A difficulty with the equivalence of all inertial reference frames

[JuanGC]

Although I am not a physicist, I am interested in physics, and recently I've been reading about special relativity. I have a doubt about it, a difficulty I see in the equivalence of all inertial reference frames which I haven't found solved anywhere, and I've thought perhaps you in this Forum could solve it.

Let's imagine two observers moving relative to each other. One of them has a laser pen, and he turns it on in a direction perpendicular to the direction of motion. In the reference frame in which the observer with the laser pen is at rest, as the light travels in a straight line from the emitter, the laser ray would be always perpendicular to the direction of motion.

But in the reference frame in which the other observer is at rest (and the one with the laser pen is in motion), as the light travels in a straight line from the emitter and the emitter is moving, the laser ray wouldn't be perpendicular to the direction of motion, but would travel with an angle from the laser pen. So it seems that the two reference frames aren't equivalent.

 

[FactChecker]

Your next conclusion should be that, in the space-time geometry of the reference frame where the light pen is moving, the pen is pointing in a different direction. That is true. [EDIT: apparently, this is wrong] There is an apparent rotation of moving objects.
You should be aware that every aspect of Special Relativity was scrutinized by skeptical physicists before they were forced to accept it. The space-time geometry of SR has a lot of effects that make it all consistent in any inertial reference frame. That is what is meant by saying that inertial reference frames are all equivalent, not that all processes are identical, but that they are all consistent within each IRF.

 

[Orodruin]

Your next conclusion should be that, in the space-time geometry of the reference frame where the light pen is moving, the pen is pointing in a different direction. That is true. There is an apparent rotation of moving objects.

However, the pen that is perpendicular to the direction of motion does not tilt.

 

[FactChecker]

However, the pen that is perpendicular to the direction of motion does not tilt.

Now I am really confused. Doesn't that mean that the flash of light does not go in the direction that a moving laser pen is pointing?

 

[Orodruin]

Doesn't that mean that the flash of light does not go in the direction that a moving laser pen is pointing?

Yes. But this is as it should be.

 

[DrGreg]

Maybe this animation helps (from an old post of mine):

Top image: ball bouncing up and down
Bottom image: ball moving diagonally
But they're both the same thing depicted in different frames.

 

[Sagittarius A-Star]

Although I am not a physicist, I am interested in physics, and recently I've been reading about special relativity. I have a doubt about it, a difficulty I see in the equivalence of all inertial reference frames which I haven't found solved anywhere, and I've thought perhaps you in this Forum could solve it.

Where did you read it? The word "equivalence" may be misleading here. Maybe you mean this: According to the first postulate of SR, the laws of physics are the same in all inertial reference frames. Your example is not contradicting to this first postulate of SR.

 

[FactChecker]

Maybe this animation helps (from an old post of mine):
View attachment 298681

Top image: ball bouncing up and down
Bottom image: ball moving diagonally
But they're both the same thing depicted in different frames.

I see. Very good. Thanks! This really helps to visualize how a light beam through a tube would work. I suppose laser light is the same -- the coherence would depend on the motion and direct it laterally to match the motion. I don't really know anything about lasers.

 

[PeroK]

So it seems that the two reference frames aren't equivalent.

It's fairly obvious that different reference frames are not equivalent. The surface of the Earth in London, England and the surface of the Earth in Wellington, New Zealand are not equivalent.

What they have in common, however, is that the laws of physics are the same in both. You can always tell by looking at the stars whether you are in the northern or southern hemisphere. Or by measuring the Earth's magnetic field. But, you don't have one set of Maxwell's equations of Electromagnetism for England and a different set of Maxwell's equations for New Zealand. Physicists in England and New Zealand use the same laws of electromagnetism. And, in that sense, they are "the same".

What matters in your example is whether the same laws of physics for the laser apply in both reference frames.

 

[JuanGC]

It's fairly obvious that different reference frames are not equivalent. The surface of the Earth in London, England and the surface of the Earth in Wellington, New Zealand are not equivalent.

What they have in common, however, is that the laws of physics are the same in both. You can always tell by looking at the stars whether you are in the northern or southern hemisphere. Or by measuring the Earth's magentic field. But, you don't have one set of Maxwell's equations of Electromagnetism for England and a different set of Maxwell's equations for New Zealand. Physicists in England and New Zealand use the same laws of electromagnetism. And, in that sense, they are "the same".

What matters in your example is whether the same laws of physics for the laser apply in both reference frames.

Then it seems that the laws of physics are not the same for both reference frames.

