Relativity and the "Twin Paradox"

bonepile · Aug 18, 2005

I have had trouble understanding how an absolute frame of reference CANNOT exist for some time now. I figured that a lot of people who can explain this to me probably visit this message board, so here goes: The "Twin Paradox" usually is explained where one "twin" accelerates away from the other "twin", and then accelerates back. I realize that acceleration has implications in motion through the space-time continuum, so allow me to present an alternate example that does not deal with acceleration. Let's say we have three probes in deep space: probes X, Y, and Z. X and Y are touching, and Z is moving towards them at near c velocity. X and Y each have an atomic clock, which they synchronize. Then, X and Y move apart from each other along the line shared with Z. After some time, X and Y both stop. At this point, X and Y's clocks are still synchronized, right? Now X and Y are both broadcasting their time into space. Z has a receiver and can take into account the Doplar affect of the transmissions from X and Y and all that. As Z flies by X (the first probe), it synchronizes its own internal clock to the value of X's broadcast. Then, as Z flies by Y, it compares Y's broadcast time with its own internal clock. Which clock is ahead, Z's or Y's? Remember, velocity terms like "c" and "stop" are all relative. From Earth's perspective, it could just as well be X and Y that are moving at near c velocity.

lunatio_gordin · Aug 18, 2005

Hm. As long as X and Y are moving at the same speed, i suppose they should still be synchronized. And Z synchronises with X, it will take some time passing at whatever speed to pass in front of Y. So it might be mere trillionths of a second behind Y. maybe... If I understood that right.

bonepile · Aug 18, 2005

You understood it correctly. At face value, it seems that whether Z is ahead or behind Y depends on their motions relative to Earth. If X and Y are moving at near c, then Z would be ahead. If Z is moving at near c, then Z would be behind, as you suggested. However, motion relative to Earth should have nothing to do with this. So does this imply that there IS an absolute frame of reference? Or is there another explanation beyond basic time dilation that would make Z be behind Y?

Saiph · Aug 18, 2005

we'll look at the reference frame thing. I stick you in a box, in space, and wander off. You feel no acceleration/gravity. Are you moving? Can you tell? How? <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

lunatio_gordin · Aug 18, 2005

He must be moving. he has to be orbiting something...

Saiph · Aug 18, 2005

no acceleration, no gravity. This is a though experiment, so he doesn't have to be orbiting anything. If you really must be technical, yes, he does. but the influence is so minor it doesn't matter (he's between galaxies, no nearby stars). <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

lunatio_gordin · Aug 18, 2005

I guess then we come to the statement that Motion is relative, huh? (wish i could find my book...)

Saiph · Aug 18, 2005

correct, as you cannot perform any experiment to show that you are moving, and not the observer, motion is relative. As such, two people in space are completely correct in stating, "I'm stationary, he's moving". As such, applying the special relativity time dilation equations shows that both see the other persons clock as slowed down, as long as both are in a uniform inertial frame (nobody turns, or accelerates). However, the only way to check who is right, is to compare clocks, side by side, in the same location. The problem: someone has to turn around. By doing so we know that someone moved (the other guy didn't feel the acceleration) and when clocks are compared, the guy who turned around had the slower clock. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

bonepile · Aug 18, 2005

It depends. If I am in a position to conduct this X, Y, Z experiment, then I could tell how fast I was moving based on the time difference between Y and Z. If motion is truly relative, then Y and Z should be synchronous, but that doesn't seem to jive with time dilation.

bonepile · Aug 18, 2005

In that experiment, Y and Z's clocks are effectively compared side by side, in the same location, though not in the same frame of reference. Neither Y nor Z has experienced any acceleration since their clocks were set. I can see how acceleration could play a part, but what happens when acceleration is taken out of the picture, as in the X, Y, Z experiment?

