless case gives an effective space-time metric for waves that corresponds to a patch of de Sitter space, as discussed in ref. [@bus12]. In that work, Busch and Parentani studied dispersive fields in de Sitter space and in a cosmological context considered the dispersive wave equation used here. These authors derived the scattering amplitudes (\[S12\])–(\[S42\]) below by a somewhat different method. Here we provide more details on the exact wave solution and its asymptotics. Also, our focus here is on laboratory systems rather than cosmology.
The linear flow profile is not the example one would most like to solve exactly because it does not model current experiments, but nevertheless it gives some interesting lessons. The results highlight the importance of *reflection*, the familiar conversion of right-moving waves to left-moving and vice versa, where left/right motion here refers to the velocity relative to the fluid. Scattering due to reflection in inhomogeneous flows has until recently been of secondary interest in studies of Hawking amplification, but reflection occurs in the experimental systems and its magnitude affects the spectrum of quantum Hawking emission [@euv15]. For theoretical purposes, dispersion can be implemented in a manner that does not give reflection, so that the left- and right-moving waves do not mix [@sch08], but this possibility is not realised in the experimental systems.
For wave equations in spatially inhomogeneous media, the exact solution for a linearly changing profile is the basis for the WKB approximation for arbitrary profiles [@heading]. In optics and quantum mechanics the solutions in question are the two Airy functions, valid for a linearly changing permittivity or potential. For the wave equation in a linearly inhomogeneous flow we will obtain four solutions, rather than two, and importantly they depend on the dispersion. Our solutions do not therefore have the same universal significance for inhomogeneous flows as the Airy functions have in optics and quantum mechanics, rather they are specific to the dispersion of the BEC model.
The model equation used together with a general description of the propagating modes is given in section \[sec:wave\]. In section \[sec:ksol\] the wave equation is Fourier transformed and solved exactly in $k$-space. Section \[sec:4sols\] presents the lengthiest part of the analysis, in which four independent solutions of the wave equation are defined and their meaning in terms of mode scattering is elucidated. In section \[sec:scat\] solutions are constructed that represent the scattering of single incident modes and the exact scattering coefficients are calculated. The final results are contained in (\[S12\])–(\[S42\]) and (\[S2S3a\])–(\[S2S3b\]), and are visually represented in Figs. \[fig:2in\] and \[fig:3in\].
Wave equation {#sec:wave}
=============
An approximate (1+1)-dimensional wave equation for sound in a flowing BEC takes the form [@bar05] $$\label{wave}
\partial_t (\partial_t+v\partial_x )\psi+\partial_x (v\partial_t+v^2\partial_x )\psi-\left(\partial_x^2-\frac{1}{k_c^2}\partial_x^4\right)\psi=0.$$ Here velocity is dimensionless, $v(x)$ is the (time-independent) flow velocity, $k_c$ quantifies a dispersive term, and $\psi(x,t)$ is the wave field; for the BEC $\psi(x,t)$ is the phase fluctuation in the field operator of the bose gas [@bar05]. A more accurate equation for sound in a BEC can be derived [@mac09b], but (\[wave\]) is often considered in studies of dispersion and the Hawking effect. In the absence of the fourth-order derivative term there is no dispersion of the wave and (\[wave\]) is the equation of a scalar field in a curved (1+1)-dimensional space-time [@unr81; @bar05]. In the non-dispersive case ($k_c\to\infty$) the waves have speed $1$ relative to the fluid and points where $v(x)=\pm1$ are in strict analogy to event horizons for the waves [@unr81].
We take (\[wave\]) as our model wave equation in a dispersive fluid. The central problem is to solve (\[wave\]) as a classical wave equation, as this determines the mode expansion for $\psi(x,t)$ as a classical or quantum field. In particular, the Hawking effect in this system is at root a classical scattering effect for waves satisfying (\[wave\]). The monochromatic wave equation follows from the substitution $\psi(x,t)=e^{-i\omega t} \phi(x)$: $$\label{mono}
\left[\omega^2 +i\omega v'+2 v (i\omega-v')\partial_x +(1-v^2) \partial_x^2 -\frac{1}{k_c^2}\partial_x^4 \right]\phi=0,$$ where a prime denotes a derivative with respect to $x$. The dispersion relation is $$\label{disprel}
(\omega-vk)^2=k^2+\frac{k^4}{k_c^2},$$ which can be solved exactly for four roots $k(\omega)$ that can all describe propagating modes in the fluid for $\omega>0$ (the expressions for $k(\omega)$ are too cumbersome to reproduce here). The frequency $\omega$ in the laboratory frame is conserved for monochromatic waves because of the time-independence of (\[mono\]). The quantity $\omega-v(x)k$ is the frequency in a frame locally co-moving with the fluid, which is the relevant frame for characterising the material dispersion. We see that (\[disprel\]) gives anomalous dispersion at all frequencies, so the group and phase speeds relative to the fluid exceed their non-dispersive values of $1$, except in the limit of zero $k$. As long as the dispersion is entirely anomalous or entirely normal for all frequencies, there will still be at most four propagating modes with $\omega>0$. If the dispersion is normal for some frequency ranges and anomalous for others, there are in general more than four $\omega>0$ propagating modes (water waves provide an example [@rou10]).
![Top: the linear velocity profile (\[flow\]) for $\alpha=1/2$. At $x=\pm1/\alpha$ the flow speed matches the speed relative to the fluid of low-$k$ waves. Middle: ray plots for waves in the flow with $\omega=1$, and with dispersion removed ($k_c\to\infty$). Red (blue) rays are for waves moving left (right) relative to the fluid. Solid (dotted) rays have positive (negative) frequency $\omega-v(x)k$ in a frame co-moving with the fluid. This corresponds to a white-hole binary. Bottom: the ray plots when dispersion is included ($k_c=5$). The mode labels 1, 2, 3 and 4 refer to the four solutions $k(\omega)$ of the dispersion relation (\[disprel\]); these solutions are shown graphically in Fig. \[fig:disp\]. []{data-label="fig:rays"}](figure1.pdf){width="\linewidth"}
![Plots of $\omega$ versus $k$ for the dispersion relation (\[disprel\]) in the linear profile (\[flow\]), for three values of $x$ (with $\alpha=1/2$, $k_c=5$). The horizontal magenta line is $\omega=1$ and its intersection with the dispersion plots gives the propagating (real $k$) modes at this frequency; these mode solutions are labeled 1 to 4. Modes from an intersection with a red (blue) dispersion curve travel left (right) relative to the fluid. In addition, if the intersection is with a solid (dotted) dispersion curve, the mode has positive (negative) co-moving frequency $\omega-v(x)k$. In the region between the horizons, such as in the middle plot for $x=0$, there are two propagating modes. Beyond the horizons ($x\ll 0$ and $x\gg0$) modes 3 and 4 can propagate. []{data-label="fig:disp"}](figure2.pdf){width="\linewidth"}
We consider the linear flow profile $$\label{flow}
v(x)=-\alpha x, \qquad \alpha>0$$ (see Fig. \[fig:rays\]). This flow exceeds the small-$k$ (non-dispersive) speed of the waves relative to the fluid at $x=\pm1/\alpha$. In the non-dispersive limit $k_c\to\infty$ the flow provides a sharp horizon for left-going waves at $x=-1/\alpha$, and for right-going waves at $x=1/\alpha$. This is shown in Fig. \[fig:rays\], where the rays for $\omega>0$ waves are plotted in the non-dispersive case. The two rays for left movers (relative to the fluid) are shown in red and are funnelled to the horizon at $x=-1/\alpha$, undergoing an infinite blue-shifting ($k\to\infty$) in the process. The solid (dotted) ray has a positive (negative) frequency $\omega-v(x)k$ in the co-moving frame. Similarly the two right-movers (relative to the fluid), one of which has a negative co-moving frequency, are funnelled to the horizon at $x=1/\alpha$ and infinitely blue-shifted. This flow profile corresponds to a white-hole binary, i.e. two 1+1-dimensional white holes facing each other. By reversing the direction of time (or reading the ray plot from top to bottom) a black-hole binary is obtained. The anomalous dispersion in the wave equation (\[mono\]) limits the blue-shifting at the horizons and redirects the blue-shifted left- and right-movers into the respective white holes, as also shown in Fig. \[fig:rays\]. This behaviour can be understood by looking at the dispersion relation at different point in the flow, shown in Fig. \[fig:disp\]. As noted above, there are four propagating modes $k(\omega)$ in the flow, labeled 1 to 4 in the bottom ray plot in Fig. \[fig:rays\] and in the graphical solutions for these modes shown in Fig. \[fig:disp\]. Modes 1 and 2 are the usual left- and right-moving $\omega>0$ waves that would be present in a nonmoving fluid ($v(x)=0$); these modes have positive frequency in a frame co-moving with the fluid. The motion of the fluid has the remarkable effect of making modes 3 and 4 into propagating $\omega>0$ waves, despite their having a negative co-moving frequency $\omega-v(x)k$. Modes 3 and 4 have complex wave-vectors in the region between the horizons, with mode 3 exponentially increasing with $x$ in this region and mode 4 exponentially decreasing with $x$. It is not visually apparent from Figs. \[fig:rays\] and \[fig:disp\] how to follow modes 3 and 4 through the region where their wave-vectors are complex and identify which is mode 3 and which is mode 4 when they again become propagating. The identification follows from the four exact roots of the quartic dispersion relation (\[disprel\]) and a choice of branch cuts in these roots. Modes 3 and 4 coincide at two points in the flow, as is clear from the bottom ray plot in Fig. \[fig:rays\]; these modes are thus converted into each other in a continuous manner by the flow. Rays move at the group velocity $d\omega/dk$, and Fig. \[fig:disp\] explains the reversal of group velocity as modes 3 and 4 are converted into each other. Note also that when mode 4 propagates in the left-hand region of the flow its phase ($\omega/k$) and group ($d\omega/dk$) velocities are in opposite directions, and similarly for mode 3 in the right-hand region.
Mixing of modes 3 and 4 occurs at the level of ray tracing and is therefore visible in the bottom ray plot in Fig. \[fig:rays\]. In addition, there is also wave scattering of each of the four modes into the other three modes because of a breakdown of geometrical optics near the horizons. Our goal is to calculate the exact scattering coefficients. The scattering of modes 1 and 2, with positive co-moving frequency, into modes 3 and 4, which have negative co-moving frequency, is accompanied by wave amplification and is the underlying mechanism of the Hawking effect [@unr95]. Modes 3 and 4 scatter into modes 1 and 2 with a similar amplification effect. The wave amplification can be partly understood in terms of the conserved quantities associated with the wave equation (\[wave\]). As detailed in Appendix \[app:conserved\], there is a conserved norm, associated with $U(1)$ symmetry of the action giving (\[wave\]), and a conserved pseudo-energy associated with its time-translation invariance. The sign of the norm and pseudo-energy of each mode is given by the sign of its co-moving frequency (see Appendix \[app:asymptotic\]). When a mode with positive co-moving frequency scatters into a mode with negative co-moving frequency, the total norm can only be conserved if the original mode *increases* its norm. There is thus an amplification of the wave excitation in the system. This occurs with conservation of pseudo-energy, but in the real physical system the energy of the wave has increased while energy is removed from the fluid motion (this corresponds to evaporation of black holes [@haw74; @bro95]). A complete treatment of the physics involved would require a full account of back-reaction on the fluid together with the resulting energy transfer from the flow to the wave, but in (\[wave\]) the fluid appears simply as an external field $v(x)$. In actual experiments the amplification effect is so small that ignoring back-reaction in theoretical predictions is justified.
In order for the notion of scattering coefficients to make sense it is necessary for the waves in the asymptotic regions $|x|\to\infty$ to reduce to non-interacting superpositions of waves associated with the four roots of the dispersion relation and their corresponding rays. In discussing the “modes" labeled 1 to 4 above, we have implicitly assumed the validity of such a picture. If the flow has constant velocity in the asymptotic regions $|x|\to\infty$, it is clear the waves will asymptotically become superpositions of non-interacting plane waves given by the dispersion relation. For the linear profile (\[flow\]), however, it is not immediately obvious that the breakdown of geometrical optics is confined to the horizon regions, since the profile has the same rate of change at all points $x$ and there are rays whose wavelengths grow longer as $|x|$ increases (mode 2 on the left and mode 1 on the right). Nevertheless, waves in the linear profile in the regions $|x|\to\infty$ do reduce to non-interacting superpositions whose components are associated with the roots of the dispersion relation and their corresponding rays. This is shown in Appendix \[app:asymptotic\], where it is also found that the low-$k$ asymptotic modes do not have the same dependence on $v(x)$ as they do for asymptotically constant flows. (This last fact contrasts with the optical and quantum mechanical case, where the WKB solutions for the linear profile have exactly the same dependence on the profile function as for asymptotically constant profiles [@heading].) The flow regions on the far left and far right are thus places where we can legitimately speak of input and output waves and compute well-defined scattering coefficients.
Solution of the wave equation in $k$-space {#sec:ksol}
==========================================
We express the monochromatic wave $\phi(x)$ by a Fourier representation $$\label{four}
\phi(x)=\int_C dk \, \tilde{\varphi}(k) e^{ik x},$$ where we allow any contour $C$ in the complex $k$-plane such that the integral converges and $\tilde{\varphi}(k) e^{ik x}$ vanishes at the endpoints (in practice our contours will run to infinity in the complex plane). The wave equation (\[mono\]) for the linear profile (\[flow\]) then gives $$\begin{aligned}
\left[k^2\left(1+\frac{k^2}{k_c^2}\right)-i\alpha\omega-\omega^2 \right] \tilde{\varphi}(k) & + 2k\alpha(\alpha-i\omega)\tilde{\varphi}'(k) \nonumber \\
& +k^2\alpha^2\tilde{\varphi}''(k)=0,\end{aligned}$$ where a prime denotes a derivative. Extracting a factor through $\tilde{\varphi}(k)=k^{-1+i\omega/\alpha}\exp[-ik^2/(2\alpha k_c)]f(k)$ and rescaling the variable $k$ leads to a differential equation of the well-studied form $$f''(z)-2 z f'(z)+af(z)=0.$$ We thereby obtain the general solution $$\begin{aligned}
\tilde{\varphi}(k)=k^{-1+\frac{i\omega}{\alpha}} & \exp\left(- \frac{ik^2}{2\alpha k_c} \right) \left[ c_1 H_{-\frac{1}{2}-\frac{ik_c}{2\alpha}}\left(\sqrt{\frac{i}{\alpha k_c}} \, k \right) \right. \nonumber \\
& \qquad \left. + c_2\,_1\!F_1\left(\frac{1}{4}+\frac{ik_c}{4\alpha};\frac{1}{2};\frac{ik^2}{\alpha k_c} \right) \right], \label{ksol}\end{aligned}$$ where $H_a(z)$ is the Hermite function (which is a polynomial for non-negative integer $a$), $_1\!F_1(a;b;z)$ is the Kummer confluent hypergeometric function, and $\{c_1,c_2\}$ are arbitrary constants.
Through (\[four\]) and (\[ksol\]) we now have the general solution for the wave $\phi(x)$, expressed as an integral representation. Although there are only two linearly independent solutions in (\[ksol\]) for $ \tilde{\varphi}(k)$, there are four independent solutions for $\phi(x)$ because of the freedom to choose a branch cut in the integrand in (\[four\]). We choose two independent solutions for $ \tilde{\varphi}(k)$ by taking just the Hermite term, or just the hypergeometric term, in (\[ksol\]). For each of these $ \tilde{\varphi}(k)$ we will obtain two independent solutions for $\phi(x)$ by the choice of the contour and branch cut in (\[four\]).
