Download hwk03ans

[ kd ]
hwk03ans
Ast 4001, Fall 2015
Answers for homework set 3
________________________________________________________________________________________
1. Imagine a sphere with radius R , surrounded by empty space. Inside the sphere we have
scattering coefficient k sc = k = constant and k abs = 0 ; i.e., radiation is scattered but is
not absorbed. The radiation emission rate per unit volume,  , has a uniform constant value
in r < R . Let’s estimate the trapped radiation density U ( r ) via the diffusion approximation.
(a) Since there’s no absorption, all the emitted radiation eventually escapes, usually after
being scattered repeatedly. Evaluate L , the total luminosity or radiated power produced
by the sphere. (This isn’t a trick question; the obvious answer is correct.)
Answer: Simply L = volume ×  = 4  R 3  / 3 . This question was asked because
it might be suggestive for part (b).
(b) Deduce a formula for F ( r ), the local radiation flux at radius r inside the sphere, in two
different ways. Explain. (Hint: One way is mathematically formal, the other isn’t.)
Answers: First the formal way: solve F = With spherical symmetry, the flux
is entirely radial and its divergence F is r d ( r F ) / d r . So the equation
becomes d ( r F ) / d r = r which has solution r F = r
Of course the general solution has an additive constant, which is zero in this
problem because we need r F = 0 at r = 0. A non-zero value at r = 0
would imply a singularity in F . A point source would produce such a singularity,
but there’s no point source in this problem.)
Therefore F ( r ) =
r
based on the F equation.
Informally (and more quickly) we get the same result by visualizing where the
radiation is coming from. The luminosity produced inside radius r is 4  r 3  / 3
like question (a) above. Therefore F ( r ) = (luminosity) / 4  r 2 = r
(c) Find two approximate formulae for U ( r ) inside the sphere. Use two different boundary
conditions:
* First, F ( R )  av c U ( R ) with av  2 / 3 , which is semi-realistic.
* Then a simpler rough approximation: U ( R )  0 . This is obviously wrong
near the surface, but turns out to be useful deep inside the sphere.
{ U ( R )  0 may appear absurd at first sight, but it simplifies the
math. It is mathematically feasible because F is related to d U / d r ,
not U . }
Compare these two solutions at the center, r = 0 . What is the condition for
their difference to be relatively small?
----- answers for part (c) are on next page -----
4001 hwk03ans - p2
(problem 1, continued:)
Answer for part (c): Knowing F ( r ) = r from part (b) , we can estimate
U ( r ) from the diffusion equation: d U / d r =  3 k F / c =  k r / c .
Since k and  are constants inside the sphere, we integrate to find
U ( r )   k r 2 / 2 c + (constant) .
Deducing the constant from the first boundary condition that was suggested above,
we find
U ( r )  ( k  / 2 c ) ( R 2  r 2 ) + R / 2 .
(The boundary parameter av  2 / 3 was chosen partly because it leads to a simple
formula here. Remember that any av in the range 0.5 to 0.75 can be a more or less
acceptable approximation in most problems with well-defined boundaries.)
With a simplified boundary condition U ( R )  0 , the last term disappears:
U ( r )  ( k  / 2c ) ( R2  r2 ) .
At r = 0, the difference between these two solutions is much smaller than U
if R / 2 << k R 2 / 2 c , which is true if k R >> 1 . Note that k R is
the optical depth  corresponding to radius R . This is one instance of a useful
generality: The choice of boundary condition has little effect in deep regions
where  >> 1.
(d) If that condition is satisfied, i.e., if the simpler form is a tolerable approximation,
integrate it over volume to estimate the total radiation energy trapped inside the
sphere, E = U d V .
Answer:
 4 r2 U ( r ) d r 

4 k R 5 / 15 c .
(e) Comparing E to L , deduce the average radiation escape time t esc . Explain.
It’s OK to omit any term that is relatively small.
Answer: If we recall part (a) of this problem, the result in part (d) can be written as
E  ( k R 2 / 5 c ) ( 4  R 3  / 3 ) = (  R / 5 c ) L .
Since { trapped radiation E }  L  t esc , this formula implies
t esc k R 2 / 5 c   R / 5 c .
This fact is significant in the following way. If the radiation could escape without
scattering, then t esc would be of the order of R / c . The general rule, therefore,
is that scattering increases the escape time by a factor of the order of  , or anyway
proportional to  . This is usually true for any shape of configuration, provided
we evaluate  along the shortest escape route. (Imagine a cylinder, for instance.)
