Ast 4001
Answers for mid-semester exam

1. (16 points) Answer any four of these five small questions. Each should take only
about two minutes.
(a) [calculate] A particular binary star normally has combined apparent magnitude
m (A+B) = 9.3, and its two stars A and B are normally equal in brightness.
Suppose that star A suddenly brightens by a factor of 4 - - i.e., its apparent
radiation flux f  increases by that factor. Calculate the resulting combined
magnitude m (A+B) . Express your answer to the nearest 0.1 magnitude.
Answer: Since we can use arbitrary units for f , let's say the radiation flux of each
star was initially 1. Hence the original total flux was f (A+B) = 1 + 1 = 2 units.
Then, after star A brightened, it became f (A+B) = 4 + 1 = 5 units. Evidently the
combined flux increased by a factor of 5 / 2 = 2.5, which corresponds to  m =
practically 1.0 magnitude. (We should remember that 1 magnitude  a flux factor
of exactly 10 0.4 = 2.512.)
Therefore m (A+B) = 9.3  1.0 = 8.3 after the brightening.
(b) [calculate] A particular star has n He = 0.1 n H (these n 's are numbers of
helium and hydrogen nuclei), and let's ignore heavier elements. In its photosphere
where T  4500 K , the gas constant is   6.6  10 7 c.g.s. units.
Quickly evaluate  at a location deep inside the star where T  300,000 K,
without using the values of physical constants k , m H , and m He .
Answer: Since P =   T = n k T , the macroscopic gas constant  is proportional
to the number of free particles per unit mass. (You might remember  = k / m av .)
In this problem the number of particles increases because the gas becomes ionized.
Imagine, for example, 10 hydrogen atoms plus 1 helium atom. At 4500 K they are
not ionized (you should remember this from other information in class), so we have
simply 11 particles. At 300,000 K they are fully ionized so we have
10 hydrogen nuclei + 1 helium nucleus + 12 electrons = 23 free particles
for the same amount of mass. Therefore  increases by a factor of 23 / 11 , leading
to   1.38  10 8 c.g.s. units at high temperature.
The above explanation fills 9 lines, but it takes only a few seconds to think the same
thoughts. For a hydrogen + helium mixture, the transition from non-ionized to fully
ionized occurs gradually from about 6000 K to roughly 30,000 K.
----- continued on next page -----
2
----- answers for question 1, continued -----
(c) [explain] In a typical stellar spectrum, ultraviolet absorption lines tend to appear
much stronger (deeper, darker) than those in the infrared. Why?
Answer: This was described in class. The figure shows two Planck-formula curves
for an idealized atmosphere with effective temperature T eff  10000 K.
At most wavelengths or frequencies, we "see" gas at the photosphere level,
for example 10000 K in the figure. That's the atmosphere depth where typical
escaping photons were created. At the wavelength of a spectral line, however,
the gas is more opaque; and this fact impedes escape from the photosphere level.
Therefore, in a spectral line we "see" mainly the cooler gas that's above the
photosphere. In simple models the lowest (outermost) temperature is roughly 20 %
cooler than the photosphere -- shown above by the 8000 K curve.
In the infrared, we see the Rayleigh-Jeans side of the Planck formula, so 8000 K gives
almost 80 percent as much light as 10000 K: 'A' in the figure. In the ultraviolet we
see the Wien-law, exponential part of the spectrum where 8000 K gives much less light
than 10000 K: 'B' in the figure. This is why UV absorption lines tend to be much
deeper (stronger, darker) than those in the infrared.
For an exam answer, it would be OK to give a shorter explanation -- just enough to
show that you understand what's going on. BTW, most real spectral lines are "weak,"
so in the figure their centers would have temperatures like 9000 K or 9500 K or even
9900 K. Another complication: The figure is inaccurate for real stars around 10000 K,
because the continuum opacity depends strongly on wavelength -- i.e., the photosphere
depth depends on wavelength. But this quibble doesn't affect the basic nature of the
absorption lines.
