text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
almost-perfect completeness, 1 (cid:0) (cid:14) for small (cid:14) > 0, since these are more
natural: if the proof is correct we usually want the veri(cid:12)er to always, or almost always, accept.
We typically want the soundness error to be as small as possible. Soundness error 0, or error smaller
than 1=2r, correspon... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
< 1.
Proof. (i) Suppose Max-3Sat1;s is N P -hard for some s < 1. Let us construct a probabilistic
veri(cid:12)er V for 3Sat with completeness 1 and soundness error s.
Input: a formula φ.
1. Apply the reduction from 3Sat to Max-3Sat1;s, to transform the formula φ to clauses
C1; : : : ; Cm over variables x1; : : : ; xn.
... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
(although may be slightly larger than
the randomness of V ).
2. The reduction will produce all the 2O(log n) = nO(1) clauses/tests that V ′ produces.
Note that the reduction is efficient.
Note that the veri(cid:12)er we construct from a Max-3Sat1;s N P -hardness uses its randomness only to
decide which test to make out o... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
constant can be 1, and the best randomness known is log n +
O(log log n).
Corollary 3.3. The following follow from Theorem 4:
1. Two queries: There are a (cid:12)xed s < 1 and a (cid:12)xed alphabet (cid:6), such that N P (cid:18) P CP1;s[O(log n); 2](cid:6).
2. Low error: For any (cid:12)xed ϵ > 0, there is q = q(ϵ), ... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
45(1):70{122, 1998.
[FGL+96] U. Feige, S. Goldwasser, L. Lovasz, S. Safra, and M. Szegedy. Interactive proofs and
the hardness of approximating cliques. Journal of the ACM, 43(2):268{292, 1996.
[GW95] M. Goemans and D. Williamson.
Improved approximation algorithms for maximum
cut and satis(cid:12)ability problems using... | https://ocw.mit.edu/courses/18-405j-advanced-complexity-theory-spring-2016/1b2c1d36df1e2a9cfa79d6531ffb8d29_MIT18_405JS16_IntroPCP.pdf |
3.15
Optical Fibers and Photonic Deviccs
C.A. Ross, DMSE, MIT
References:
Braithwnite and Weaver chapter 6.4 (fibers)
Sanger. How fiber optics works, The Industrial Physicist p l 8 . FebIMar 2002
Savage. Linking lvith Light, IEEE Spectrum p32. Aug. 2002
Saleh and Teich, Furldamentals of' Photonics. Wiley 199 I .... | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/1b318e5a5a7d1ba5a21d9ddf6edc5c63_lecture15_16.pdf |
Total internal reflection when angle of incidence exceeds $<: sin $, = n/n'
Fibers rely on total internal reflection, with a very small An. e.g.
core n = 1.53, cladding n = 1.50 gives $, = 78.6'
Step index fibers can have rrtohl dispersion: different modes of light traveling at
different angles traverse different p... | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/1b318e5a5a7d1ba5a21d9ddf6edc5c63_lecture15_16.pdf |
)
detcclcd (PIN diodes)
Example: Comrnrlnicntion system
Local: inexpensive rnultimode fibers, directly modulated LEDs, typically 850 nm
waveletlgth.
Long distance: loss and dispersion are important. Single mode fiber, distributed feedback
laser, external ~notlulation. amplifiers every 100 km, dispersion compensat... | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/1b318e5a5a7d1ba5a21d9ddf6edc5c63_lecture15_16.pdf |
(drop) one channel from a fiber.
Couplers: different wavelength channels can be added to the fiber by bringing another
fiber (or waveguide) close, allowing light to couple between the fibers.
e . 5 a tantalum oxide/silica rnultilayer. By designing the multilayer. it
Thir~filt~~filtcr,
can have a photonic band gap ... | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/1b318e5a5a7d1ba5a21d9ddf6edc5c63_lecture15_16.pdf |
6.867 Machine learning, lecture 7 (Jaakkola)
1
Lecture topics:
• Kernel form of linear regression
• Kernels, examples, construction, properties
Linear regression and kernels
Consider a slightly simpler model where we omit the offset parameter θ0, reducing the
model to y = θT φ(x) + � where φ(x) is a particular fe... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
terms of prediction differences αt and the
feature vectors:
θ =
n1 �
λ
t=1
αtφ(xt)
(3)
The implication is that the optimal θ (however high dimensional) will lie in the span of the
feature vectors corresponding to the training examples. This is due to the regularization
Cite as: Tommi Jaakkola, course materials... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
)
(7)
(8)
(9)
Note that finding the estimates ˆαt requires inverting a n × n matrix. This is the cost of
dealing with inner products as opposed to handing feature vectors directly. In some cases,
the benefit is substantial since the feature vectors in the inner products may be infinite
dimensional but never needed ... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
= 1, 2, . . .
K(x, x�) = exp −
�x − x��2
, β > 0
�
β
2
�
3
(11)
(12)
We have already discussed the feature vectors corresponding to the polynomial kernel. The
components of these feature vectors were polynomial terms up to degree p with specifically
chosen coefficients. The restricted choice of coefficients was... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
1. K(x, x�) = f (x)K1(x, x�)f (x�) for any function f (x),
2. K(x, x�) = K1(x, x�) + K2(x, x�),
3. K(x, x�) = K1(x, x�)K2(x, x�)
Cite as: Tommi Jaakkola, course materials for 6.867 Machine Learning, Fall 2006. MIT OpenCourseWare
(http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YY... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
use a
double index i, j to index the components of φ(x) where i ranges over the components of
φ(1)(x) and j refers to the components of φ(2)(x). Then
It is now easy to see that
(1)
φi,j (x) = φi (x)φj (x)
(2)
K(x, x�) = φ(x)T φ(x�)
φi,j (x)φi,j(x�)
�
i,j
�
i,j
�
[
=
=
=
i (x)φ(2)
φ(1)
j (x)φ(1)
j (x�) ... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
67 Machine learning, lecture 7 (Jaakkola)
is a valid kernel.