Four independent solutions of the wave equation {#sec:4sols}
===============================================
Both the Hermite function $H_a(z)$ and the confluent hypergeometric function $_1\!F_1(a;b;z)$ are entire functions of $z$ [@dlmf]. The $k$-space solution (\[ksol\]) therefore has one singularity and one branch cut in the complex $k$-plane due to the factor $k^{-1+\frac{i\omega}{\alpha}}$. In the contour integral (\[four\]), which gives us the exact solutions to the wave equation, we must avoid the singularity at $k=0$ and negotiate the branch cut. We will choose the branch cut to run either along the positive or the negative imaginary $k$-axis. Then we choose the contour in (\[four\]) to run from $k=-\infty$ to $k=\infty$, as in the usual Fourier transform, but avoiding $k=0$ by running below $k=0$ (branch cut along positive imaginary axis) or above $k=0$ (branch cut along negative imaginary axis). These two choices of branch cut and contour will give two independent solutions for $\phi(x)$.
Two solutions with $H_a(z)$ in their integral representations
-------------------------------------------------------------
With $c_1=1$ and $c_2=0$ in (\[ksol\]), the integrand in (\[four\]) contains the Hermite function $H_a(z)$. This integrand is plotted in the complex $k$-plane in Figs. \[fig:Hermpos\] and \[fig:Hermneg\], with the branch cut running along the positive or negative imaginary axis, respectively. Also shown in each figure is a contour that that defines a solution (\[four\]). The contours are as described above and labeled $C'$ (branch cut along positive imaginary axis) or $C''$ (branch cut along negative imaginary axis).
![The integrand in (\[four\]) plotted in the complex $k$-plane, with $c_1=1$ and $c_2=0$ in (\[ksol\]). Colour indicates the phase while brightness indicates the absolute value (brighter means bigger). The branch cut is placed along the positive imaginary axis. Parameter values are $\omega=1$, $\alpha=1/2$ and $k_c=5$. The top plot is for $x=4$, the bottom for $x=-4$. We take $C'$ as the contour in (\[four\]) and thereby define an exact solution of (\[mono\]). []{data-label="fig:Hermpos"}](figure3a.pdf "fig:"){width="\linewidth"} ![The integrand in (\[four\]) plotted in the complex $k$-plane, with $c_1=1$ and $c_2=0$ in (\[ksol\]). Colour indicates the phase while brightness indicates the absolute value (brighter means bigger). The branch cut is placed along the positive imaginary axis. Parameter values are $\omega=1$, $\alpha=1/2$ and $k_c=5$. The top plot is for $x=4$, the bottom for $x=-4$. We take $C'$ as the contour in (\[four\]) and thereby define an exact solution of (\[mono\]). []{data-label="fig:Hermpos"}](figure3b.pdf "fig:"){width="\linewidth"}
![The same as Fig. \[fig:Hermpos\], but with the branch cut along the negative imaginary axis and a different contour $C''$. This defines another independent solution of (\[mono\]). []{data-label="fig:Hermneg"}](figure4a.pdf "fig:"){width="\linewidth"} ![The same as Fig. \[fig:Hermpos\], but with the branch cut along the negative imaginary axis and a different contour $C''$. This defines another independent solution of (\[mono\]). []{data-label="fig:Hermneg"}](figure4b.pdf "fig:"){width="\linewidth"}
Our main goal is to quantify the wave scattering in the flow and for this we need the asymptotic expansions of the exact solutions (\[four\]) in the large $|x|$ regions. Every exact solution must reduce as $|x|\to\infty$ to a superposition of the asymptotic wave components derived in Appendix \[app:asymptotic\]. These asymptotic waves do not interact and correspond to the four ray solutions of Sec. \[sec:wave\], propagating in the large $|x|$ regions.
If the contour in (\[four\]) ran along the real $k$-axis we would have a Fourier transform. For large $|x|$, the rapid oscillation in $k$ of the $e^{ikx}$ factor tends to make the Fourier transform zero as $|x|\to\infty$, provided the function being transformed is well behaved. The Riemann-Lebesgue lemma gives the technical requirements [@titchmarsch]. In our case there is a singularity at $k=0$ and we do not integrate along the real $k$-axis, so we need not expect the integral (\[four\]) to vanish as $|x|\to\infty$. Indeed, we know that the low-$k$ asymptotic wave components do not vanish as $|x|\to\infty$ (see Appendix \[app:asymptotic\]). Hence the integral (\[four\]) will be nonzero as $x\to-\infty$ ($x\to\infty$) if and only if the solution $\phi(x)$ contains any of the low-$k$ asymptotic wave components in the far left (far right) regions. The high-$k$ asymptotic wave components, on the other hand, vanish as $|x|\to\infty$ (see Appendix \[app:asymptotic\]) and we must also extract these components from the solution (\[four\]). We thus require not simply the leading-order part of the solution (\[four\]) for large $|x|$, but rather the leading-order contributions to all asymptotic wave components that are present.
To find the asymptotic expansions for $x\to\pm\infty$ of the integral (\[four\]), we must consider the behaviour of the integrand along the contour. The contours $C'$ and $C''$ in Figs. \[fig:Hermpos\] and \[fig:Hermneg\] run mostly along the real $k$-axis. For large $|x|$, the rapid oscillation of $e^{ikx}$ gives net cancellation for the parts of the contours on the real axis and so their contribution to the integral will vanish as $|x|\to\infty$. The leading-order term in these contributions is determined by whether or not there are points of stationary phase in the integrand that lie close to the contour. One can see two patches in the top plots in Figs. \[fig:Hermpos\] and \[fig:Hermneg\] where the phase has an extremum in the complex plane, while in the lower plots the integrand appears to be the sum of two parts, one with an extremum of the phase that is situated to the left and another part that has no extrema. Moreover all these extrema move out along the real $k$-axis as $|x|$ increases, so to investigate them analytically we must consider the form of the integrand for large $|k|$. We will find below that there are terms in the integrand with points of stationary phase that are precisely the points we have just identified visually. Moreover each of these stationary-phase points is the wave-vector of one of the high-$k$ asymptotic waves and the resulting contributions to the integral will give all the high-$k$ asymptotic wave components. There remains the parts of the contours $C'$ and $C''$ that do not lie on the real $k$-axis. For $x>0$ the factor $e^{ikx}$ decreases exponentially along the positive imaginary $k$-axis, whereas for $x<0$ it decreases along the negative imaginary axis. All other factors in the integrand are dominated by this behaviour, as is visible in Figs. \[fig:Hermpos\] and \[fig:Hermneg\]. We can deform the parts of the contours $C'$ and $C''$ that leave the real axis so that as much as possible they lie along the half of the imaginary axis where the integrand is exponentially small. If there is no branch cut along the relevant half of the imaginary axis (as in the bottom plot of Fig. \[fig:Hermpos\] and the top plot of Fig. \[fig:Hermneg\]), then the contribution of this part of the contour is arbitrarily small. If however the branch cut lies along the half of the imaginary axis where the integrand is exponentially decreasing (as in the top plot of Fig. \[fig:Hermpos\] and the bottom plot of Fig. \[fig:Hermneg\]), then the contour gets wrapped around $k=0$ and runs along both sides of the branch cut. The latter situation will give a contribution to the integral that does not vanish as $|x|\to\infty$ and corresponds to low-$k$ asymptotic wave components present in the solution. This outlines how the asymptotic wave components are encoded in the exact integral representation (\[four\]). A similar identification of wave components occurs in [@unr05], where the techniques used here were applied to an approximate treatment of the wave equation in an arbitrary flow profile.
To identify the points of stationary phase discussed above, we require the asymptotic expansions for $k\gg 0$ and $k\ll 0$ of the Hermite function that appears in (\[ksol\]). Asymptotic expansions of the Hermite function for large variable are given in [@dlmf]; in our case the variable is $\sqrt{i/(\alpha k_c)}k$, giving the following leading-order expansions:
$$\begin{aligned}
H_{-\frac{1}{2}-\frac{ik_c}{2\alpha}}\left(\sqrt{\frac{i}{\alpha k_c}}k \right) \sim & \, 2^{-\frac{1}{4}-\frac{ik_c}{4\alpha}} \left( \sqrt{\frac{2i}{\alpha k_c}} \, k \right)^{ -\frac{1}{2}-\frac{ik_c}{2\alpha} }, \quad -\pi < \text{arg}(k) <\frac{\pi}{2}, \label{Hposk} \\
H_{-\frac{1}{2}-\frac{ik_c}{2\alpha}}\left(\sqrt{\frac{i}{\alpha k_c}}k \right) \sim & \, 2^{-\frac{1}{4}-\frac{ik_c}{4\alpha}} \left[ \left( \sqrt{\frac{2i}{\alpha k_c}} \, k \right)^{ -\frac{1}{2}-\frac{ik_c}{2\alpha} } \right.
\nonumber \\
& \left. - \frac{i \sqrt{2\pi}}{\Gamma\left( \frac{1}{2}+\frac{ik_c}{2\alpha} \right)} \left( \sqrt{\frac{2i}{\alpha k_c}} \, k \right)^{ -\frac{1}{2}+\frac{ik_c}{2\alpha} } \exp\left( \frac{i k^2}{\alpha k_c} - \frac{\pi k_c}{2\alpha} \right) \right] , \quad -2\pi < \text{arg}(k) < -\frac{\pi}{2}. \label{Hnegk}\end{aligned}$$
Here the branch cuts in the asymptotic expressions are rotated out of the range of $\arg(k)$ for which they are valid. This amounts to analytically continuing $\ln z$ from its principal branch $-\pi<\arg(z)\leq \pi$.
The points of stationary phase in the integrand in (\[four\]) can now be identified (where here we take just the Hermite-function term in (\[ksol\])). In the region where (\[Hposk\]) is valid, which includes the positive real $k$-axis, the integrand has an exponential factor containing $k$ given by $\exp\left[ikx- ik^2/(2\alpha k_c) \right]$. This factor has a point of stationary phase at $$\label{stat1}
k=\alpha k_c x, \qquad x>>0,$$ where the condition on $x$ occurs because (\[Hposk\]) is not valid for negative real $k$. This point of stationary phase is visible on the right of the top plots in Figs. \[fig:Hermpos\] and \[fig:Hermneg\]. In the region where (\[Hnegk\]) is valid, which includes the negative real $k$-axis, we have two terms in the asymptotic expansion of the Hermite function, the second of which has an exponential involving $k$. We must therefore treat the two terms in (\[Hnegk\]) separately as they have different phase factors. The first term in (\[Hnegk\]) gives a contribution to (\[four\]) that has a phase factor $\exp\left[ikx- ik^2/(2\alpha k_c) \right]$, giving a point of stationary phase at $$\label{stat2}
k=\alpha k_c x, \qquad x<<0,$$ where now the condition is for $x$ to be negative because (\[Hnegk\]) is not valid on the positive real $k$-axis. The second term in (\[Hnegk\]) leads to a contribution to (\[four\]) with a phase factor $\exp\left[ikx+ ik^2/(2\alpha k_c) \right]$, with stationary phase at $$\label{stat3}
k= - \alpha k_c x, \qquad x>>0.$$ The significance of the point (\[stat2\]) is partly visible on the left of the lower plots in Figs. \[fig:Hermpos\] and \[fig:Hermneg\] but the integrand also has the contribution of the second term in (\[Hnegk\]), for which there are no points of stationary phase for negative $x$. The point (\[stat3\]) is similarly a stationary-phase point for only one term in the integrand but its significance is nevertheless apparent on the left of the top plots in Figs. \[fig:Hermpos\] and \[fig:Hermneg\].
The points (\[stat1\])–(\[stat3\]) lie on the contours $C'$ and $C''$. For large $|x|$, we use the method of steepest descent to compute the the leading-order contribution to the integral of the terms which have a stationary phase at these points. Note that the points (\[stat1\])–(\[stat3\]) are the local wave-vectors of asymptotic wave components (see Appendix \[app:asymptotic\]) whose amplitudes fall off as $1/|x|^{3/2}$. We will find that the stationary-phase contributions give exactly these asymptotic wave components in a particular superposition. Contributions to this part of the integral that are not stationary-phase contributions fall off faster in $|x|$ than the stationary-phase contributions and therefore they are not needed to compute the asymptotic wave components. In the method of steepest descent [@ablowitz] we expand the phase around the stationary point to second order and find the directions in the complex plane in which the quadratic term in the expansion is real and negative. We deform the contour to run along this line of steepest descent. Other factors in the integrand are evaluated at the stationary point and the resulting Gaussian integral along the steepest-descent line gives the leading-order contribution to the integral. In our case the phase factors are $\exp\left[ikx\mp ik^2/(2\alpha k_c) \right]$, which are simple to work with and give steepest-descent lines running through the stationary-phase points at angles of $\pm45^\circ$. When evaluating the other factors in the integrand at the stationary-phase point we must remember to choose the branch cut in the $k^{-1 + \frac{i\omega}{\alpha}}$ factor from (\[ksol\]) to lie on the positive or negative imaginary axis, depending on which of the two solutions we are evaluating. This will complete the leading-order contributions to the solution from the parts of the contours $C'$ and $C''$ that lie on the real $k$-axis. There remains the portions of the contours that run around the singularity at $k=0$.
### First solution: branch cut along positive imaginary $k$-axis
The branch cut in the integrand is first chosen to lie along the positive imaginary $k$-axis (Fig. \[fig:Hermpos\]) and the solution is defined by the contour $C'$. We denote this solution by $\phi^{(a)}(x)$. There are two expansions of $\phi^{(a)}(x)$ in terms of the asymptotic waves, one for $x\ll 0$ and one for $x\gg 0$.
