In Einstein's book "Relativity" there is a wonderful derivation of the Lorentz transformation, requiring no more than high school algebra (pp. 117 - 121). It is quite clear but I do not understand one early step.

Equation (1) is $$x - ct = 0$$ Equation (2) is $$x' - ct' = 0$$

I don't see how (1) and (2) imply

Equation (3) $$(x - ct) = \lambda (x' - ct')$$

This seems to be saying $0 = \lambda 0$ mathematically, which makes no sense.

Other questions in this forum have dealt with this, and one commenter said that (3) follows because the transformation between the two coordinate systems is linear.

Linear transformations do take straight lines through the origin of one coordinate system to straight lines through the origin of another, but are (1) and (2) enough to imply that the transformation is linear, and if so does that make them imply (3)?

  • 2
    $\begingroup$ Hi, you should use MathJax to format your equations :) $\endgroup$ Jan 9, 2019 at 20:22
  • $\begingroup$ In the book edition printed by Crown Publishers Inc., New York, 1961 this is pp 115-120. $\endgroup$ Jan 9, 2019 at 21:19
  • $\begingroup$ Perhaps you should try reading a different book. $\endgroup$
    – safesphere
    Jan 10, 2019 at 3:47

3 Answers 3


Equations (1) and (2) relate to light signals, while equation (3) applies to any event, including a light signal.


Multiplying each side of the second equation by $\lambda$, we get:

$$\lambda (x' - ct') = \lambda × 0 =0 \space ,$$

Using this equation along with the first one, we easily get:

$$\lambda (x' - ct') = x - ct \space .$$

This seems to be saying $0 = \lambda × 0$ mathematically, which makes no sense.

Dandy! Why do you think it does not make sense?!

Second method:

The first and second equations imply:

$$x=ct, \space x'=ct'\rightarrow -x=-ct, \space \lambda x'=\lambda ct' \space.$$

Adding left sides and adding right sides gives:

$$\lambda x'-x=\lambda ct'-ct \rightarrow \lambda x'-\lambda ct'=x-ct \rightarrow$$

$$\lambda (x' - ct') = x - ct \space .$$

  • $\begingroup$ After downvoting please add a comment to show me where I am wrong. $\endgroup$ Dec 14, 2019 at 10:24
  • $\begingroup$ FYI, the downvote here wasn’t me. $\endgroup$
    – Dale
    Dec 14, 2019 at 20:19
  • $\begingroup$ @Dale Then why don't you upvote to uphold the justice?! $\endgroup$ Dec 14, 2019 at 21:21

The reason for my confusion is that I came to Einstein's derivation of the Lorentz transformation by recommendation from another source, and so I just read it in the appendix (Routledge Edition pp. 117 - 121) without reading the rest of the book.

Not explicitly stated in the appendix was that Einstein was discussing something called the standard configuration -- which I found elsewhere -- reprinted below

The assumptions of the standard configuration are as follows:

  • An observer in frame of reference K defines events with coordinates t, x, y, z
  • Another frame K' moves with velocity v relative to K, with an observer in this moving frame K' defining events using coordinates t', x', y', z'
  • The coordinate axes in each frame of reference are parallel
  • The relative motion is along the coincident xx' axes
  • At time t = t' =0, the origins of both coordinate systems are the same.

Another assumption is that at time t = t' = 0 a light pulse is emitted by R at the origin (x = y = z = x' = y' = z' = 0)

The only possible events in K and K' are observations of the light pulse. Since the velocity of light (c) is independent of the coordinate system, K' will see the pulse at time t' and x' axis location ct', NOT x'-axis location ct' + vt'. So whenever K sees the pulse at time t and on worldline (ct, t), K' will see the pulse SOMEWHERE on worldline (ct', t').

The way to express this mathematically is by (3) (x - ct) = lambda * (x' - ct')

x - ct = 0 (observation of the pulse in K) FORCES x' - ct' = 0

An event on K's worldline will be an event on a similarly constructed worldline in K'.


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