special relativity - Taylor Series in Einstein's 'On the Electrodynamics of Moving Bodies'

In his famous paper on Special Relativity, Einstein derives the Lorentz Transformations. He considers a light beam emitted at time $t$ from the origin of the system of coordinates $k$ towards a point that moves with the origin of the system $K$ such that its coordinate on the $K$ system is $x'=x-vt$ and is then reflected back. He begins with the equation

$$ \frac{1}{2}\left[\tau(0,0,0,t)+\tau\left(0,0,0,t+\frac{x'}{c-v}+\frac{x'}{c+v}\right)\right]=\tau\left(x',0,0,\frac{x'}{c-v}\right)\tag{1} $$

Then the paper says "Hence, if $x'$ be chosen infinitesimally small"

$$ \frac{1}{2} \left(\frac{1}{c-v}+\frac{1}{c+v}\right)\frac{\partial\tau}{\partial t}=\frac{\partial\tau}{\partial x'}+\frac{1}{c-v}\frac{\partial\tau}{\partial t}\tag{2} $$

which is simplified to

$$ \frac{\partial\tau}{\partial x'}+\frac{v}{c^2-v^2}\frac{\partial\tau}{\partial t}=0\tag{3} $$ I have read and know how to go from equation (1) to equation (2) using differentials and partial derivatives, but recently, I found a forum thread which stated that what Einstein means by "Making $x'$ infinitely small" is to take a Taylor Series of the components of equation (1) and reducing $x'$ to $0$. Yet, I am not sure of how to do that. Can someone help me?

Answer

It's also worth remembering that we don't strictly need a Taylor expansion in this case, because the purpose of a Taylor expansion is to linearize a function, which in this case we already know to be linear (as stated earlier in Einstein's paper, the transformation equations must be linear for space and time to be homogeneous).

So we know a priori that $\tau$ is of the form: $$ \tau\left(x',y,z,t\right) ~=~ Ax' + By + Cz + Dt + E \,,$$ and $$ \tau\left(0,0,0,t\right) ~=~ Dt+E \,,$$ where $D$ is $\frac{\partial \tau}{\partial t} .$ Likewise all the coefficients are simply the respective partial derivatives: $A=\frac{\partial \tau}{\partial x'}$, $B=\frac{\partial \tau}{\partial y}$, $C=\frac{\partial \tau}{\partial z}$, and $E$ is an additive constant.

So taking Einstein's $\frac{1}{2}(\tau_0 + \tau_2) = \tau_1$ formula (your Eqn (1)), plugging in the coordinate arguments, and performing a little algebra, you can easily obtain his partial differential equation (your Eqn (3)): $$ \frac{\partial \tau}{\partial x'} + \frac{v}{c^2-v^2} \frac{\partial \tau}{\partial t} ~=~ 0 \,.$$ Note that during this algebraic process $x'$ simply cancels out anyway.

And the same process (which he omits in the paper), can be used to show algebraically that the partial derivatives with respect to $y$ and $z$ are zero, by considering a light ray that moves vertically (along $y$). It departs the origin at $\tau\left(0,0,0,t_0\right)$, reflects at $\tau\left(0,L,0,t_0+\frac{L}{\sqrt{c^2-v^2}}\right) ,$ and returns at $\tau\left(0,0,0,t_0+\frac{2L}{\sqrt{c^2-v^2}}\right) ,$ where $L$ is the vertical length in question. Same thing for $z$.