Datasets:
vnpthack
/
vnpt_data

Dataset card Data Studio Files
xet
 Community
1
text
stringlengths
0
6.73k
the moment, the integral we wish is
0=œo eiaur/o
J ——— jpdịp. (30.13)
To do this integral we need to use the relation bebween r and ø:
r?ˆ= p?+ 2. (30.14)
Since z is independent of ø, when we take the diferential of this equation, we get
2r dr = 2p dp,
which is lucky, since in our integral we can replace øđø by r dr and the z will
cancel the one in the denominator. 'Phe integral we want is then the simpler one
?=CC -
J e~⁄/$ dự, (30.15)
To integrate an exponential is very easy. We divide by the coeflicient oŸ r in the
exponent and evaluate the exponential at the limits. But the limits of z are not
the same as the limits of . When ø = 0, we have r = z, so the limits oŸ z are z
to infinity. We get for the integral
— C —i¡œ __ „—(iœ/c)z 30.16
¬"m.“nnh (30.16)
where we have written oo for (œ/c)oo, since they both just mean a very large
numberl -
NÑow e"??° is a mysterious quantity. Its real part, for example, is cos (—o©),
which, mathematically speaking, is completely indefnite (although we would
--- Trang 532 ---
expect i% to be somewhere—or everywhere (?)—between +1 and —1l). But in a
phụs?cal situation, 1t can mean something quite reasonable, and usually can Jjust
be taken to be zero. 'To see that this is so In our case, we go back to consider
again the original integral (30.15).
W©e can understand (30.15) as a sum of many small complex numbers, each
of magnitude Az, and with the angle Ø = —œr/c in the complex plane. We can
try to evaluate the sum by a graphical method. In Eig. 30-11 we have drawn the
first five pieces of the sum. Each segment of the curve has the length Az and is
placed at the angle AØ = —ưw Ar/c with respect to the preceding piece. The sum
for these first five pieces is represented by the arrow from the starting point to
the end of the fifth segment. As we continue to add pieces we shall trace out a
polygon until we get back to the starting point (approximately) and then start
around once more. Adding more pieces, we just go round and round, staying
close to a circle whose radius is easily shown to be c/œ. We can see now why the
integral does not give a definite answerl
ạ= —, lmaginary Axis
A0 = — âr
¬—- Real Axis
1" “¿0
lã \-Aø
cụ Sum >¬A0
Fig. 30-11. Graphical solution of J" e—ðr/€ qr,
But now we have to go back to the øñh#s¿cs of the situation. In any real
situation the plane of charges cœnno‡ be infnite in extent, but must sometime
stop. lfit stopped suddenly, and was exactly circular in shape, our integral would
have some value on the cirele in Fig. 30-11. If, however, we let the number of
charges in the plane gradually taper off at some large distance from the center
(or else stop suddenly but in an irregular shape so for larger ø the entire ring
--- Trang 533 ---
lmaginary Axis
_g—®>‹j¿Start;r =z_ Real Axis
Fig. 30-12. Graphical solution of J" re #/£ dự,
of width đø no longer contributes), then the coefficient r in the exact integral
would decrease toward zero. Since we are adding smaller pieces but still turning
through the same angle, the graph of our integral would then become a curve
which is a spiral. The spiral would eventually end up at the center of our original
circle, as drawn in Eig. 30-12. 'Phe ph¿#/s/call correct integral is the complex
number 4 in the figure represented by the interval from the starting point to the
center of the circle, which is just equal to
¬... (30.17)
as you can work out for yourself. This is the same result we would get from
E4q. (30.16) if we set e”?% = 0.
(There is also another reason why the contribution to the integral tapers of
for large values of z, and that is the factor we have omitted for the projection of
the acceleration on the plane perpendicular to the line P@Q.)
W© are, of course, interested only in physical situations, so we will take e—”%
cqual to zero. Returning to our original formula (30.12) for the ñeld and putting
back all of the factors that go with the integral, we have the result
Total fñeld at P= —- T zoe«ứ=#/2) (30.18)
(remembering that 1/2 = —)).
It is interesting to note that (2#oe”““) is just equal to the œelociy of the
charges, so that we can also write the equation for the fñeld as
Total fñeld at P = _¬ [veloeity of charges]a ¿ _ ;/e, (30.19)
--- Trang 534 ---
which is a little strange, because the retardation is just by the distance z, which
is the shortest distance from ? to the plane of charges. But that is the way
1t comes out—fortunately a rather simple formula. (We may add, by the way,
that although our derivation is valid only for distances far from the plane of
oscillatory charges, it turns out that the formula (30.18) or (30.19) is correct at
any distance z, even for z < À.)
--- Trang 535 ---
Tho €)riqgirt of tho lHofretcfftco InăiÏox
31-1 The index of refraction
WS have said before that light goes slower in water than in air, and slower,
slightly, in air than in vacuum. 'This effect is described by the index of refraction 0ø.
Now we would like to understand how such a slower velocity could come about. In
particular, we should try to see what the relation is to some physical assumptions,
or statements, we made earlier, which were the following:
(a) That the total electric field in any physical circumstance can always be
represented by the sum of the fñelds rom all the charges in the universe.
(b) That the fñeld from a single charge is given by its acceleration evaluated
with a retardation at the speed œ, aøa¿/s (for the radiafion feld).
But, for a piece of glass, you might think: “Oh, no, you should modify all
this. You should say it is retarded at the speed c/w” That, however, is not right,
and we have to understand why it is not.
lt 7s approximately true that light or any electrical wave đoes øppear to travel
at the speed c/n through a material whose index of refraction is nø, but the
fñelds are still produced by the motions oŸ øÏ/ the charges——including the charges