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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

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

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)?

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  • $\begingroup$ It seems like this question could be of help for you. $\endgroup$ – mrtaurho Dec 30 '18 at 16:18
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    $\begingroup$ The question is strictly about the derivation of the differential equation's solution. Hadn't the OP mentioned physics at all, it would still be a perfectly fine question for this site. $\endgroup$ – Niki Di Giano Dec 30 '18 at 16:26
  • $\begingroup$ @NikiDiGiano I guess I overreacted. I retracted my close vote but the question I found on physics stackexchange could be useful nevertheless. $\endgroup$ – mrtaurho Dec 30 '18 at 16:28
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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 $\tau$: $$\tau = - \frac{v}{c^2 - v^2}Dx' + Dt = D\bigg(t - \frac{v}{c^2 - v^2}x'\bigg)$$ 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: $$ \tau = a\bigg(t - \frac{v}{c^2 - v^2}x'\bigg) + E$$ 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.

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  • $\begingroup$ Thank you for your help so far. I am just thinking through your answer at the moment, trying to get my head round it. I am just trying to see how you got from $A=-\frac{v}{c^2-v^2}D$ to $\tau=a\left(t-\frac{v}{c^2-v^2}x'\right)+E$. Could you perhaps extend your answer a little to show every step? $\endgroup$ – Rational Function Dec 30 '18 at 16:39
  • $\begingroup$ I have added the missing intermediate step. It's just a matter of plugging the results in the original expression. $\endgroup$ – Niki Di Giano Dec 30 '18 at 16:45
  • $\begingroup$ Thank you very much, that makes perfect sense - brilliant answer! $\endgroup$ – Rational Function Dec 30 '18 at 16:52
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There is really no need to assume that $\tau$ is linear or affine to derive the general form of $\tau$. Write the equation as $$ \frac{\partial\tau}{\partial x'}+k\frac{\partial\tau}{\partial t} = 0 , $$ where $k={v}/({c^2-v^2})$. Change coordinates from $(x',t)$ to $(\xi,u)$ by $$ \begin{cases}\xi= x'\\ u=t-k x' \end{cases} \qquad \text{or equivalently,}\qquad \begin{cases}x'=\xi\\ t=u+k\xi. \end{cases} $$ This gives $$ \frac{\partial\tau}{\partial\xi} = \frac{\partial\tau}{\partial x'} \frac{\partial x'}{\partial\xi}+ \frac{\partial\tau}{\partial t}\frac{\partial t}{\partial\xi} = \frac{\partial\tau}{\partial x'}+k\frac{\partial\tau}{\partial t} = 0 , $$ 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).

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    $\begingroup$ Thank you, this is another interesting approach. I especially like that it finds a general form before invoking the linearity of $\tau$, because it would allow us to proceed in a different way if for some reason we decided that the universe wasn't homogeneous (which forces $\tau$ to be linear/affine). If only I could tick multiple answers! $\endgroup$ – Rational Function Dec 30 '18 at 17:18
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    $\begingroup$ Indeed this answer is much more elegant than mine in my opinion. Props! $\endgroup$ – Niki Di Giano Dec 30 '18 at 17:35
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Making the change of variables

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

we have

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

so choosing

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

we get

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

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