A paradox?

[Glom]

Probably not but anyway.

The C-ship Muppet leaves the planet Pigeon and heads away from it at speeds approaching c. The observation post on Pigeon shows that time has dilated and everything on Muppet is running slow. But, relative to Muppet, Pigeon is moving away at near c. Doesn't this mean that Muppet observes slowing on Pigeon as well. How can it be both ways?

More importantly, what would happen if Muppet were to slow, turn around, and return to Pigeon at near c. Would the same dilation effect be observed, albeit a bit bluer?

This Twins Paradox that Stephen Hawking loves so much, why would one twin be older? If one twin goes away near c, turns around and returns near c and as a result of dilation has taken years to do a short journey, wouldn't it be that the twin left at home has, relative to the other one, done the same thing?

[Grey]

The C-ship Muppet leaves the planet Pigeon and heads away from it at speeds approaching c. The observation post on Pigeon shows that time has dilated and everything on Muppet is running slow. But, relative to Muppet, Pigeon is moving away at near c. Doesn't this mean that Muppet observes slowing on Pigeon as well. How can it be both ways?

Each of them does observe the other to be slowing down as they move away. Since time is only relative to the observer, there's no reason that the two time frames have to follow common sense expectations. As long as the clocks are far from each other, you cannot compare them to see which one is "really" slowing down, so it's not fair to ask.

More importantly, what would happen if Muppet were to slow, turn around, and return to Pigeon at near c. Would the same dilation effect be observed, albeit a bit bluer?

Pigeon will continue to see Muppet's clock slowed. Muppet will actually see Pigeon's clock begin to move faster, and it will overtake Muppet's. Both observers would be able to predict these effects from general relativity, and the end result (that less time will have passed for Muppet) will be exactly what they would expect. See my next comment below.

This Twins Paradox that Stephen Hawking loves so much, why would one twin be older? If one twin goes away near c, turns around and returns near c and as a result of dilation has taken years to do a short journey, wouldn't it be that the twin left at home has, relative to the other one, done the same thing?

The crucial thing to be aware of is that the experiences of the two observers are not the same. Muppet accelerates at the beginning of the trip to a velocity near c, and then halfway there has to decellerate and then re-accelerate in the other direction. Pigeon experiences no such acceleration. Although it may seem like these accelerations last a short time relative to the flight, they are crucial.

In Muppet's rest reference frame, for example, those periods of acceleration would involve large g-forces. If the frame is considered at rest, those forces are indistinugishable from an intense gravitational field which, according to general relativity, should cause enormous time dilation, even from Muppet's point of view. If you do the math, it all works out, amazingly enough.

[SeanF]

More importantly, what would happen if Muppet were to slow, turn around, and return to Pigeon at near c. Would the same dilation effect be observed, albeit a bit bluer?

Pigeon will continue to see Muppet's clock slowed. Muppet will actually see Pigeon's clock begin to move faster, and it will overtake Muppet's. Both observers would be able to predict these effects from general relativity, and the end result (that less time will have passed for Muppet) will be exactly what they would expect.

Not really, no. While Muppet is traveling back towards Pigeon, Muppet will still see Pigeon's clock as moving slower, not faster. It's only during the actual changing of directions that Pigeon's clock would seem to "jump forward."

[BigJim]

Although primarily about string theory, the book "The Elegant Universe" by Brian Greene contains unsurpassed explanations of special and general relativity, as well as quantum theory. It would explain the question you ask amazingly well- consider picking it up from the library.

[daver]

This took longer than i expected to type (long enough that the session timed out). If this explanation isn't clear, you might try the sci.relativity FAQ.

There's a problem with reference frames when you're talking relativistic velocities. There are three reference frames involved--the planet, the outbound Muppet, and the inbound Muppet.

Let's make this a bit more explicit--say the outbound muppet is Honeydew, the inbound muppet is Beaker. And we have two planets, one light year apart, named Pigeon and Canary. Honeydew passes Pigeon at time point 0, after a bit Honeydew and Beaker pass Canary simultaneously, and eventually Beaker passes Pigeon. Canary and Pigeon think both Honeydew and Beaker are moving at .8 c, Honeydew and Beaker think both Pigeon and Canary are moving at .8 c, Honedew and Beaker think the other is moving at 1.6/1.64 ~= .976 c (relativistic velocity addition). Honeydew and Beaker think that Canary and Pigeon are moving slowly--at 60% normal rate. Pigeon and Canary think the same of Beaker and Honeydew. Honeydew and Beaker think the other is moving particularly slowly--at 22% normal rate. OK so far? You should crank the numbers through if you know the formulae.

