Cfrgtkky

In his paper "On the Electrodynamics of Moving Bodies", Einstein writes the equation

$$dfrac{partial tau}{partial x'}+dfrac{v}{c^2-v^2}dfrac{partial tau}{partial t}=0$$

where

• 
  $tau=tau(x',y,z,t)$ is a linear function (i.e. $tau=Ax'+By+Cz+Dt$)


• $x'=x-vt$


• 
  $dfrac{partial tau}{partial y}=0$ (i.e. $B=0$)


• 
  $dfrac{partial tau}{partial z}=0$ (i.e. $C=0$)


• 
  $c$ is a constant


• 
  $x,x',y,z,t,v$ are variables

and he derives that

$$tau=aleft(t-dfrac{v}{c^2-v^2}x'right)$$

where $a=a(v)$

Could someone please walk me through step-by-step how he derived this? I am not very familiar with integrals invovling partial derivatives, so I would be appreciative if any answers are quite explicit.

Additionally, if I modified the question to say that $tau$ was an affine function (i.e. $tau=Ax'+By+Cz+Dt+E$), would it make any difference to the result (I suspect it wouldn't)?

From the definitions given:
$$partial_{x'}tau = A, quad partial_t tau = D$$
Also:
$$partial_{y}tau = B = 0, quad partial_z tau = C = 0$$
From the differential equation we get:
$$A + frac{v}{c^2 - v^2}D = 0 \
implies A = - frac{v}{c^2 - v^2}D$$
Now using the definition given for $tau$:
$$tau = - frac{v}{c^2 - v^2}Dx' + Dt = Dbigg(t - frac{v}{c^2 - v^2}x'bigg)$$
So if you define $D=a$ you get the final expression for $tau$. As you have noticed, requiring $tau$ to be affine doesn't change the results at all - your $tau$ would be:
$$ tau = abigg(t - frac{v}{c^2 - v^2}x'bigg) + E$$
The reason $a=a(v)$ is because $D$, who is actually $a$ in disguise, does not depend on $x', y, z, t$ but is assumed to depend on $v$. Otherwise, the relation wouldn't be linear. Without further information, $a$ is a function of potentially anything except those four variables.

There is really no need to assume that $tau$ is linear or affine to derive the general form of $tau$. Write the equation as
$$
frac{partialtau}{partial x'}+kfrac{partialtau}{partial t} = 0 ,
$$
where $k={v}/({c^2-v^2})$. Change coordinates from $(x',t)$ to $(xi,u)$ by
$$
begin{cases}xi= x'\
u=t-k x'
end{cases}
qquad
text{or equivalently,}qquad
begin{cases}x'=xi\
t=u+kxi.
end{cases}
$$
This gives
$$
frac{partialtau}{partialxi} = frac{partialtau}{partial x'} frac{partial x'}{partialxi}+ frac{partialtau}{partial t}frac{partial t}{partialxi} = frac{partialtau}{partial x'}+kfrac{partialtau}{partial t} = 0 ,
$$
meaning that in the new coordinates, $tau$ is a function of only $u$. Hence
$$
tau=a(u)=a(t-k x'),
$$
for some function $a$. At this point, you can use the assumption that $tau$ is linear (or affine) to deduce that $a$ is linear (or affine).

Making the change of variables

$$
x'=x- vt\
t'=alpha t
$$

we have

$$
(v^2+alpha(c^2-v^2))frac{partialtau}{partial x'}+vfrac{partial tau}{partial t'}=0
$$

so choosing

$$
alpha = frac{v^2}{v^2-c^2}
$$

we get

$$
frac{partial tau}{partial t'} = 0Rightarrow tau(x',t') = f(x') = f(x-v t)
$$