 

[PeroK]

Then it seems that the laws of physics are not the same for both reference frames.

Give me an example of a law of physics!

 

[Sagittarius A-Star]

I don't really know anything about lasers.

For this you don't need to know anything about lasers, only about relativity of simultaneity.

Assume, a horizontal electromagnetic wavefront in the vertical laser crosses the horizontal x-axis. In the primed rest frame of the laser, the left and right side of the wavefront cross the x'-axis simultaneously, that means
$\Delta t' = 0$.

With inverse LT and length contraction follows with reference to the unprimed frame, that the left and right side of the wavefront cross the x-axis with the following time difference:
$\Delta t = \gamma (\Delta t' + v/c^2 * \Delta x') = \gamma (0 + v/c^2 * \Delta x') = v/c^2 * \Delta x$.

For small angles: When the right side of the wavefront crosses the x-axis, the left side has already moved further across the x-axis by
$\Delta d \approx c * \Delta t = c * v/c^2 * \Delta x = v/c * \Delta x$.

That means the wavefront is tilted by

$\Delta d / \Delta x \approx v/c$.

In both frames, the Poynting vector $\vec S = \vec E \times \vec H$ is oriented perpendicular to the wavefront.

 

[phinds]

Then it seems that the laws of physics are not the same for both reference frames.

Huh? How do you conclude that?

 

[JuanGC]

Huh? How do you conclude that?

Because depending on the reference frame, the laser ray goes in different directions, and reaches different places.

 

[Dale]

But in the reference frame in which the other observer is at rest (and the one with the laser pen is in motion), as the light travels in a straight line from the emitter and the emitter is moving, the laser ray wouldn't be perpendicular to the direction of motion, but would travel with an angle from the laser pen. So it seems that the two reference frames aren't equivalent

It is correct that the path of a pulse of light is perpendicular in one frame and not in the other; this is called relativistic aberration. It is incorrect that this implies any inequivalence between the frames.

In this context the equivalence between reference frames means that the laws of physics are the same in all reference frames. In order to produce a collimated pulse of light requires a device designed according to the laws of physics to do so. If you analyze any such device you will see that the same laws produce the different results in the different frames.

The easiest example to analyze is a spherical light flash in a box with a pinhole. The flash goes in all directions, but only the ray that goes from the flash to the hole leaves the box. In the frame where the box is moving, this ray is angled, using only the same laws of physics.

 

[Orodruin]

If you analyze any such device you will see that the same laws produce the different results in the different frames.

Just to clarify: This means that the light signal can travel at different angles relative to the relative velocity between the frames, not that it can hit a hole in one frame and not in the other. The same set of events will always occur in all frames.

 

[JuanGC]

In order to produce a collimated pulse of light requires a device designed according to the laws of physics to do so. If you analyze any such device you will see that the same laws produce the different results in the different frames.

The easiest example to analyze is a spherical light flash in a box with a pinhole. The flash goes in all directions, but only the ray that goes from the flash to the hole leaves the box. In the frame where the box is moving, this ray is angled, using only the same laws of physics.

Does the same happens with a laser pen, or a flashlight?

 

[PeroK]

Because depending on the reference frame, the laser ray goes in different directions, and reaches different places.

Would the same be true of an apple? If you are on a moving train and throw an apple out so that it goes perpendicular to your motion. Let's say you throw it out as you enter a platform and it smashes a window half way along the platform (just as you are passing the smashed window).

What you are saying is that to someone standing on the platform, the apple will go out in a straight line at the end of the platform and miss the windows?

The apple goes in two different paths depending on your frame of reference? In one frame of reference it smashes a window and in the other it misses the window?

 

[FactChecker]

For this you don't need to know anything about lasers, only about relativity of simultaneity.

Assume, a horizontal electromagnetic wavefront in the vertical laser crosses the horizontal x-axis. In the primed rest frame of the laser, the left and right side of the wavefront cross the x-axis simultaneously, that means
$\Delta t' = 0$.

With inverse LT and length contraction follows with reference to the unprimed frame, that the left and right side of the wavefront cross the x-axis with the following time difference:
$\Delta t = \gamma (\Delta t' + v/c^2 * \Delta x') = \gamma (0 + v/c^2 * \Delta x') = v/c^2 * \Delta x$.

For small angles: When the right side of the wavefront crosses the x-axis, the left side has already moved further across the x-axis by
$\Delta d \approx c * \Delta t = c * v/c^2 * \Delta x = v/c * \Delta x$.

