WS 9.2: ALL of the Relationships

The point of this worksheet, as seen in the title of this blog post, was to find all of the relationships between a star's mass, temperature, and luminosity and its radius.

In question 1, we were asked to recall two equations from previous classes.

$\frac{dP}{dr}=-\frac{GM_r\rho_r}{r^2}$

$\frac{dT}{dr}=-\frac{L_rk_B\rho_r}{16\pi r^2acT^3}$

Pretty easy, right? Well, it gets harder.

Question 2 asked us to derive the equation for the relationship between mass and radius, $\frac{dM}{dr}$.  

We wrote down the equation for the mass of an object, which is actually the rearranged equation for the density of an object.  

$M=\frac{4\pi r^3\rho_r}{3}$

Then we found its derivative 

$\frac{dM}{dr}=4\pi r^3\rho_r$

Still pretty easy to understand

Question 3 asked us to find the relationship between density and gas pressure.

Obviously, I saw the words "gas" and "pressure" and immediately thought of the ideal gas law

$P=nk_BT$

But, since the question wanted to include density, we had to find a way to put a $\rho$ in the equation.  Since $\rho=n\bar{m}$, the ideal gas law can be written as 

$P=\frac{\rho}{\bar{m}}k_BT$

Question 5 asked us to rewrite the all of the relationships.  

We were given that $P_C\sim \frac{M^2}{R^4}$.

$\frac{dP}{dr}\sim \frac{P}{r}=-\frac{GM_r\rho_r}{r^2}$*

If you substitute in the full equation for $\rho$ and rearrange for P, you do in fact find that 

$P_C\sim \frac{M^2}{R^4}$

Next we'll do the temperature-radius relationship.

$\frac{dT}{dr}\sim \frac{T}{r}=-\frac{L_rk_B\rho_r}{16\pi r^2acT^3}$

Again, if you substitute in for $\rho$ and rearrange, you find 

$T^4\sim \frac{LM}{R^4}$

Now, it's mass' turn.

$\frac{dM}{dr}\sim \frac{M}{r}=4\pi r^3\rho_r$

Substituting and rearranging gives you 

$\frac{M}{R}\sim\frac{M}{R}$

which is so obvious that it's not actually helpful.  So, the trick is not to substitute for $\rho$ and you get 

$\rho\sim\frac{M}{R^3}$

Question 6 asks us to summarize all the relationships found above by using power laws

a)  $R\sim M$ which you can find by simplifying the relationship found in Question 5

b) $L\sim \frac{R^4}{M}$ but because of the relationship in part (a), we can rewrite this as $L\sim M^3$

c) $L\sim \frac{M}{\rho}$ which makes sense because if a star isn't particularly massive, a dense star will be really small, and therefore not very luminous.  If we do the substitution for $\rho$ one last time, and we use the relationship from part (a), we find that $L\sim M^5$

e) For low-mass stars, we start with 

$L\sim M^2T^4$

which is the from equation for the surface luminosity of a star.  

But we can substitute using the mass-luminosity relation found in part (c) and get

$L\sim L^{2/5}T^4$

When you rearrange, you get 

$L\sim T^{20/3}$

For massive stars, we once again start with 

$L\sim M^2T^4$

but now we substitute the mass-luminosity relation in part (b)

$L\sim M^{2/3}T^4$

$L\sim T^12$

See? I told you it would get harder.  

* You can say that $\frac{dP}{dr}\sim\frac{P}{r}$ because we assume that the initial pressure is 0 and the final pressure (the pressure at the surface) is not.  So since we assume that dP=P(final)-P(initial), we can posit make that leap.