1
exp{−
2
�x − x��2} = exp{−
x T x + x T x� −
1
2
f (x)
��
�
1
= exp{−
2
�
1
2
x�T x�}
�
f (x�)
��
1
2
�
x�T x�}
x T x} · exp{x T x�} · exp{−
5
(25)
(26)
Here exp{xT x�} is a sum of simple products xT x� and is therefore a kernel ba... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
, x contains the elements of u in
positions i1 < i2 <
< ik. If the elements of u are found in successive positions in x,
then ik − i1 = k − 1. A simple string kernel corresponds to feature vectors with counts of
occurences of length k subsequences:
· · ·
�
φu(x) =
δ(ik − i1, k − 1)
i:u=x[i]
(27)
In other word... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
as l(i) = ik − i1. The feature vectors in a
weighted gapped substring kernel are given by
φu(x) =
λl(i)
(30)
�
i:u=x[i]
where the parameter λ ∈ (0, 1) specifies the penalty for non-contiguous matches to u. The
resulting kernel
K(x, x�) =
�
u∈Ak
φu(x)φu(x�) =
⎛
� �
⎝
λl(i)
⎞ ⎛
⎠ ⎝
�
λl(i)
⎞
⎠
u∈Ak
i:... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
θ0 and θ to zero gives the following optimality
conditions:
dJ(θ, θ0)
dθ0
dJ(θ, θ0)
dθ
n
�
�
= −2
yt − θT φ(xt) − θ0 = 0
�
t=1
= 2λθ − 2
n
� ��
αt
��
��
yt − θT φ
φ(xt) = 0
(xt) − θ0
t=1
(34)
(35)
Cite as: Tommi Jaakkola, course materials for 6.867 Machine Learning, Fall 2006. MIT OpenCourseWare
(ht... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
be interpreted as prediction
differences:
αi = yi − θT φ(xi) − θ0
= yi −
= yi −
= yi −
n1 �
λ
t�=1
n
1 �
λ
t�=1
n
1 �
n
t=1
αt� φ(xt� )T φ(xi) − θ0
�
n
1 �
n
t=1
yt −
αt� φ(xt� )T φ(xi) −
�
n
1 �
λ
t�=1
yt −
αt� φ(xt� )T φ(xi) −
�
n
1 �
λ
t�=1
n
1 �
n
t=1
αt� φ(xt� )T φ(xt)
�
φ(xt� )T φ... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
CKCa
1
λ
(43)
Cite as: Tommi Jaakkola, course materials for 6.867 Machine Learning, Fall 2006. MIT OpenCourseWare
(http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].(cid:13)(cid:10)
6.867 Machine learning, lecture 7 (Jaakkola)
8
where we have introduced an additional ce... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1b55295adbca700abfa7356a4f832e33_lec7.pdf |
Global Illumination and Monte Carlo
MIT EECS 6.837 Computer Graphics
Wojciech Matusik
with many slides from Fredo Durand and Jaakko Lehtinen
© ACM. All rights reserved. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.
1
Today
• Lots of r... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
integral there!
• Recursive!
x
10
The Rendering Equation
• Where does Lin come from?
– It is the light reflected towards x from the surface point in
direction l ==> must compute similar integral there!
• Recursive!
– AND if x happens
to be a light source,
we add its contribution
directly
x
11
... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
Cast random rays from the visible point
• Recurse
19
“Monte-Carlo Ray Tracing”
• Cast a ray from the eye through each pixel
• Cast random rays from the visible point
• Recurse
20
“Monte-Carlo Ray Tracing”
• Systematically sample light sources at each hit
– Don’t just wait the rays will hit it by chance
21 ... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
acing Results: Glossy Scene
• 100 paths/pixel
n
e
s
n
e
J
n
n
a
W
k
i
r
n
e
H
Courtesy of Henrik Wann Jensen. Used with permission.
28
Importance of Sampling the Light
Without explicit
light sampling
With explicit
light sampling
1 path
per pixel
4 paths
per pixel
✔
✔
29
Why Use Random Numbers?
•... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
cached values
• But do full calculation for direct lighting
40
Irradiance Caching
• Yellow dots:
indirect diffuse sample points
The irradiance cache tries to
adapt sampling density to
expected frequency content of
the indirect illumination (denser
sampling near geometry)
Courtesy of Henrik Wann Jensen. Used... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
Courtesy of Henrik Wann Jensen. Used with permission.
49
More Subsurface Scattering
6
0
0
2
.
l
a
t
e
h
c
i
r
y
e
W
Photograph
Rendering
© ACM. All rights reserved. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.
50
That... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
© ACM. All rights reserved. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.
53
Motivational Eye Candy
• Rendering glossy reflections
• Random reflection rays around mirror direction
– 1 sample per pixel
© source unknown. All rights reser... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
S
• {xi} are independent uniform random points in S
• The integral is the average of f times the volume of S
• Variance is proportional to 1/N
– Avg. error is proportional 1/sqrt(N)
– To halve error, need 4x samples
61
Monte Carlo Computation of
• Take a square
• Take a random point (x,y) in the square
• T... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
Images by Veach and Guibas, SIGGRAPH 95
Naïve sampling strategy
Optimal sampling strategy
© ACM. All rights reserved. This content is excluded from our Creative Commons
license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.
67
Hmmh...
• Are uniform samples the best we can do?