Saiph · Aug 18, 2005

If you compare yourself to Y and Z, you've just done everything relative to Y and Z. There's no reason everybody else has to use those two clocks. Now, if Y and Z are moving the same way, relative to you (both going say 30 m/s), they'll read the same. Doesn't matter which direction they're going, just that the speed is the same. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

bonepile · Aug 19, 2005

Y and Z are not moving at the same speed (or more precisely, they do not share the same frame of reference). X and Y are moving at the same speed, and Z is moving at near c velocity relative to X and Y. Allow me to change the experiment. I've thought some more about it, and I've realized that the experiment is too theoretical because Z had to accelerate at some point. This means that Z will be behind Y. If X and Y did the acceleration to set up the experiment, then Z will be ahead of Y. So here is a new experiment, which is actually more simple and also realistic in the sense that we could some day go out and try it out. 1) We start with three probes again: U, V, and W (so as not to confuse with the first experiment). They are all together, in the same frame of reference. All three probes synchronize their clocks. 2) V and W accelerate away from U to near c velocity (relative to U). 3) V decelerates until its velocity is 0 relative to U. 4) W decelerates until its velocity is 0 relative to U. 5) Now, with all three probes in the same frame of reference, they each transmit a signal with their time stamp to each probe. When a probe receives a signal, it echoes back with its own time stamp at the time it received the signal. In this way, each probe can precisely determine the current time in all probes by calculating the trip delay and comparing with the received time stamp. With the above experiment, we would expect the ordering of the clocks to be U /> V > W, right? Ok, no problem so far. But let's do one more simple experiment to illustrate the problem. 1) Start with two probes, S and T. They synchronize their clocks. 2) T accelerates away from S to near c velocity. 3) S accelerates to T's frame of reference. 4) T and S compare clocks as in the first experiment. In this second experiment, S /> T, right? Well, herei

Saiph · Aug 20, 2005

Once you throw in acceleration, you're dealing with GR, and that can make things complicated. Because it also depends on how large the acceleration is IIRC, to determine total dime dilation. Assuming identical accelerations: The first experiment U />V & V=W. Second: S=T Assuming identical accelerations. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

bonepile · Aug 29, 2005

I apologize for taking so long to reply. I was out of town for a week and without internet. If this is the case, then my understanding of relativity must be even worse than I thought. It seems to me that V must be greater than W if time dilation applies even when you are no longer accelerating. W spends more time in a near c frame of reference (relative to U) than V does. If time dilation is only occurs during acceleration, it seems that there are other paradoxes out there.

Saiph · Aug 29, 2005

oh, it works during times of constant velocity. But the only time it becomes "fixed" and permanent, is during acceleration. During times of uniform motion, you'll only have relative time dilation. Both travelers will be able to look at the other and say they are running slow. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

mrmux · Aug 29, 2005

The universal 'axis' suggested (discovered?) by researchers from the Universities of Rochester and Kansas (1997) could mean the Universe is 'spinning'. I don't know whether that helps GR or hinders it.

bonepile · Aug 30, 2005

Wierd stuff! Thanks for the help. That explains the paradox, though it also means that I need to do some serious reading up on GR before I try to think about this anymore...

emperor_of_localgroup · Aug 30, 2005

The universal 'axis' suggested (discovered?) by researchers from the Universities of Rochester and Kansas (1997) could mean the Universe is 'spinning'. DUH!!!! From atoms to mega galaxies are spinning and now iin 1997 the astro people are guessing the universe may be spinning?? If something, even an obvious conclusion, doesn't fit their models they never do any research on a topic. <div class="Discussion_UserSignature"> Earth is Boring </div>

mrmux · Aug 30, 2005

Sorry, Emperor, I don't know if you're telling me off for being pointless and dumb or for stating the already well-known and accepted. I just remembered that article from 1997, that's all. It stuck in my mind at the time and I always wondered what became of it. Apologies if it wasn't relevant.

emperor_of_localgroup · Aug 30, 2005

@ MrMux: Sorry, I didn't make it clear. I always bash Astrophysicists for not doing research on new subjects that can cast new lights on this mystery, rather than squeezing the last drop of juice out of 'black holes' and 'big bang'. I had this idea for sometime that since everything in the universe is in circular motion, the universe itself may be in circular motion. Since we are not sure about the shape of the univ, it may be the 'space' that is spinning. This spinning theory may even lead to an explanation of the expansion of the univ. Who knows. So now, as you can see, I am mad at the Astronomers/Astrophysicists who didn't pay attention to this scenario until 1997. Next time I'll be more clear about who Im displeased with. <div class="Discussion_UserSignature"> Earth is Boring </div>

bonepile · Aug 31, 2005

I was thinking about this a little bit more, and I realized that it has some crazy implications. Consider the U, V, W experiment, except this time let's say that we're starting from a point close to Earth. V makes it to the edge of the solar system, but W makes it all the way to Alpha Centauri. You're saying that V and W's clocks will be synchronized after the experiment is over? And can we predict how much their clocks differ from U, given the curve of their velocities relative to U (in U meters/ U second)?

