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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?

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    $\begingroup$ Link to paper ? $\endgroup$
    – Qmechanic
    Apr 5, 2018 at 20:19

2 Answers 2

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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 $z$.

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  • $\begingroup$ I wanted to add that I appreciate this original question. I was trying to follow the 1905 paper derivations myself, and had run into a brick wall trying to a brick wall trying to go from (1) to (2), but this helped point me in the right direction. $\endgroup$
    – RC_23
    Jul 22, 2019 at 4:43
  • $\begingroup$ Shouldn't it be $\partial$ \partial instead of $\nabla$ \nabla? $\endgroup$ Jul 22, 2019 at 7:18
  • $\begingroup$ Holy cow. It's been two years since I asked this. $\endgroup$ Jul 22, 2019 at 15:15
  • $\begingroup$ Appreciate the user who formatted my original math to look good. I'm not a pro at this particular coding language. $\endgroup$
    – RC_23
    Aug 1, 2019 at 0:00
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I have my attempt but it needs to be checked for logical faults.

  • $\tau$(0,0,0,t) = $\tau$(0,0,0,0) + [$\frac{\partial\tau}{\partial x'}$] * (0-0) + [$\frac{\partial\tau}{\partial y}$] * (0-0) + [$\frac{\partial\tau}{\partial z}$] * (0-0) + [$\frac{\partial\tau}{\partial t}$] * (t-0) = $\tau$(0,0,0,0) + t$\frac{\partial\tau}{\partial t}$

  • $\tau$(0,0,0,t+$\frac{x'}{c-v}$+$\frac{x'}{c+v}$) = $\tau$(0,0,0,0) + [$\frac{\partial\tau}{\partial x'}$] * (0-0) + [$\frac{\partial\tau}{\partial y}$] * (0-0) + [$\frac{\partial\tau}{\partial z}$] * (0-0) + [$\frac{\partial\tau}{\partial t}$] * (t+$\frac{x'}{c-v}$+$\frac{x'}{c+v}$-0) = $\tau$(0,0,0,0) + t$\frac{\partial\tau}{\partial t}$ + $\frac{x'}{c-v}$$\frac{\partial\tau}{\partial t}$ + $\frac{x'}{c+v}$$\frac{\partial\tau}{\partial t}$

  • $\tau$(x',0,0,t+$\frac{x'}{c-v}$) = $\tau$(0,0,0,0) + [$\frac{\partial\tau}{\partial x'}$] * (x'-0) + [$\frac{\partial\tau}{\partial y}$] * (0-0) + [$\frac{\partial\tau}{\partial z}$] * (0-0) + [$\frac{\partial\tau}{\partial t}$] * (t+$\frac{x'}{c-v}$-0) = $\tau$(0,0,0,0) + * x'$\frac{\partial\tau}{\partial x'}$ + t$\frac{\partial\tau}{\partial t}$ + $\frac{x'}{c-v}$$\frac{\partial\tau}{\partial t}$

Plug those Taylor approximations back into equation (1):

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

Distribute and combine

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

Subtract $\tau$(0,0,0,0) and t$\frac{\partial\tau}{\partial t}$ from both sides

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

Take the derivative with respect to x' of both sides

$\frac{d}{dx'}$[$\frac{x'}{2(c-v)}$$\frac{\partial\tau}{\partial t}$ + $\frac{x'}{2(c+v)}$$\frac{\partial\tau}{\partial t}$] = $\frac{d}{dx'}$[x'$\frac{\partial\tau}{\partial x'}$ + $\frac{x'}{c-v}$$\frac{\partial\tau}{\partial t}$]

Factor out $\frac{\partial\tau}{\partial t}$ and $\frac{1}{2}$ on the left side

$\frac{1}{2}$[$\frac{1}{c-v}$ + $\frac{1}{c+v}$]$\frac{\partial\tau}{\partial t}$ = $\frac{\partial\tau}{\partial x'}$ + $\frac{1}{c-v}$$\frac{\partial\tau}{\partial t}$

I assume you know how to get from (2) to (3)

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  • $\begingroup$ So I took the Taylor approximation of each component at the origin (0,0,0,0), which I assumed Einstein would've chosen the origin, to me it just makes sense to choose the origin. And instead of using Tau(x',y,z,t) for the approximations I treated each Tau within equation (1) as separate functions rather than different values of the same function (that function again being Tau(x',y,z,t)). This could be problematic when simplifying being that this would mean Tau(o,o,o,o) in each approximation has distinct values and therefore cannot be combined when you plug them back into equation (1) $\endgroup$
    – Tjow
    Mar 2, 2017 at 20:04
  • $\begingroup$ But I think this can be overcame by the fact that these components of Tau(x',y,z,t) are still operating under that same function, they may only be treated as separate functions within Tau(x',y,z,t) so long as they contain variables in the input (e.g. Tau[x',0,0,t+(x'/c-v)]). So when the three components contain Tau(0,0,0,0) in the approximations, I treat each Tau(0,0,0,0) as equal to the other two Tau(0,0,0,0) in the other approximations since the variables are removed and so the value is no longer a composite and may go back to being considered a value of Tau(x',y,z,t) $\endgroup$
    – Tjow
    Mar 2, 2017 at 20:19

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