68
Smarter Samp... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
.
• The problem is designing ps that are easy to sample
from and mimic the behavior of f
74
Monte Carlo Path Tracing
Video removed due to copyright restrictions – please see the link below for further details.
http://www.youtube.com/watch?v=mYMkAnm-PWw 75
Questions?
1200 Samples/Pixel
Traditional imp... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
References
Images of the following book covers have been removed due to copyright restrictions:
-Advanced Global Illumination by Philip Dutre, Philippe Bekaert, and Kavita Bala
-Realistic Ray Tracing by Peter Shirley and R. K. Morley
-Realistic Image Synthesis Using Photon Mapping by Henrik Wann Jensen
Please check th... | https://ocw.mit.edu/courses/6-837-computer-graphics-fall-2012/1b5985f78c68379e15543fe27d41b72c_MIT6_837F12_Lec18.pdf |
6.096 Lecture 3:
Functions
How to reuse code
Geza Kovacs
#include <iostream>
using namespace std;
int main() {
int threeExpFour = 1;
for (int i = 0; i < 4; i = i + 1) {
threeExpFour = threeExpFour * 3;
}
cout << "3^4 is " << threeExpFour << endl;
return 0;
}
Copy-paste
coding
#include <i... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
10; i = i + 1) {
twelveExpTen = twelveExpTen * 12;
}
cout << "12^10 is " << twelveExpTen << endl;
return 0;
}
With a
function
#include <iostream>
using namespace std;
// some code which raises an arbitrary integer
// to an arbitrary power
int main() {
int threeExpFour = raiseToPower(3, 4);
co... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
: Lets other people use algorithms
you’ve implemented
Function Declaration Syntax
Function name
int raiseToPower(int base, int exponent)
{
int result = 1;
for (int i = 0; i < exponent; i = i + 1) {
result = result * base;
}
return result;
}
Function Declaration Syntax
Return type
int raiseToPower(i... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
int result = 1;
for (int i = 0; i < exponent; i = i + 1) {
result = result * base;
}
return result;
}
Function Declaration Syntax
int raiseToPower(int base, int exponent)
{
int result = 1;
for (int i = 0; i < exponent; i = i + 1) {
result = result * base;
}
return result;
}
Return statement
F... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
return 0;
}
Returning a value
• Return statements don’t necessarily need to be at the end.
• Function returns as soon as a return statement is executed.
void printNumberIfEven(int num) {
if (num % 2 == 1) {
cout << "odd number" << endl;
return;
}
cout << "even number; number is " << num << endl;
}
int... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
<< x << endl;
}
• printOnNewLine(3) prints “Integer: 3”
• printOnNewLine(“hello”) prints “String: hello”
Function Overloading
void printOnNewLine(int x)
{
cout << "1 Integer: " << x << endl;
}
void printOnNewLine(int x, int y)
{
cout << "2 Integers: " << x << " and " << y << endl;
}
• printOnNewLine(3... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
}
• Function prototypes should match the signature of the
method, though argument names don’t matter
int square(int z);
function prototype
int cube(int x)
{
return x*square(x);
}
int square(int x)
{
return x*x;
}
• Function prototypes are generally put into separate
header files
– Separates s... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
variable to determine this.
– Can be accessed from any function
int numCalls = 0;
Global variable
void foo() {
++numCalls;
}
int main() {
foo(); foo(); foo();
cout << numCalls << endl; // 3
}
int numCalls = 0;
Scope
• Scope: where a
variable was declared,
determines where it
can be accessed fr... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
determines where it
can be accessed from
int raiseToPower(int base, int exponent) {
numCalls = numCalls + 1;
int result = 1;
for (int i = 0; i < exponent; i = i + 1) {
result = result * base;
}
return result;
}
• numCalls has global
scope – can be
accessed from any
function
result has function
scop... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
result = result * base;
}
// A
return result;
}
int max(int num1, int num2) {
numCalls = numCalls + 1;
int result;
if (num1 > num2) {
result = num1;
}
else {
result = num2;
}
// B
return result;
}
Global scoperaiseToPower function scopemax function scopeint baseint exponentint resultint nu... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
ops and if/else
statements also have
their own scopes
– Loop counters are in the
same scope as the body of
the for loop
squareRoot function scopefor loop scopeIf statement scopeelse statement scopedouble lowdouble highdouble numdouble estimateint idouble newHighdouble newLow
double squareRoot(double num) {
do... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
for (int i = 0; i < 30; i = i + 1) {
estimate = (high + low) / 2;
if (estimate*estimate > num) {
double newHigh = estimate;
high = newHigh;
} else {
double newLow = estimate;
low = newLow;
}
}
return estimate; // A
}
• Cannot access variables
that are out of scope
• Solution 2: declare the
vari... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
cout << "q in main " << q << endl;
}
Output
a in increment 4
q in main 3
main function scopeq=3increment function scopea=3
Pass by value vs by reference
// pass-by-value
void increment(int a) {
a = a + 1; // HERE
cout << "a in increment " << a << endl;
}
int main() {
int q = 3;
increment(q); // doe... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
cout << "q in main " << q << endl;
}
Output
a in increment 4
q in main 4
main function scopeincrement function scopeq=3a
Pass by value vs by reference
// pass-by-value
void increment(int &a) {
a = a + 1; // HERE
cout << "a in increment " << a << endl;
}
int main() {
int q = 3;
increment(q); // work... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
r 3
}
main function scopeswap function scopeq=3r=5ab
Implementing Swap
void swap(int &a, int &b) {
int t = a; // HERE
a = b;
b = t;
}
int main() {
int q = 3;
int r = 5;
swap(q, r);
cout << "q " << q << endl; // q 5
cout << "r " << r << endl; // r 3
}
main function scopeswap function scopeq=3r=5... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
4;
int rem;
int result = divide(num, den, rem);
cout << result << "*" << den << "+" << rem << "=" << num << endl;
// 3*4+2=12
}
Libraries
• Libraries are generally distributed as the
header file containing the prototypes, and a
binary .dll/.so file containing the (compiled)
implementation
– Don’t need... | https://ocw.mit.edu/courses/6-096-introduction-to-c-january-iap-2011/1b954ce3fc96fd1168f8cd123c5748e4_MIT6_096IAP11_lec03.pdf |
Massachusetts Institute of Technology
Department of Materials Science and Engineering
77 Massachusetts Avenue, Cambridge MA 02139-4307
3.205 Thermodynamics and Kinetics of Materials—Fall 2006
October 31, 2006
Kinetics Lecture 2: Mass Diffusion and Heat Conduction
Lecture References
1. Porter and Easterling, Phase ... | https://ocw.mit.edu/courses/3-205-thermodynamics-and-kinetics-of-materials-fall-2006/1baa318e2b9bd0996e61c1e5017c1dbf_lecture02_review.pdf |
materials and in many alloy crystals occurs by the vacancy mecha
nism.