Taking first $x\ll 0$, we see from (\[stat1\])–(\[stat3\]) that there is just one stationary-phase contribution, from the point (\[stat2\]). This point is the wave-vector of mode 1 in the far-left region, whose normalized asymptotic form is $\phi_1^-(x)$ given by (\[-1\]). The part of the contour $C'$ that leaves the real $k$-axis can be pushed down along the negative imaginary $k$-axis (see bottom plot in Fig. \[fig:Hermpos\]) where the integrand is exponentially small and gives no contribution. The only asymptotic wave component for $x\ll 0$ is thus mode 1 with a constant complex amplitude that we obtain from the method of steepest descent. The expansion of the solution is $$\begin{gathered}
\phi^{(a)}(x) \stackrel{x\ll 0}{\sim} a_1^- \phi_1^-(x) , \label{H1left} \\
a_1^-=(-1)^\frac{1}{8}\sqrt{\pi}2^{\frac{1}{2}-\frac{ik_c}{2\alpha}}e^{\frac{\pi\omega}{\alpha}-\frac{3\pi k_c}{8\alpha}}\alpha (\alpha k_c)^{\frac{i\omega}{\alpha}-\frac{ik_c}{4\alpha}-\frac{1}{4}}. \label{a1-}\end{gathered}$$
![A portion of the upper plot in Fig. \[fig:Hermpos\], with the contour $C'$ deformed to wrap around the singularity at $k=0$. The integrand decays exponentially along the positive imaginary axis. For the portion of the contour shown here, only the part running along both sides of the branch cut and around the singularity contributes to the integral for large $x$. []{data-label="fig:branch"}](figure5.pdf){width="7cm"}
For $x\gg 0$ we will find contributions from all four asymptotic wave components and the expansion of the solution is $$\label{H1right}
\phi^{(a)}(x) \stackrel{x\gg 0}{\sim} a_1^+ \phi_1^+(x) +a_2^+ \phi_2^+(x) +a_3^+ \phi_3^+(x) +a_4^+ \phi_4^+(x),$$ for constant coefficients $a_n^+$. For large positive $x$ there are contributions from the two stationary-phase points (\[stat1\]) and (\[stat3\]). These points are the local wave-vectors of modes 2 and 3, respectively, in the far-right region, where they have the normalized asymptotic expressions (\[+2\]) and (\[+3\]). The steepest descent method gives the coefficients $a_2^+$ and $a_3^+$ of these modes: $$\begin{aligned}
a_2^+= & (-1)^\frac{13}{8} \sqrt{\pi} 2^{\frac{1}{2}-\frac{ik_c}{2\alpha}} e^{\frac{\pi k_c}{8\alpha}} \alpha (\alpha k_c)^{\frac{i\omega}{\alpha}-\frac{ik_c}{4\alpha}-\frac{1}{4}}, \label{a2+} \\
a_3^+= & \frac{ (-1)^\frac{9}{8} 2 \pi e^{\frac{\pi\omega}{\alpha} - \frac{\pi k_c}{8\alpha}} \alpha (\alpha k_c)^{\frac{i\omega}{\alpha}+\frac{ik_c}{4\alpha}-\frac{1}{4}} }{\Gamma \left( \frac{1}{2}+\frac{ik_c}{2\alpha} \right) } . \label{a3+} \end{aligned}$$ There is also a contribution from the part of the contour $C'$ that runs around the singularity at $k=0$. For positive $x$ the integrand exponentially decreases along the positive imaginary $k$-axis (see top plot in Fig. \[fig:Hermpos\]) and we move the contour upwards so that it runs along both sides of the branch cut, as shown in Fig. \[fig:branch\]. The exponential decrease along the positive imaginary $k$-axis is very rapid for large $x$ so only the part of the contour in Fig. \[fig:branch\] close to $k=0$ is significant for the asymptotic integral. We can therefore compute this part of the integral by expanding the integrand around $k=0$. The Taylor expansion of the Hermite-function factor around $k=0$ is $$\begin{aligned}
H_{-\frac{1}{2}-\frac{ik_c}{2\alpha}} & \left(\sqrt{\frac{i}{\alpha k_c}} \, k \right) = \frac{\sqrt{\pi} 2^{-\frac{1}{2}-\frac{ik_c}{2\alpha}} }{\Gamma \left( \frac{3}{4}+\frac{ik_c}{4\alpha} \right) } \nonumber \\
& \qquad\qquad\quad - \frac{\sqrt{i\pi} 2^{\frac{1}{2}-\frac{ik_c}{2\alpha}} k}{\sqrt{\alpha k_c}\,\Gamma \left( \frac{1}{4}+\frac{ik_c}{4\alpha} \right) } +O(k^2). \label{Htaylor}\end{aligned}$$ Using the first two terms in this series we compute the integral along the contour in Fig. \[fig:branch\] for $x\gg 0$, where we can take the ends of the contour to run to positive imaginary infinity along both sides of the branch cut. We make the substitution $k=i s/x$ in the integral and expand the integrand to find the two leading-order terms for large $x$. This gives the integral $$\begin{aligned}
\sqrt{\pi} 2^{-\frac{1}{2}-\frac{ik_c}{2\alpha}} \int ds \,\left(\frac{is}{x}\right)^{\frac{i\omega}{\alpha}}
& e^{-s} \left[ \frac{1}{s \, \Gamma \left( \frac{3}{4}+\frac{ik_c}{4\alpha} \right) } \right. \nonumber \\
& \left. + \frac{2\sqrt{-i}}{x \sqrt{\alpha k_c}\, \Gamma \left( \frac{1}{4}+\frac{ik_c}{4\alpha} \right) } \right] , \label{sintH}\end{aligned}$$ where our original choice of contour and branch cut in the complex $k$-plane means that in (\[sintH\]) the branch cut in the integrand runs along the positive real $s$-axis and the contour runs in from infinity above the branch cut, around $s=0$, then out to infinity below the branch cut. Both terms in the integral (\[sintH\]) are related to Hankel’s integral representation of the gamma function [@whittaker], which states $$\label{Han}
\Gamma(z)=\frac{i}{2\sin(\pi z)} \int_{C_\Gamma} dt \, (-t)^{z-1} e^{-t}, \quad z\notin \text{Integers},$$ where the branch cut in the integrand is along the positive real $t$-axis and the contour $C_\Gamma$ runs in from infinity above the branch cut, around $t=0$, then out to infinity below the branch cut. The branch cut of the log function is chosen differently in (\[sintH\]) compared to (\[Han\]), and we must bear this in mind when employing the latter. The result for (\[sintH\]) follows from (\[Han\]): $$\begin{aligned}
-\sqrt{\pi} 2^{\frac{1}{2}-\frac{ik_c}{2\alpha}} e^{\frac{\pi \omega}{2 \alpha}} \Gamma \left( \frac{i \omega}{\alpha} \right) & \sinh\left( \frac{\pi \omega}{\alpha} \right) x^{-\frac{i\omega}{\alpha}}
\left[ \frac{1}{ \Gamma \left( \frac{3}{4}+\frac{ik_c}{4\alpha} \right) } \right. \nonumber \\
& \left. + \frac{2\sqrt{i} \,\omega}{x \alpha \sqrt{\alpha k_c}\, \Gamma \left( \frac{1}{4}+\frac{ik_c}{4\alpha} \right) } \right] . \label{Hlowk1}\end{aligned}$$ Referring to (\[+1\]) and (\[+4\]), we see that (\[Hlowk1\]) is a superposition of the two low-$k$ asymptotic wave components on the far right (modes 1 and 4). Solving for the coefficients in this superposition we find $a_1^+$ and $a_4^+$ in (\[H1right\]): $$\begin{aligned}
a_1^+ = - & \frac{ \sqrt{\omega} \, e^{\frac{\pi \omega}{2 \alpha}} \Gamma \left( \frac{i \omega}{\alpha} \right) \sinh\left( \frac{\pi \omega}{\alpha} \right) }{ \sqrt{2} \, \Gamma \left( \frac{1}{2}+ \frac{i k_c}{2 \alpha} \right) }
\left[ \Gamma \left( \frac{1}{4}+\frac{ik_c}{4\alpha} \right) \right. \nonumber \\
& \qquad\qquad\qquad \left. + 2i \sqrt{\frac{i \alpha}{ k_c}}\ \Gamma \left( \frac{3}{4}+\frac{ik_c}{4\alpha} \right) \right] , \label{a1+} \\
a_4^+ = - & \frac{ \sqrt{\omega} \, e^{\frac{\pi \omega}{2 \alpha}} \Gamma \left( \frac{i \omega}{\alpha} \right) \sinh\left( \frac{\pi \omega}{\alpha} \right) }{ \sqrt{2} \, \Gamma \left( \frac{1}{2}+ \frac{i k_c}{2 \alpha} \right) }
\left[ \Gamma \left( \frac{1}{4}+\frac{ik_c}{4\alpha} \right) \right. \nonumber \\
& \qquad\qquad\qquad \left. - 2i \sqrt{\frac{i \alpha}{ k_c}}\ \Gamma \left( \frac{3}{4}+\frac{ik_c}{4\alpha} \right) \right] . \label{a4+}\end{aligned}$$ Here the expressions have been simplified a little by use of Legendre’s duplication formula [@htf1] for the gamma function, which gives $$\Gamma \left( \frac{1}{2}+\frac{ik_c}{2\alpha} \right) = \frac{ 2^{- \frac{1}{2}+\frac{ik_c}{2\alpha}} }{ \sqrt{\pi} } \Gamma \left( \frac{1}{4}+\frac{ik_c}{4\alpha} \right) \Gamma \left( \frac{3}{4}+\frac{ik_c}{4\alpha} \right) .$$
This completes the decomposition of the solution $\phi^{(a)}(x)$ into the asymptotic modes on both sides of the flow. The content of this solution in terms of mode scattering can be seen from (\[H1left\]) and (\[H1right\]), together with the lower ray plot in Fig. \[fig:rays\]. The input wave is incident from the right and is a superposition of the low-$k$ modes 1 and 4 with different amplitudes (\[a1+\]) and (\[a4+\]) (see Fig. \[fig:rays\]). Mode 1 propagates through the horizon regions and continues to the left, while mode 4 reverses its group velocity and is transformed to mode 3, as in the ray plots in Fig. \[fig:rays\]. In addition, there is scattering of the input modes 1 and 4 into mode 2 on the right. In fact there is also scattering of input mode 1 into mode 3 on the right and of input mode 4 into mode 1 on the left, but this will become clearer when we calculate the scattering coefficients. Note that there is no scattering into mode 4 on the left. Although both of the incident modes 1 and 4 from the right scatter into mode 4 on the left, the weighting of the incident modes in this solution causes cancellation of the scattering into mode 4 on the left.
The norm flux of any monochromatic solution is constant throughout the flow (see Appendix \[app:conserved\]). It must therefore be the case that the norm flux of the asymptotic wave decomposition (\[H1left\]) on the far left of the flow is equal to that of the decomposition (\[H1right\]) on the far right. As discussed in Appendix \[app:asymptotic\], the norm flux of the normalized asymptotic wave components is $\pm1$, with a sign given by the product of the signs of the co-moving frequency and the group velocity. Also, a superposition of asymptotic wave components has a norm flux which is the sum of the fluxes of the individual components. For the solution (\[H1left\]) and (\[H1right\]), constancy of the norm flux therefore implies $$-\left| a_1^- \right|^2 = - \left| a_1^+ \right|^2 + \left| a_2^+ \right|^2 - \left| a_3^+ \right|^2 + \left| a_4^+ \right|^2.$$ One can verify that this relation indeed holds for the coefficients (\[a1-\]), (\[a2+\]), (\[a3+\]), (\[a1+\]), and (\[a4+\]). The calculation of the absolute values of the coefficients requires the gamma-function identities [@htf1] $$\begin{aligned}
\Gamma(z)\Gamma(-z) = & \, - \frac{\pi}{z\sin(\pi z)}, \label{gammaid1} \\
\Gamma(z)\Gamma(1-z) = & \, \frac{\pi}{\sin(\pi z)}. \label{gammaid2}\end{aligned}$$
### Second solution: branch cut along negative imaginary $k$-axis
We now take the branch cut in the integrand to lie along the negative imaginary $k$-axis (Fig. \[fig:Hermneg\]), with the solution defined by the contour $C''$. We denote this solution by $\phi^{(b)}(x)$ and calculate its two expansions in terms of the asymptotic wave components, one for $x\ll 0$ and one for $x\gg 0$.
For $x\ll 0$ we have a stationary-phase contribution to the integral from the point (\[stat2\]), as in the first solution. But now there is also a branch-cut contribution because in deforming the contour $C''$ into the region along the negative imaginary axis (where the integrand decreases exponentially) it gets wrapped around the singularity at $k=0$. This last contribution will give low-$k$ wave components corresponding to modes 2 and 3 on the far left. The decomposition of $\phi^{(b)}(x)$ for $x\ll 0$ will consequently have three asymptotic wave components: $$\label{H2left}
\phi^{(b)}(x) \stackrel{x\ll 0}{\sim} b_1^- \phi_1^-(x) +b_2^- \phi_2^-(x) +b_3^- \phi_3^-(x).$$ The calculation of $b_1^-$ by the steepest descent method is almost identical to that of $a_1^-$ in the first solution. The different position of the branch cut in the two cases means that $b_1^-$ differs from $a_1^-$ by a simple factor: $$\label{b1-}
b_1^- = e^{-\frac{2\pi\omega}{\alpha}} a_1^-.$$ The branch-cut contribution gives the coefficients $b_2^-$ and $b_3^-$, and the calculation here has only minor differences from that represented in Fig. \[fig:branch\]. We again use the Taylor expansion (\[Htaylor\]) of the Hermite function and Hankel’s integral (\[Han\]). The final results are simply related to the low-$k$ coefficients (\[a1+\]) and (\[a4+\]) in the previous solution: $$\label{b2-}
b_2^- = - e^{-\frac{\pi\omega}{\alpha}} a_4^+, \qquad b_3^- = - e^{-\frac{\pi\omega}{\alpha}} a_1^+.$$
For $x\gg 0$ there is no branch-cut contribution to the integral as the part of the contour $C''$ near $k=0$ can be moved upwards along the positive imaginary axis where the integrand is exponentially small (see Fig. \[fig:Hermneg\]). There are stationary-phase contributions from the points (\[stat1\]) and (\[stat3\]), as in the previous solution, but here the new position of the branch cut in the integrand gives some minor differences. We obtain an expansion $$\label{H2right}
\phi^{(b)}(x) \stackrel{x\gg 0}{\sim} b_2^+ \phi_2^+(x) +b_3^+ \phi_3^+(x)$$ in which the coefficients $b_2^+$ and $b_3^+$ are simply related to (\[a2+\]) and (\[a3+\]): $$\label{b2+}
b_2^+ = a_2^+, \qquad b_3^+ = e^{-\frac{2\pi\omega}{\alpha}} a_3^+.$$
The expansions (\[H2left\]) and (\[H2right\]) reveal how the modes scatter in the solution $\phi^{(b)}(x)$. Referring to Fig. \[fig:rays\], we see that the incident wave is a superposition of modes 2 and 3 from the left. Mode 2 propagates through into the right-hand region and there is also scattering into mode 3 on the right and mode 1 on the left. Note that there is no mode-4 component on the left even though the incident mode-3 component on the left would normally convert to a blue-shifted mode-4 wave on the same side. As in the solution $\phi^{(a)}(x)$, here the different contributions to mode 4 on the left exactly cancel.
The constancy of the norm flux for $\phi^{(b)}(x)$ implies the following relation between the “$b$" coefficients for the modes on the left and the right: $$-\left| b_1^- \right|^2 + \left| b_2^- \right|^2 - \left| b_3^- \right|^2 = \left| b_2^+ \right|^2 - \left| b_3^+ \right|^2.$$ The coefficients (\[b1-\]), (\[b2-\]), and (\[b2+\]) indeed satisfy this relation.
Two solutions with $_1\!F_1(a;b;z)$ in their integral representations
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We now take $c_1=0$ and $c_2=1$ in (\[ksol\]), so that the integrand in (\[four\]) contains the confluent hypergeometric function $_1\!F_1(a;b;z)$. This integrand is plotted in the complex $k$-plane in Figs. \[fig:Hyperpos\] and \[fig:Hyperneg\], with the branch cut running along the positive or negative imaginary axis, respectively. A solution of (\[mono\]) is obtained for each choice of branch cut and corresponding contour $C'$ or $C''$. We proceed to find the expansions of these two solutions in terms of the asymptotic waves. The derivation follows exactly the steps described above when the integrand contained the Hermite function.
