# Lorentz triangle and length contraction perpendicular to propagation

1. May 4, 2014

### ANvH

Consider a pole of 1 light second long in the $y$ direction (the vertical line(s) in the enclosed figure). It is moving in the $-x$ direction. According SR, the pole's length is not contracted because its length is not parallel to the propagation direction. However, given the time of flight of light, a signal from the pole's end will arrive 1 second later than a light signal from the pole's start. Given the pole is moving, one would expect an aberration of the pole's end. In the figure I have drawn a triangle, with the hypothenuse indicating the aberration angle $\phi$. The horizontal line depicts the speed.

Both triangles in the figure are identical, the labeling of the sides is according to Lorentz's triangle that defines the $\gamma$ factor. The labeling of the sides of the lower triangle are obtained by dividing the labels of the upper triangle by the $\gamma$ factor.

When observing the moving pole in a rest frame one would expect that the hypothenuse is slanted to the left (think of mirror images of these figures). Choosing a frame where the pole is at rest, one would observe (by taking a picture) an aberration as shown in the figure. But SR does not predict this. It would predict no aberration.

Further, a peculiar issue is shown in the figure. The bottom triangle indicates that the vertical line is contracted by the $\gamma$ factor, which is not predicted by SR. The upper triangle does not suggest length contraction of the vertical line (the pole), but shows that the hypothenuse is dilated by the $\gamma$ factor.

I am trying to figure Lorentz's assertion of length contraction that explained the null result of MM and the above is a representation of one the light-arms involved.

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2. May 4, 2014

3. May 4, 2014

### ANvH

Thanks for the heads up, but don't you agree that the Lorentz transformation fails to predict this? After all, when moving with the pole, i.e., observing a pole at rest does not require a correction of the time of flight of light, because we can measure the pole's length at will. If we were to allow such a correction then we would define
$t'=t - c^{-1}y$ ​
$y'=y-ct$​
with a determinant equal to zero. Combining this transformation with the usual Lorentz transformation requires a matrix with two time components and two ($x, y$) spatial components. Reading up on such matrix scenario's is apparently a no, no.

So does the Penrose-Terrell rotation has a place in SR, or is it considered as visual effects extraneous to SR, while SR has nothing to do with optical phenomena?

4. May 4, 2014

### Staff: Mentor

Fails to predict what? It certainly doesn't fail to predict Penrose-Terrell rotation. Of course, in order to correctly predict that, you need to take into account the fact that the light reaching your eyes at a given instant from different parts of a moving object will have left the object at *different* instants.

First of all, if you are at rest relative to the pole, Penrose-Terrell rotation doesn't come into play. Neither do any of the aberration effects you're talking about. So I'm not sure why you're considering that case.

Given that you're at rest relative to the pole, however, you haven't stated things correctly. The correct statement is: you can measure the pole's length if it is at rest relative to you, without having to make corrections for light travel time (for example, by seeing where its ends fall relative to a measuring rod that's at rest relative to you and the pole). But measuring the pole's length in this way is *not* the same as "observing" the pole in the sense of actually looking at the image produced by the light entering your eyes from the pole at a particular instant. The light entering your eyes at a particular instant is always time-delayed if it's coming from a part of the pole that is spatially separated from you: even if you are at rest relative to the pole, you can't be co-located with more than one point along the pole, so light from any other part of the pole is time-delayed when it reaches your eyes.

I have no idea what you're talking about here. Do you have a reference you can give for this transformation, and why it's a "no, no" to read up on it?

Certainly. If you want to know what an object moving relativistically actually looks like, as in, what the images entering your eyes at a given instant will look like, Penrose-Terrell rotation is what tells you that.

No. See above.

5. May 4, 2014

### ANvH

When at rest relative to the pole, a light signal at the end of the pole is delayed and therefore there is an aberration if the pole and I are moving with respect to some absolute rest frame. This means that in such a scenario I would know that I am moving with the pole, thanks to the finite speed of light. Is that not correct?

The matrix I constructed for an object at rest. Was trying to combine this with the Lorentz transformation, realizing that a) the determinant is zero, that b) in the particular scenario two time components are required. A while ago I was reading up on multidimensional matrices on one of the wikipedia pages, where it was stated that a 1:3 matrix is stable (1 component for time, 3 components for space). Other combinations involving more than one time components were not stable; can't find the specific page. Don't know why it would be unstable, it was merely expressed as such.

6. May 4, 2014

### Staff: Mentor

No. Aberration is relative to the actual observer seeing the light, not relative to some arbitrary frame. If you are at rest relative to the pole, you observe no aberration in light from the pole. An observer that is moving with respect to the pole will see aberration. But if there's no actual observer observing the light. "aberration" is a meaningless concept.

Also, there is no such thing as an "absolute rest frame" in SR.