• Interdiffusion occurs in an alloy with composition gradients. The motion of each species in a labora
tory frame fixed to the crystal follows Fick’s first law, with a proportionality constant known as the
intrinsic diffusivity. The intrinsic diff... | https://ocw.mit.edu/courses/3-205-thermodynamics-and-kinetics-of-materials-fall-2006/1baa318e2b9bd0996e61c1e5017c1dbf_lecture02_review.pdf |
Chapter 3
Grand Unified Theory
3.1 SU(5) Unification
Gauge bosons:
SU (3)
�
SU (2)
�
U (1): (commuting with SU (3)
SU (2))
×
Or Lie algebra:
2iλ
e
2iλ
e
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
2iλ
e
3iλ
e−
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
3iλ
e−
2
2
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
2
3
−
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
3
−
(3.1)
(3.2)
(3.3)
One gets breaking SU (5)
... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
tensor
����
+ 5
vector
Y = 0
����
�
Altogether?
6
1
× 6
+ 2
× −
1
2
+ 3
× −
2
3
+ 3
1
× 3
v
e
�
L
+ 1
=
0
make 5
1
2
−
(dC
R) +
1
3
�
Actually (using
αβ )
�
Note:
dC
R
e
−
v
⎛
⎜
⎝
⎞
⎟
⎠
=
¯5
2
2
g˜
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
2
3
−
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
3
−
11
(3.4)
(3.5)
(3.6)
(3.7)
(3.8)
... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
−
u C
R
Ψij, color: singlet, SU (2) : 2
Y =
1
6
(
3
−
−
−
3) = 1
e C
R
⇒
It clicks.
Normalizing ˜g:
(3.14)
(3.15)
(3.16)
SU (3) generators
:
gun.
SU (2) generators
:
gun.
fab
trΓaΓb =
1
2
U (1) generators
:
g˜
1
−
2
1
2
−
0
0
0
0
0
1
2
1
2−
0
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
etc.
(3.17)
⎞
⎟
⎟... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
3
−
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
1
2
1
2
g
S = g
2
2
ω
=
5
2
g
3 un.
sin 2 θω =
g
�2
g
�2 + g2
ω
=
3
5
1 +
3
5
=
3
8
13
(3.20)
(3.21)
(3.22)
(3.23)
(3.24)
(3.25)
Expt.
0.22.
≈
Coupling Constant
Energy
Q 0
Figure 3.1: Coupling Constant
Need substantial remaining. Q0 = observation point (e.g. M2).
Constra... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
):
real
weyl
1
2
× 1
× 2
β(3) =
β(2) =
4
3 ×
2 +
11 +
11
3 ×
−
−
1
6
7
=
× 2 −
4
3 ×
12
1
1
× 2 ×2
weyl
+
1
19
1
=
3 × 2 − 6
β(1) =
3 4
5 {3 ×
3
f amilies
[6
����
×
1
( )2 + 3
6
2
( )2 + 3( )2 + 2
3
1
3
1
( )2 + 12]
2
×
×
1
× 2
weyl
41
����
10
����
1
( )2 =
2 }
+
1
3 ×
2
·
MSSM:
Δβ(3)... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
36)
3.1. SU(5) UNIFICATION
=
25
6
Higgsions:
Δβ(1) =
3 4
5{3 ×
2
(2
×
×
1
( )2)
2
1
( )2 + 3
6
×
1
3 ×
1
3 ×
3[6
×
1
( )2
2 }
+
+
5
2
=
×
1
2
weyl
2
����
( )2 + 3
3
1
( )2 + 2
3
×
1
( )2 + 1]
2
×
Leaving out Higgs:
MSM:
MSSM:
β(3) =
β(2) =
7
−
10
− 3
β(1) = 4
Δβ(3) = 4
=
Δβ(2) =
10
... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
6)
(3.47)
6 effective
⇒
16
CHAPTER 3. GRAND UNIFIED THEORY
ln
MU
1
βj
Q ∝ βi −
1
βi −
1016.5
)SU SY �
(
1012 11 +2
9·
βj
βj
�
(
1
βi −
1017
102 →
(3.48)
)M SM (1
2
11
)
�
−
11
1
(
9 βi −
βj
)M SM
(3.49)
(3.50)
3.2. SU (5)
17
3.2 SU (5)
Fermion Multiplets
•
Needs U (1) traceless.
