I Is this another "paradox" or a veritasium mistake?

[mgkii]

TL;DR

Laser pointer shot across a spaceship in deep space hits the wall directly across from it. When the spaceship accelerates the beam curves and hits lower down. However, an outside observer will (according to the video source) observe no change in light path.

This Veritasium video https : / / youtu.be / XRr1kaXKBsU?si=zNqHwmCTq-cmcsbW (spaces added in case there's an issue with posting URLs) at 13:04 makes a statement which I can't get my head around. Veritasium videos always seem to be well researched so my working assumption is it's my lack of brain power rather than their mistake! However, I can't see a direct reference to this one on the forum.

The video describes a spaceship in deep space under no acceleration. A laser pointer it set up to point across the ship (side to side as opposed to nose to engine) and as expected, the pointer highlights a spot directly across from the source. The spaceship then fires up an goes into constant acceleration and the laser point moves down the wall slightly in the same way it would if the acceleration were due to gravity. So far so good.

The video then goes on to say that an outside observer in an inertial FOR would see no bend in the light path. My understanding breaks here, because they couldn't see the light hitting a different part of the wall, so how would they see something different? I would have reasoned that an outside observer would see exactly the same effect as the accelerating observer; the movement of the light beam is perpendicular to the direction of acceleration.

Am I missing something or was it an error in (as far as I can tell) an otherwise excellent video?

Many thanks in advance
Matt

 

[Dale]

In the inertial frame the light goes straight. The accelerating rocket’s path is a parabola in spacetime. So they see the laser hit lower because the rocket’s path has curved up.

 

[PeroK]

The video then goes on to say that an outside observer in an inertial FOR would see no bend in the light path. My understanding breaks here, because they couldn't see the light hitting a different part of the wall, so how would they see something different?

The inertial observer would see the same emission and impact points, but that impact point would have moved (relative to the point of emission) after the light ray was emitted.

I would have reasoned that an outside observer would see exactly the same effect as the accelerating observer; the movement of the light beam is perpendicular to the direction of acceleration.

A straight line in an inertial reference frame may be curved in an accelerating reference frame and vice versa.

 

[PeroK]

PS note that this scenario in itself is not peculiar to special relativity. In classical physics, a light beam could bend in an accelerating reference frame and be a straight line in an IRF. As could the motion of any particle.

 

[mgkii]

So it is a mistake in the video. Many Thanks Indeed everyone! Happy to close the thread if that's the normal protocol.

 

[PeroK]

So it is a mistake in the video. Many Thanks Indeed everyone! Happy to close the thread if that's the normal protocol.

I didn't watch the video, but I assume it's correct. Light moves in a straight line in an IRF.

 

[256bits]

TL;DR: Laser pointer shot across a spaceship in deep space hits the wall directly across from it. When the spaceship accelerates the beam curves and hits lower down. However, an outside observer will (according to the video source) observe no change in light path.

Am I missing something or was it an error in (as far as I can tell) an otherwise excellent video?

The video is somewhat lax in the explanation.
Good catch!

We can take as true for the spacecraft thought experiment in that light travels in a straight line for a non-accelerating observer.
With the coasting rocket ( no acceleration ) the light beam hits the wall a a particular point on the opposite wall - ie the 0 ( zero ) mark as seen in the video. Same observation for the outside inertial astronaut.

With the accelerating rocket the light beam hits a spot lower down, say spot B.
Both observers should see that.
Astronaut in the accelerating ship concludes that light has curved.
Inertial astronaut outside the ship concludes the the ship has moved up ( ie ship has different velocity at time of absorption ( at B ) from the velocity at previous time of emission.

His Veritasium statement that the inertial astronaut sees the same thing - ie light is moving in a straight line - but offered little else on what the inertial astronaut sees on the opposite wall ( spot B )

 

[PeroK]

I watched the relevant part of the video. The video is correct, in what it says about light as measured in inertial and accelerating reference frames. Note that, as above, that is as expected.

The difference in this context between classical physics and relativity is that the light would bend twice as much when calculated using the equations of GR as it would using the classical acceleration of gravity.

 

[256bits]

Shoot. I was writing and did not see all the other responses coming in.

 

[A.T.]

The difference in this context between classical physics and relativity is that the light would bend twice as much when calculated using the equations of GR as it would using the classical acceleration of gravity.

This is true for the total global deflection over a large region around a mass, where the effects of spatial curvature become non-negligible. For the local experiment in the accelerating space-ship, where the effects of spatial curvature are negligible, the deflection predicted by GR matches the one by classical acceleration of gravity.

See:
https://www.mathpages.com/rr/s8-09/8-09.htm

 

[A.T.]

The video is correct, in what it says about light as measured in inertial and accelerating reference frames.

I'm not so sure about the end of the video though, where it claims that charges under constant proper acceleration (e.g. resting on the Earth's surface) would be "radiating" according to relativity, but we could not test this so far.

I rather think that charges under constant proper acceleration have a distorted but static E-field in their rest frame, according to relativity. You need a changing proper acceleration of a charge to radiate energy away, in a frame invariant manner.

 

[PeroK]

I'm not so sure about the end of the video though,

I just watched a bit of it. I must admit, I'm not predisposed to watch a video where someone claims everyone else is wrong!

It's quite ironic that Veritasium found corroboration here. That might even be considered a paradox.