As Honeydew passes Pigeon, he starts a stopwatch. This is going to be a very robust stopwatch--a Timex, possibly. Pigeon thinks the stopwatch is running 60% normal speed. Eventually, Honeydew passes Canary, and tosses the stopwatch to Beaker. An observer on Canary notes that the stopwatch was running slow (60% normal) with Honeydew, and slow (again 60%) when it was passed off to Beaker. The Canarinier noticed as well that .75 years had elapsed on the stopwatch. He's not surprised--he thinks that it should have taken 1.25 years to get there, and the clock is moving at 60% rate (1.25 * .6 = .75). Eventually Beaker passes Pigeon; an observer on Pigeon notices that the stopwatch is still running slow (again 60%) and that 1.5 years had elapsed on it. The Pigeoneer thinks that 2.5 years had elapsed since Honeydew passed. No surprises, right?

Alright. From Honeydew's perspective, it is only .6 light years between Pigeon and Canary (relativistic distance contraction). It takes .75 years for Canary to pass him. He tosses the watch to Beaker, and notices that as soon as the watch has accelerated to .976 c it starts moving at only 22% its normal speed. He thinks it is going to take 0.6/(.976 - 0.8 ) = .6/.176 = 3.417 years for Beaker to reach Pigeon. However, he sees the stopwatch is moving slowly, so he thinks that it will only register another .22 * 3.417 = .75 years. He thinks that 3.417+.75 = 4.17 years will have elapsed on Pigeon since it passed him, however, since their clocks are running slow, they will think only 4.17 * .6 = 2.5 years will have passed.

You could work the same thing out from Beaker's perspective. It is pretty much a mirror image of Honeydew's--3.417 years for the watch to get from Pigeon to Canary, .75 years more to get back to Pigeon.

Let's say someone with real good eyesight on Pigeon were watching the stopwatch all the way. As the stopwatch passes Pigeon the first time, he sees it moving at 60% normal speed. However, as it passes, Doppler rears its ugly head, and the stopwatch seems to move even slower (if he compensated for Doppler, it would still be running at 60% c). Anyway, 2.25 years after it passed, he sees the handoff from Honeydew to Beaker at Canary (1.25 years to get to Canary, and another 1 year for the light to travel from Canary back to Pigeon). The stopwatch has registered .75 years in 2.25 actual years. Now, however, the stopwatch is coming back. It is severly blue-shifted, and appears to be moving quite a bit faster. It apparently gains another .75 years in only .25 years. However, if he compensated for the Doppler, it would still seem to be running slowly.

So, to (finally) answer your questions. If you can ignore Doppler, the Piegeonese and Canarians think the others are moving at normal speed, everybody else thinks the others are moving slower then normal.

The Twins paradox crops up because the situations aren't symmetrical--the Pigeons might think that the Muppets are experiencing similar time distortions, but the outbound Muppet and the inbound Muppet think that the other is moving much more quickly, and experiencing a more severe time distortion than the planets.


Oh. By the way, this is all Special Relativity. You don't need General Relativity for this.

[Glom]

Thanks for all the effort, Daver. I'll have to read your post a few times before I fully comprehend it, but it seems pretty concise.

Thanks to the others as well.

[daver]

Thanks for all the effort, Daver. I'll have to read your post a few times before I fully comprehend it, but it seems pretty concise.

Thanks to the others as well.

You're welcome.

I think you should skip the section on Doppler shift in my explanation--if you use the relativistic formula (sqrt((1-beta)/(1+beta))) the time distortion is already factored in. If you use the classical formula i think it works out ok.

One of the hard parts in explaining relativity problems is keeping track of the reference frames. Another is knowing under what circumstances you can compare events in one reference frame to those in another. However, to me, once you have the situation set up properly, it's almost magical the way all the numbers work out.