LH fiel... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
1,
2
3
1−
6
1
•
Symmetry Breaking
Adjoint (traceless)
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
→
SU (5)
2
2
2
3
−
3
−
SU (2)
×
×
SU (3)
SU (3)
⎞
M
⎟
⎟
⎟
⎟
⎟
⎟
⎠
U (1)
�
SU (2)
�
Still need SU (2)
U (1)
×
⇒
U (1).
(3.52)
(3.53)
(3.54)
(3.55)
(3.56)
(3.57)
18
CHAPTER 3. GRAND UNIFIED THEORY
Minimal:
0
0
0
0
υ
√... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
η
η
−
�
�
�
where, a and b are Dirac indices.
Therefore,
η(i)�
a �abη(j)
b
Note symmetric in i
⇔
j, due to Fermi statistics.
(3.58)
(3.59)
(3.60)
(3.61)
(3.62)
(3.63)
(3.64)
(3.65)
(3.66)
(3.67)
(3.68)
(3.69)
3.2. SU (5)
Link between texture and representation at unification: (gab symmetric)
all... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
(3.75)
(3.76)
(3.77)
Single appearance of �αβγ.
Phenomenology
M large.
⇒
•
Implementing SB; Hierarchy problem ϕ+ϕ, ϕ+Aϕ, ϕ+A2ϕ, trA2ϕ+ϕ, trA2 ,
trA3 , trA4 , tr(A2)2 . Need big vev for A, small for ϕ. Heavy ϕα not by decou
pling, but by conspiracy. Not inconsistent, but ugly.
20
CHAPTER 3. GRAND UNIFIED THEOR... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
(3.78)
(3.79)
(3.80)
(3.81)
(3.82)
(3.83)
(3.84)
3.3. SO(10) UNIFICATION
3.3 SO(10) Unification
SU (6)?
SU (5) in SO(10): 5 complex components
5
6
×
2
, F ab , T µν
Zj = Xj + iYj
< Z �
Z > =
|
X �Xj + Yj�Yj
j
+i
SO(10) leaves this part invariant
�
Xj�Yj −
Yj�Xj
SP (10) leaves this part invariant
�
��... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
.)
1[Γk, Γl] satisfy the SO(10) commutators.
4
Γk, Γl}
{
= 2δkl
[Γk, Γl] = 2(ΓkΓl −
δkl)
1
16
[[Γk, Γl], [Γm, Γn]] =
=
1
4
1
4
[ΓkΓl, ΓmΓn]
(ΓkΓlΓmΓn −
ΓmΓnΓkΓl)
Claim:
−
So
Now use
21
(3.85)
(3.86)
(3.87)
(3.88)
(3.90)
(3.91)
(3.92)
(3.93)
(3.94)
(3.95)
22
CHAPTER 3. GRAND UNIFIED THEORY
ΓaΓb... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
δnk[Γm, Γl] + δmk[Γn, Γl]
(
−
−
δnl[Γk, Γm] + δml[Γk, Γn])
(3.99)
1
4
Compare to
QED.
δnkΓlm
δnlΓkm
−
−
δmkΓln + δmlΓkn
Construction of Γ matrices:
U −
1(R)T µU (R) = RµT ν
ν
Γ1 = σ1 ⊗
Γ2 = σ2 ⊗
Γ3 = σ3 ⊗
Γ4 = σ3 ⊗
Γ5 = σ3 ⊗
. . .
1
1
1
⊗
⊗
⊗
1
1
1
⊗
⊗
⊗
σ1 ⊗
1
1
⊗
⊗
σ2 ⊗
1
1
⊗
⊗
σ1 ⊗
σ2 ⊗
1
⊗
1 ... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
⊗
⊗
SO(2)
⊗
SO(2)
⊂
SO(10)
(3.109)
This gives us a 25 = 32 – dimensional representation of SO(10) by
R(e
iθabTab)
iθab(
= e
1
− 4 [Γa,Γb])
(3.110)
It is not quite irreducible.
Note
K =
iΓ1Γ2
−
Γ10
· · ·
(3.111)
anticommutes with all the Γi.
Also K ∗ = K and K is Hermitean (exercise).
K 2 =
iΓi
−
· · ·
Γ1... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
σ3 ⊗
σ3
(3.114)
(3.115)
24
CHAPTER 3. GRAND UNIFIED THEORY
J
=
υ
Δυ�Jυ
→
=
=
⇒spinor
Similarly we identify
0
1
1 0
−
⎛
.
.
.
−
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
υ + �Gυ
�υ�(GT J + JG)υ
GJ + JG)υ
�υ�(
−
σ2 invariant
0
1
1 0
�
SU (3)
SU (2)
⊂
⊂
SO(6)
SO(4)
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
T12 +
· · ·
+ T910
(3.116)
(3.1... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
0 ∝
U (1)Y
1 state
+ + + ++ >
U (1) singlet
|
×
SU (3)
SU (2)
×
1 + signs
(3.123)
(3.124)
(3.125)
(3.126)
5 state, 2 types
>
+
| − −
−−
3.3. SO(10) UNIFICATION
25
+
− −−
− − −−
>
>
| −
+
|
Y˜ =
1
6
1)
(
−
+
−
1
4
2) =
(
−
SU (3) (3)
��
�
�
SU (2) singlet
1
3
>
−
+ >
| − − −
| − − − −
SU (3) sin... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
C
R
=
−
−
−
|
|
|
�
|
|
1
Y = (
6
1)
−
−
1
4
(2)
(3.127)
(3.128)
(3.129)
(3.130)
(3.131)
26
CHAPTER 3. GRAND UNIFIED THEORY
+ + +
−
+ +
+
−
− −
+
−
−
+ +
−
|
|
|
|
>
−
+ >
+ >
+ >
SU (3) 3
SU (2) 2
u
d
�
�
L
Y =
1
6
(1) =
1
6 ≡
Comments:
(3.132)
1. The construction of Γ –matrices – anticommut... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
Higgs doublet, in the form
SU (2)
×
×
µij N¯i Lµjϕ∗
R
µ
(3.135)
where, i, j are formly indices and µ = SU (2) index.
