Struggling with a special relativity "paradox"

[KipIngram]

The argument goes like this:

We imagine Bob going in a very large circle, so large that its curvature can be neglected. There is a clock on the circumference of the circle, synchronised to a clock at the centre.

As Bob passes this clock, both that clock and his get synchonised to 0. Sometime later he passes that clock again. According to the dodgy version of SR (adopted by some) where time dilation applies equally to Bob and the clock on the circle, there is a paradox. According to each the other should have recorded less time during Bob's circular orbit. Which is physically impossible.

The resolution is that circular motion, no matter how large the circle, is not linear inertial motion. Bob's clock in non-inertial and records less time during a circular orbit than any clock at rest relative to this orbit.

Yes, I do think that having Bob go all the way around a full circle would raise problems. I do think, however, that you could approximate a tiny angle of a huge circle as a straight line and have the error go to zero as you made the circle big enough.

 

[PeroK]

Yes, I do think that having Bob go all the way around a full circle would raise problems. I do think, however, that you could approximate a tiny angle of a huge circle as a straight line and have the error go to zero as you made the circle big enough.

It's not enough to think that. You have to prove it mathematically. That's the acid test. I can assure you that in the case the mathematics does not cooperate with your beliefs.

 

[Ibix]

Your problem is still the relativity of simultaneity. Alice's camera aperture has finite extent; Bob does not see it open all at once. So he sees any point on the aperture open less than 30 times while he is in shot, although every point opens 30 times overall, and some part of the aperture is open 30 times while he's in shot (just not the same part).

 

[SiennaTheGr8]

But see my latest comment above - I think I saw through to the problem with my reasoning late last night. Alice can see the signals coming to her as approaching straight on along a path perpendicular to Bob's motion, but Bob cannot say the same - he will see the signals follow an angled path to Alice, because he sees Alice moving between emission and reception. Alice sees Bob moving between emission and reception as well, but it doesn't matter - once Bob "emits" it's done - his motion after emission is irrelevant. My originally presented "Bob analysis" of what Alice would see also assumed "straight shot" signals, and that was the invalid assumption. Even though Alice is at closest approach when Bob emits the signals, she's not when she receives them.

I'm going to work the algebra on this later today, but my gut tells me that all is well now.

I haven't followed this thread closely, but it seems to me that you're on the right track here.

I'd like to point out, though, that the "straight shot" assumption would be wrong even in Galilean relativity. Two inertial observers in standard configuration will disagree on the angle that some third party's velocity-vector makes with their coincident $x$/$x^\prime$ axis (unless that angle is $0$ or $\pi$), and this is true in both classical and relativistic mechanics. That angle is given by:

• $\cos \theta = v_x / v$ in the unprimed frame, and
• $\cos \theta^\prime = v_x^\prime / v^\prime$ in the primed frame,

where $v^{[\prime]}$ is the third party's speed and $v_x^{[\prime]}$ is the signed $x$-component of the third party's velocity. To find the relationship between the unprimed quantities and their primed counterparts, you need the appropriate transformation formulas (Galilean or Lorentz).

I suggest you do the algebra. That the "third party" is a light wave makes your life easier: $v = v^\prime = c$. Work out how $\cos \theta$ and $\cos \theta^\prime$ are related.

 

[KipIngram]

Ok, guys - I just did the math on this and it works out perfectly. The images from Bob that will line up with Alice's telescope are the ones he emits at closest approach. In Alice's frame, those just make a straight shot to her and everything is easy to see. In Bob's frame, though, he sees Alice traverse a 30 degree angle (for v=0.5) before those signals reach her. If Alice's speed is 0.5, then the hypotenuse of that triangle is lengthening at the speed of light. But Alice has a radially outward velocity component (in Bob's frame) of 0.25.

Bob sees 38.49 ms elapse between Alice's shutter events, but since she's moving radially away from him at 0.25, then only 3/4 of that 38.49 ms has passed her as far as the data stream she's seeing from Bob goes when her shutter fires again. 0.75*38.49 is 28.87 ms, which is exactly the right answer.

I think part of what made this hard for me to see what that my first knee jerk impression was that signals arriving from Bob at an angle like that wouldn't line up with her telescope. But we know that the signals Bob emits at closest approach will go into her scope - that's obvious by looking at it in Alice's frame. So they'll make it in viewed from Bob's frame too. As soon as I chased the timing of those signals down it all fell into place.

PeroK, I believe it's standard knowledge that if you zoom in enough on a circle or a spherical surface you get a manifold that looks more and more flat. In the limit you can treat it as flat. I'm happy with the resolution I outlined above.

