Thread: Einstein's mathematics in his derivation of the Lorentz transformation.

I found his book (on Sp. Rel.) on the internet at this page

Appendix 1. Simple Derivation of the Lorentz Transformation. Einstein, Albert. 1920. Relativity: The Special and General Theory

Over a period of days, weeks I have more or less come to terms with it but there is a little piece that is puzzling me still.

I should start by posting the two equations(5) he refers to a little earlier

So start quote

We should thus have the solution of our problem, if the constants a and b were known. These result from the following discussion.	5
For the origin of k' we have permanently x' = 0, and hence according to the first of the equations (5) If we call v the velocity with which the origin of k' is moving relative to K, we then have

This is the point where I puzzle over -coming up

The same value v can be obtained from equation (5), if we calculate the velocity of another point of k' relative to K, or the velocity (directed towards the negative x-axis) of a point of K with respect to K'. In short, we can designate v as the relative velocity of the two systems.

8

Furthermore, the principle of relativity teaches us that, as judged from K, the length of a unit measuring-rod which is at rest with reference to k' must be exactly the same as the length, as judged from K', of a unit measuring-rod which is at rest relative to K. In order to see how the points of the x'-axis appear as viewed from K, we only require to take a “snapshot” of k' from K; this means that we have to insert a particular value of t (time of K), e.g. t = 0. For this value of t we then obtain from the first of the equations (5) x' = ax. Two points of the x'-axis which are separated by the distance x'=1 when measured in the k' system are thus separated in our instantaneous photograph by the distance But if the snapshot be taken from K'(t' = 0), and if we eliminate t from the equations (5), taking into account the expression (6), we obtain From this we conclude that two points on the x-axis and separated by the distance 1 (relative to K) will be represented on our snapshot by the distance But from what has been said, the two snapshots must be identical; hence x in (7) must be equal to x' in (7a), so that we obtain OK I do follow his logic but I am not quite clear why we cant just use x' = ax. and to solve for "x". I do see that this will not give the correct answer and "(a^-1)xprime=xprime*(a(1-v^2/c^2))^-1) just gives me a = a(1-v^2/c^2) Am I just going down a wild goose chase path or is my mathematics also faulty? As an aside , this relationship between x and xprime does it arise in other circumstances (in mathematics or physics)? I mean can you find other pairs of functions where one functions "view" of the other is practically a mirror image of the other's although they are not inverse functions? I hope I made some sense (it was difficult for me to copy out the equations in my browser)