By 2nd order perturbation theory we induce Majorana Masses for the νL.
<Φ>
<Φ>
M
Figure 3.3: Majorana Masses
m
2µ
∼ M
(3.136)
3.3. SO(10) UNIFICATION
Breaking Scheme: Higgs ϕ in 16
< ϕN >... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
(3.145)
�
28
CHAPTER 3. GRAND UNIFIED THEORY
Ce :R
α
3
+
γ
6
+
2
6
=
5
3
= 1
+ γ = 1
β
−
αB + βL + γY = 5B
There are 45adjoint and 10vectro as before.
Fermion masses: bilinears in ϕ
5L
4Y
−
−
Γ1 =
Γ2 =
. . .
σ1 ⊗
σ2 ⊗
1
1
⊗ · · · ⊗
⊗ · · · ⊗
1
1
ϕ∗ transforms as e[Γµ,Γν ]∗
(Cϕ∗)� = Ce[Γµ,Γν ]∗
ϕ... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
16
16
×
×
16 = 10vector + 1203
S
−
A
tensor + 1265
−
tensor:self
dual
−
S
16 = 1scale + 452
�
� �� �
−
tensor + 2104
�
��
�
−
tensor
��
�
(3.155)
Totally
(3.156)
(3.157)
3.3. SO(10) UNIFICATION
29
Final comments on SO(10) vs. SU (5): 3 RH neutrinos vs. SU (5)
quantization.
�(B
Q
∝
−
L)
U (1). Charg... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/1bb0ca3337af13703289df734a424e7c_chap3.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.005 Elements of Software Construction
Fall 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Today
conclusion
¾ take-away messages
¾ what to do next
project 3 awards
6.005 quiz game
HKN evaluations
HKN evaluations... | https://ocw.mit.edu/courses/6-005-elements-of-software-construction-fall-2008/1bb173ac0fc980981d33851b7de8af6b_MIT6_005f08_lec23.pdf |
6.720J/3.43J - Integrated Microelectronic Devices - Spring 2007
Lecture 3-1
Lecture 3 - Carrier Statistics in Equilibrium
(cont.)
February 9, 2007
Contents:
1. Equilibrium electron concentration
2. Equilibrium hole concentration
3. np product in equilibrium
4. Location of Fermi level
Reading assignment:
del Ala... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1bccffbfa7b4487be99b8af8b46186ef_lecture3.pdf |
to answer question:
1. derive relationship between no and EF
2. derive relationship between po and EF
3. derive expressions for nopo and ni
4. figure out location of EF from additional arguments (such as
charge neutrality)
Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Sp... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1bccffbfa7b4487be99b8af8b46186ef_lecture3.pdf |
:
no =
(cid:2) ∞
Ec
no(E) dE
At a certain energy, no(E) is the product of CB density of states and
occupation probability:
no(E) = gc(E) f (E)
Then:
⎛
no = 4π
⎝
√
2m ∗
de
h2
⎞
3/2 ∞
E − Ec
(cid:2)
⎠
Ec 1 + exp E−
EF
kT
dE
Refer energy scale to Ec and normalize by kT . That is, define:
η =
E − Ec
k... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1bccffbfa7b4487be99b8af8b46186ef_lecture3.pdf |
0 1 + eη−ηc
dη
F1/2(x) is Fermi integral of order 1/2.
Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.720J/3.43J - Integrated Microelectronic Devi... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1bccffbfa7b4487be99b8af8b46186ef_lecture3.pdf |
− EF (cid:11) kT , or no (cid:10) Nc:
no (cid:7) Nc exp
EF − Ec
kT
Simple exponential relationship when Fermi level is well below con
duction band edge.
f(E)
gc(E-Ec)
no(E)
Ec
E
EF
Ec
E
Can obtain same result with Maxwell-Boltzmann statistics for f (E).
Cite as: Jesús del Alamo, course materials for 6.720J... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1bccffbfa7b4487be99b8af8b46186ef_lecture3.pdf |
concentration
• Q: How many holes are there in a semiconductor in TE?
A: It depends on location of EF .
• Why? Because EF defines probability that states are occupied by
electrons.
• The closer EF is to the valence band edge, the more holes there
are in the valence band.
1-f(E)
gv(Ev-E)
po(E)
Ev E
Ev
EF
E
•... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1bccffbfa7b4487be99b8af8b46186ef_lecture3.pdf |
19 cm−3
Then:
po = NvF1/2(ηv)
Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.720J/3.43J - Integrated Microelectronic Devices - Spring 2007
Lecture... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1bccffbfa7b4487be99b8af8b46186ef_lecture3.pdf |
n
o
Ev
Ec
EF
Fermi-Dirac for electrons
Maxwell-Boltzmann for holes
Maxwell-Boltzmann for electrons
Maxwell-Boltzmann for holes
Maxwell-Boltzmann for electrons
Fermi-Dirac for holes
Ec
Ev
Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Spring 2007.
