No, because that statement isn’t invariant under Lorentz transformations. Light rays are solutions of the wave equation so all these statements are exercises about the wave equation.

The two ends of the rod are separated by a space-like distance, that's why they can't be in causal relation.

It suffices to draw the light cones centered on each end to find that they can only intersect on points that lie on the straight line that bisects the rod...

Dear Professor Nicolis,

Thanks for your comment.

"No, because that statement isn’t invariant under Lorentz transformations"

Do you mean that light-ray lengths are not invariant under Lorentz tranformations? and this somehow precludes the above as a possibility? I'm not sure I catch your drift?

Dear Professor Nicolis,

"The two ends of the rod are separated by a space-like distance, that's why they can't be in causal relation."

This is a very interesting comment. Are you suggesting the above would indicate some kind of causal relation, if the above cicumstance did exist?

Hi Gary

We do not need to talk about Lorentz transformations to answer your question.

First, it is correct, and, I'm glad you understand that clearly, that if two events happen simultaneously at the same place, that is an absolute fact. That means they are simultaneously happening at the same place in every reference frame.

What you seem to be missing is that if two events happen simultaneously at different places for one observer, they will not be simultaneous for observers moving w.r.t to the first one.

Now, let's go back to your initial example. If the observer sees the light emitted simultaneously from the two extremes, they will meet at the center otherwise we have a logical contradiction. However, other observers in relative motion will not see the light meeting at the center of their frames. To see this you have to understand two things

1) In their reference frames, M' is not at the center of the segment determined by places where the light rays were emitted. The point M' is in the center of the stationary observer but cannot be at the center of the other reference frame that is moving w.r.t. the stationary one.

2) In the moving frames, the light rays are not emitted simultaneously, otherwise the point M' will also be at their center.

Dear Professor Lambare,

Thank you for your excellent comment, clearly and concisely made.

I'm with you down to here, when you say :

"What you seem to be missing is that if two events happen simultaneously at different places for one observer, they will not be simultaneous for observers moving w.r.t to the first one."

If we look at Norton's first animation, a man is moving to the left. We are not with that man. The location of the man on the plank of wood is M.

https://sites.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/Special_relativity_rel_sim/index.html

If we look at Norton's second animation, we are now with that man moving to the left. This is exactly equivalent to the carriage moving to the right, by the Relativity of Motion (for uniform motion). The man on the plank of wood is still standing at the same place. This place we now say is M'.

But, M and M' are "at the same place".

This is why in both frames, the rays -have to- arrive simultaneously to the same place, as you right pointed out in your piece.

If what you said in the latter part was right, (and M and M' were not at the same place) then the man on the plank of wood could say he was in two places at once. He never is. He is only at the one place, and that place is in the middle of the plank of wood.

Dear Gary

If we say that M' is the "place" where the man on the train is, and M is the "place" where the man on plank wood is, then M and M' cannot be the same place in space. Why are you saying "But, M and M' are "at the same place"?

Maybe you are confused by the following:

- M is stationary and always in the same place in the plank wood but not in the train.

- Reciprocally, M' is always in the same place in the moving train but not in the plank wood.

- M and M' coincide at a given moment but not when the light beams hit those places.

Dear Professor Lambere,

You'd be right if we were talking about Absolute motion (if there is absolute motion.)

Then every point in space would have a unique position. And the train carriage would be objectively at different points in space on its journey.

But, instead, we are talking only about Relative motion.

The two are antipodes of each other.

So, when I say the man on the train at M is "at the same place" -- you will come back and say, -- I don't think so-- the carriage was in Amsterdam, now it is in Brussels-- hardly the same place!

To which I would respond. No the carriage is at the same place, with the man on it at M. It is Brussels that has moved to where Amsterdam was.

Gary, you are not understanding relative motion.

The man on the train is at the same point M' always. What happens is that M' is stationary in his own refernce frame (the train) but in motion w.r.t. the wood plank.

The two frames are objectively different, when we say they are equivalent wnat we mean is that the same physical laws apply to both.

Dear Professor Lambere,

"The two frames are objectively different"

How are the two frames "objectively different" ?

So in the case of the clock-paradox, Dingle would say "which clock is doing the moving ?" -- which clock is the one that is supposed to "run slow".

Dear Gary Stephens

They are objectively different because they are two different physical systems. You and I are governed by the same physical laws, but we are objectively different persons, aren't we?

You either are on the train or in the wood plank. The observer in the train is one person, and the one on the plank wood is another.

How could they be the same? The principle of relativity means that both "different" frames are governed by the same physical laws, it does mean they are the same indistinguishable things.

In the case of Dingle, clock A in frame X1 run slow with respect the clock B in frame X2 and vice versa, it does not mean that A and B are the same thing.

Why that is not a contradiction? To understand that you have to understand how time is measured according to SR and Dingle simply did not understand that.

Returning to the original question points M and M' are different points in space. M is at rest in one frame but is moving w.r.t the other and the same goes for M'. When M and M' coincide crossing each other, two light rays are emitted from two points that are equidistant from M and M' in their respective reference frames. Do the rays are emitted simultaneously in both frames, the answer according to relativity is a definitive "NO" because if they are simultaneous in one frame they cannot be simultaneous in the other one.

That is what Einstein and Norton were trying to explain.