![The integrand in (\[four\]) plotted in the complex $k$-plane, with $c_1=0$ and $c_2=1$ in (\[ksol\]). Colour indicates the phase while brightness indicates the absolute value (brighter is bigger). The branch cut is placed along the positive imaginary axis. Parameter values are $\omega=1$, $\alpha=1/2$ and $k_c=5$. The top plot is for $x=4$, the bottom for $x=-4$. With $C'$ as the contour in (\[four\]) this defines an exact solution of (\[mono\]). []{data-label="fig:Hyperpos"}](figure6a.pdf "fig:"){width="\linewidth"} ![The integrand in (\[four\]) plotted in the complex $k$-plane, with $c_1=0$ and $c_2=1$ in (\[ksol\]). Colour indicates the phase while brightness indicates the absolute value (brighter is bigger). The branch cut is placed along the positive imaginary axis. Parameter values are $\omega=1$, $\alpha=1/2$ and $k_c=5$. The top plot is for $x=4$, the bottom for $x=-4$. With $C'$ as the contour in (\[four\]) this defines an exact solution of (\[mono\]). []{data-label="fig:Hyperpos"}](figure6b.pdf "fig:"){width="\linewidth"}
![The same as Fig. \[fig:Hyperpos\], but with the branch cut along the negative imaginary axis and a different contour $C''$. This defines another independent solution of (\[mono\]). []{data-label="fig:Hyperneg"}](figure7a.pdf "fig:"){width="\linewidth"} ![The same as Fig. \[fig:Hyperpos\], but with the branch cut along the negative imaginary axis and a different contour $C''$. This defines another independent solution of (\[mono\]). []{data-label="fig:Hyperneg"}](figure7b.pdf "fig:"){width="\linewidth"}
Figures \[fig:Hyperpos\] and \[fig:Hyperneg\] indicate that there are points of stationary phase on the real $k$-axis, but only for parts of the integrand. In the region near the positive real $k$-axis, the integrand appears to be the sum of a part with a stationary-phase point and a part with no extrema of the phase, and similarly for the region near the negative real $k$-axis. These stationary-phase features in the plots are present for both positive and negative $x$ and they move out along the real $k$-axis as $|x|$ increases. We confirm this picture by looking at the asymptotic expansion of the hypergeometric function in (\[ksol\]) for $k\gg0$ and $k\ll0$. We refer to [@dlmf] for asymptotic expansions of the hypergeometric function for large variable. In our case the form $ik^2/(\alpha k_c)$ of the variable means there is an asymptotic expansion that is valid for both $k\gg0$ and $k\ll0$, which to leading order is $$\begin{aligned}
& _1\!F_1\left(\frac{1}{4}+\frac{ik_c}{4\alpha};\frac{1}{2};\frac{ik^2}{\alpha k_c} \right) \sim \frac{\sqrt{i\pi} \,e^{-\frac{\pi k_c}{4\alpha}}}{\Gamma\left( \frac{1}{4}-\frac{ik_c}{4\alpha} \right)} \left( \frac{ik^2}{\alpha k_c} \right)^{ -\frac{1}{4}-\frac{ik_c}{4\alpha} }
\nonumber \\
& \qquad \quad\ + \frac{\sqrt{\pi}}{\Gamma\left( \frac{1}{4}+\frac{ik_c}{4\alpha} \right)} \left( \frac{ik^2}{\alpha k_c} \right)^{ -\frac{1}{4}+\frac{ik_c}{4\alpha} } \exp\left( \frac{i k^2}{\alpha k_c} \right) , \nonumber \\
& \qquad\quad -\frac{\pi}{2} < \arg(k) < \frac{\pi}{2}\ \ \text{or}\ \ \frac{\pi}{2} < \arg(k) < \frac{3\pi}{2}. \label{Fk}\end{aligned}$$ The two terms in the expansion (\[Fk\]) give terms with different phase factors in the integrand in (\[four\]) (we take just the hypergeometric-function term in (\[ksol\])). The first term in the resulting integrand has a phase factor $\exp\left[ikx- ik^2/(2\alpha k_c) \right]$, which has a point of stationary phase at $$\label{stat4}
k=\alpha k_c x.$$ As (\[Fk\]) is valid for both $k\gg0$ and $k\ll0$, the first term in the integrand thus has a stationary-phase point (\[stat4\]) on the positive real $k$-axis for $x\gg0$ and on the negative real $k$-axis for $x\ll0$. The second term in (\[Fk\]) gives a term in the integrand in (\[four\]) with a phase factor $\exp\left[ikx+ ik^2/(2\alpha k_c) \right]$ and a point of stationary phase at $$\label{stat5}
k= -\alpha k_c x.$$ The second term in the integrand thus has a stationary-phase point (\[stat5\]) on the negative real $k$-axis for $x\gg0$ and on the positive real $k$-axis for $x\ll0$. These results confirm what is already apparent from the plots in Figs. \[fig:Hyperpos\] and \[fig:Hyperneg\].
The asymptotics (\[Fk\]) is not valid on the imaginary $k$-axis, and we have not given an expansion of the hypergeometric function valid in this region of the complex $k$-plane. We do not include it here because it does not give any points of stationary phase for terms in the integrand in (\[four\]).
The stationary-phase points (\[stat4\]) and (\[stat5\]) lie on both contours $C'$ and $C''$ and give contributions to the asymptotic expansions of the integral (\[four\]) for large $|x|$. These contributions are calculated by the method of steepest descent. There may also be a branch-cut contribution, which is evaluated as above for the solutions generated by the Hermite function.
### First solution: branch cut along positive imaginary $k$-axis
With the branch cut in the integrand chosen to lie along the positive imaginary $k$-axis (Fig. \[fig:Hyperpos\]) the solution is defined by the contour $C'$. We denote this solution by $\phi^{(c)}(x)$ and find its expansions in terms of the asymptotic waves for $x\ll 0$ and for $x\gg 0$.
For $x\ll0$ there are stationary-phase contributions from the points (\[stat4\]) and (\[stat5\]), which give asymptotic waves corresponding to modes 1 and 4, respectively. There is no branch-cut contribution and so the expansion is $$\phi^{(c)}(x) \stackrel{x\ll 0}{\sim} c_1^- \phi_1^-(x) + c_4^- \phi_4^-(x) . \label{F1left}$$ The steepest descent method gives the coefficients $c_1^-$ and $c_4^-$:
$$\begin{aligned}
c_1^-= & -\frac{2\pi (-1)^\frac{7}{8} e^{\frac{\pi\omega}{\alpha}-\frac{\pi k_c}{8\alpha}} \alpha (\alpha k_c)^{\frac{i\omega}{\alpha}-\frac{ik_c}{4\alpha}-\frac{1}{4}} }{ \Gamma\left( \frac{1}{4}-\frac{ik_c}{4\alpha} \right) } , \label{c1-} \\
c_4^-= & -\frac{2\pi (-1)^\frac{1}{8} e^{-\frac{\pi k_c}{8\alpha}} \alpha (\alpha k_c)^{\frac{i\omega}{\alpha}+\frac{ik_c}{4\alpha}-\frac{1}{4}} }{ \Gamma\left( \frac{1}{4}+\frac{ik_c}{4\alpha} \right) } . \label{c4-} \end{aligned}$$
For $x\gg 0$, the solution $\phi^{(c)}(x)$ has contributions from all four asymptotic wave components: $$\phi^{(c)}(x) \stackrel{x\gg 0}{\sim} c_1^+ \phi_1^+(x) + c_2^+ \phi_2^+(x) + c_3^+ \phi_3^+(x) + c_4^+ \phi_4^+(x) . \label{F1right}$$ The two stationary-phase points (\[stat4\]) and (\[stat5\]) give rise to the mode-2 and mode-3 waves on the far right, with coefficients $$\begin{aligned}
c_2^+= & -\frac{2\pi (-1)^\frac{7}{8} e^{-\frac{\pi k_c}{8\alpha}} \alpha (\alpha k_c)^{\frac{i\omega}{\alpha}-\frac{ik_c}{4\alpha}-\frac{1}{4}} }{ \Gamma\left( \frac{1}{4}-\frac{ik_c}{4\alpha} \right) } , \label{c2+} \\
c_3^+ = & -\frac{2\pi (-1)^\frac{1}{8} e^{\frac{\pi\omega}{\alpha} -\frac{\pi k_c}{8\alpha}} \alpha (\alpha k_c)^{\frac{i\omega}{\alpha}+\frac{ik_c}{4\alpha}-\frac{1}{4}} }{ \Gamma\left( \frac{1}{4}+\frac{ik_c}{4\alpha} \right) } . \label{c3+} \end{aligned}$$ There is also a branch-cut contribution, which gives the low-$k$ asymptotic wave components on the right (modes 1 and 4). The relevant contour is as in Fig. \[fig:branch\], but here the integrand is different. As before, we only require the integrand in the region near $k=0$ in order to evaluate the branch-cut contribution. We employ the Taylor expansion of the hypergeometric function around $k=0$: $$_1\!F_1\left(\frac{1}{4}+\frac{ik_c}{4\alpha};\frac{1}{2};\frac{ik^2}{\alpha k_c} \right) = 1- \frac{ (k_c-i\alpha)k^2 }{ 2\alpha^2k_c } +O(k^4). \label{Ftaylor}$$ As previously, we make the substitution $k=i s/x$ in the resulting integral and expand the integrand to find the first two leading-order terms for large $x$. This gives the integral $$\int ds \,\left(\frac{is}{x}\right)^{\frac{i\omega}{\alpha}}
e^{-s} \left[ \frac{1}{s }
+ \frac{ s }{ 2 x^2 \alpha^2 } \right] , \label{sintF}$$ where the branch cut in the integrand runs along the positive real $s$-axis and the contour runs in from infinity above the branch cut, around $s=0$, then out to infinity below the branch cut. Evaluation of (\[sintF\]) using Hankel’s integral (\[Han\]) produces $$\begin{aligned}
- 2 e^{\frac{\pi \omega}{2 \alpha}} \Gamma \left( \frac{i \omega}{\alpha} \right) & \sinh\left( \frac{\pi \omega}{\alpha} \right) x^{-\frac{i\omega}{\alpha}}
\left[ 1 + \frac{ i \omega ( \alpha +i \omega) }{2 x^2 \alpha^4 } \right] . \label{Flowk1}\end{aligned}$$ This is a superposition of the two low-$k$ asymptotic wave components (\[+1\]) and (\[+4\]), but note that (\[Flowk1\]) does not have a term falling off as $1/x$, whereas both (\[+1\]) and (\[+4\]) contain such a term. It must therefore be the case that (\[Flowk1\]) is an equal superposition of (\[+1\]) and (\[+4\]), so that the terms containing $x^{-i\omega/\alpha-1}$ cancel out. This would require the ratio of the two terms in (\[Flowk1\]) to be equal to the ratio of the first and third terms in (\[+1\]) and (\[+4\]), as is indeed the case. Solving for the coefficients $c_1^+$ and $c_4^+$ in the superposition we obtain $$c_1^+=c_4^+ = - \sqrt{2\omega} \, e^{\frac{\pi \omega}{2 \alpha}} \Gamma \left( \frac{i \omega}{\alpha} \right) \sinh\left( \frac{\pi \omega}{\alpha} \right) . \label{c1+}$$
The expansions (\[F1left\]) and (\[F1right\]), together with (\[c1+\]), show that the solution $\phi^{(c)}(x)$ is an incident wave from the right that is an equal superposition of modes 1 and 4 (see Fig. \[fig:rays\]). These incident modes scatter into all outgoing modes on the left and right. The “$c$" coefficients obey the relation that follows from constancy of the norm flux: $$-\left| c_1^- \right|^2+\left| c_4^- \right|^2 = - \left| c_1^+ \right|^2 + \left| c_2^+ \right|^2 - \left| c_3^+ \right|^2 + \left| c_4^+ \right|^2.$$
### Second solution: branch cut along negative imaginary $k$-axis
For the second solution generated by the hypergeometric function, we take the branch cut in the integrand to lie along the negative imaginary $k$-axis (Fig. \[fig:Hyperneg\]). The solution is defined by the contour $C''$ and we denote it by $\phi^{(d)}(x)$. The calculation of the expansions of the solution in terms of the asymptotic wave components has by now been well rehearsed and here we quote the results.
For $x\ll0$ the expansion contains all four asymptotic wave components: $$\label{F2left}
\phi^{(d)}(x) \stackrel{x\ll 0}{\sim} d_1^- \phi_1^-(x) +d_2^- \phi_2^-(x) +d_3^- \phi_3^-(x) +d_4^- \phi_4^-(x),$$ where the coefficients are given by $$d_1^- = e^{-\frac{2 \pi\omega}{\alpha}} c_1^-, \quad d_2^- = d_3^- = - e^{-\frac{\pi\omega}{\alpha}} c_1^+, \quad
d_4^- = c_4^-.$$ For $x\gg0$ the expansion has two wave components, for modes 2 and 3: $$\label{F2right}
\phi^{(d)}(x) \stackrel{x\gg 0}{\sim} d_2^+ \phi_2^+(x) +d_3^+ \phi_3^+(x) ,$$ with coefficients $$\begin{gathered}
d_2^+ = c_2^+, \qquad d_3^+ = e^{-\frac{2\pi\omega}{\alpha}} c_3^+ .\end{gathered}$$ The “$d$" coefficients obey the norm-flux constancy condition $$-\left| d_1^- \right|^2 + \left| d_2^- \right|^2 - \left| d_3^- \right|^2 + \left| d_4^- \right|^2 = \left| d_2^+ \right|^2 - \left| d_3^+ \right|^2.$$
The solution $\phi^{(d)}(x)$ is an incident wave from the left that is an equal superposition of modes 2 and 3 (see Fig. \[fig:rays\]). The incident wave is scattered into all outgoing modes on the left and right.
Scattering coefficients {#sec:scat}
=======================
The four independent solutions derived in the previous section contain all the information about scattering in the flow. To compute the scattering coefficients however, it is convenient to have solutions that contain just one incident mode. These are straightforwardly obtained by superposing our four solutions in an appropriate manner.
Scattering of mode 2 incident from the left
-------------------------------------------
We first construct a solution that corresponds to mode 2 incident from the left (see Fig. \[fig:2in\]). In this solution the asymptotic wave components on the far left do not include mode 3, and on the far right there are no asymptotic waves for modes 1 and 4. We also choose the mode-2 component on the left to be normalized so that it is exactly (\[-2\]). Two of our four solutions have a wave incident from the left only, namely $\phi^{(b)}(x)$ and $\phi^{(d)}(x)$. We superpose these solutions so that on the far left the incident mode-3 component is removed and the mode-2 component is normalized. From (\[H2left\]) and (\[F2left\]) the required wave is $$\label{2in}
\phi^{\text{2in}}(x)= \frac{ d_3^- \phi^{(b)}(x) - b_3^- \phi^{(d)}(x) }{ d_3^- b_2^- - b_3^- d_2^-}.$$
![ Heuristic ray picture for the wave solution (\[2in\]), whose only incident asymptotic wave component is the normalized mode-2 wave (\[-2\]) on the left. The incident wave scatters into all outgoing modes, some of which have negative norm (modes 3 and 4) and some propagate to the left relative to the fluid (modes 1 and 4). As in Fig, \[fig:rays\], the flow corresponds to a white-hole binary. Reversing $t$ (i.e. reading the plot top to bottom) corresponds to a black-hole binary, in which several modes in the past combine to give an outgoing low-$k$ mode 2 on the left. []{data-label="fig:2in"}](figure8.pdf){width="\linewidth"}
A qualitative picture of the solution (\[2in\]) is given in Fig. \[fig:2in\]. The incident mode 2 is right-moving relative to the fluid and has positive norm. This incident mode propagates through to the right where it scatters into mode 3, which is also a right-mover but has negative norm. As a result of the scattering, the outgoing mode 2 on the right has been amplified (otherwise norm would not be conserved). This part of the scattering process occurs even in the limit of no dispersion, where it corresponds exactly to the Hawking effect at an event horizon. Because of dispersion there is also *reflection* of mode 2 into modes which are left-moving relative to the fluid (modes 1 and 4 on the left). Moreover the reflection also involves further scattering into a negative-norm mode, namely mode 4 on the left, and therefore additional amplification of positive-norm components.