Thanks again, everyone - this one scrambled my brain for a while, but it was still fun. Cheers!

 

[PeroK]

PeroK, I believe it's standard knowledge that if you zoom in enough on a circle or a spherical surface you get a manifold that looks more and more flat. In the limit you can treat it as flat. I'm happy with the resolution I outlined above.

A circle is not a straight line and a sphere is not a plane. You could have a polygon that approximates a circle, but a polygon isn't a straight line either.

 

[KipIngram]

I'm not going to argue with you about it. No, they're not exactly the same thing, but you can reduce the error by so assuming to as small a value as you wish if you look at a small enough part of the circle / sphere. Is the error zero? No. But it becomes less significant as you go. This is how numerical integration works - break the region into small enough pieces and you can consider them bounded by straight lines, or parabolas, or whatever your particular integration method happens to impose.

 

[Mister T]

I'm not going to argue with you about it. No, they're not exactly the same thing, but you can reduce the error by so assuming to as small a value as you wish if you look at a small enough part of the circle

So you have a segment of a circle with a radius much much larger than the length of the segment. You place a clock at each end of the segment and synchronize them in the rest frame of someone at the center of the circle. In the rest frame of someone traversing that segment those two clocks will not be synchronized.

 

[jbriggs444]

but you can reduce the error by so assuming to as small a value as you wish if you look at a small enough part of the circle / sphere

If you are looking at only a small part, you might as well make it a straight line or plane. The circular or spherical shape is not buying you anything. You still have mutual and symmetric time dilation (or a non-inertial coordinate system).

If you look at the whole thing, global topology enters in. Now the symmetry will be broken. Regardless of how large you make the circle or sphere.

 

[PeroK]

No, they're not exactly the same thing, but you can reduce the error by so assuming to as small a value as you wish if you look at a small enough part of the circle / sphere. Is the error zero? No. But it becomes less significant as you go.

Let's look at what we know. If Bob travels in a circle, then the rate of differential ageing depends only on the velocity (relative to the centre of the circle, where Alice is). That amounts to a certain differential ageing per unit time, hence per unit distance travelled. This is independent of the radius of the circle. No matter how large the circle, the differential ageing per unit time is the same.

The explanation for this, vis-a-vis the approximation argument, is that the larger the circle, the further away Alice is, so the reduction of angle is compensated for by the larger distance to the centre. These cancel out to leave the same differential ageing per unit time, as it must be.

The moral is that you have to be careful what aspects of the approximation to use. The distance traveled by a straight line or a chord can be made arbitrarily close, so you can approximate the distance travelled. But, the effects of following a curved path cannot be neglected, no matter how large the circle.

 

[PeterDonis]

I was trying to create a situation where time dilation was the only applicable effect

You can't. Not because of length contraction, but because of relativity of simultaneity. It is impossible to create a scenario in relativity in which time dilation is significant but relativity of simultaneity is not.

 

[Mister T]

I was trying to create a situation where time dilation was the only applicable effect,

IIRC Einstein was struggling with this puzzle when he was formulating his special theory of relativity. He knew that time dilation had to be symmetrical, and the notion that clocks synchronized in one rest frame will not necessarily be synchronized in a different rest frame appeared as the solution in his mind, it caused him to sit upright while lying in bed.

There is no getting around it. Simultaneity is relative, one of the hardest things for learners to grasp.

 

[PAllen]

I would like to point out the case of signals between a circular moving body and the center of rotation can be analyzed using the scheme of my post #27 as long as one crucial caveat about MCIFs (momentarily comoving inertial frames) is recognized. This caveat is that given a body undergoing general (not inertial) motion, an MCIF for one event on it must make the same observations as the non-inertial body for physics local to the comoving event. This explains the asymmetry between the center of circular motion and the circular moving body. The center is generally taken to be actually inertial. The revolving body is not. Thus you can validly use an MCIF to predict the result of local physics at event on the revolving body, but certainly not distant physics.

The upshot of this is that in scheme of post #27, the revolving body must be taken to be emitter for signals received by the center, and the receiver for signals emitted by the center. The reason is that this is the only way that has the MCIF matching local physics. If, instead, you treat a signal emitted by the center as 'emitted at closest approach', you have an MCIF for the circular moving body purporting to describe distant physics (the center).

Once this is accepted, the logic in #27 may be rigorously used to compute that signals emitted by the revolver are received at the center red shifted by the speed time dilation factor, while those emitted by the center are received by the revolver blue shifted by the time dilation factor.

Of course, there are much more elegant, direct ways to compute the circular situation, but the above shows the unique way to match it to transverse Doppler.