----- continued on next page -----
3
----- multi-part question 1, continued -----
(d) [explain] Two stars X and Y both have surface temperature T eff = 8000 K .
Hydrogen is mostly ionized in the photosphere of star X , but mostly un-ionized
in the photosphere of Y. They differ because one star is a main-sequence dwarf
while the other is a giant star with much weaker surface gravity.
Which is the giant, X or Y ? Briefly explain in terms of ionization equilibrium.
Answer: X, because it is more ionized. At any given temperature, lower gas densities
imply higher ionization.
(Why? Mainly because higher densities produce higher recombination rates.)
(e) [explain] The Planck formula for radiation energy density has its maximum
value u(max) at photon energy h  2.82 k T . Without using the actual
Planck formula, deduce the form of the relation between u(max) and T .
In other words, evaluate exponent n in u(max) = (constant)  T n .
Answer: u(max)  T 3 . We know that U =  u d   T 4 , and this
integral must scale like "height  width" where "height" is u(max).
The "width" of the curve is obviously proportional to the value of  at the peak,
which is proportional to T . Thus we have
(height)  (width) 
u(max)  T

T4
which implies
u(max)  T 3 .
The figure for answer (c), on the preceding page, shows an example. We expect
u(max at 10000 K) / u(max at 8000 K) = 1.25 3  1.95, which is confirmed
by the figure.
In most branches of science and technology, professionals habitually think in
terms of scaling laws like the above case, a.k.a. proportionality relations. It's not
just good skill to develop; it's almost a necessary one.
___________________________________________________________________________
4
_________________________________________________________________________
2. (11 points) The Sun, with effective temperature 5800 K, has an average density
of 1.4 g cm . Star X has spectral type B8 V, effective temperature 11600 K,
luminosity L = 256 L sun , and mass M = 4 M sun .
Calculate the following quantities for star X . Clearly indicate what formulae
you're using.
(a) Relative radius R / R sun .
(b) Average mass density .
(c) Relative gravitational potential energy, E G (X) / E G (Sun) . Assume that
star X's structure is similar ("homologous") to that of the Sun.
Answers:
(a) Let L , R , and T represent quantities relative to the Sun. Then L = R 2 T 4 ,
and we know that T = 2 and L = 256. Therefore
R = sqrt ( L / T 4 ) = sqrt ( 256 / 16 ) = 4 ,
meaning that the R (X) = 4 R (Sun) if we switch back to normally-defined radii.
(You probably noticed that every input and output quantity is a power of 2.)
Caution: Later in this course we'll be concerned with the central temperatures
of stars. Always keep the distinction between surface temperature and
central temperature very clear in your mind.
(b) Without any serious assumptions at all, we know that average density is
proportional to M / R 3. Relative to the Sun this is 4 / 4 3 = 1 / 16 .
Referring to the quoted average density of the Sun, we thus find about
0.09 g cm for star X, which is interestingly low. (A factoid worth
remembering: massive stars tend to be less dense than low-mass objects.)
Remember this too: If we assume that the interior structure is similar
("homologous"), then the M / R 3 scaling relation also applies to central
densities of stars. Be very clear about when this is true and when it isn't.
(c) E G =   ( G m / r ) d m  G M 2 / R  M 2 / R . With both M and
R four times that of the Sun, we find E G (X) = 4 E G (Sun) .
_________________________________________________________________________
5
_________________________________________________________________________
3. (10 points)
[ Caution: Don’t calculate any numbers that you don’t need! ]
Two samples of pure hydrogen gas are each in thermodynamic equilibrium at the same
temperature, about 8000 K. Both are partly ionized, and n ( e  ) = n ( H  ) because
every free electron has to come from a hydrogen atom. H 2 and H  are negligible.
Sample 1 has total hydrogen density n H = n ( H 0 ) + n ( H + ) = 9  10 12 cm3 .
It’s 50% ionized: n ( H + ) = 4.5  10 12 cm3 .
Sample 2 is 75% ionized, n ( H + ) = 0.75 n H , at the same temperature.
(a) Before calculating anything, which sample has the higher total density n H ?