The scattering coefficients for this process are just the coefficients of the asymptotic wave components of (\[2in\]) on the far left and far right. These are easily obtained from (\[H2left\]), (\[H2right\]), (\[F2left\]), (\[F2right\]) and (\[2in\]). We denote the coefficient for scattering of incident mode 2 into outgoing mode $n$ by $S_{n,2}$, so that the expansions of (\[2in\]) into asymptotic waves read $$\begin{aligned}
\phi^{\text{2in}}(x) \stackrel{x\ll 0}{\sim} & \, \phi_2^-(x) + S_{1,2} \phi_1^-(x) + S_{4,2} \phi_4^-(x), \label{2inleft} \\
\phi^{\text{2in}}(x) \stackrel{x\gg 0}{\sim} & \, S_{2,2} \phi_2^+(x) + S_{3,2} \phi_3^+(x). \label{2inright}\end{aligned}$$ The most important information is given by the absolute values of the scattering coefficients. In computing $|S_{n,2}|^2$ we make use of the identities (\[gammaid1\]) and (\[gammaid2\]) to obtain
$$\begin{aligned}
\left| S_{1,2} \right|^2 = & \, \left( e^{ \frac{2\pi\omega}{\alpha} } -1 \right)^{-1} \left[ - \frac{1}{2}
+ e^{ -\frac{\pi k_c}{4\alpha} } \left( \frac{ \pi \sqrt{ k_c } } { 4 \sqrt{\alpha} } \left| \Gamma\left( \frac{3}{4}+\frac{ik_c}{4\alpha} \right) \right|^{-2}
+ \frac{ \pi \sqrt{ \alpha } } { \sqrt{k_c} } \left| \Gamma\left( \frac{1}{4}+\frac{ik_c}{4\alpha} \right) \right|^{-2} \right) \right] , \label{S12} \\
\left| S_{2,2} \right|^2 = & \, \left( 1 - e^{ -\frac{2\pi\omega}{\alpha} } \right)^{-1} \left[ \frac{1}{2}
+ e^{ -\frac{\pi k_c}{4\alpha} } \left( \frac{ \pi \sqrt{ k_c } } { 4 \sqrt{\alpha} } \left| \Gamma\left( \frac{3}{4}+\frac{ik_c}{4\alpha} \right) \right|^{-2}
+ \frac{ \pi \sqrt{ \alpha } } { \sqrt{k_c} } \left| \Gamma\left( \frac{1}{4}+\frac{ik_c}{4\alpha} \right) \right|^{-2} \right) \right] , \label{S22} \\
\left| S_{3,2} \right|^2 = & \, \left( e^{ \frac{2\pi\omega}{\alpha} } -1 \right)^{-1} \left[ \frac{1}{2}
+ e^{ -\frac{\pi k_c}{4\alpha} } \left( \frac{ \pi \sqrt{ k_c } } { 4 \sqrt{\alpha} } \left| \Gamma\left( \frac{3}{4}+\frac{ik_c}{4\alpha} \right) \right|^{-2}
+ \frac{ \pi \sqrt{ \alpha } } { \sqrt{k_c} } \left| \Gamma\left( \frac{1}{4}+\frac{ik_c}{4\alpha} \right) \right|^{-2} \right) \right] , \label{S32} \\
\left| S_{4,2} \right|^2 = & \, \left( 1- e^{- \frac{2\pi\omega}{\alpha} } \right)^{-1} \left[ - \frac{1}{2}
+ e^{ -\frac{\pi k_c}{4\alpha} } \left( \frac{ \pi \sqrt{ k_c } } { 4 \sqrt{\alpha} } \left| \Gamma\left( \frac{3}{4}+\frac{ik_c}{4\alpha} \right) \right|^{-2}
+ \frac{ \pi \sqrt{ \alpha } } { \sqrt{k_c} } \left| \Gamma\left( \frac{1}{4}+\frac{ik_c}{4\alpha} \right) \right|^{-2} \right) \right] . \label{S42} \end{aligned}$$
These are exact amplitudes of scattering coefficients for the linear flow profile. They were previously obtained by Busch and Parentani [@bus12] by a different approach as part of a study of dispersive fields in de Sitter space. We refer to [@bus12] for the application of these results to cosmological particle creation and black-hole thermodynamics when Lorentz invariance is broken.
Similarly to the four wave solutions in the previous section, (\[2inleft\]) and (\[2inright\]) imply that $|S_{n,2}|^2$ are (up to a sign) equal to the norm fluxes of the outgoing modes. Constancy of the norm flux explains the following identity: $$1- \left| S_{1,2} \right|^2 + \left| S_{4,2} \right|^2 = \left| S_{2,2} \right|^2 - \left| S_{3,2} \right|^2 .$$
The scattering amplitudes (\[S12\])–(\[S42\]) demonstrate the importance of dispersion in laboratory analogues of event horizons. Note the following simple relation that is clear from (\[S12\])–(\[S42\]): $$\frac{ \left| S_{3,2} \right| }{ \left| S_{2,2} \right| } = \frac{ \left| S_{1,2} \right| }{ \left| S_{4,2} \right| } = e^{ -\frac{\pi\omega}{\alpha} } . \label{ratio}$$ The first ratio in (\[ratio\]) has the same value as in the non-dispersive case (see below) but dispersion leads to non-zero values for $S_{1,2}$ and $S_{4,2}$ whose ratio matches that of $S_{3,2}$ and $S_{2,2}$. An interesting question is whether the elementary relation (\[ratio\]) is a property of the linear flow profile with *arbitrary* anomalous dispersion, but our results allow no conclusions on this point.
The non-dispersive result for the scattering amplitudes is very simple and was first derived by Hawking [@haw74]. To extract the non-dispersive case from (\[S12\])–(\[S42\]) we can employ asymptotic expansions of the gamma functions for large $k_c$. The following complete asymptotic expansion for large $z$ will allow us to take the required limit [@htf1] : $$\begin{aligned}
\ln \left[\Gamma(a+z) \right] \sim & \, \left(a+z-\frac{1}{2}\right) \ln z - z + \frac{1}{2} \ln (2\pi) \nonumber \\
&+ \sum_{n=1}^{\infty} \frac{ (-1)^{n+1} B_{n+1} (a) }{ n(n+1) z^n } , \label{lnGamma}\end{aligned}$$ where $B_n(z)$ are the Bernoulli polynomials. Employing this expansion in (\[S12\])–(\[S42\]) we can compute the leading-order scattering amplitudes as $k_c\to\infty$: $$\begin{aligned}
\left| S_{1,2} \right|^2 \sim & \, \left( e^{ \frac{2\pi\omega}{\alpha} } -1 \right)^{-1} \frac{ \alpha^4 }{ 64 k_c^4 } \left( 1+ \frac{5 \alpha^2 }{k_c^2 } \right) , \label{S12as} \\
\left| S_{2,2} \right|^2 \sim & \, \left( 1 - e^{ -\frac{2\pi\omega}{\alpha} } \right)^{-1} \left[ 1+ \frac{ \alpha^4 }{ 64 k_c^4 } \left( 1+ \frac{ 5\alpha^2 }{k_c^2 } \right) \right] , \label{S22as} \\
\left| S_{3,2} \right|^2 \sim & \, \left( e^{ \frac{2\pi\omega}{\alpha} } -1 \right)^{-1} \left[ 1+ \frac{ \alpha^4 }{ 64 k_c^4 } \left( 1+ \frac{5 \alpha^2 }{k_c^2 } \right) \right] , \label{S32as} \\
\left| S_{4,2} \right|^2 \sim & \, \left( 1 - e^{ -\frac{2\pi\omega}{\alpha} } \right)^{-1} \frac{ \alpha^4 }{ 64 k_c^4 } \left( 1+ \frac{5 \alpha^2 }{k_c^2 } \right) . \label{S42as} \end{aligned}$$ These expansions should be treated with caution. The appearance of exponentials containing $\pm k_c/\alpha$ in the asymptotic expansions of the gamma functions in (\[S12\])–(\[S42\]) means it is not possible to develop complete asymptotic expansions of the scattering amplitudes in powers of $\alpha/k_c$. For $k_c\to\infty$, we obtain from (\[S12as\])–(\[S42as\]) the Hawking result $$\begin{gathered}
\left| S_{1,2} \right|^2 = \left| S_{4,2} \right|^2 = 0 , \label{S12haw} \\
\left| S_{2,2} \right|^2 = \left( 1 - e^{ -\frac{2\pi\omega}{\alpha} } \right)^{-1} , \quad
\left| S_{3,2} \right|^2 = \left( e^{ \frac{2\pi\omega}{\alpha} } -1 \right)^{-1} , \label{S32haw} \end{gathered}$$ in which there is no scattering of the incident mode into modes left-moving relative to the fluid. The squared amplitude $ \left| S_{3,2} \right|^2$ in (\[S32haw\]) for scattering into the right-moving negative-norm mode has the form of the Planck distribution.
The frequency dependence of the scattering amplitudes (\[S12\])–(\[S42\]) factors out neatly from the dependence on dispersion, the latter being a function of $k_c/\alpha$. This factorization may be unique to the linear flow profile. In heuristic terms, each frequency in the linear profile experiences the same profile shape in the region around a blocking point (where the group velocity changes sign), even though the blocking point is at a different position in the flow for each frequency. For a curving flow profile each frequency will experience a different flow shape near a blocking point and the dependence of the scattering coefficients on frequency, profile shape and dispersion will be very complicated [@rob12; @mac09; @leo12; @fin12; @cou12; @mic14; @rob14; @euv15].
The fact that classical waves in the flow experience scattering into modes of opposite norm implies spontaneous emission in the quantum theory, provided an appropriate quantum description exists. The derivation of quantum emission from classical scattering into negative-norm modes is well described elsewhere [@rob12; @bro95; @unr95] and is not repeated here. Recall that we chose the sign of $t$ so that the flow corresponds to a white-hole binary, whereas a black-hole binary is obtained by $t\to-t$. For the black-hole binary, the scattering of mode 2 given by the solution (\[2in\]) occurs backward in time, so that time runs from the top to the bottom in Fig. \[fig:2in\]. In this case the positive-norm mode 2 on the left contains negative-norm components in the past. As a consequence, the annihilation operator for mode-2 quanta is a sum of annihilation and creation operators for modes in the past. If all modes were in their ground state in the past then mode 2 will now contain quanta and there will be emission of low-$k$ waves to the left (see Fig. \[fig:2in\], reading top to bottom). The expectation value for the number of quanta in mode 2 is $ \left| S_{3,2} \right|^2 + \left| S_{4,2} \right|^2 $, and is thus determined by the scattering into negative-norm modes. In the dispersionless case this gives the familiar thermal spectrum of quanta, as is seen from (\[S12haw\]) and (\[S32haw\]). When dispersion is included the spectrum is obtained from (\[S32\]) and (\[S42\]) and is no longer thermal.
Note that $\left| S_{4,2} \right|^2$ does not go to zero at large frequencies, but rather approaches a value given by the quantity in square brackets in (\[S42\]). This means the result for quantum emission does not vanish at large frequency, in contrast to the dispersionless case. But we cannot of course employ the model assumed here at arbitrarily high frequencies. The missing ingredient is dissipation. When waves propagate in any medium there is a limit to the size of the wavelength that can be supported and one manifestation of this is the loss of energy from the wave into the medium. This is very clear in the case of water waves where it is readily observed how dissipation increases as the wavelength decreases into the capillary-wave regime. The influence of dissipation on the Hawking effect has been explored in [@ada13; @rob15].
Scattering of mode 3 incident from the left
-------------------------------------------
We now construct a solution that corresponds to mode 3 incident from the left (see Fig. \[fig:3in\]). Here the asymptotic wave components on the far left do not include mode 2, and on the far right there are no asymptotic waves for modes 1 and 4. The mode-3 component on the left is normalized to be exactly (\[-3\]). This solution is given by a superposition of $\phi^{(b)}(x)$ and $\phi^{(d)}(x)$ that removes the mode-2 wave on the left and normalizes the mode-3 component on the left. From (\[H2left\]) and (\[F2left\]) we find the required wave: $$\label{3in}
\phi^{\text{3in}}(x)= \frac{ d_2^- \phi^{(b)}(x) - b_2^- \phi^{(d)}(x) }{ d_2^- b_3^- - b_2^- d_3^-}.$$
![ Heuristic ray picture for the wave solution (\[3in\]), whose only incident asymptotic wave component is the normalized mode-3 wave (\[-3\]) on the left. The incident wave has negative norm and scatters into all outgoing modes, some of which have positive norm (modes 1 and 2) and some propagate to the right relative to the fluid (modes 2 and 3 on the right). []{data-label="fig:3in"}](figure9.pdf){width="\linewidth"}
The incident mode-3 wave on the left is left-moving relative to the fluid and has negative norm. It is scattered into all four outgoing modes, including those right-moving relative to the fluid (modes 2 and 3 on the right). The outgoing mode 1 on the left and mode 2 on the right have positive norm, so conservation of norm again implies amplification of the incident wave. We denote the coefficient for scattering of the incident mode 3 into outgoing mode $n$ by $S_{n,3}$, The expansions of (\[3in\]) into asymptotic waves then take the form $$\begin{aligned}
\phi^{\text{3in}}(x) \stackrel{x\ll 0}{\sim} & \, \phi_3^-(x) + S_{1,3} \phi_1^-(x) + S_{4,3} \phi_4^-(x), \label{3inleft} \\
\phi^{\text{3in}}(x) \stackrel{x\gg 0}{\sim} & \, S_{2,3} \phi_2^+(x) + S_{3,3} \phi_3^+(x), \label{3inright}\end{aligned}$$ with scattering coefficients that follow from (\[H2left\]), (\[H2right\]), (\[F2left\]), (\[F2right\]) and (\[3in\]). We calculate the absolute values of the scattering coefficients and find expressions that already appear in (\[S12\])–(\[S42\]): $$\begin{gathered}
\left| S_{1,3} \right|^2 = \left| S_{3,2} \right|^2 , \qquad \left| S_{2,3} \right|^2 = \left| S_{4,2} \right|^2 , \label{S2S3a} \\
\left| S_{3,3} \right|^2 = \left| S_{1,2} \right|^2 , \qquad \left| S_{4,3} \right|^2 = \left| S_{2,2} \right|^2 . \label{S2S3b} \end{gathered}$$ The scattering amplitudes for mode 3 obey the identity that follows from constancy of the norm flux: $$-1- \left| S_{1,3} \right|^2 + \left| S_{4,3} \right|^2 = \left| S_{2,3} \right|^2 - \left| S_{3,3} \right|^2 .$$ There is also a simple relation analogous to (\[ratio\]): $$\frac{ \left| S_{1,3} \right| }{ \left| S_{4,3} \right| } = \frac{ \left| S_{3,3} \right| }{ \left| S_{2,3} \right| } = e^{ -\frac{\pi\omega}{\alpha} } . \label{ratio2}$$
By means of (\[S12as\])–(\[S42as\]) we find immediately the leading-order scattering amplitudes (\[S2S3a\]) and (\[S2S3b\]) as $k_c\to\infty$. The scattering amplitudes thus reproduce the non-dispersive Hawking result $$\begin{gathered}
\left| S_{2,3} \right|^2 = \left| S_{3,3} \right|^2 = 0 , \label{S23haw} \\
\left| S_{4,3} \right|^2 = \left( 1 - e^{ -\frac{2\pi\omega}{\alpha} } \right)^{-1} , \quad
\left| S_{1,3} \right|^2 = \left( e^{ \frac{2\pi\omega}{\alpha} } -1 \right)^{-1} , \label{S43haw} \end{gathered}$$ in which there is no scattering into modes right-moving relative to the fluid and $\left| S_{1,3} \right|^2$ has the form of the Planck distribution.
