Hello everyone, hope you all are having a good day. I am trying to derive Lorentz transformations from the principle of relativity. I read a few papers on the topic and one of the papers I read was this.

The paper is about how only the principle of relativity can be used to derive the general form of transformation between inertial frames of reference. However, I am getting confused over the part where the author shows such a transformation is a linear one. Their argument goes like this:

Suppose you have two inertial frames of reference A and B and B is moving at a relative velocity v with respect to A, say in the positive x-direction. In general, the transformation between any two arbitrary frames of reference will be of the form-



The author then says that if we have one end of a rod (placed along the x-axis) of some length l, in B's frame of reference, at point xB1 then the other end of the rod will be at xB1+l at that instant. Thus, the the position of the first end in A's frame of reference will be X(xB1,tB,v) and the position of the other end will be X(xB1+l,tB,v). The author then says that the length of the rod in A's frame of reference will be

l'= X(xB1+l,tB,v)-X(xB1,tB,v) .....1

And this where I am confused. How is this the length of the rod in A's frame of reference? Even though the positions of the two ends of the rod are known, these positions are taken at different times in A's frame of reference in general. The time in A's frame of reference for the position of the first end will be T(xB1,tB,v) and the time when the position of the other end is taken is T(xB1+l,tB,v). Unless the rod is stationary in A's frame of reference, l' won't be the length of the rod in A's frame of reference in general, right? The rod could be in motion with respect to A and as far as I know, there shouldn't be any restrictions on whether the length being measured is stationary or moving or accelerating. Shouldn't the position of both ends be taken at a fixed time in A's frame of reference to measure the length? This is a crucial step in the derivation because later the author displaces the rod in B's frame of reference and homogeneity of space implies as long as the rod is identical, the length of the rod doesn't depend on where it is kept along the x-axis. Then in A's frame of reference the new positions of the ends are calculated and using the above equation 1, the new length is equated to the original length to show that xA varies linearly with xB.

At the start I was convinced about this method's validity and I only came up with this doubt when I was trying to find what argument can be used to show that tA varies linearly with xB and I was stuck. So far I've only been able to show that xA and tA vary linearly with tB using homogeneity of space and time.

I have seen identical derivation in other papers but I am still confused about the method the authors use. Am I just missing something simple? Can someone give a physical explanation of whether the method used by the author is correct or not?

  • $\begingroup$ Note we use Mathjax for mathematical expressions. It's the site standard. $\endgroup$ Commented Jul 4, 2021 at 5:07

1 Answer 1


I cannot speak in detail about that paper (I have it myself and have only glossed over it, but it seems valid - it gets the right answer for a start!).

The reason I did not persevere is that I already had this article, which is (to me) much simpler. The derivation itself starts with the paragraph beginning "For a typical axiomatic derivation of the Lorentz transformation, . . .", but it is well worth looking at the whole thing.

Reading this might help in understanding the "Nothing but Relativity" paper, if you find you still need to.

  • $\begingroup$ Thanks for the reference. I am convinced that the arguments the author has used are correct. I am just confused about the way they have used those arguments. $\endgroup$
    – A.G.47
    Commented Jul 4, 2021 at 11:43
  • $\begingroup$ So am I, hence my answer! $\endgroup$
    – m4r35n357
    Commented Jul 5, 2021 at 9:58

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