Special relativity - Einstein's transformations I am reading "On the electrodynamics of moving bodies" and have got to page 6 and become stuck.  Is anyone able to please help explain how:

*

*Einstein went from the first line of workings to the second line (I can see how the first line is created but not what or how the second line is created by which rule of differentiation)

$$\frac{1}{2}\left[\tau(0,0,0,t)+\tau\left(0,0,0,t+\frac{x'}{c-v}+\frac{x'}{c+v}\right)\right]=\tau\left(x',0,0,t+\frac{x'}{c-v}\right).$$
Hence, if $x'$ be chosen infinitesimally small,
$$\frac{1}{2}\left(\frac{1}{c-v}+\frac{1}{c+v}\right)\frac{\partial\tau}{\partial t}=\frac{\partial\tau}{\partial x'}+\frac{1}{c-v}\frac{\partial\tau}{\partial t},$$



*We can then deduce (I've missed out the simplification of the above equation from the original document but it can all be seen on page 6)

Since $\tau$ is a linear function, ir follows from these equations that
$$\tau=a\left(t-\frac{v}{c^2-v^2}x'\right)$$

 A: 1) How to get from equation 1 to equation 2
If we assume that $x'$ (defined as $x'=x-vt$ in the paper) is infinitesimally small, i.e., $ d x' $, then we can Taylor expand $\tau(x',y,z,t)$ and retain only terms up to first order (higher-order terms in $x'$ can be neglected because they're small). The Taylor expansion for multi-variable function $f(x+dx,y+dy,z+dz,t+dt)$ is $$f(x+dx,y+dy,z+dz,t+dt)=f(x,y,z,t)+dx \dfrac{\partial f}{\partial x}+dy \dfrac{\partial f}{\partial y} +dz \dfrac{\partial f}{\partial z}+dt \dfrac{\partial f}{\partial t}$$
Thus $\tau(0,0,0,t+dt)$ in the LHS becomes
$$\tau(0,0,0,t+dt)=\tau(0,0,0,t)+dt \dfrac{\partial \tau}{\partial t}=\tau(0,0,0,t)+dx'[\dfrac{1}{c-v} +\dfrac{1}{c+v}] \dfrac{\partial \tau}{\partial t}$$
where $$dt=dx'(\dfrac{1}{c-v} +\dfrac{1}{c+v})$$
Therefore, the LHS of the first equation becomes
$$\dfrac{1}{2}[2\tau(0,0,0,t)+dx' (\dfrac{1}{c-v} +\dfrac{1}{c+v})\dfrac{\partial \tau}{\partial t}]$$
Similarly, by Taylor expanding the RHS, i.e., $\tau(dx',0,0,t+dt')$,  we get
$$\tau(dx',0,0,t+dt')=\tau(0,0,0,t)+dx'\dfrac{\partial \tau}{\partial x'}+dt'\dfrac{\partial \tau}{\partial t} =\tau(0,0,0,t)+dx'\dfrac{\partial \tau}{\partial x'}+\dfrac{dx'}{c-v}\dfrac{\partial \tau}{\partial t}$$
where $$dt'=\dfrac{dx'}{c-v}$$
The first terms in the RHS and LHS cancel, and by dividing both sides by $dx'$, we finally get the desired equation
$$\dfrac{1}{2}(\dfrac{1}{c-v}+\dfrac{1}{c+v})\dfrac{\partial \tau}{\partial t}= \dfrac{\partial \tau}{\partial x'}+ \dfrac{1}{c-v}\dfrac{\partial \tau}{\partial t}$$
2) How to get from equation 2 to equation 3
Assuming $\tau$ is a linear function of $x'$ and $t$ (in addition to his assertions in the paper that $\partial \tau /\partial y= \partial \tau /\partial z=0$), by definition we have
$$\tau=a_1 t+ a_2 x' $$
where $a_1$ and $a_2$ are constants. Note that equation 2 can be rewritten (through simple algebra) as
$$-\dfrac{v}{c^2-v^2}\dfrac{\partial \tau }{\partial t}=\dfrac{\partial \tau }{\partial x'} $$
Plugging in $\tau$ in the above equation we get
$$-\dfrac{v}{c^2-v^2} a_1= a_2 $$
Which finally gives us
$$\tau=a_1( t -\dfrac{v}{c^2-v^2}x')  $$
as desired.
