For a while I've been trying to write a simple geometric derivation of the Lorentz transformation, based only on the second postulate. Although I'm able to successfully derive the Lorentz transformation, the geometry of that derivation is always a combination of relativistic effects (ie. Lorentz contraction and time dilation) together with observational effects related to observing an event through the medium of light.

Is observation a component of the Lorentz transformation?

From what I've understood previously, the Lorentz transformation has always been communicated as a transformation of space-time for a single event from one inertial frame to the same event in another inertial frame. So I'm guessing there's something wrong with the geometry that I'm deriving.

Here's what it looks like:

geometry of the Lorentz transformation

In words, given two inertial frames (Amy and Henry), for any given photon emission event $(x_{\small{e}}, t_{\small{e}})$ from a stationary torch in Amy’s inertial frame:

  • calculate where the photon from that event, travelling at the speed of light relative to Amy, intercepts Henry’s path,
  • calculate where, in Henry’s time-dilated and Lorentz-contracted coordinates, a photon observed at the same point must have originated from the torch if the photon was travelling at the speed of light relative to Henry.

The result is the Lorentz transformation. If it helps to find where I'm going wrong, the visualisations that I've included on the geometry of the Lorentz transformation are interactive SVG's so you can test any point you want. Thanks for any help!

  • $\begingroup$ Closely related: physics.stackexchange.com/questions/59985/… $\endgroup$
    – user10851
    Feb 17, 2014 at 5:02
  • $\begingroup$ I don't think that question is closely related - the geometry above implies the normal lack of simultaneity, regardless of observations. But I think I've asked the wrong question - maybe: "Is it just a coincidence that you can derive the Lorentz transformation without assuming the Minkowski space-time metric?" would be better. It seems quite neat and interesting (to me), and still exhibits all the expected physical L-contraction, time-dilation, lack of simultaneity, but does imply observation is involved in the complete transform (baulk). See the link for interactive examples. $\endgroup$ Feb 17, 2014 at 9:37

1 Answer 1


No, the Lorentz transformations do not include the travel time of the photon from the transformed event to an observer.

The Lorentz transformations do indeed derive from a geometrical property, but the property in question is the invariance of the spacetime interval:

$$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $$

I don't really understand your diagram, but it looks as if you've assumed the $S'$ frame is merely the $S$ frame rotated about the origin. If so, you've got off on the wrong foot. If you draw the original and primed coordinates together they look more like:


  • $\begingroup$ Thanks John. Yes, I can derive the L-transform if I start with the Minkowski space-time metric, but I was interested to see if I can derive the L-transform starting only with the 2nd postulate. It turns out it is indeed possible to do this, but from what you're saying, just a coincidence, or simply an incorrect derivation of a correct result - not sure. If you have time, I'd really appreciate feedback from someone who had gone quickly through the derivation (see the link in the question), but I understand if that's not possible. $\endgroup$ Feb 16, 2014 at 16:29

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