MIT OpenCourseWare... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1bccffbfa7b4487be99b8af8b46186ef_lecture3.pdf |
• T ↑ ⇒ ni
• Eg ↑ ⇒ ni
• Nc ↑, Nv ↑ ⇒ ni
Remember: in Si at RT: ni (cid:7) 1010 cm−3 ((cid:10) Nc, Nv)
Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1bccffbfa7b4487be99b8af8b46186ef_lecture3.pdf |
extrinsic semiconductor
Require no (cid:7) ND. If non-degenerate (ND (cid:10) Nc):
EF − Ec (cid:7) kT ln
ND
Nc
Evolution of EF with doping:
non-degenerate
degenerate
intrinsic
extrinsic
Ec
Ei
Ev
EF
ni
Nc
log ND
� p-type extrinsic semiconductor
Require po (cid:7) NA. If non-degenerate (NA (cid:10) Nv):
... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1bccffbfa7b4487be99b8af8b46186ef_lecture3.pdf |
– p-type non-degenerate semiconductor:
po (cid:7) NA,
EF − Ev (cid:7) kT ln
ND
Nc
Nv
NA
• Order of magnitude of key parameters for Si at 300 K:
– effective density of states of CB and VB: Nc, Nv ∼ 1019 cm−3
Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Spring 2007.
M... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1bccffbfa7b4487be99b8af8b46186ef_lecture3.pdf |
3.032 Mechanical Behavior of Materials
Fall 2007
shear bands (red) forming in
polycrystalline elemental metal
with many line and point defects
Images removed due to copyright restrictions.
Please see:
http://www-geol.unine.ch/03_france/granites/Granites-Thumbnails/3.jpg
shear bands in granite
(complex crystal) form... | https://ocw.mit.edu/courses/3-032-mechanical-behavior-of-materials-fall-2007/1bd633873707143d7607ffb3b6d08fb6_lec20.pdf |
(CalTech). www.csem.caltech.edu/Facilities/tem.html
Data sources: Applied Physics Letters, 82: 1030–1032, (2003); TG Nieh, Bulk Metallic Glasses, Ch. 6.
Lecture 20 (10.26.07)
3.032 Mechanical Behavior of Materials
Fall 2007
fibrils of polymer
hydrocarbon chains
aligned within fibril
crazing in amorphous polymer
r... | https://ocw.mit.edu/courses/3-032-mechanical-behavior-of-materials-fall-2007/1bd633873707143d7607ffb3b6d08fb6_lec20.pdf |
20.110/5.60 Fall 2005
Lecture #10
page
1
Chemical Equilibrium
Ideal Gases
Question: What is the composition of a reacting mixture of ideal
gases?
e.g.
½ N2(g, T, p) + 3/2 H2(g, T, p) = NH3(g, T, p)
What are
p p
,
H
2
N
2
, and
p at equilibrium?
NH3
Let’s look at a more general case
νA A(g, T, p) + νB B(g... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/1bdff1fc0dbacbe7a759e6ca10d918d3_l10.pdf |
W. Field
20.110/5.60 Fall 2005
Lecture #10
page
2
where ε is an arbitrary small number that allows to let the reaction
proceed just a bit.
We know that
µ
i
(
)
T p
,
g,
=
o
µ
i
(
T RT p
i
ln
+
)
⎡
⎢
⎣
p
i
1 bar
implied
⎤
⎥
⎦
where
o
(
i Tµ
)
is the standard chemical ... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/1bdff1fc0dbacbe7a759e6ca10d918d3_l10.pdf |
rxn
or
∆
o
o
= ∆
G G
form
(
products
)
− ∆
o
G
form
(
reactants
)
If
ε∆
G
<
( ) 0
then the reaction will proceed spontaneously to form
more products
ε∆
G
>
( ) 0
then the backward reaction is spontaneous
ε∆
G
=
( ) 0
No spontaneous changes ⇒
Equilibrium
20.110J / 2.772J / 5.601JThermodynamics of Biomolecular ... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/1bdff1fc0dbacbe7a759e6ca10d918d3_l10.pdf |
by 1 bar, so Kp and KX are
both unitless.
________________________________________________
Example: H2(g) + CO2(g) = H2O(g) + CO(g)
T = 298 K
p =1 bar
H2(g) CO2(g) H2O(g) CO(g)
a
b
a-x
b-x
0
x
0
x
Initial #
of moles
# moles
at Eq.
Total # moles at Eq. = (a – x) + (b – x) + 2x = a + b
Mole fraction
... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/1bdff1fc0dbacbe7a759e6ca10d918d3_l10.pdf |
H
2
CO
2
=
2
x
)(
(
a x b
−
)
−x
Let’s take a = 1 mol and b = 2 mol
We need to solve
(
1
−
2
x
)(
2
x
=
9.7 10
x
−
6
−
x
)
A) Using approximation method:
K << 1, so we expect x << 1 also.
Assume
1
−
x
≈
1, 2
−
x
≈
2
⇒
x ≈
0.0044 mol (indeed
1
(
<<
2
x
)(
2
−
x
≈
2
x
2
−
x
)
=
9.7 10
x
−
6
1)
B)
Exactly:
2
x
−
... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/1bdff1fc0dbacbe7a759e6ca10d918d3_l10.pdf |
.110J / 2.772J / 5.601JThermodynamics of Biomolecular SystemsInstructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #10
page
5
Effect of total pressure: example
N2O4(g) = 2 NO2(g)
I... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/1bdff1fc0dbacbe7a759e6ca10d918d3_l10.pdf |
2
α
=
⎛
+⎜
1
⎜
⎝
p
4
K
p
⎞
⎟
⎟
⎠
∴
If p increases, α decreases
Le Chatelier’s Principle, for pressure:
An increase in pressure shifts the equilibrium so as to decrease the
total # of moles, reducing the volume.
In the example above, increasing p shifts the equilibrium toward the
reactants.
