Flux and Luminosity

The objective of this problem was to introduce us to the concept of blackbodies and the method to finding one's flux and/or luminosity.

Problem 2(a) dealt with bolometric flux, which is energy per area per time.  To do this, one integrates the equation for blackbody flux, $F_\upsilon (T)$ (found in the last part of problem 1), over all frequencies.  Because the bolometric flux is found by adding up the flux at all frequencies from 0 to $\infty$, it is independent of frequency.

$F(T)=\int_{0}^{\infty}F_\upsilon (T)d\upsilon =\frac{2\pi h\upsilon ^3}{c^2(e^{\frac{h\upsilon }{kT}}-1)}d\upsilon $

$u=\frac{h\upsilon }{kT}\rightarrow \upsilon =\frac{ukT}{h}$

$d\upsilon =\frac{kT}{h}du$

$F(T)=T^4\int_{0}^{\infty}\left ( \frac{2\pi h}{c^2} \right )\left ( \frac{k}{h} \right )^4\left ( \frac{u^3}{e^u-1} \right )$

$\sigma =\int_{0}^{\infty}\left ( \frac{2\pi h}{c^2} \right )\left ( \frac{k}{h} \right )^4\left ( \frac{u^3}{e^u-1} \right )$

$F(T)=T^4\sigma$


In part (b), we were asked to convert the equation $B_\upsilon (T) =\frac{2\pi h\upsilon ^3}{c^2(e^{\frac{h\upsilon }{kT}}-1)}$ so that it was in terms of wavelength instead of frequency.  The tricky part was that this was not just a simple matter of substituting $\upsilon=\frac{c}{\lambda}$ into the equation.  Simple substitution would not conserve the amount of energy during a conversion.

$B_\upsilon (T) =\frac{2\pi h\upsilon ^3}{c^2(e^{\frac{h\upsilon }{kT}}-1)}$

$\upsilon=\frac{c}{\lambda}$

$d\upsilon=\frac{-c}{\lambda^2}d\lambda$

$B_\lambda(T) =\frac{2\pi h\left ( \frac{c}{\lambda } \right ) ^3c}{c^2(e^{\frac{hc}{\lambda kT}}-1)\lambda ^2}$
$B_\lambda(T)=\left ( \frac{2c^2h}{\lambda ^5} \right )\left ( \frac{1}{e^\frac{hc}{\lambda kT}-1} \right )$


Part (c) asked us derive an expression for $\lambda_{max}$, corresponding to the peak of the intensity distribution at a given temperature T.  To do this, we first derived the final equation we got for $B_\lambda(T)$ with respect to $\lambda$.  After many failed attempts at deriving this by myself and some help from WolframAlpha, I found that the derivative of $B_\lambda(T)$ is:

$B_\lambda'(T)=\left ( 2c^2h \right )\frac{\left ( \frac{hc}{kT} \right )\left ( e^\frac{hc}{\lambda kT} \right )-5\lambda \left ( e^\frac{hc}{\lambda kT}-1 \right )}{\lambda ^7\left ( e^\frac{hc}{\lambda kT}-1 \right )^2}$

To find $\lambda_{max}$, one has to set this derivative equal to 0 and solve for $\lambda$.

$5=\left ( \frac{hc}{\lambda kT} \right )\left ( \frac{e^\frac{hc}{\lambda kT}}{e^\frac{hc}{\lambda kT}-1} \right )$

Substituting $u=\frac{hc}{\lambda kT}$ leads to the following equation:

$5=\frac{ue^u}{e^u-1}$

I solved for u by graphing this equation on WolframAlpha.  I got the following graph,

and found a u-value of 4.965.

By solving $4.965=\frac{hc}{\lambda kT}\rightarrow \lambda T=\frac{hc}{4.965k}$ for $\lambda$, I found that $\lambda_{max}=2.85\times10^{-3}mK$

Part (d) of the problem asked us to find a simplified version of $B_{\lambda}(T)$ by using a first order Taylor expansion on $e^\frac{hc}{kT}$.

In a first order Taylor expansion, $e^x=1+x$.  Therefore, $e^\frac{hc}{kT}=1+\frac{hc}{kT}$.

The simplified version of  $B_{\lambda}(T)$ becomes $B_\lambda (T)= \frac{2h\upsilon ^3}{c^2\frac{h\upsilon }{kT}}=\frac{2kT\upsilon ^2}{c^2}$.

Part (e) asked us to find an equation for the bolometric luminosity, L, for a blackbody with radius, R.  This quantity should have units of energy per time.  We were told to start with the equation for bolometric flux, $F(T)=T^4\sigma$.  From there, because we know that blackbodies emit blackbody radiation isotropically, we can simply multiply the flux by the surface area of the blackbody.  

$L=4\pi \sigma R^2T^4$

This works out units-wise because bolometric flux is in units of $\frac{energy}{area\cdot time}$.  multiplying by area will give units of energy per time.  

The final part of the problem posed a situation in which two stars are gravitationally bound.  One is blue and one is yellow; the yellow one is a lot brighter than the blue one.  We were first asked to qualitatively compare the temperatures and radii of the two stars.  The blue star is hotter, because blue corresponds to higher energy, which corresponds to higher temperature.  The yellow star has a bigger radius because it appears to give off more light than the blue star.  The only way this would happen (assuming the two stars are roughly the same distance from Earth) is if the yellow star were just a lot bigger than the blue star.  

We were then asked to quantitatively compare the radii of the two stars.  To answer this question, I randomly assigned relative values to the luminosities and temperatures of the two stars.  I said that the blue star was twice as hot as the yellow star, which was twice as luminous.  

$8\pi \sigma R_B^2T_B^4 = 4\pi \sigma R_Y^2T_Y^4$

$R_B^2T_B^4 = R_Y^2\left ( \frac{1}{2}T_B \right )^4$

$\left ( \frac{R_B}{R_Y} \right )^2=\frac{1}{32}$

$\frac{R_B}{R_Y}=.2$