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hwk04ans
Ast 4001, Oct 2015
Answers for homework set 4
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1. In an ordinary gas, we expect the average kinetic energy of a particle to be E av  1.5 k T .
This is untrue, however, for a thermal gas of photons -- i.e., the Planck distribution,
energy density u = A  / { exp ( h  / k T )  1 } .
(a) Devise an easy way to calculate the peak of f ( x ) = x 3 / { e x  1 } to an accuracy
better than 5 decimal places. In other words, the value of x where f ( x ) is maximum,
and the value of that maximum.
Simplest solution: Of course the maximum occurs at the frequency where d f / d x = 0.
As often happens for multifactor expressions, it’s easier to differentiate the logarithm,
d ln f / d x = 0. A good way to write the resulting equation is x = 3 ( 1  e –x ) .
By inspection (or by trying two or three sample values), x will be fairly close to 3 .
Therefore the exponential term is only a small perturbation, so a stable iteration formula
is x n +1 = 3 [ 1  exp (  x n ) ] . This is well adapted for easy calculating.
Starting with a guess of 3.0 (for example), we find successive values:
2.85,
2.827,
2.8224,
2.82160,
2.82147,
2.82145,
2.82144.
This convergence is slow by formal numerical-analysis standards, but it’s good enough
for a problem that we only need to do once. It minimizes the amount of thinking!
The maximum value of f ( x ) turns out to be 1.4214…
Note: Often, when you try to solve an equation this way, the iterations diverge. In
such cases, sometimes you can make it converge by rearranging the terms. Try to find
a form like ( relatively sensitive expression in x ) = ( less sensitive expression ) .
Of course Newton’s Method is better for mass production or for elaborate equations.
Many software packages do this stuff automatically, but check the results because
they can fail for various reasons.
(b) For temperature T = 6000 K, evaluate the photon wavelength  which corresponds to
the peak described in part (a). Express your answer in m , nm , and Å .
Answer: A handy fact or number worth remembering is
x = h  / k T  ( 14390 m K ) /  T .
If T = 6000 K , then evidently x = 2.82… at  0.85 m  850 nm  8500 Å .
Note that this is near-infrared, not visible light as one might have expected.
u, on the other hand, peaks at x  4.965 and   0.483 m, a blue wavelength.
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hwk04ans - p2
(c) Explain why the average photon energy, E av , is directly related to the quantity
X =

f (x) dx
/
x f ( x ) d x .
Calculate the value of X to at least five decimal places.
value of x found in part (a).
Compare X to the special
Answer: Since f ( x ) is related to energy density, and a photon has energy h = x k T ,
the x f ( x ) integral is proportional to the number density, i.e., density of photons.
Hence the ratio of integrals is proportional to ( energy density ) / (photon density ) =
energy per photon. These two definite integrals are pretty famous and they have
values 6.493939402… and 2.404114… ; so X = 2.70117… This factoid may be
worth remembering: the average photon energy in a Planck distribution is 2.7 kT .
(d) For a photon gas, what value does C have in E av = C k T ? Compare this to the
value of C that applies for an ordinary ideal gas of non-relativistic particles.
(Here E av refers only to the kinetic energies, excluding "internal degrees of freedom"
that atoms and molecules usually have.)
Answers: For photons, C  2.701 as noted above at the end of the part (c) answer.
For an ideal, non-relativistic, monatomic gas, the average kinetic energy per particle
is famously 3 × ( k T / 2 ), i.e. k T / 2 for each degree of freedom; so C = 1.5.
Thermal photons evidently have considerably more energy per particle than an ordinary
gas with the same temperature.
----- problem 2 is on the next page -----
hwk04ans - p3
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2. Concerning the Saha equation for ionization. Imagine a sample of pure hydrogen gas in
LTE, but there are no H 2 molecules. We’ll use n to denote density of particles per unit
volume:
n ( H 0 ) + n ( H + ) = n H , the total gas density;
n ( e ) = n ( H + ) , because in pure hydrogen every free electron must be supplied
by a hydrogen atom;
 = m H n H , the mass density.