The scattering of the incident mode 3 into modes with opposite norm again implies spontaneous emission in the quantum theory [@rob12; @bro95; @unr95]. In the black-hole binary, the expectation value for the number of quanta in mode 3 emitted to the left (reading Fig. \[fig:3in\] from top to bottom) is $ \left| S_{1,3} \right|^2 + \left| S_{2,3} \right|^2 $. From (\[S2S3a\]) we see that this is equal to the result we obtained for the number of quanta in mode 2 emitted to the left by the black-hole binary. The symmetry of the problem shows that this is also the number of quanta in the low-$k$ modes 1 and 4 emitted to the right (see Fig. \[fig:rays\]). There is thus emission of the same non-thermal spectrum of radiation in all the modes, both left-moving and right-moving relative to the fluid.
Our results were derived for the strictly linear flow profile (\[flow\]), but our analysis showed that there is no wave scattering in the far-left and far-right regions of the flow. The scattering coefficients will therefore be the same for a flow profile that flattens out far from the horizons, provided the change in the flow velocity with distance is slow enough not to induce further scattering.
Concluding remarks
==================
We chose the fourth-order wave equation (\[wave\]) because it is relatively simple while still being applicable to a physical system (the flowing BEC). The same equation with $k_c\to ik_c$ has dispersion that is normal rather than anomalous, but this gives a fourth-order normal dispersion relation $\omega(k)$ that is not monotonic in $k$ in the fluid frame. A monotonic normal dispersion relation leads to singular wave propagation in the linear flow profile, as some modes are infinitely blue-shifted as they approach any point where the flow speed is zero [@bar05]. In reality such modes would be heavily damped as their wavelengths go to zero. For these reasons we have not treated the case of normal dispersion in the linear flow profile.
Our motivation was to obtain an exact solution for the Hawking effect in the presence of dispersion. The linear flow profile has two horizons but it can be solved exactly and the scattering amplitudes (\[S12\])–(\[S42\]) and (\[S2S3a\])–(\[S2S3b\]) are our final results. They demonstrate in exact formulas how dispersion changes the Hawking effect in one particular flow profile.
For the wave equation (\[wave\]), the new qualitative feature introduced by the dispersive term is the reflection of waves, i.e. the scattering of right-movers relative to the fluid into left-movers and vice versa. In the absence of dispersion there is no reflection because equation (\[wave\]) is then exactly the 1+1-dimensional wave equation in curved space-time, and conformal flatness of the metric tensor leads to a strict separation of left- and right-movers. Dispersion causes coupling between the left- and right-movers and this is why reflection occurs in our example. The modification of the Hawking effect due to dispersion is related to the amount of refection because the total scattering into all channels must conserve the norm. The precise relationship between the various scattering channels will depend on the flow profile, even when the dispersion is fixed. For more complicated flow functions $v(x)$ than the one considered here, the scattering coefficients for the wave equation (\[wave\]) will also be more complicated, if indeed exact results can be found.
Instead of trying to solve the wave scattering in a given flow profile, an alternative possibility is to design profiles that give a desired scattering. This approach has been fruitfully pursued in optics and quantum mechanics [@ber90; @lek07; @hor15; @lon15; @phi16; @hor16a; @hor16b; @lon16]. An important lesson from this work is that a breakdown of the geometrical-optics approximation does not necessarily imply scattering. In fact several infinite classes of inhomogeneous profiles are known in optics that have strictly zero scattering, even when the geometrical-optics approximation is violated arbitrarily badly. If these techniques can be extended to the wave equation in a moving medium, then dramatic differences in the spectrum of spontaneous quantum emission may be achieved by careful control of the flow profile.
Acknowledgements {#acknowledgements .unnumbered}
================
I am indebted to R. Parentani for informing me of the close connection between this work and ref. [@bus12], and also for commenting on the manuscript. I also thank S.A.R. Horsley, C. G. King and R. J. Churchill for many helpful discussions.
Conserved quantities {#app:conserved}
====================
An action giving the wave equation (\[wave\]) can be written in a general form that allows for arbitrary dispersion [@unr95]: $$\begin{aligned}
S=\int \int dt \, dx\left[\frac{1}{2}(\partial_t\psi^*+v\partial_x\psi^*) (\partial_t\psi+v\partial_x\psi) \right. \nonumber \\
\left. -\frac{1}{2}F^*(-i\partial_x)\psi^*F(-i\partial_x)\psi \right], \label{act} \\
F(-i\partial_x)=\sum_{n=0}^\infty (-1)^{n+1} i b_{2n+1} \partial_x^{2n+1}. \qquad\end{aligned}$$ This gives the wave equation $$\label{wavegen}
\partial_t (\partial_t+v\partial_x )\psi+\partial_x (v\partial_t+v^2\partial_x )\psi+F^2(-i\partial_x)\psi=0,$$ with the dispersion relation $$\label{disprelgen}
(\omega-vk)^2=F^2(k).$$ The general dispersive equation (\[wavegen\]) has spatial derivatives of $\psi$ of all even orders (terms in the wave equation with an odd number of derivatives would give dissipation). The fourth-order equation (\[wave\]) corresponds to $F^2(-i\partial_x)=-\partial_x^2+\frac{1}{k_c^2}\partial_x^4$, which gives an $F(-i\partial_x)$ that is defined by the power series $$\label{F}
F(k)=k\sqrt{1+\frac{k^2}{k_c^2}}=\sum_{n=0}^\infty \binom{\frac{1}{2}}{n} \frac{k^{2n+1}}{k_c^{2n}}.$$ In (\[act\]) we allow $\psi(x,t)$ to be complex to see better the quantities conserved by (\[wavegen\]). We derive the conservation laws for arbitrary dispersion and then specialise to the fourth-order equation (\[wave\]).
The action (\[act\]) is invariant under the $U(1)$ transformation $\psi\to e^{i\theta}\psi$ and also under time translation (since $v(x)$ is time independent). The conserved quantities associated with these symmetries are the norm and the pseudo-energy, respectively. To construct the conservation laws we must apply Noether’s theorem to an action with an (in general) unbounded number of terms containing derivatives of arbitrarily high order. The method for applying Noether’s theorem to such actions is described in [@phi11], with examples from dispersive optics. We refer to [@phi11] for the technicalities of how to construct the conservation laws and here quote the results for the norm and pseudo-energy. The norm density $\rho_N(x,t)$ and norm flux $s_N(x,t)$ are $$\begin{aligned}
\rho_N = \, & i\psi^*(\partial_t\psi+v\partial_x\psi) +\text{c.c.}, \label{rhoN} \\
s_N = \, & iv\psi^*(\partial_t\psi+v\partial_x\psi) \nonumber \\
& +\sum_{n=0}^\infty \sum_{m=0}^{2n} (-1)^{n+m} b_{2n+1}[\partial_x^mF(-i\partial_x)\psi ]\partial_x^{2n-m}\psi^* \nonumber \\
& +\text{c.c.}, \label{sN}\end{aligned}$$ where c.c. means complex conjugate. It is straightforward to verify that the norm conservation law $$\label{conlaw}
\partial_t \rho_N (x,t)+ \partial_x s_N (x,t)=0$$ holds for waves satisfying the general dispersive equation (\[wavegen\]). The pseudo-energy density $\rho_E(x,t)$ and pseudo-energy flux $s_E(x,t)$ are $$\begin{aligned}
\rho_E = \, &\frac{1}{2}\left( \partial_t\psi^* \partial_t\psi - v^2\partial_x\psi^* \partial_x\psi \right) \nonumber \\
&+\frac{1}{2} F^*(-i\partial_x)\psi^* F(-i\partial_x)\psi , \label{rhoE} \\
s_E = \, & \frac{1}{2} v \partial_t\psi^*(\partial_t\psi+v\partial_x\psi) \nonumber \\
- & \frac{i}{2} \sum_{n=0}^\infty \sum_{m=0}^{2n} (-1)^{n+m} b_{2n+1}[\partial_x^mF(-i\partial_x)\psi ]\partial_x^{2n-m}\partial_t\psi^* \nonumber \\
& +\text{c.c.} \label{sE}\end{aligned}$$ These also obey the conservation law of form (\[conlaw\]), because of (\[wavegen\]).
For monochromatic waves $\psi(x,t)=e^{-i\omega t} \phi(x)$, the densities (\[rhoN\]) and (\[rhoE\]), and fluxes (\[sN\]) and (\[sE\]), are clearly time-independent. The conservation law (\[conlaw\]) thus gives for monochromatic waves $$\label{sNconst}
\partial_x s_N =0,$$ so that the flux is the same at each point in the inhomogeneous flow. This constant-flux condition is of great importance in analysing wave propagation in the fluid. The fourth-order wave equation (\[wave\]) corresponds to $F$ given by (\[F\]), and in this case the norm flux (\[sN\]) reduces to a finite number of terms: $$\begin{aligned}
s_N (x,t) = \, & iv\psi^*(\partial_t\psi+v\partial_x\psi) -i \psi^* \partial_x \psi \nonumber \\
& + i k_c^{-2} \left( \psi^* \partial_x^3 \psi - \partial_x \psi^* \partial_x^2 \psi \right) +\text{c.c.}. \label{sN4th} \end{aligned}$$ The constancy of this flux for monochromatic waves will be referred to throughout.
Asymptotics of the wave equation {#app:asymptotic}
================================
The main aim here is to understand the wave equation (\[wave\]) in the asymptotic regions $|x|\to\infty$ of the linear flow profile. We show that waves in the linear profile must reduce, as $|x|\to\infty$, to non-interacting wave components associated with the dispersion relation. The norm flux (\[sN4th\]) for these wave components is then calculated.
The issue addressed here could be phrased as that of finding the WKB solutions of the wave equation in the flow. But it is worth pointing out some significant differences between asymptotic solutions for sound in a flowing fluid and the familiar WKB solutions in optics and quantum mechanics. For the Helmholtz equation (equivalently the time-independent Schrödinger equation) in an inhomogeneous medium, there is a simple criterion for the WKB solutions to be good approximations, namely that the fractional change in the refractive index must be very small over a local wavelength. This criterion is clearly satisfied as $|x|\to\infty$ if the permittivity profile approaches constant values. But the Helmholtz equation has the property that the WKB criterion is satisfied for large $|x|$ even for profiles that diverge as $|x|\to\infty$, because the local wavelength goes to zero. This means that general WKB solutions can be written which are functionals of an arbitrary permittivity profile and these will always be valid as $|x|\to\infty$. The wave equation for sound in a moving fluid shows important differences. One can compute leading-order asymptotic solutions in this case also, but the criterion for their validity is not very simple. For all flow profiles with regions where $v'(x)$ is very small, leading-order asymptotic solutions that are accurate in these regions can be derived as functionals of $v(x)$. These functionals, however, do *not* give the leading-order asymptotic solutions for the linear profile in the regions $|x|\to\infty$. This is because $v'(x)$ stays constant as $|x|\to\infty$ for the linear profile while the wavelengths of some roots of the dispersion relation get larger.
For completeness, we first give the leading-order asymptotic solutions valid in any regions where $v'(x)$ is “small" and allowing for arbitrary dispersion. Then we treat the leading-order asymptotic solutions of (\[wave\]) for the linear profile (\[flow\]), in the regions $|x|\to\infty$.
Asymptotics in regions of slowly varying flow velocity
------------------------------------------------------
The wave equation (\[wavegen\]) in the case of arbitrary dispersion gives the monochromatic equation $$\label{monogen}
\left[\omega^2 +i\omega v'+2 v (i\omega-v')\partial_x -v^2 \partial_x^2 -F^2(-i\partial_x) \right]\phi=0.$$ Following a standard approach to the WKB approximation in quantum mechanics [@merz], we substitute $\phi(x)= e^{i\chi(x)}$ and arrange (\[monogen\]) as $$\begin{aligned}
(\omega-v\chi')^2 - & F^2(\chi') = -i\omega v'+2ivv'\chi'+iv^2\chi'' \nonumber \\
&+e^{-i\chi}\left[F^2(-i\partial_x) -F^2(\chi') \right] e^{i\chi}. \label{wkb1}\end{aligned}$$ In regions where $v'$, and therefore $\chi'$, are nearly constant, terms on the right-hand side of (\[wkb1\]) are small compared to terms on the left-hand side. We therefore iterate (\[wkb1\]) as follows. To lowest order the solution of (\[wkb1\]) is $\chi_0'$, satisfying $$\label{wkb2}
(\omega-v\chi_0')^2 - F^2(\chi_0') = 0,$$ i.e. the branches of the dispersion relation (\[disprelgen\]). The lowest order correction $\chi_1'$ to $\chi_0'$ is found by inserting $\chi'=\chi_0'+\chi_1'$ into (\[wkb1\]), applying (\[wkb2\]), and keeping only terms linear in small quantities of the same order as $\chi_1'$, i.e. $\chi_0''$, $\chi_1'$ and $v'$. This gives $$\begin{aligned}
-2(\omega-v\chi_0')v\chi_1' - & 2F(\chi_0')F'(\chi_0')\chi_1' = -i\omega v'+2ivv'\chi_0' \nonumber \\
& +iv^2\chi_0'' -i \frac{d}{dx}\left[F(\chi_0') F'(\chi_0') \right] , \label{wkb3}\end{aligned}$$ which yields the following solution for $\chi_1'$: $$\begin{aligned}
\chi_1' &=\frac{i}{2}\frac{d}{dx} \ln \left|(\omega-v\chi_0')v+F(\chi_0') F'(\chi_0') \right| \nonumber \\
&= \frac{i}{2}\frac{d}{dx} \ln \left|F(\chi_0') V_g(\chi_0') \right|.\end{aligned}$$ The last expression contains $V_g(\chi_0')$, the group velocity in the laboratory frame of the mode given by the root $\chi_0'$ of the dispersion relation (\[wkb2\]), i.e. $$V_g(k)=v\pm F'(k),$$ where the sign depends on the branch $k$. The solution for $\phi(x)= e^{i\chi(x)}$ to order $\chi_1'$ is thus $$\label{wkbsoln}
\phi(x)\sim \frac{1}{2\sqrt{\left|F(\chi_0') V_g(\chi_0') \right|}} e^{i\chi_0},$$ where a normalization factor of $1/2$ is inserted. This result, valid for arbitrary dispersion, can also be derived from an analysis of the wave equation (\[monogen\]) in $k$-space [@rob12].