---------------
20.11... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/1bdff1fc0dbacbe7a759e6ca10d918d3_l10.pdf |
3 -
≈x
2
K
p
≈
1
p
2
−
x
3
)
(
1
(
or 1
−
3
)
x
≈
2
pK
p
x
=
1
−
⎛
⎜
⎜
⎝
2
pK
p
1 3
⎞
⎟
⎟
⎠
In this case, if p↑ then x↑ as expected from Le Chatelier’s principle.
20.110J / 2.772J / 5.601JThermodynamics of Biomolecular SystemsInstructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Fie... | https://ocw.mit.edu/courses/20-110j-thermodynamics-of-biomolecular-systems-fall-2005/1bdff1fc0dbacbe7a759e6ca10d918d3_l10.pdf |
2.160 Table of Contents
1. Introduction
Physical modeling vs. Black-box modeling
System Identification in a Nutshell
Applications
Part 1 ESTIMATION
2. Parameter Estimation for Deterministic Systems
2.1 Least Squares Estimation
2.2 The Recursive Least-Squares Algorithm
2.3 Physical meanings and properties of ma... | https://ocw.mit.edu/courses/2-160-identification-estimation-and-learning-spring-2006/1bf206dfb15f096397afc979ba28bddb_updtd_table_cont.pdf |
6.1 Model Sets
6.2 A Family of Transfer Function Models
6.2.1 ARX Model Structure
6.2.2 Linear Regressions
6.2.3 ARMAX Model Structure
6.2.4 Pseudo-linear Regressions
6.2.5 Output Error Model Structure
6.3 State Space Model
6.4 Consistent and Unbiased Estimation: Preview of Part 3, System ID
6.5 Times-Series D... | https://ocw.mit.edu/courses/2-160-identification-estimation-and-learning-spring-2006/1bf206dfb15f096397afc979ba28bddb_updtd_table_cont.pdf |
13 Asymptotic Distribution of Parameter Estimates
13.1 Overview
13.2 Central Limit Theorems.
13.3 Estimate Distribution
13.4 Expression for the Asymptotic Variance
13.5 Frequency-Domain Expressions for the Asymptotic Variance
14 Experiment Design
14.1 Review of System ID Theories for Experiment Design
Key Requi... | https://ocw.mit.edu/courses/2-160-identification-estimation-and-learning-spring-2006/1bf206dfb15f096397afc979ba28bddb_updtd_table_cont.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
(cid:10) 6.642 Continuum Electromechanics
Fall 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
(cid:13)
6.642, Continuum Electromechanics
Prof. Markus Zahn
Lecture 4: Continuum Electromechanics (Melcher) – Sections 2.18... | https://ocw.mit.edu/courses/6-642-continuum-electromechanics-fall-2008/1bf674b3da895a32ac2d670975ead983_lec04_f08.pdf |
ε
⇒ Φ
=
∫
V
4
π
ρ
ε
dV
r - r '
∇
2f = 0
⇒
f = 0
⇒
C = A
C. Vector Poisson’s Equation Solutions
2
∇
A = - J
μ ⇒
( )
A r =
μ
π ∫
4
V
)
(
J r ' dV
r - r '
6.642, Continuum Electromechanics Lecture 4
Prof. Markus Zahn Page 1 of 6
_
Courtesy of M... | https://ocw.mit.edu/courses/6-642-continuum-electromechanics-fall-2008/1bf674b3da895a32ac2d670975ead983_lec04_f08.pdf |
∂ ⇒
A
∂
r
∂
-
A
∂
r
∂
dr +
A
∂
θ
∂
dθ = dA = 0
⇒
3. Axisymmetric Cylindrical
A =
Λ
(
r, z
r
)
−
i
θ
1
B = × A = -
r
∇
∂Λ
z
∂
1
−
i +
r
r
∂Λ
r
∂
−
i
z
dz
dr
=
B
B
z
r
=
1
r
1
r
-
∂Λ
r
∂ ⇒
∂Λ
z
∂
∂Λ
r
∂
dr +
∂Λ
z
∂
dz = d = 0
Λ
Λ
(
r, z = constant
)
4. Axisymmetric Spherical
A =
Λ
r, θ
)
(
r sin θ
−
i
φ
B =
∇ ×
... | https://ocw.mit.edu/courses/6-642-continuum-electromechanics-fall-2008/1bf674b3da895a32ac2d670975ead983_lec04_f08.pdf |
= 0
[Vector Laplace’s Equation]
A. Cartesian Coordinates
2
∇
A =
2
∇
⎡
⎢
⎣
−
−
A i + A i + A i
y
∇
∇
2
2
x
−
z
x
y
⎤
⎥
⎦
z
A = i R e A x e
(cid:105) (
⎡
⎣
)
-jky
⎤
⎦
−
z
(cid:105) (
)
A x =
α
(cid:105)
(cid:105)
A sinh kx - A sinh k x -
β
(
Δ
)
sinh k
Δ
(cid:105)
H = -
y
(cid:105)
1 A
∂
x
μ ∂
= -
k
μ
⎡
⎢
⎢
⎣
α
(cid:... | https://ocw.mit.edu/courses/6-642-continuum-electromechanics-fall-2008/1bf674b3da895a32ac2d670975ead983_lec04_f08.pdf |
= -jk
α
β
(cid:105)
A
(cid:105)
A
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
6.642, Continuum Electromechanics Lecture 4
Prof. Markus Zahn Page 4 of 6
Courtesy of MIT Press. Used with permission.
B. Polar Coordinates
2
∇
A
2
= ∇
(
A r,
θ
)
⎛
⎜
⎝
−
i
z
⎞
⎟
⎠
= ⇒ ... | https://ocw.mit.edu/courses/6-642-continuum-electromechanics-fall-2008/1bf674b3da895a32ac2d670975ead983_lec04_f08.pdf |
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