Drag force on mirror immersed in blackbody radiation

Question

In Einstein's "On The Present Status of the Radiation Problem" he mentions a thought experiment where A mirror of mass $M$ and area $A$ that is perfectly reflective in the range of frequencies $\nu$ to $\nu+d\nu$ and transparent in all others is restrained to movement in the $x$-axis, to which it is perpendicular, and immersed in an isotropic bath of gas in thermal equilibrium and black body radiation, which has energy density $$\rho=\frac{8\pi\nu^2}{c^3}\frac{h\nu}{e^{h\nu/kT}-1}d\nu$$ The mirror will experience a resistive force proportional to its velocity $-Rv$. Over that time, it'll also be imparted some extra momentum $\Delta$ by the small and random variations in the energy density and direction of the radiation. So, if at a certain time its momentum is $Mv$ its momentum after a very short interval of time $\tau$ will be $$Mv-Rv\tau+\Delta$$ Once in equilibrium, the average value of the square of the momentum won't change, so we have $$\langle M^2v^4\rangle=\langle(Mv-Rv\tau+\Delta)^2\rangle$$ Expanding, neglecting terms of order $\tau^2$ or higher and noting that the average of $v\Delta$ is 0 $$\langle \Delta^2\rangle=2MR\tau\langle v^2\rangle$$ Since the mirror is in thermal equilibrium with the gas, we obtain $$\langle \Delta^2\rangle=2RkT\tau$$ Calculating the force on the mirror (neglecting terms of order $\left (\frac{v}{c}\right )^2$ or higher, we obtain $$R=\frac{3}{2c}\left(\rho-\frac{1}{3}\nu\frac{\partial\rho}{\partial\nu}\right)Ad\nu$$ So $$\frac{\langle\Delta^2\rangle}{\tau}=3kT\left(\rho-\frac{1}{3}\nu\frac{\partial\rho}{\partial\nu}\right)Ad\nu$$ Einstein then says that $$\frac{\langle\Delta^2\rangle}{\tau}=\frac{1}{c}\left(h\rho\nu+\frac{c^3}{8\pi}\frac{\rho^2}{\nu^2}\right)Ad\nu$$ I don't understand how the last follow from the previous.

Answer 1

First off, the final two equations differ by a factor of $c$, which might be due to a misprint or a transcription error. What we actually want to show is that $$k T \left( 3 \rho - \nu \frac{\partial \rho}{\partial \nu} \right) = h \rho \nu + \frac{c^3}{8\pi} \frac{\rho^2}{\nu^2}.$$ To do this, we just evaluate the derivative on the left-hand side directly. We have $$\rho(\nu) = \frac{8 \pi h \nu^3}{c^3} \frac{1}{e^{h \nu / k T} - 1}.$$ When you differentiate $\rho$, there are two terms, because of the two separate appearances of $\nu$. The first term comes from the differentiation of $\nu^3$. Since the derivative of $\nu^3$ is $(3/\nu) \nu^3$, this contributes $$\frac{\partial \rho}{\partial \nu} \supset \frac{3}{\nu} \rho$$ which precisely cancels the first term on the left-hand side. So what remains on the left-hand side is the term you get from differentiating the $\nu$ in the denominator, $$- k T \nu \left( \frac{8 \pi h \nu^3}{c^3} \frac{\partial}{\partial \nu} \frac{1}{e^{h \nu / kT} - 1} \right) = - k T \nu \, (-h/kT)\, \left(\frac{8 \pi h \nu^3}{c^3} \frac{e^{h \nu/kT}}{(e^{h \nu / kT} - 1)^2} \right)$$ where the new factors are from the chain rule. The term in parentheses can be simplified as $$h \nu \left(\frac{8 \pi h \nu^3}{c^3} \frac{e^{h \nu / kT} - 1 + 1}{(e^{h \nu / kT} - 1)^2} \right) = h \nu \rho + \frac{h \nu \rho}{e^{h \nu /kT} - 1} = h \nu \rho + \frac{c^3}{8 \pi h \nu^3} h \nu \rho^2$$ which is exactly the right-hand side.