The norm density (\[rhoN\]) and pseudo-energy density (\[rhoE\]) for the asymptotic solutions (\[wkbsoln\]) is positive or negative according to the sign of the co-moving frequency $\omega-v\chi_0'$. Norm and pseudo-energy are transported at the group velocity of the mode, i.e. $s_N/\rho_N=s_E/\rho_E=V_g(\chi_0')$.
The norm flux (\[sN\]) of the asymptotic solutions (\[wkbsoln\]) is equal to $\pm1$ (to the same order of approximation). The sign of the norm flux is given by the product of two signs: the sign of the norm being transported by the mode (given by the sign of the co-moving frequency) and the sign of the mode’s group velocity. A superposition of asymptotic solutions (\[wkbsoln\]) for different roots $\chi_0'$ of the dispersion relation has the important property that its norm flux is just the sum of the fluxes of the individual components in the superposition, i.e. all cross terms in (\[sN\]) involving different components cancel out.
Asymptotics of the wave equation in the linear profile
------------------------------------------------------
Here we confine attention to the fourth-order equation (\[mono\]) in the linear profile (\[flow\]), and find its asymptotic solutions in the regions $|x|\to\infty$.
As noted in Sec. \[sec:wave\], the four roots of the dispersion relation (\[disprel\]) in the linear profile have complicated expressions, but here we require only their asymptotic expansions for large $|x|$. We denote the roots by $k_n(x)$, where $n$ labels the ray solutions 1 to 4 discussed in Sec. \[sec:wave\]. For large positive $x$ the first few terms of the expansions of $k_n(x)$ are $$\begin{aligned}
k_1(x) & \sim \frac{\omega}{\alpha x}\left( -1+\frac{1}{\alpha x} -\frac{1}{\alpha^2 x^2} +\frac{2k_c^2+\omega^2}{2k_c^2\alpha^3 x^3} \right) , \label{k1} \\
k_2(x) & \sim \alpha k_c x - \frac{k_c-2\omega}{2\alpha x} - \frac{k_c^2-8k_c\omega+8\omega^2}{8k_c\alpha^3 x^3} , \label{k2} \\
k_3(x) & \sim -\alpha k_c x + \frac{k_c+2\omega}{2\alpha x} + \frac{k_c^2+8k_c\omega+8\omega^2}{8k_c\alpha^3 x^3}, \label{k3} \\
k_4(x) & \sim \frac{\omega}{\alpha x}\left( -1-\frac{1}{\alpha x} -\frac{1}{\alpha^2 x^2} -\frac{2k_c^2+\omega^2}{2k_c^2\alpha^3 x^3} \right) . \label{k4} \end{aligned}$$
The invariance of the dispersion relation under $x\to-x,\ k\to-k$ leads to the following (exact) relations: $k_1(-x)=-k_2(x)$, $k_3(-x)=-k_4(x)$ (this can be seen in the dispersion plots in Fig. \[fig:disp\]). We can thus easily obtain from (\[k1\])–(\[k4\]) the asymptotic expansions of the wave-vectors $k_n(x)$ for large negative $x$.
We first seek asymptotic solutions to (\[mono\]) for large positive $x$. In view of (\[wkbsoln\]), we make the Ansatz $$\label{asym1}
\phi^+_n(x)\sim A^+_n(x) \exp\left[i\int^x dx\, k_n(x)\right],$$ where $A^+_n(x)$ are unknown amplitude functions for the modes 1 to 4, and the superscript $+$ labels the region (large positive $x$) in which the expansion (\[asym1\]) is to be valid. We insert (\[asym1\]) for $n=1,\dots,4$ into the wave equation (\[mono\]) and demand that it be satisfied for $x\to\infty$. For modes 1 and 4 the wave equation is satisfied by (\[asym1\]) as $x\to\infty$ even with constant $A^+_1$ and $A^+_4$. By demanding that $A^+_1(x)$ and $A^+_4(x)$ increase the accuracy of the asymptotic solutions (\[asym1\]), so that the wave equation is satisfied to higher orders of $1/x$, we can build up the required amplitudes $A^+_1(x)$ and $A^+_4(x)$ as asymptotic series. For modes 2 and 3 we perform the same procedure, but here the wave equation is not satisfied by (\[asym1\]) to any order of $1/x$ without $x$-dependent amplitudes $A^+_2(x)$ and $A^+_3(x)$. The amplitudes $A^+_n$, to the orders consistent with the accuracy of the expansions (\[k1\])–(\[k4\]), are found to be $$\begin{aligned}
A^+_1(x) & \sim \frac{1}{\sqrt{2\omega}}\left( 1+\frac{i\omega^3}{6k_c^2\alpha^4 x^3} \right) , \label{A1} \\
A^+_2(x) & \sim \frac{1}{\sqrt{2k_c\alpha^3x^3}}\left( 1+ \frac{12 k_c-27i\alpha-32 \omega}{16k_c\alpha^2 x^2} \right) , \label{A2} \\
A^+_3(x) & \sim \frac{1}{\sqrt{2k_c\alpha^3x^3}}\left( 1+ \frac{12 k_c+27i\alpha+32 \omega}{16k_c\alpha^2 x^2} \right) , \label{A3} \\
A^+_4(x) & \sim \frac{1}{\sqrt{2\omega}}\left( 1-\frac{i\omega^3}{6k_c^2\alpha^4 x^3} \right) . \label{A4} \end{aligned}$$ where convenient constant normalization factors have been included. Using these and (\[k1\])–(\[k4\]) in (\[asym1\]) we obtain asymptotic solutions for the four modes. We will only need the leading order of (\[asym1\]) for modes 2 and 3, but we require the first three orders for modes 1 and 4, as follows: $$\begin{aligned}
\phi^+_1(x) & \sim \frac{ x^{-i\omega/\alpha}}{\sqrt{2\omega}}\left( 1-\frac{i\omega}{\alpha^2 x}+\frac{i\omega(\alpha+i\omega)}{2\alpha^4 x^2} \right) , \label{+1} \\
\phi^+_2(x) & \sim \frac{ x^{-\frac{3}{2}+i\frac{2\omega-k_c}{2\alpha}}}{\sqrt{2k_c\alpha^3}} \exp\left( \frac{i \alpha k_c x^2}{2} \right) , \label{+2} \\
\phi^+_3(x) & \sim \frac{ x^{-\frac{3}{2}+i\frac{2\omega+k_c}{2\alpha}}}{\sqrt{2k_c\alpha^3}} \exp\left(- \frac{i \alpha k_c x^2}{2} \right) , \label{+3} \\
\phi^+_4(x) & \sim \frac{ x^{-i\omega/\alpha}}{\sqrt{2\omega}}\left( 1+\frac{i\omega}{\alpha^2 x}+\frac{i\omega(\alpha+i\omega)}{2\alpha^4 x^2} \right) . \label{+4} \end{aligned}$$
One can show that the amplitudes (\[A2\]) and (\[A3\]) for modes 2 and 3 are the same as would be obtained by using the result (\[wkbsoln\]) for a slowly varying flow velocity. But the amplitudes (\[A1\]) and (\[A4\]) for modes 1 and 4 are not given correctly by (\[wkbsoln\]).
The four expressions (\[+1\])–(\[+4\]) are also asymptotic solutions for large negative $x$, but the identification of each with one of the four modes is different in the region $x\to-\infty$. It is straightforward to find the corresponding mode in each case, and the four asymptotic solutions $\phi^-_n(x)$ for $x\to-\infty$ take the form $$\begin{aligned}
\phi^-_1(x) & \sim \frac{ \left|x\right|^{-\frac{1}{2}+i\frac{2\omega-k_c}{2\alpha}}}{x\sqrt{2k_c\alpha^3}} \exp\left( \frac{ i \alpha k_c x^2}{2} \right) , \label{-1} \\
\phi^-_2(x) & \sim \frac{ \left|x\right|^{-i\omega/\alpha}}{\sqrt{2\omega}}\left( 1+\frac{i\omega}{\alpha^2 x}+\frac{i\omega(\alpha+i\omega)}{2\alpha^4 x^2} \right) , \label{-2} \\
\phi^-_3(x) & \sim \frac{ \left|x\right|^{-i\omega/\alpha}}{\sqrt{2\omega}}\left( 1-\frac{i\omega}{\alpha^2 x}+\frac{i\omega(\alpha+i\omega)}{2\alpha^4 x^2} \right) , \label{-3} \\
\phi^-_4(x) & \sim \frac{ \left|x\right|^{-\frac{1}{2}+i\frac{2\omega+k_c}{2\alpha}}}{x\sqrt{2k_c\alpha^3}} \exp\left( - \frac{i \alpha k_c x^2}{2} \right) . \label{-4} \end{aligned}$$
The norm flux (\[sN\]) of each of the asymptotic solutions (\[+1\])–(\[-4\]) is equal to $\pm1$, to leading order. Modes 2 and 4 have norm flux equal to $+1$ while modes 1 and 3 have norm flux of $-1$. The sign of the norm flux is the product of the sign of the norm (positive for modes 1 and 2, negative for modes 3 and 4) and the group velocity (positive for modes 2 and 3, negative for modes 1 and 4). A superposition of the asymptotic solutions has a norm flux that is the sum of the fluxes of the individual mode components, i.e. all cross terms in (\[sN\]) between different modes cancel out, to leading order.
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Q:
Catch errors from validation when saving product or category ($product->validate() or $category->validate())
I would like to validate the data before saving a product programatically because ->save() saves a product even if it doesn't have an sku.
If I do ->validate() I get the expected error:
[Magento\Eav\Model\Entity\Attribute\Exception]
The value of attribute "sku" must be set
But I would like to catch that error and log it somewhere and let the script continue on another product.
This doesn't work:
try{
$product->validate();
} catch(Exception $e){
}
If I look at the validate method it doesn't supply much info about where can I catch the error (except for that @todo):
\module-catalog\model\Product.php
/**
* Validate Product Data
*
* @todo implement full validation process with errors returning which are ignoring now
*
* @return array
*/
public function validate()
{
$this->_eventManager->dispatch($this->_eventPrefix . '_validate_before', $this->_getEventData());
$result = $this->_getResource()->validate($this);
$this->_eventManager->dispatch($this->_eventPrefix . '_validate_after', $this->_getEventData());
return $result;
}
I imagine it is quite a common task. Does anyone have any solutions?
A:
You can check like this:
$errors = $product->validate();
//validate returns true if everything is ok and an array with the errors otherwise
if (is_array($errors)) {
//log your errors
} else {
//proceed to save
}
The errors array should look like this:
array(
'attribute_code1_here' => 'Error message 1 here',
'attribute_code2_here' => 'Error message 2 here',
....
)
There is a catch also.
If the product is not assigned to an attribute set, you will get only this error
array('attribute_set' => 'Invalid attribute set entity type');
all the other errors will not be shown for that product.
For the categories it works the same without the restriction on the attribute set because there is only one attribute set for the categories.
[EDIT]
before calling $product->validate() call this:
$product->setCollectExceptionMessages(true);
This will collect all the validation error messages inside an array and it will not throw an exception.
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BUENA PARK, Calif (KABC) -- Knott's Berry Farm has partnered with Heritage Auctions to hold a special auction featuring more than 200 items from the park's archives.Many of the items were once displayed inside the 75-year-old park but have since spent years in storage.Pieces include coin-operated player pianos, ride animatronics and western paintings.The park hopes much of the memorabilia will not only go to the highest bidder, but to owners who will find a good home for a slice of Knott's history.The auction date is set for March 31 with a preview day the week prior.
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Webb managed to pitch in a few rehab games at Double-A Frisco before undergoing season-ending rotator cuff surgery. He showed decreased velocity and showcased pitches with "Hit Me" on them instead of "Rawlings" during his rehab appearances, allowing 21 hits in 12 innings. Although he has expressed no desire to retire, one has to wonder if he'll ever toe the rubber on a major league mound again after what will now be at least a three-year absence.
2011
Webb has made one start the last two seasons, while all of 2010 was lost following shoulder surgery. On multiple occasions, it appeared as though he was going to return in August or September and log innings out of the Arizona bullpen, but multiple setbacks in his rehab sessions prevented him from making it back as far as a minor league rehab assignment. After signing with the Rangers in January, Webb is a lottery ticket at this point as the days of inking him in for 225-plus innings appear well in the past.
2010
The D-Backs picked up their $8.5 million option on Webb after he made just one start in 2009 and underwent a season-long attempt to rehab back from shoulder stiffness following his Opening Day start. His week-to-week status quickly turned into month-to-month, before culminating in a shoulder debridement operation in September. As spring training approached, Webb had successfully completed his rehab work and was on track to prepare for the start of spring training as he normally would. Getting him back into the rotation for another 30-plus starts would provide a huge boost to an Arizona pitching staff looking to fill a few holes on the free-agent market this winter. Fortunately for Webb, tests and the subsequent operation didn't reveal any tears of his labrum, so there's reason to believe that he'll bounce back in 2010. Be sure to carefully monitor his health and performance during spring training before an attempt to draft him on the cheap this spring.
2009
Webb had his best major league campaign yet, finishing with 22-7 in 34 starts. You can chalk it up as another great season in what is turning into a very productive career for Webb. He's still generating plenty of groundballs (2.93 G/F), while maintaining a healthy strikeout rate (7.27 K/9IP) and limiting his walks (2.58 BB/9IP). He's proven capable of handling a heavy workload on a yearly basis, rolling up over 225 innings in each of the last four seasons, while Webb has finished with an ERA below 3.60 in all six of his seasons in Arizona. A long-term contract extension appears to be in the offing, while Arizona may need to get a hometown discount to keep him beyond an $8 million team option in 2010. Expect him to come off the board as one of the high-priced elite pitching options on draft day.
2008
Webb had another excellent season in 2007, posting numbers again worthy of Cy Young consideration. His G/F ratio slid from an extreme 3.53 in 2006 to 2.57 last season, but his 194 strikeouts were a career high and he cut back on homers allowed for the third straight season. Webb is in his prime and with the D-Backs poised to be competitive again in 2008, expect another year of Top-10 production from the Arizona ace.
2007
Webb improved a bit, and the defense improved a bit (.310 BABIP dropped to .293), and that's how you win a Cy Young Award. He's terrific, and likely to stay this good for a while. Mid-season shoulder issues didn't seem to bother him down the stretch.
2006
Webb had a big comeback in 2005. Forget his ERA, which barely budged from 2004, and check the walk rate instead. Yes, he really did cut his walks in half while keeping his strikeout rate constant and pitching more innings to boot. The sinkerball artist also led the majors in groundball-flyball ratio last year (by a good margin), so the acquisition of the slick-fielding Orlando Hudson could bode well for Webb's stats.
2005
In his rookie year Webb walked roughly one out of every twelve batters he faced. In 2004 hitters learned to lay off Webb's sinker and that walk rate skyrocketed to roughly one batter out of nine, which led to a lot of other numbers getting worse as well. Don't be fooled by that relatively pleasant 3.59 ERA, Webb's component ERA (component ERA is a statistic that estimates what a pitcher's ERA should have been, based on his hits, walks, and home runs allowed.) last year was a much higher 4.32, which indicates that, as bad as Webb's walk rate was, he was actually lucky in terms of how his walks and hits translated to runs. There's still much of talent here, and a new pitching coach can help, so if he slips to the second half of your draft, he'd be a good buy.