 

[PeroK]

This is true for the total global deflection over a large region around a mass, where the effects of spatial curvature become non-negligible. For the local experiment in the accelerating space-ship, where the effects of spatial curvature are negligible, the deflection predicted by GR matches the one by classical acceleration of gravity.

See:
https://www.mathpages.com/rr/s8-09/8-09.htm

I've just been out for a bike ride and I was wondering about that as I was cycling round!

 

[Orodruin]

TL;DR: Laser pointer shot across a spaceship in deep space hits the wall directly across from it. When the spaceship accelerates the beam curves and hits lower down. However, an outside observer will (according to the video source) observe no change in light path.

My understanding breaks here, because they couldn't see the light hitting a different part of the wall, so how would they see something different

Yes, they would, but this would be attributed to the spaceship accelerating rather than the light following a curved path.

 

[Dale]

So it is a mistake in the video. Many Thanks Indeed everyone! Happy to close the thread if that's the normal protocol.

Yes, I would say the video is wrong in that spot. The two mistakes in that clip are:

(13:18) From the perspective of the external inertial observer when he correctly says that the light travels in a straight line they show a graphic of it hitting the original location on the wall (it will actually hit lower because the light goes straight and the rocket accelerates)

(13:29) In the rocket’s accelerating frame the rocket is always at rest. So in the accelerating frame it is not true that the rocket has sped up. So that cannot be the reason in that frame that the light’s path curves.

 

[FactChecker]

Good catch. You are correct. What the man says from 13:04 to 13:20 seems all correct, but the video illustration is bad. Everyone sees the light hit the same spot on the spaceship wall. Relative to the ship's accelerating frame, the light path curved down and hit the wall lower. Relative to the non-accelerating observer's reference frame, the light goes straight and the ship wall accelerates up, so the light hits the wall lower. They both see the light hit the same spot on the ship wall. The video illustration does not show that correctly, but all the statements are correct.

 

[PeroK]

(13:18) From the perspective of the external inertial observer when he correctly says that the light travels in a straight line they show a graphic of it hitting the original location on the wall (it will actually hit lower because the light goes straight and the rocket accelerates)

I missed that. I assumed he was still talking about inertial rocket motion at that point.

This is the problem with spending all the effort dressing up in a rocket suit. Instead of showing us the two perspectives of the two cases drawn on a blackboard!

 

[PeterDonis]

The difference in this context between classical physics and relativity is that the light would bend twice as much when calculated using the equations of GR as it would using the classical acceleration of gravity.

As @A.T. pointed out, this is not correct. The factor of 2 you describe here is global, not local; the local light bending in relativity, which is what is being described in the video, is the same as it would be in Newtonian physics. The global factor of 2 comes from the contribution of spacetime curvature to the global path of the light; in the local context spacetime curvature is not detectable.

 

[PeroK]

As @A.T. pointed out, this is not correct. The factor of 2 you describe here is global, not local; the local light bending in relativity, which is what is being described in the video, is the same as it would be in Newtonian physics. The global factor of 2 comes from the contribution of spacetime curvature to the global path of the light; in the local context spacetime curvature is not detectable.

I already read and replied to his response.

 

[Dale]

This is the problem with spending all the effort dressing up in a rocket suit. Instead of showing us the two perspectives of the two cases drawn on a blackboard!

And making the animations appealing rather than accurate. The rocket doesn’t appear to be accelerating, or at least if it is accelerating it is too subtle to be visible.

 

[A.T.]

And making the animations appealing rather than accurate.

I'm not sure if this is the case here, but I think it's common to have the graphics done by someone else. And at some point, one might simply give up explaining how it is still not quite accurate. Just like when you contribute a chapter to a technical book, and the publisher has the diagrams redone by their graphics department, which is clueless about what they mean.

I never had a disconnect between audio and visuals when watching 3blue1brown videos, because he makes everything himself.

 

[FactChecker]

IMO, Veritasium tries hard to be correct. It might be nice to leave a comment on the video about the mistake.
UPDATE: I see that the video already has an appropriate comment by Rafael Silva.

PS. Proofreading your own work is a hard, endless, thankless, job.

 

[PeroK]

IMO, Veritasium tries hard to be correct. It might be nice to leave a comment on the video about the mistake.

The video was posted 5 years ago and has already been described in the comments as a "masterpiece"!

That said, it would be interesting to see the reaction to a proposed correction at this stage.

 

[pervect]

I watched the relevant part of the video. The video is correct, in what it says about light as measured in inertial and accelerating reference frames. Note that, as above, that is as expected.

The difference in this context between classical physics and relativity is that the light would bend twice as much when calculated using the equations of GR as it would using the classical acceleration of gravity.

It would be bent twice as much by a mass. But I don't think not on an elevator. I don't have a reference for this statement, but I believe it to be true. If you want to be absolutely sure, you'll need to either do your own calculation of the path of light in a Rindler frame, look up such a calculation (and hope it's right), or find some other reference.

 

[PeterDonis]

It would be bent twice as much by a mass.

To be more precise: the global bending of light when passing close by a massive object, i.e., the angle between its incoming path at infinity, and its outgoing path at infinity, is twice what you would expect from a naive Newtonian calculation

But I don't think not on an elevator.

To be more precise: the local path of light through an accelerating elevator is "bent" downwards by the same amount as you would expect from a naive Newtonian calculation.