Some of the formulae will be a little easier to write if we measure temperature T in units of
10000 K: define t = T / ( 10 4 K ). Then, if I H = 13.6 eV is the ionization potential
of hydrogen, the ratio I H / kT is equal to 15.78 / t . It is also convenient to denote
the ionization fraction as x ;
x = n ( H + ) / n H = n ( H + ) /{ n ( H0 ) + n ( H + ) } .
The Saha equation for hydrogen atoms can be written as
n ( H + ) n ( e ) / n ( H 0 ) = C t 1.5 exp (  15.78 / t ) .
(a) If we regard t as a dimensionless parameter, what dimensions or units does the constant
C have?
Answer: The left side has dimensions (density) × (density) / (density) = density,
and on the right side C is the only factor that is not dimensionless.
Therefore C must have the dimensions of a number density, units m or cm .
(b) Rewrite the equation so the left side is an expression containing the ionization fraction x
while n H and t appear only on the right side.
Answer:
x 2 / (1  x ) = ( C / n H ) t 1.5 exp (  15.78 / t ) .
(c) Now assume that the gas has a particular fixed value of the total density n H . At this
density the gas is 50% ionized, n ( H 0 ) = n ( H + ) , at temperature T = 8000 K,
t = 0.8. (This is a reasonable value in many stellar atmospheres.) Accurately
calculate the ionization fraction at temperatures of 7000 and 9000 K.
Answer: First, adjust C / n H so that it gives the stated 50% ionization at 8000 K.
If x 2 / ( 1  x ) = 0.5 at t = 0.8, then we must have C / n H = 2.575 × 10 8 .
Next, plugging into the formula found in part (b), we calculate x 2 / ( 1  x ) =
0.02445 for t = 0.7 and 5.340 for t = 0.9. …
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hwk04ans - p4
----- part (c) continued: -----
… The solution to quadratic equation
x 2 / ( 1  x ) = Q can be written this way:
x = 2 Q / { Q + sqrt ( Q 2 + 4 Q ) } .
(This isn’t the form that most people remember from high school algebra,
but it is equivalent and it has no minus signs.) Plugging in Q = 0.02445 and
5.34 respectively, we find x = 0.1446 at 7000 K and x = 0.861 at 9000 K.
Thus, in this case a 29-percent increase in T raises the ionization fraction by
a factor of 6.
(d) Next, suppose we wish to calculate the temperature that gives some particular
ionization fraction x . Invent, test, and describe a simple iterative procedure that
will do this.
Solution: (This is much like the attitude expressed in problem 1a. For mass-production
calculations we’d develop an algorithm based on Newton’s method, but here we use a
technique that converges slower but is much simpler to prepare.)
Temperature appears in two factors, t 1.5 and exp ( -15.78 / t ). The first of these is
far less sensitive to t , so put it on the right side of the equation:
exp ( 15.78 / t ) = ( C / n H ) t 1.5 { x 2 / (1  x ) } ,
or
t = 15.78 / ln [ ( C / n H ) t 1.5 { (1  x ) / x 2 } ] .
Now suppose that n H and x are known quantities and we wish to calculate t . The
right side is comparatively insensitive to t . Start by putting a reasonable guess for t
into the right-side expression, use the formula to calculate an improved value of t ,
and then reiterate. As an example, here’s one of the cases quoted in part (c).
C / n H = 2.575 × 10 8 and x = 0.5, so we wish to solve
t = 15.78 / ln ( 5.15 × 10 8 t 1.5 ).
To demonstrate the method, let’s intentionally use a poor starting guess, t = 2.0 which
is far too high. Successive iterations give the following values.
t = 2.0, 0.748, 0.8041, 0.79969, 0.80003, 0.80000, 0.80000.
As we said earlier: If you’re writing, say, a computer code that will be used many times,
then it’s usually best to adopt a professional-grade technique such as Newton’s method
because each individual case converges faster. . On the other hand, sometimes we don’t
mind if an algorithm takes 5 iterations to converge instead of 3. Try to minimize
overall effort, and meanwhile aim for whatever precision you really need.
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