Drag force on mirror immersed in blackbody radiation 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.
 A: 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.
