2008W

Complete two of the following four problems. You can consult your textbook, notes, calculator but not your neighbours.

Problem 1

Use the blackbody equations to find the approximate temperature of a dust grain that radiates mainly at 150 μm. Show that a large dust grain 1 pc from an O star with $L=10^{6}L_{\rm {Sun}}$ will be heated to roughly this temperature.

Using Wien's law we have

$\lambda _{\rm {max}}=\left[{\frac {2.9{\rm {K}}}{T}}\right]{\rm {mm}}=0.150{\rm {mm}}$

so $T=19.33{\rm {K}}$ . The dust grain will radiate according to the black-body formula from its entire surface and absorb radiation from the side facing the star so

$4\pi R^{2}\sigma T^{4}={\frac {L}{4\pi d^{2}}}\pi R^{2}$

and

$d^{2}={\frac {L}{16\sigma T^{4}}}=\left(1.02{\rm {pc}}\right)^{2}$

or

$T^{4}={\frac {L}{16\sigma d^{2}}}=\left(19.59{\rm {K}}\right)^{4}$

Problem 2

How does the metallicity of the gas evolve with the mass of long-lived stars if you assume that the gas is refreshed as stars form? You should assume that the yield is $p$ and that the initial metallicity of the gas is zero.

We have

$\Delta M_{h}=\left(p-Z\right)\Delta M_{*}$

and

$\Delta Z={\frac {1}{M_{g}}}\left[p\Delta M_{*}-Z\left(\Delta M_{*}+\Delta M_{g}\right)\right]$

but $\Delta M_{g}=0$ yielding

$M_{g}{\frac {dZ}{dM_{*}}}=p-Z.$

Let us define <m>U=Z - p</m> to give

$M_{g}{\frac {dU}{dM_{*}}}=-U$

so

$U=A\exp(-M_{*}/M_{g}),Z=p-p\exp(-M_{*}/M_{g})~$

where I have applied the initial conditions at the last step.

Problem 3

Derive the relationship between the orbital frequency for a circular orbit <m>\Omega</m> and the epicyclic frequency for an object in a slightly non-circular orbit around the Galaxy, assuming that the rotation curve is flat.

One can get at this answer is several ways. A quick way is the formula for the epicyclic frequency in terms of the derivative of the rotation frequency. We have

$\kappa ^{2}={\frac {1}{R^{3}}}{\frac {d}{dR}}\left[\left(R^{2}\Omega \right)^{2}\right]={\frac {1}{R^{3}}}{\frac {d}{dR}}\left[R^{2}\left(R\Omega \right)^{2}\right]=2\Omega ^{2}$

where we used the fact that the value of <m>\Omega R</m> is constant with respect to <m>R</m>.

We could have used the definition of the epicyclic frequency in terms of the effective potential

$\kappa ^{2}=\left.{\frac {d^{2}\Phi _{\rm {eff}}}{dr^{2}}}\right|_{r_{g}}$

where

$\Phi _{\rm {eff}}=\Phi +{\frac {L^{2}}{2r^{2}}}.$

Let's take the first derivative

${\frac {d\Phi _{\rm {eff}}}{dr}}={\frac {d\Phi }{dr}}-{\frac {L^{2}}{r^{3}}}=0$

but we know that

${\frac {d\Phi }{dr}}={\frac {v^{2}}{r}}$

Now take the second derivative keeping $v=\Omega r$ and $L=\Omega r^{2}$ fixed to give

$\kappa ^{2}=-{\frac {v^{2}}{r^{2}}}+3{\frac {L^{2}}{r^{4}}}=-\Omega ^{2}+3\Omega ^{2}=2\Omega ^{2}.$

Problem 4

Derive the specific heat of a bound system of $N$ particles interacting under their mutual gravitational attraction. You can assume that the mean kinetic energy of a particle is $(3/2)kT$ where $T$ is the temperature. What happens if you bring two such systems with different temperatures into contact? Does energy flow? In which direction? Do the temperatures approach each other?

Here we use the virial theorem. We have

$2\langle {\rm {KE}}\rangle +\langle {\rm {PE}}\rangle =0$

so

$E=-\langle {\rm {KE}}\rangle =-{\frac {3}{2}}NkT$

so the specific heat is

${\frac {dE}{dT}}=-{\frac {3}{2}}Nk$

or $-{\frac {3}{2}}k$ per particle. Energy will flow from the hotter to the colder system but as the hotter system loses energy it gets hotter and the cold system gets colder as it gains energy, so equilibrium is impossible.

2009W

Complete one of the following two problems. You can consult your textbook, notes, calculator but not your neighbours.

Problem 1

At the centres of disk galaxies, the circular velocity increases linearly with distance from the centre. Let's model the circular velocity as $v_{c}=v_{\infty }{\rm {tanh}}(r/a)$ .

What is the mass enclosed within a given radius assuming that it is spherically symmetric? Sketch the density as a function of radius on a log-log plot and include the transition radius.

Sketch the epicyclic frequency as a function of the distance from the centre. Derive the epicyclic frequency in terms of the local orbital frequency near the centre and at large distances as well as giving an approximate value of the transition radius. What range of pattern speeds for a two-arm spiral are acceptable near the centre?

The centripetal acceleration is given by

${\frac {v^{2}}{r}}={\frac {GM(<r)}{r^{2}}}$

so

$M(<r)={\frac {v^{2}r}{G}}={\frac {v_{\infty }^{2}r{\rm {tanh}}^{2}(r/a)}{G}}.$

At small radii $M(<r)\propto r^{3}$ so $\rho$ is constant. At large radii, $M(<r)\propto r$ , so $\rho \propto r^{-2}$ . The transition is at $a$ .

We have $\kappa ^{2}={\frac {1}{R^{3}}}{\frac {d}{dR}}\left[\left(R^{2}\Omega \right)^{2}\right]={\frac {1}{R^{3}}}{\frac {d}{dR}}\left[\left(RV\right)^{2}\right].$

Near the centre, $\Omega$ is constant and on the outside $v$ is constant ( $\Omega \propto r^{-1}$ ) so we have at the centre

$\kappa ^{2}={\frac {1}{R^{3}}}\Omega ^{2}4R^{3}=4\Omega ^{2}$

and at large distances we have

$\kappa ^{2}={\frac {1}{R^{3}}}V^{2}2R=2{\frac {V^{2}}{R^{2}}}=2\Omega ^{2}.$

The transition radius is $a$ . We know that $m\left|\Omega -\Omega _{p}\right|\leq \kappa$ , so $\left|\Omega -\Omega _{p}\right|\leq \Omega$ and $0\leq \Omega _{p}\leq 2\Omega$ .

Problem 2

Consider an attractive force between particles that is proportional to the distance between particles and the product of their masses. Prove a relationship between the mean kinetic energy and the mean potential energy of a group of particles that remains bounded.

Let

$K=\sum _{i}{\frac {1}{2}}m_{i}x_{i}^{2}$

and calculate the second derivative of $K$ with time. We have

${\frac {dK}{dt}}=\sum _{i}m_{i}x_{i}{\dot {x}}_{i}$

and

${\frac {d^{2}K}{dt^{2}}}=\sum _{i}m_{i}\left({\dot {x}}_{i}^{2}+x_{i}{\ddot {x}}_{i}\right)=2T+\sum _{i}\sum _{j}x_{i}F_{ij}.$

Let's pair up the particles

${\frac {d^{2}K}{dt^{2}}}=2T-\sum _{i<j}\left(x_{i}-x_{j}\right)^{2}m_{i}m_{j}k=2T-2V.$

When averaged over a long time the left-hand side vanishes, so $<T>=<V>$ .