Dear Professor Lambare,

To say that one frame is definitely still, and the other frame is definitely moving, is an Absolutist view, is it not?

So if I said, you're right, the carriage is definitely moving, and the embankment is definitely still, would you say, yes, that is correct?

Or would you say, no, the carriage can just as easily be said to be still, and the embankment moving.

I accept we have two frames, but I don't accept that one can be said to be moving, and the other can be said to be still.

Each frame is both still and moving.

Dear Gary,

I am sorry but can't figure out how are thinking too.

Nothing is definitely still or moving either in Newtonian Mechanics or SR. In both theories only acceleration is absolute. It is so because we can physically measure acceleration but not uniform motion w.r.t. absolute space.

The train is actually moving w.r.t. the embankment and at the same time is acually at rest w.r.t. to itself. Both frames absolutely do not accelarate w.r.t. absolute space if we assume both are inertial systems.

I am sorry if that does not enlighten you.

By the way Gary, what is your educational background. Just want to know why you are interested in understanding SR.

Gary Stephens ,

it is obvious for example in air that relative moving objects having the same position when a sound is emitted, will not detect the sound at the same time and affected by Doppler effect.

Dear Professor Quattrini,

"it is obvious for example in air that relative moving objects having the same position when a signal is emitted, will not detect the sound at the same time and with a Doppler effect."

But it is light that has the extraordinary properties-- not sound. (Confer the Michelson Morley experiment). Or I suppose you could say sound has very much less of the extraordinary properties -- perhaps so much less, that to all intents and purposes it has no extraordinary properties.

"Each frame is both still and moving."

And since in the still frame the light-rays always meets in the middle at M. It follows logically that in the moving frame the light rays must also meet in the middle at M'.

This is also true for the off-middle setup-- described under this head : since in the still frame the light-rays always meet off-middle at M, then it follows logically that in the moving frame the light rays must also meet off-middle at M'.

(Dear Professor Lambare, I have a background in Low Temp Phys lab. experiments - ie. superconductivity.)

Dear Gary Stephens

You said:

- And since in the still frame the light-rays always meets in the middle at M. It follows logically that in the moving frame the light rays must also meet in the middle at M'

No, that is incorrect because the light rays meet in the middle only in the frame where they are emitted simultaneously. They can be emitted simultaneously only in one frame.

To put it in perspective, in the Einstein and Norton examples, they assume that the rays are emitted simultaneously in one frame and then show that in the other frame, they do not meet in the middle, therefore they could not have been emitted simultaneously if light speed is invariant.

(Thanks for your answer, I just was curious. You are on the right track, many people go directly to the math without understanding the basics and end up understanding nothing.)

Gary Stephens ,

if you consider that air is at rest with the plaform the same applies.

Stefano Quattrini

Your analogy could be confusing because in this case, the speed of sound is different in both frames so you cannot conclude that if "simultaneous in one frame is not simultaneous in the other".

Justo Pastor Lambare ,

considering the platform, where the air is at rest, the result is the same if the observer travels in the air of the platform.

Stefano Quattrini

I do not understand, are you saying that both observers in relative motion measure the same speed of sound in the air? Do you have air at rest within the two frames in relative motion?

Justo Pastor Lambare ,

• considering that the air is stationary with the embankment
• considering that I can synchronize two sources of sound at rest in the embankment using sound up to a good precision with ultrasounds

The observer M stationary in the middle between the sources will detect the signals at once

if a moving observer M' is at the same location of M when the emission is performed, M' will not detect the signals at once.

Stefano Quattrini

Yes, that's right. But I said the example is confusing because the velocity of the signals is not the same for both observers as is the case with light.

Justo Pastor Lambare,

the velocity of light does not sum up with the speed of the source, this can be certainly said also for light. That the velocity of signals is the same for both observers is something synchronization dependent since there is no way to determine the one-way speed of light.

I agree, it is not necessary even to mention the velocity of the source.

But, certainly, the speed of signals is not the same for both observers.

For the observer stationary w.r.t. the air the velocity of signals is the same from both sides. However, for the one that is moving w.r.t. to the air, the velocity is higher for the signal to which he is approaching and lesser for the signal from which is receding. That does not happen with light because it is forbidden by principle.

Justo Pastor Lambare , In (non-flying) temporal light theory (TLT), light creates relative time and inertia, so within one inertial frame (flying) rays must intersect at the center.

https://note.com/s_hyama/n/n36da9fb827df

Justo Pastor Lambare ,

the time that the signal takes to reach the moving target is shorter both for sound and for light.

In the case of sound it is straightforward to find the speed of the body in air (which is the medium). Also the Doppler effect (only first order) allows to know the speed in the medium, thanks to the two distinct formulas for moving source and moving observer. In the case of light, exchanged in longitudinal motion it is not possible to that, there is no distinction between moving source and moving observer, it is just one, thanks to the gamma factor and this has been experimentally verified.

The pure transverse Doppler absent in sound, at least in that form, makes a difference in detecting who is moving more or moving less (circular motion)

Gary Stephens ,

if you bend that meter, making the extremes match, and then move M' you will have a similar configuration as the Sagnac experiment being sure that the rays depart at once. The Sagnac effect is due specifically to that difference in arrival time of the light waves at the interferometer which is the common observer.