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I am currently working my way through "Special Relativity and Classical Field Theory (The Theoretical Minimum)" by Susskind and Friedman, and I'm a little bit perplexed by the interpretation of this picture being used to describe Lorentz Contraction:

enter image description here

The idea is that an observer in the x', t' frame observes a 1-unit length OQ in the x, t frame as they move past. The author plugs the 1 into the Lorentz transformation equations and we see that OP is $\sqrt{1-v^2}$ where $v$ is in units of light-speeds. The authors say that the unit length therefore looks shorter.

It seems to me that an alternate interpretation is that the moving observer at point P is just witnessing the future of the end of the stationary meterstick at point Q. This is, I think, easy to see from the diagram, that P is in Q's future from the PoV of the stationary frame. Observing the future of Q, the moving observer does not see the end of the meterstick, but instead sees that they are somewhat past it (having passed it, since they are in a moving frame).

To illustrate, imagine that, to signal the measurement-taking, the moving observer flashes a light signal at both the origin and at point P simultaneously (for them) at time t'=0. They stop and come back to chat with the stationary observer and say "hey, I measured your meterstick to be less than 1." The stationary observer might reply "well no duh, I saw your signals and, even accounting for the speed of light I saw that you took your second measurement much later than the first."

Is this a valid alternate interpretation of the situation, or am I missing some crucial point here? Is length contraction just the manifestation of our divergent time axes?

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  • $\begingroup$ Remember, O, P, Q aren't just points in space. They're events in spacetime. $\endgroup$
    – PM 2Ring
    Commented Apr 10, 2023 at 22:58

2 Answers 2

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The length of a stationary object is the distance between the positions of its endpoints. With moving objects, the distance between the two endpoints only gives the length if you pin them down at the same time. If I take the position of the front of a moving train and compare that with the position of the rear a few second later, the rear will have moved forward in the intervening few seconds, so the distance between the two points will be less than the true length of the train. That is essentially what is happening with length contraction. To measure the length of a moving object you have to determine the distance between its two endpoints simultaneously, and the problem is that two moments that are simultaneous in one frame are not simultaneous in another.

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Your core concept is correct, but some of the specific details you state are not.

Any segment from the $x=0$ line to the $x=1$ which is parallel to OQ represents the length in the unprimed frame. And any segment parallel to OP represents the length in the primed frame. The fact that they are not parallel indeed stems from the divergent time axes, and also implies length contraction.

However, none of the comments hold regarding future or past. The distance is constant in each frame. It doesn’t change over time in either frame. So regardless which segment parallel to OP you choose, there are segments parallel to OQ which are entirely earlier, segments which are entirely later, and segments which “straddle” the chosen OP segment. None of those pairs of segments are priveliged over any other.

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  • $\begingroup$ "there are segments parallel to OQ which are entirely earlier, segments which are entirely later." Sure. The authors chose the segments they did because the frames coincide at the origin (i.e. (0,0) in frame 1 is (0,0) in frame 2.) It makes the math easier because you don't have to fiddle with both endpoints, but only one. $\endgroup$
    – Him
    Commented Apr 11, 2023 at 0:39
  • $\begingroup$ "none of the comments hold regarding future or past" so point P is not, actually, in Q's future? Possibly I'm misunderstanding the diagram. $\endgroup$
    – Him
    Commented Apr 11, 2023 at 0:43
  • $\begingroup$ P is indeed in the causal future of Q, but that fact has nothing to do with length contraction. $\endgroup$
    – Dale
    Commented Apr 11, 2023 at 1:16
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    $\begingroup$ @Him The rotation of the time axis is not reflected in the fact that P is in the future of Q. That is true in both frames. What they disagree on is P and Q’s relationship to O. For the unprimed frame P is in the (non-causal) future of O, while in the primed frame they are simultaneous. And the reverse for Q. That is what shows the rotation of the time axis and what they disagree on. In contrast “P is in Q's future” is not just “from the PoV of the stationary frame”, but from both frames. That fact does not depend on the rotation of the time axis $\endgroup$
    – Dale
    Commented Apr 11, 2023 at 11:31
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    $\begingroup$ @Him FYI, it is discouraged to alter a question after it has been answered. At least in a way that makes the answers no longer make sense. You don’t need to revise the question here, it is a good question as is $\endgroup$
    – Dale
    Commented Apr 11, 2023 at 12:52

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