(b) Calculate the total hydrogen density n H in sample 2 . Briefly explain.
Answers:
(a) Sample 1 must have higher density, since it is less ionized. Same as question
1(d), and the Saha equation, see below. Remember this qualitative relation.
(b) For this problem we do not need any physical constants, and we do not need
the temperature dependence of the Saha equation. Instead let's merely write
n (H+ ) n (e ) / n (H0 ) = f ( T ) .
In terms of ionization fraction x , the above formula becomes
x 2 / ( 1  x ) = f ( T ) / n H .
(There was a homework problem like this.)
The quantity on the left side is 0.5 for sample 1, and 2.25 for sample 2.
Of course they both have the same value of f ( T ) . Comparing expressions
for the two samples, we thereby find
n H (sample 2) = n H (sample 1) / 4.5 = 2  10 12 cm3 .
.
_________________________________________________________________________
6
_______________________________________________________________________
4. (13 points) Reflectivity of a translucent slab ...
A plane parallel region 0 < z < z 1 has isotropic scattering coefficient k ( z ), with
no absorption or emission. Everywhere else is transparent, k = 0.
Radiation density is U ( z ) and the net radiation flux in the z-direction is F ( z ) .
An extended radiation source below the slab creates a diffuse upward-moving flux
F + which enters the bottom side of the slab. Here "diffuse" implies that
F +  0.5  ( radiation density of the upward radiation )  c .
As the sketch shows, some radiation will be reflected and some will escape through
the top boundary. We'll use the diffusion approximation. Instead of z , let's refer to
optical depth  : d  = k d z , with  = 0 at z = 0 and  1 at z = z 1 .
(a) What does the F equation tell us about F (  ) in this case ?
Answer: Since there is no absorption or emission, F = 0 so d F / d z = 0.
Therefore F is a constant for all values of z or  .
(b) What does the U equation (diffusion) tell us about U (  ) ?
Answer: In this plane-parallel case, it's d U / d z =  3 k F / c which is equivalent
to d U / d  =  3 F / c = constant. Evidently U is a linear function of  .
(We can say this even if we don't know anything about k ( z ).)
(c) Find the general solution U  ( known function of  ) + ( constant ) .
Answer: Based on part (b) above, U  ( constant )  ( 3 F / c )  .
----- concluded on next page -----
7
----- last part of problem 4 -----
(d) A good upper boundary condition is F (  )  0.5 c U (  ) . Write a good boundary
condition for  = 0 , relating the quantities U ( 0 ) , F + , and F (  ) .
Answer: Both F + and F  contribute to the radiation density at  = 0 . Let's
assume that the average direction cosine is 0.5 for each, because this is simple and
it's like the stated boundary condition at  . Let's arbitrarily define F  to be a
positive quantity (pointed in the negative z-direction). Then we could write
F + = 0.5 c U + and F  = 0.5 c U  at  = 0 ; but what really matters is
the sum U + + U  = total radiation density there:
F + + F  = 0.5 c U (  ) .
This isn't useful by itself, since we don't know enough about F  yet. But we
do know that the total net flux in the z-direction at  = 0 is
F = F+  F .
(If we had defined F  to be a negative quantity, then the sign in each equation
would be reversed.) This equation was worth writing because part (a) showed us
that F is a constant. In particular, it's equal to F (  ) which appears in the
other boundary condition. Eliminating F  , the two above equations give us
the overall condition that was requested:
0.5 c U (  ) + F (  ) = 2 F + .
This is obviously useful because U (  ) is simply the constant in the solution
we found in part (c) , and F (  ) = the constant net flux F .
The exam specifically did not request the completed solution, because the last step
is prone to algebraic errors. The completed solution includes this result:
Q = fraction that passes through the slab = F (  ) / F + = 1 / ( 1 + 0.75  ) .
The "reflectivity" or reflected fraction is of course 1  Q .
This is only an approximation, and may be doubtful for small  . (Why? Can you
imagine a more accurate way to estimate the reflectivity for small  ?)
_______________________________________________________________________