# How did Einstein integrate $\frac{\partial \tau}{\partial x'}+\frac{v}{c^2-v^2}\frac{\partial \tau}{\partial t}=0$?

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)?

• It seems like this question could be of help for you. – mrtaurho Dec 30 '18 at 16:18
• 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. – Niki Di Giano Dec 30 '18 at 16:26
• @NikiDiGiano I guess I overreacted. I retracted my close vote but the question I found on physics stackexchange could be useful nevertheless. – mrtaurho Dec 30 '18 at 16:28

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.

• 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? – Rational Function Dec 30 '18 at 16:39
• I have added the missing intermediate step. It's just a matter of plugging the results in the original expression. – Niki Di Giano Dec 30 '18 at 16:45
• Thank you very much, that makes perfect sense - brilliant answer! – Rational Function Dec 30 '18 at 16:52

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).

• 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! – Rational Function Dec 30 '18 at 17:18
• Indeed this answer is much more elegant than mine in my opinion. Props! – Niki Di Giano Dec 30 '18 at 17:35

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

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