That means the wavefront is tilted by

$\Delta d / \Delta x \approx v/c$.

In both frames, the Pointing vector is oriented perpendicular to the wavefront.

"only about relativity of simultaneity" Ha! As usual. That, and looking at the wavefront. Thanks.

 

[PeroK]

Maybe this animation helps (from an old post of mine):
View attachment 298681

Top image: ball bouncing up and down
Bottom image: ball moving diagonally
But they're both the same thing depicted in different frames.

@JuanGC it would be interesting to know what you think of this animation? Is it right? Or, wrong? Does it apply to a little red ball but not to light?

 

[phinds]

Because depending on the reference frame, the laser ray goes in different directions, and reaches different places.

Wowzers, you really have missed the boat on this one. The laser doesn't give a hoot what various observers see it doing, it just does what it does and if the second observer has any brains he can figure that out. I assume you get this now, after looking at post #6

 

[FactChecker]

Does the same happens with a laser pen, or a flashlight?

Yes. @Sagittarius A-Star 's post #12 explains how the wavefront of the light will be tilted due to the relativity of simultaneity. To the frame moving with the pen, the wavefront seems to go in the direction that the pen points. But to the frame that sees the pen moving, the pen's time has been distorted and the wavefront is tilted to keep up with the moving pen. So the light pulse does not travel in the direction that the pen was pointing. That will be true for any light source.

 

[JuanGC]

@JuanGC it would be interesting to know what you think of this animation? Is it right? Or, wrong? Does it apply to a little red ball but not to light?

I thought it wouldn't apply to light, but for what I'm reading here it seems I was wrong.

 

[PeroK]

I thought it wouldn't apply to light, but for what I'm reading here it seems I was wrong.

It must apply to light. Otherwise you have inconsistent experimental data.

Personally, I would never say all IRF's are "equivalent" - for the reasons I gave above. Instead, they are all equally valid. This means you may analyse a physical scenario in any IRF. The same set of physical events must take place in each IRF. You may then transform the coordinates of those events to any other IRF.

This is true in both classical and relativistic physics. The difference is the way you transform coordinates: In classical, Newtonian physics you use the Galilean transformation; and in SR, you use the Lorentz Transformation.

 

[Dale]

Does the same happens with a laser pen, or a flashlight?

Yes. It is just that the relevant laws for those are not as simple as for a pinhole. If the laser or flashlight beams line up with a pinhole beam in one frame, then they will line up in all frames.

 

[Dale]

Just to clarify: This means that the light signal can travel at different angles relative to the relative velocity between the frames, not that it can hit a hole in one frame and not in the other. The same set of events will always occur in all frames.

Yes, thanks for the clarification

 

[vanhees71]

One should mention that the phenomenon described in #1, which is known as the aberration of light, has already been treated in Einstein's very first paper on Special Relativity in 1905. In modern terms it uses the fact that the wave vector $\vec{k}$ together with the (angular) frequency $\omega$ builds a Minkowski four-vector, $(k^{\mu})=(\omega/c,\vec{k})$. Since Maxwell's theory of classical electrodynamics is a relativistic field theory that implies that when transforming a plane-wave em. field from one inertial reference frame to another using the corresponding Lorentz transformation, leads again to a plane wave, and the wave-fourvector is transforming as a fourvector. This leads to the prediction of both the Doppler effect of light (including the transverse Doppler effect which goes beyond the analogous non-relativistic Doppler effect of, e.g., sound waves) and the aberration of light. For a complete treatment see Sect. 4.2.1 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[JuanGC]

Thank you very much for your answers. Now I understand.

 

[Mister T]

Let's imagine two observers moving relative to each other. One of them has a laser pen, and he turns it on in a direction perpendicular to the direction of motion. In the reference frame in which the observer with the laser pen is at rest, as the light travels in a straight line from the emitter, the laser ray would be always perpendicular to the direction of motion.

But in the reference frame in which the other observer is at rest (and the one with the laser pen is in motion), as the light travels in a straight line from the emitter and the emitter is moving, the laser ray wouldn't be perpendicular to the direction of motion, but would travel with an angle from the laser pen. So it seems that the two reference frames aren't equivalent.

Suppose A is the observer with the laser pen and B is the observer who sees the ray as not perpendicular.

If instead B has the laser pen and points it perpendicular to the direction of travel, A will see it as not perpendicular.

Thus you cannot use this experiment to determine which observer is stationary and which is moving. This is the essence of what is meant by the equivalence of inertial reference frames. There is no one frame that can be determined to be absolutely at rest.