Einstein's Derivation of Lorentz transformations

Question

I have been reading the derivation of Lorentz transformations in the appendix of relativity:the special and general theory by Einstein and am stuck in the following spot: He states $$ x - c t = 0\tag{1} $$ implies $$ x' - c t' = 0\tag{2} $$ and this implies $$ x - c t = k (x' - c t')\tag{3} $$ where $k$ is a constant. I am not able to see where the third equation comes from. Will be grateful for any help.

Answer 1

The issue is less trivial than it may appear. Let $(x,t)$ and $(x',t')$ be the Cartesian coordinates of two inertial reference frames whose origins coincide, but where the primed frame is moving with velocity $v\hat x$ with respect to the unprimed frame. For an event with coordinates $(x,t)$ in the primed frame, the Galilean transformation equations give that the coordinates of the event in the primed frame are $$\pmatrix{x'\\t'} = \pmatrix{x-vt\\t} = \pmatrix{1 & -v \\ 0 & 1}\pmatrix{x\\t}\tag{1}$$

However, this transformation is incompatible with the requirement that the speed of light be the same in both frames. To see this, observe that the trajectory of a light ray moving in the $+\hat x$ direction in the unprimed frame is given by $x = ct$. If we plug in the results of $(1)$, we find that $$x = ct \implies (x' + vt') = ct' \implies x' = (c-v) t'$$ so the observer in the primed frame sees the light ray traveling with velocity $(c-v)\hat x$.

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Einstein sought to modify $(1)$ to encode the invariance of the speed of light. He demanded that if $x=ct$, then $x'=ct'$ - which simply means that if an object is moving with velocity $c\hat x$ in one frame, it must be moving with velocity $c\hat x'$ in any other frame.

If we assume that the relationship between coordinates is still linear, then we can write

$$\pmatrix{x'\\t'} = \pmatrix{Ax + Bt \\ Dx + Ft} = \pmatrix{A&B\\D&F}\pmatrix{x\\t}\tag{2}$$ for some $A,B,D,F$. If we let $x = ct$, then we find that $x' = (Ac+B)t$ and that $t' = (Dc+F)t$. If we now require that this implies that $x' = ct'$, then we have that $$Ac+B = c(Dc+F)\tag{3}$$

Now consider the expression $x-ct$, which does not necessarily equal zero. It is zero if we're talking about the trajectory of a light ray, but now we could be talking about the trajectory of a car moving with velocity $u_0\hat x$, in which case $x=u_0 t \implies x-ct = (u_0-c)t \neq 0$.

If we apply $(2)$, we find that

$$x'-ct' = (Ax+Bt)-c(Dx+Ft) = (A-cD)x + (B-cF)t$$

However, if we rearrange $(3)$ then we find that $B-cF = -c(A-cD)$, implying that $$x'-ct' = (A-cD)(x-ct) \equiv k(x-ct)$$ where $k\equiv (A-cD)$ is a constant which we do not yet know.

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The main takeaway is that if we require that something moving with velocity $c\hat x$ in one frame must be moving with velocity $c\hat x'$ in every other frame and that the transformation between coordinates be linear, then we can conclude that $x-ct$ and $x'-ct'$ (which are zero for light rays, but are generally nonzero) are proportional to one another.

Answer 2

He states $$ x - c t = 0\tag{1} $$ implies $$ x' - c t' = 0\tag{2} $$ and this implies $$ x - c t = k (x' - c t')\tag{3} $$ where $k$ is a constant. I am not able to see where the third equation comes from. Will be grateful for any help.

The third equation equation follows almost trivially from the fact that any number $k$ times zero is zero. Starting from Eq (2): $$ x' - c t' = 0 $$ Multiply both sides by $k$: $$ k(x'-ct') = k*0 = 0 $$

Two things that are both equal to a third thing are equal to each other. In this case we have two things that are equal to zero. So since we also have: $$ (x-ct)=0 $$ we have: $$ x-ct = k(x' -ct') $$

Answer 3

Actually, his argument is not watertight, and conformal transformations are a counter-example. If you also require the transform functions to be bounded functions, then the light-speed preservation condition is enough to derive their linearity; hence the constancy of the extra coefficient.

The condition $x' - c t' = 0$ if $x - c t = 0$ is actually a functional equation, expressed as $U(0,t) = 0$, where $U(u,t) = X(u + c t, t) - c T(u + c t, t)$, with $x' = X(x, t)$, $t' = T(x, t)$ being the transform functions. By the order 0(!) Taylor's Theorem, it follows that $u$ factors out of $U(u,t)$ and $U(u, t) = u V(u, t)$ for some continuous function $V(u, t)$, and in fact $V(u, t)$ is the average value of the partial derivative $U_u(⋯, t)$ taken over the interval $[0,u]$ ... provided the first-order derivative of $U$ is absolutely continuous (the usual assumption made for Taylor's Theorem). So, $$X(x,t) - c T(x,t) = (x - ct) V(x - ct, t).$$ The linearity condition makes the function $V$ a constant.

Linearity already follows by deep geometric arguments. It turns out that it is possible to build up the entire infrastructure of Minkowski geometry solely from the "light-connection" relation, which may be defined as: $(x,t) ⇔ (x',t')$ iff $|x - x'| = c |t - t'|$. This works at least for spatial dimensions of 2 or more. I'm not sure it can be done for 1+1 dimensional Minkowski geometry.

From the light-connection relation - and from that alone(!) - it is possible to define orthogonality, collinearity, co-planarity, parallelism, congruency for angles and segments, ratios of lengths and durations, and so on. The geometry is uniquely defined up to a choice of length or time unit, and an orientation of the time axis.

That's a purely optical axiomatization of Minkowski geometry. Here's one reference

Optical Axiomatization of Minkowski Space-Time Geometry
Cambridge University Press: 01 April 2022
Brent Mundy

but axiomatizations of this type go back to the 1950's. They are all reifications of the construction of Minkowski Geometry by A. A. Robb in 1914

A Theory Of Time And Space
https://archive.org/details/theoryoftimespac00robbrich

as a spatial extension of temporal logic, using as its sole primitive, the "before-after" relation, defined by: $(x,t) < (x',t')$ iff $t < t'$ and $|x - x'| < c |t - t'|$.

These are synthetic geometries. There are no coordinate systems assumed at the outset. They are derived and constructed.

Any transform that preserves the light-connection relation - by virtue of transforming lines to lines - must, therefore, be linear ... or else unbounded.

Without doubt, Einstein was aware of some of these developments, so he was probably piggy-backing on it, so as to be able to use linearity as a folklore theorem and have cover for his assertion of linearity.

Answer 4

Not sure what the context is but I will assume we are talking about travelling light beams. Light travels at c in both reference frames. Suppose both axes are lined up at the start.

A light beam travels at $c$ in the first frame $x=ct$

In the other frame, the light beam is also travelling at c. An observer in that frame measuring with their time is $x'=ct'$.

Now you have two equations $$x-ct=0$$ $$x'-ct'=0$$

Both equations are zero but they don't necessarily have to be equal. The equations could differ by multiplication of a constant or a function. From this you can construct your third equation for arbitrary $k$.