2004
Here he is, the best rookie pitcher of 2003. (No, not the award winner, but what do THEY know?) In 2003, Webb posted a quality start in 21 of his 28 starts (75 percent, the best rate in the league for anyone with 20 or more starts). He held NL hitters to a .605 OPS (or, in other words, he made everyone look like Tony Womack), second in the league only to Jason Schmidt. Yes, he only got 10 wins, but his run support (3.2 runs per game) was among the worst in the majors. According to Baseball Prospectus' stats, with average run support Webb would have gone 14-5, not 10-9. Should we expect a downturn in 2004? It could always happen, but we wouldn't bet on it. He'll be the No. 2 man behind Randy Johnson in Arizona's rotation this year, and he'd be worth a high draft pick in your league.
2003
Webb went 10-6, 3.14 in 25 starts at Double-A in 2002 (122 K's in 152 innings), then wowed ‘em in the Arizona Fall League, posting a 0.55 ERA in eight games (16 innings). He'll turn 24 in May; probably starts 2003 at Triple-A, but could crack the big-league bullpen with great spring outings and a break or two.
|
Contact Information
Job Description
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Fun, energy filled atmosphere. State of the art
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|
Local civilian military workers worry about sequestration
Published: Wednesday, February 27, 2013 at 10:15 AM.
About 3,700 civilian employees work at Eglin Air Force Base and another 1,400 work at Hurlburt Field.
The final implementation plan hasn’t been released, but the Secretary of the Air Force said in a memo Monday that affected employees would be subject to furloughs, or unpaid leave, for up to 22 discontinuous workdays.
That’s if Congress doesn’t agree to avert the budget cuts, or sequestration, by Thursday, the memo said.
Eglin anticipates that if the furloughs go into effect, most civilian employees will not report to work one day a week from the end of April to the end of September, said Andy Bourland, a spokesman for Eglin.
For Blount, that could mean a loss of about $800 a month, or 23 percent of his pay, he said.
The Iraq war veteran served six years in the Air Force before leaving to join the civilian workforce.
He said he feels like he has a good-paying job, but that he’s budgeted his life for that income and a sudden cutback will quickly lead to hard times for his family.
Last week the Department of Defense announced its intention to make civilian employees take unpaid leave if Congress doesn’t avert the some $85 billion in budget cuts set to take effect by the end of the week.
For Blount, a civilian carpenter at Eglin Air Force Base, that could mean he won’t be able to pay his bills. The 32-year-old is the primary earner for his wife and toddler.
“This is the only check for my family,” Blount said. “This is not only going to affect my job, but my future — probably for a long time.”
He said he already has contacted his mortgage company and credit lenders to try to get payments lowered to brace for the possible loss of income.
Even so, he knows he likely won’t be able to make all the scheduled payments.
“You’ve got to feed the child first before you pay the other bills, so that’s where the money is going to be going,” he said. “We will probably be in a lot of debt trouble. My credit is going to go to nothing.”
About 3,700 civilian employees work at Eglin Air Force Base and another 1,400 work at Hurlburt Field.
The final implementation plan hasn’t been released, but the Secretary of the Air Force said in a memo Monday that affected employees would be subject to furloughs, or unpaid leave, for up to 22 discontinuous workdays.
That’s if Congress doesn’t agree to avert the budget cuts, or sequestration, by Thursday, the memo said.
Eglin anticipates that if the furloughs go into effect, most civilian employees will not report to work one day a week from the end of April to the end of September, said Andy Bourland, a spokesman for Eglin.
For Blount, that could mean a loss of about $800 a month, or 23 percent of his pay, he said.
The Iraq war veteran served six years in the Air Force before leaving to join the civilian workforce.
He said he feels like he has a good-paying job, but that he’s budgeted his life for that income and a sudden cutback will quickly lead to hard times for his family.
Curt Kirkland, 48, is a civil engineer at Eglin. His wife is also a civil employee there.
He also is concerned the cuts will prevent his family from being able to meet their financial obligations, from house and car payments to their son’s college tuition.
His job requires security clearance that depends on a clean financial record, so if his credit takes a beating he is worried he eventually could lose his job.
Amanda Patterson, a 33-year-old civilian employee at NAS Pensacola with two children, said she may try to pick up a delivery job on the days she is required to take off to make up for the loss of income.
Civilian employees already haven’t had a cost-of-living pay raise for several years, Blount said.
“There is a lot of animosity in the local shops,” he said. “Morale has not been good.”
Local federal workers’ unions are concerned.
“We just continue to support the mission and do what we do,” said Thaddeus Wallace, president of the American Federation of Government Employees Local 1897, which represents thousands of local blue collar civilian military employees.
“This is another extremely hard hit to take,” he said.
Wallace said he’s worried not only about his workers’ pay, but also about possible impacts to the military’s mission that relies on civilian support work and how the loss of income will affect local businesses that civilian workers frequent.
Because of that, he questions whether laying civilians out on the chopping block is the best way to cut the military’s budget.
“I think overall we can look at how we spend in general,” he said. “There’s lots of ways to address this. Just looking down at the civilians at this particular point is a method that can have long-range effects on our economy.”
The civilian workforce makes up about 10 percent of the military’s total personnel budget, which includes military, civilian and contractors, according to a Department of Defense document from 2011. Contractors are the costliest, at about 50 percent of the total personnel budget.
While many employees are scrambling to try to plan for the pay cut, Rocky Tasse, president of the American Federation of Government Employees Local 1942, is bracing for a fight.
According to the Air Force, the unions have until early March to propose their own plans for implementing the furlough.
Tasse said he met with other union presidents representing Air Force civilian employees last week. He said they will likely propose that the furlough days be consecutive: up to 22 days off in a row without pay.
That would allow workers to file for unemployment benefits in most states to help make up for the loss of income, he said.
The employees could schedule their unpaid leave over a five-month period.
If implemented that way, it would have the added benefit of threatening to interrupt day-to-day Air Force base operations, which may be unpalatable enough to stave off or weaken the furlough, Tasse said.
“I want to make this so painful that they will never, ever consider doing this again.”
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Saint Patrick
There is probably no instance in the world of a saint and a country being linked so closely as Saint Patrick and Ireland. When he began his mission to that Celtic nation in the 5th century, it was almost entirely pagan, with spiritual matters firmly in the hands of the Druid. By the time he died, some decades later, a significant portion of its large and dispersed population was Christian. And the Catholic faith has burned strongly in Ireland ever since.
For so popular a saint, frustratingly little is known for sure about Patrick. He was a man of action rather than of the pen and the two short pieces of his writing that survive are, though lively and readable, quite inadequate as autobiographical material.
Fortunately, the Irish have been particularly gifted in handing down their history by word of mouth for thousands of years, and so there is a lively and extensive body of tradition linked with the man. Equally important, there has been no lack of scholars to sift this tradition for what is sound in it.
Devotion to Saint Patrick has of course accompanied the Irish in their vast emigrations, and the English speaking world is rich with churches bearing his name. Among the most magnificent surely, is our own Cathedral of Saint Patrick. Here the saint is represented with appropriate attention. His statue anchors the north end of the front of the Sanctuary. He also appears in a small statue set into the front doors of the Cathedral. Most striking of all is the immense window above the south transept door consisting of panels depicting various key moments linked, in legend or in fact with Patrick's life. Beginning in the lower left corner and reading the scenes upwards in groups of three, these are: his baptism; his capture; an angel revealing his vocation; his preaching the gospel aboard a ship, his being sold in Ireland; being set free; being made a cleric by Saint Martin of Tours; pursuing his studies at Lerins; being ordained a priest; setting out for Rome; receiving the Pope Saint Celestine l's blessing; being consecrated a bishop; visiting Saint Germain d' Auxerre; his first conversions in Ireland; giving Communion to King Laoghaire's daughters; raising a person from the dead; his own death; and angels singing his funeral hymn.
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Application of the CO2 laser to infertility surgery.
In addition to use of the CO2 laser in gynecologic surgery for the treatment of pelvic, vaginal and vulvar neoplasms, it has been more recently applied to the treatment of infertility. The advantages afforded by the laser over conventional techniques are discussed. The authors' clinical experience in the surgical treatment of infertility, with emphasis on endometriosis and microtubal surgery, are described, and use of the new electro-pulsed laser in infertility surgery is evaluated.
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Offshore verification needs to move with the times and improve its contribution to safety according to global risk management company, DNV GL.
In a paper to be presented at SPE Offshore Europe 2019 this September, DNV GL will reveal, for the first-time, details of its project to develop the next generation of verification to manage risks relating to major accident hazards.
The organisation is piloting a new verification approach that utilises digital technologies, particularly data mining techniques and smart software that were not available 30 years ago when verification was first introduced to the UK.
DNV GL believes that this approach will improve the benefits contributed by the independent verifiers to offshore safety by taking a more focused approach towards assessing management of risks without increasing cost to the duty holder and while more effectively meeting the intent of the current legislation.
Several North Sea operators have been working closely with DNV GL to review how the verifier can improve its contribution to the management of offshore safety.
Jack Downie, Head of Development and Innovation, UK Verification Services at DNV GL and author of the paper says, “Through discussion with many duty holders, we know that there is a common understanding that the current verification system has not evolved in line with industry’s new reality. Much has changed in the last 30 years; our collective understanding of major accident hazards and the contribution made by safety and environmental critical plant and equipment has improved, safe systems of work are more advanced, computerised maintenance management systems have developed and offshore communication and technology is unrecognisable compared to the 1990s.
“Our paper at SPE Offshore Europe will present a critical review of the verifier role and whether it still meets the original intent of the process that was created by Lord Cullen through the Safety Case Regulations (1996).”
The paper will give examples of deficiencies in the current verification system and launch DNV GL’s proposed more efficient and encompassing service which follows the principles of process safety and will ensure that the verifier is looking at the right risks at the right time.
The aim of this new approach is to enhance the ability of the verifier, and therefore the duty holder, to visualise signals of deterioration in performance of their hazard management.
DNV GL is to hold workshops with duty holders over the next few months and begin pilot assessments early in 2020 where technical results will be compared directly with traditional methods. The offshore workforce will be engaged through offshore meetings and requests for feedback.
The DNV GL paper “Next Generation Verification of Offshore Assets” will be presented on Tuesday 3rd September at 15:30 at SPE Offshore Europe being held for the first time at P&J Live, Aberdeen’s new GBP 333 million state-of-the-art event complex.
The conference programme includes a three parallel stream technical conference with 75 technical papers and a two-stream keynote programme with more than 70 senior executives discussing key industry issues in 12 2-hour sessions.
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include_rules = [
"+ash/sysui/public/interfaces",
"+components/arc",
"+components/drive",
"+components/user_manager",
"+mash/shelf/public/interfaces",
"+media",
]
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Q:
Firefox loads old versions of changed files from apache
So, I'm running an apache server on linux. Sometimes, Firefox decides to not load the new version of a file after I edited it. For example, right now I have a .js file wich is loaded dynamicly. It had a bug, wich I corrected (I checked with Chromium), but when the file is loaded in Firefox, it still has the bug! When looking at the response header of the ajax request, I see the code of the file BEFORE it was changed. But that code doesn't exist anymore... I had this happen with CSS files too.
When I rename the file to something else, it loads the right stuff, but as soon as I rename it to the old name, it starts loading an old version of the file again!
I restarted apache2, but that didn't change anything.
I checked for file permissions too, no problem there as far as I could tell (I changed all files' permissions to rwxrwxrwx to be sure).
When accessing with an other browser, it works fine!
In previous cases, the next day or so, the problem would have vanished, but I can't always just stop for a day in what I'm doing...
A:
This is caused by browser cache,
you can consider to use url with version parameter,
like http://yourdomain.com/js/some.js?v=$version,
and update the $version whenever you update a css/js
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Betty or Veronica
If you happen to be an accident prone, red-headed 17 year old with an unexplainable crosshatch pattern in your hair, chances are you can identify pretty closely with Archie Comics’ own Archie Andrews. For those of us who are none of those things, we can still identify with Archie due to the monumental choice that is constantly in front of him. Our faithful protagonist is constantly torn between the sweet, blond-haired Betty Cooper, and the stylish, raven-haired heiress Veronica Lodge.
Betty Cooper is the girl next-door; middle class, smart, athletic and well liked; she was also ranked at 66 in Comic Buyer’s Guide’s: 100 sexiest women in comics issue, as creepy as that may be. Veronica Lodge is her perfect foil; an aristocrat through and through, stylish, narcissistic, spoiled and popular. She also made the list in Comic Buyer’s Guide’s: 100 sexiest women in comics issue, but came in at 87th. At first glance it would appear that Betty is the clear choice for Archie, however given the recent storylines within the comics both girls have been pursuing other love interests. The girls are both operating under a similar idea: going out with another boy will incite jealousy in Archie and he will vie for their attention.
Where does this leave poor ol’ Archie? Pulling out his ridiculously orange hair by the fistful I would imagine. Betty and Veronica are frenemies and have said they only fight over things that don’t matter, namely: boys. These girls have been slugging it out over Archie since 1941 and they openly admit that at the end of the day, boys are no big deal. To me this puts both girls on even standing; after all they’ve been toying with Archie’s feelings for the past 71 years. So it ultimately comes down to what either of them can offer Archie. I could go into detail about which girl I myself would choose but that might lead into some scary territory. If he chooses Betty he resigns himself to a solid middle class life, content but ever wondering what could have been with Veronica. If he chooses Veronica he moves up the social ladder, into the high life with guaranteed financial security but married to a materialistic and vain woman. Either way Archie loses.
The clear choice is for Archie to remain a swinging bachelor, driving around in his crap car with his dopey friend, who clearly has an eating disorder. This is the Archie that everyone has come to know and I can’t see it changing anytime soon, they’ve tried making Archie a married man in some dream sequence issues, but it just wouldn’t be Archie. So, there you have it — quite an anticlimactic conclusion to such a controversial topic.
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Inactivation of a voltage-dependent K+ channel by beta subunit. Modulation by a phosphorylation-dependent interaction between the distal C terminus of alpha subunit and cytoskeleton.
Kv1.1/Kvbeta1.1 (alphabeta) K+ channel expressed in Xenopus oocytes was shown to have a fast inactivating current component. The fraction of this component (extent of inactivation) is increased by microfilament disruption induced by cytochalasins or by phosphorylation of the alpha subunit at Ser-446, which impairs the interaction of the channel with microfilaments. The relevant sites of interaction on the channel molecules have not been identified. Using a phosphorylation-deficient mutant of alpha, S446A, to ensure maximal basal interaction of the channel with the cytoskeleton, we show that one relevant site is the end of the C terminus of alpha. Truncation of the last six amino acids resulted in alphabeta channels with an extent of inactivation up to 2.5-fold larger and its further enhancement by cytochalasins being reduced 2-fold. The wild-type channels exhibited strong inactivation, which could not be markedly increased either by cytochalasins or by the C-terminal mutations, indicating that the interaction of the wild-type channels with microfilaments was minimal to begin with, presumably because of extensive basal phosphorylation. Since the C-terminal end of Kv1.1 was shown to participate in channel clustering via an interaction with members of the PSD-95 family of proteins, we propose that a similar interaction with an endogenous protein takes place, contributing to channel connection to the oocyte cytoskeleton. This is the first report to assign a modulatory role to such an interaction: together with the state of phosphorylation of the channel, it regulates the extent of inactivation conferred by the beta subunit.
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