Taylor Series in Einstein's 'On the Electrodynamics of Moving Bodies'

Question

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?

Answer 1

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 $z$.

Answer 2

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)

Answer 3

Your formula is inconsistent. You probably meant: $$½ \left(τ\left(0, 0, 0, t\right) + τ\left(0, 0, 0, t + \frac{x}{c + v} + \frac{x}{c - v}\right)\right) = τ\left(x, 0, 0, t + \frac{x}{c - v}\right),$$ and you can drop the primes here and just write $x'$ as $x$ since the prime is not relevant to the discussion here. You probably forgot that extra $t$ on the right-hand side.

A couple things you should be aware of (and that includes you, too, Einstein wherever you are!) The Taylor Series, properly done, is not an approximation formula, but an exact representation. So, when we say $$f(x, y) = f(0, 0) + x f_x(0, 0) + y f_y(0, 0) + ⋯$$ what we're actually saying is that $$f(x, y) = f(0, 0) + x f_x(0, 0) + y f_y(0, 0) + \frac{x^2}{2} A(x, y) + x y B(x, y) + \frac{y^2}{2} C(x, y),$$ for some continuous functions $A$, $B$, $C$ of $(x,y)$, provided that the second derivative of $f$ (i.e. its Hessian matrix) is absolutely continuous in some connected domain that includes both the points $(0, 0)$ and $(x, y)$. Normally, in the Physics literature, one assumes that functions are "smooth" (i.e. that they have continuous derivatives to all orders), so that the "minimum condition" technicalities can be collated together in the simpler "smoothness" condition. In that case, the functions $A$, $B$ and $C$ are also smooth in $(x, y)$. In addition, the theorem will also provide you an integral formula for these remainder functions in terms of the second order partial differentials of $f$.

Ignore the other coordinates and just write $τ$ as a function of $(x,t)$ only. The second order Taylor Theorem is $$τ(x,t) = τ(0,0) + x τ_x(0,0) + t τ_t(0,0) + \frac{x^2}{2} A(x,t) + x t B(x,t) + \frac{t^2}{2} C(x,t)$$ for all $(x,t)$ is some connected domain that includes $(0,0)$, where $A$, $B$ and $C$ are smooth functions of $(x,t)$. We assume that the domain is for all $(x,t)$ in some neighborhood of $(0,0)$.

Substitute this into your condition (in the form which I modified it as) to write: $$½ \left(τ(0,0) + t τ_t(0,0) + \frac{t^2}{2} C(0,t) + τ(0,0) + t_2 τ_t(0,0) + \frac{{t_2}^2}{2} C\left(0,t_2\right)\right) \\ = τ(0,0) + x τ_x(0,0) + t_1 τ_t(0,0) + \frac{x^2}{2} A(x,t_1) + x t_1 B(x,t_1) + \frac{{t_1}^2}{2} C(x,t_1),$$ where, for convenience, I've set $$t_1 = t + \frac{x}{c - v}, \hspace 1em t_2 = t_1 + \frac{x}{c + v} = t + \frac{x}{c - v} + \frac{x}{c + v}.$$

That's not an approximation. That's an exact formula - as they are for all applications of Taylor's Theorem, when the theorem is used right. In reality, you're supposed to pay attention to those remainder terms, because the expressions involving them contain structural information about the form of the remainder (e.g. the multipliers of $A$, $B$ and $C$) that may be relevant to the physics at hand.

Simplify to get the equation: $$\left(t - 2 t_1 + t_2\right) τ_t(0,0) + \frac{t^2}{2} C(0,t) + \frac{{t_2}^2}{2} C\left(0,t_2\right) \\ = 2 x τ_x(0,0) + x^2 A(x,t_1) + 2 x t_1 B(x,t_1) + {t_1}^2 C(x,t_1),$$ and note that $$t - 2 t_1 + t_2 = \frac{x}{c + v} - \frac{x}{c - v} = -\frac{2vx}{c^2 - v^2}.$$ Thus $$-\frac{2vx}{c^2 - v^2} τ_t(0,0) + \frac{t^2}{2} C(0,t) + \frac{{t_2}^2}{2} C\left(0,t_2\right) \\ = 2 x τ_x(0,0) + x^2 A(x,t_1) + 2 x t_1 B(x,t_1) + {t_1}^2 C(x,t_1).$$

This holds for all $(x,t)$ in the said neighborhood of $(0,0)$, which includes $t = 0$, provided $x$ is small enough. So, set $t = 0$, and assume $x$ is small enough that $(x,0)$ is in the neighborhood of $(0,0)$. The result is: $$-\frac{2vx}{c^2 - v^2} τ_t(0,0) + \frac{{x_2}^2}{2} C\left(0,x_2\right) = 2 x τ_x(0,0) + x^2 A(x,x_1) + 2 x x_1 B(x,x_1) + {x_1}^2 C(x,x_1),$$ where $$x_1 = \frac{x}{c - v}, \hspace 1em x_2 = \frac{x}{c - v} + \frac{x}{c + v}.$$

Factor out $x$: $$-\frac{2v}{c^2 - v^2} τ_t(0,0) + x \frac{{z_2}^2}{2} C\left(0,x_2\right) = 2 τ_x(0,0) + x A(x,x_1) + 2 x_1 B(x,x_1) + x {z_1}^2 C(x,x_1),$$ where $$z_1 = \frac{1}{c - v}, \hspace 1em z_2 = \frac{1}{c - v} + \frac{1}{c + v}.$$

Set $x = 0$, factor out a $2$ and move everything over to one side: $$τ_x(0,0) + \frac{v}{c^2 - v^2} τ_t(0,0) = 0.$$

What Einstein didn't realize is that you don't need any calculus at all to derive the relevant identities. The reason is that Minkowski geometry is so utterly rigid that from the relation $A ⇔ B$ defined as "space-time point $A$ is at a light-like separation from $B$", and from the primitive notion of space-time point, with a suitable set of purely geometric axioms concerning the "$⇔$" relation (and nothing more), it is possible to define all geometric relations and concepts(!) (including parallelism, orthogonality, angles, lines, planes, congruence, similarity, coordinate axes, etc.) and to characterize Minkowski space exactly up to the choice of a scale unit: the "second", and an orientation of time-direction (i.e. the selection of which direction points to the future and which points to the past).

Thus, no transformation of the desired type is possible that is not linear (or else unbounded). I'm not even sure if continuity needs to be assumed. Moreover, it must not only be linear, but must produce the result corresponding to the Lorentz transforms underlying Minkowski geometry.

For the less rigid family of curved Lorentzian (i.e. 3+1 dimensional) space-time geometries, the specification of the "$⇔$" relation lays out the structure of the light-cones. From the light-cones, alone, one has a space-time metric defined, uniquely, up to a conformal transformation.

Among other things, this uniquely specifies the Weyl Tensor in geometries of 4 or more dimensions, since the Weyl Tensor is conformally invariant. The importance of that is that the wave modes of the Weyl Tensor are the modes of gravitational radiation.

So, those who advance the idea that gravitational radiation is to be quantized are caught in the trap of also requiring the Weyl Tensor to be a quantum object - subject to quantum fluctuations. That means: light cone fluctuations, the ability to do light-cone tunneling and to violate causality and potentially to even do time travel. That's a heavy millstone to hang on anyone who wants to do quantum gravity and who thinks it can be done, at all; and, as I see it, it's one small step shy of a conditional impossibility proof: that either no quantum gravity or else causality-violation.