The factor of 1 / 5 for t esc is not surprising, if we consider that most of the
radiation is created fairly near the surface, say in r > 0.75 R . Obviously this
makes it easier for an "average" photon to escape.
4001 hwk03ans - p3
__________________________________________________________________________________________
2. Let’s use the equations F =  c k abs U +  and F   ( c / 3 k tot )  U for a case
with weak absorption and strong scattering. In rectilinear coordinates x , y , z , the half-infinite
region x > 0 has  = constant, k sc = constant, k abs = constant, and k sc >> k abs .
The region x < 0 is empty space, and some radiation escapes toward negative x through
the boundary at x = 0. For a boundary condition let’s use | F ( 0 ) | = avc U (0) with
av = 1 /  3 . (This choice of av simplifies the algebra, and is in the “credible” range
0.5 to 0.75.) Deep inside the gas, most radiation is absorbed before it can escape. Near the
surface x = 0, however, newly-created photons escape after a few scattering events.
The algebra may look better if you write k abs =  k and k sc = (1 –) k , where
k = k abs + k sc and  is a small constant.
(a) Before solving the equations, approximately guess the value of U for very large k x .
(Hint: Since almost all of the radiation produced in that region is absorbed before
it can escape, there must be an equilibrium balance of emission vs. absorption.)
Answer: The local equilibrium has absorption rate  emission rate per unit volume.
This can be expressed as c k abs U = c  k U   , so we expect U   / c  k at
large x . This is the same as saying that F  at large x .
(b) Find reasonable approximate solutions for U ( x ) and F ( x ) .
Important advice: The algebra here is a little tedious. In a case like this,
it is usually a good idea to define quantities that simplify the equations, for
example  ,  , and  in the outline below. If you don't do that, expressions
can become long and ugly. Good writers know that short paragraphs, short phrases
and short words are best; a good software coder uses compact subroutines and
functions; and a good scientist or engineer learns to streamline the math!
Solution for (b): Start with dF / d x =  c  k U +  and F   ( c / 3 k ) d U / d x .
Substituting the second into the first, we find d 2 U / d x 2 = + 3  k 2 U  3 k  / c .
Since the coefficients are all constants, the general solution for x > 0 is
U ( x ) = C 1 exp (   x ) + C 2 exp (   x ) +
 / ck ,
where  = (3 ) k .
Coefficient C 2 must be zero in order to avoid absurd behavior at large x . Meanwhile
F ( x )   ( c / 3 k ) d U / d x = + ( c  / 3 k ) C 1 exp (   x ) .
The boundary condition, F ( 0 ) =  c U (0) /  3 , gives C 1 . Be careful with +/
signs, because C 1 will be negative. (Do you see how we can predict that?) The result
is C 1 =   /  c k , where  = 1 / (1 +   ) . If  is small, then  is somewhat
less than 1 but not very much less...
----- continued on next page -----
4001 hwk03ans - p4
Solution for problem 1(c), continued:
... Therefore
U( x )  (  / c  k ) { 1   exp (  x ) } and
F ( x )   (   / ) exp (  x )   {  / (3 ) } {  / k } exp (  x ) .
Because the diffusion approximation is only an approximation and we’re not sure
about the precise value of av , we shouldn’t be surprised if the values are wrong
by 10 or 15 percent at the surface, x = 0. The accuracy improves deep inside the
gas, because the diffusion equation becomes more accurate there and the boundary
condition only affects the deep-interior solution by a small amount.
(c) Estimate the escaping radiation flux, F (0) .
Answer:
F (0)  {  / (3 ) } {  / k } .
When viewing this result, you should notice that  is fairly insensitive to  , and
it's always between 0.5 and 1.0 . Hence the other factors are more critical.
(d) What thickness  x would produce that much radiation [ i.e., F (0) ] via the
emissivity  ?
Answer: If we identify | F ( 0 ) | with   x (i.e., the total amount of radiation
flux produced in some thickness  x ), then  x  /  (3) k .
This concept is usually applied to cases with strong scattering but very weak
absorption, so  << 1. In those cases, for rough estimates we can ignore
 = 1 / (1 +   ) because it's "of order unity."
Summary: The radiation that escapes is mostly produced in a surface layer with
thickness given by (3) 1  2 k  x = (3) 1  2   roughly 1. This is
called the “thermalization depth,” because the emergent radiation has about the
same temperature as that region. (Deeper zones may be much hotter.)
In very hot stellar atmospheres and winds where k sc >> k abs , this concept
replaces the classical idea of a "photosphere."
__________________________________________________________________________________________