You would know you were moving with the pole because you can directly measure that you are moving with the pole: zero Doppler shift of light signals between you and the pole, constant round-trip light travel time from you to any point on the pole, etc. You don't need to know what any other frame is doing; in fact you don't even need to *define* a frame. The fact that you are at rest relative to the pole is a direct, frame-independent observable.

I still have no idea what you are talking about here.

Then I'm afraid I can't be much help, because none of this makes any sense to me.

7. May 4, 2014

### ANvH

Exactly, SR tells you this. However, you do know that after the first Michelson experiment in 1881, Lorentz's remark was that the transverse path should undergo aberration to account for the null result. The M&M paper of 1887 describes this modification in experiment and approach, however, again a null result was obtained.

I respect your answer, but I am not satisfied with it. I think this pole concept introduced here is incompatible with SR. I also think that Lorentz's suggestion after the 1881 experiment is similar to the pole concept. His suggestion was of course way before 1905, before SR, but given Lorentz's high standing in physics, the concept of aberration where the pole and the observer are at rest is certainly not meaningless.

I also think it is unfortunate that you dismiss aberration when at rest to the pole. I think you are missing the point when the pole and the observer are moving. The delay of light arriving at the observer under the condition described is exactly the point Lorentz made in 1881.

8. May 4, 2014

### Staff: Mentor

Yes, and all this was well before SR anyway. I'm not sure what you think this is supposed to show.

I'm not sure what "concept" you're referring to. SR explains the null result of the MM experiment perfectly well, without having to resort to the stuff Lorentz hypothesized. If you're finding something in your description of the pole that seems incompatible with SR, my money is on you misunderstanding something, not on SR being wrong. But I confess I don't understand what you're trying to describe well enough to be able to point out a specific error.

This is an argument from authority and carries no weight here.

I didn't "dismiss" it, I said that aberration is observer-dependent. An observer at rest relative to the pole will see no aberration in the light coming from it; an observer moving relative to the pole will.

Delay relative to which observer? I think you're confusing light speed time delay with aberration; they're not the same thing. At least, the standard meaning of "aberration" is not the same as light speed time delay. You may be using "aberration" to mean something other than the standard SR meaning of that word; if so, you should explain exactly what you mean by "aberration", and preferably find a different word to use for it.

9. May 5, 2014

### ghwellsjr

10. May 5, 2014

### ANvH

Ok, aberration causes objects to appear to be angled or tilted towards the direction of motion of the observer. I mentioned this in the first post and referred to the triangles you should see as the mirror image of the triangle shown.

The point I make is a reverse aberration, i.e., causes the object to be angled or tilted in the opposite direction of motion. This would occur when both observer and object move with the same speed and direction.

But I understand your point here, it is a matter of velocity addition and my scenario and that of Lorentz do not correspond to this.

Edit: The M&M experiment of 1887 to account for this suggestion of Lorentz did not make sense.

Last edited: May 5, 2014
11. May 5, 2014

### ANvH

12. May 5, 2014

### Staff: Mentor

Yes, I understand that, but "motion of the observer" here means motion relative to the source. If the observer is at rest relative to the source, there is no aberration.

I don't understand what the triangle is representing or why you should see a mirror image, under what circumstances?

This makes no sense to me (unless it's based on a belief in an "absolute rest frame"--see below), but more importantly, it's inconsistent with actual experimental observations, which show, as I said above, that there is no aberration when the observer and the source are at rest relative to each other.

With regard to "observer and object moving with the same speed and direction", you used the term "absolute rest frame" in an earlier post, and I responded that there is no such thing as an absolute rest frame in SR. I might add that there is also no experimental evidence for any such absolute rest frame. Lorentz apparently believed that there was such a thing, but every proposal he made for experimentally detecting it (of which, IIRC, his proposal about aberration being present when both observer and object were moving with the same speed and direction was one) turned out not to work, i.e., the proposed experiments, when done, gave null results. So if you're trying to argue that Lorentz was right somehow, you aren't going to make much progress here without being able to point at any experiments justifying his view.

13. May 5, 2014

### ANvH

In essence I already agreed with this when I responded to your previous post.

14. May 5, 2014

### ghwellsjr

I'm not sure it has been solved. In that thread, you were trying to define a Doppler with a moving source and no observer and I see you trying to do the same thing with regard to aberration in this thread. Otherwise, why are you insisting on the source moving relative to the frame?

EDIT: To clarify: Aren't you trying to define aberration as something that happens with a moving source with regard to the frame and then you define a reverse aberration that applies from the frame to the moving observer so that the two cancel out leaving the observer with the same observation that he would get if both the source and observer were not moving?

Last edited: May 5, 2014
15. May 5, 2014

### Staff: Mentor

Agreed with what? That Lorentz's proposals were disproved by experiment? Then I don't understand what the point of this thread is, since as far as I can tell, your "triangle diagram" in the OP was based on Lorentz's disproved proposals.