Saiph · Aug 31, 2005

Assuming no acceleration (they synchronize clocks when the all pass U at the same instant): V is traveling slower than W, wrt U. So V's time dilation is less than W, and it's clock will be closer to U than W's. So say W takes 8 years to get to alpha centauri according to U. Say W experience 6 years of time pass themselves. V, traveling slower, experiences somewhere between that, lets say 7 years (also means it got further than edge of solar system, but not as far as alpha centauri). however, as nobody has changed direction or speed, W can say that U traveled away very fast, and see's U's clock at 6 years time, V at 7, and W has the 8 year time lapse. heck, V could really do a number, and say U and W are moving, V see's 8 years, but U and W both saw 7. Ack! the mess!. Okay, lets get them all together again, in the same frame, and check out who's right. W stops at alpha centauri (accelerates) making itself stationary wrt U. Vstops on the way to alpha centauri (accelerates) making itself stationary wrt U. U does nothing. Now, by accelerating W and V have "broken reciprocity". You can now tell who moved, absolutely definitively tell who moved. U doesn't feel any acceleration, W and V do however. Now all the clocks read exactly as U said they would. If you arrange that any other way (U accelerates to be stationary wrt W for instance) the clocks all match the guy who didn't accelerate (the one they all became stationary wrt). Now, lets look at the final situation: W says it took 6 years to get to alpha centauri, according to it's clocks. U says W took 8 years to get there (at .5c). Who's right? Ans: Both! U will measure the distance to alpha centauri to be 4 light years. W will disagree however, and say they only traveled ~3 light years. 4 ly / 8 years = .5 ly/year 3 ly/ 6 yeras = .5 ly/ year Well...that w <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

bonepile · Sep 1, 2005

I intended for V and W to be traveling at the same velocity, only for different lengths of time. The experiment is internally consistent regardless of the velocities of V and W if you look at the whole thing from start to finish, so this doesn't affect your explanation. However, the confusion occurs when you look at the S/T experiment and recognize its parallel to what V and W are doing in the second part of the U, V, W experiment (and this is why I wanted V velocity = W velocity). I'll continue with your numbers. W takes 8 years to get to Alpha Centauri, as you said. Let's say that V, travelling at the same rate as W, only goes 3/4 of the way, so it travels (with W) for 6 years and then stops. We'll say that W experienced 6 years total, as you suggested. This means that V experiences (3/4)*6 + (1/4)*8 years, the first term occurring while moving and the second term occurring while stationary and waiting for W to reach Alpha Centauri. So V experiences 6.5 years. So far so good. When we compare the clocks at the end of the experiment, U = 8 /> V = 6.5 > W = 6. This is exactly what we expect. But now let's look at the parallel S/T experiment. S and T are stationary, and Alpha Centauri is coming towards them at relativistic speeds. It will take 6 years for Alpha Centauri to reach them at its current velocity. When Alpha Centauri has bridged 3/4 of the distance, S accelerates so as to keep its distance to Alpha Centauri constant. This occurs at (3/4)*6 = 4.5 years. 1.5 (T) years later, as Alpha Centauri reaches T, T also accelerates so as to keep up with Alpha Centauri. Now S, T, and Alpha Centauri are all moving along at the same speed. For T, the experiment takes 6 years. However, for S, the experiment takes (3/4)*6 + (1/4)*6*(6/8), or 5.625 years. Do you see the dilemma? The first term, (3/4)*6, is the same for both experiments, but the second term depends entirely on your initial fra

bonepile · Sep 1, 2005

I apologize for dragging this on, and I really appreciate the effort you guys are making to help me understand this. T is W, just with a different starting frame of reference. W takes 2.6 years to get to Alpha Centauri, so T must wait for 2.6 years for Alpha Centauri to reach it (using the new numbers). I like the notion of using 0.866c with a factor of 2 for time dilation, as using real numbers is less confusing. At any rate, if we actually managed to accelerate Alpha Centauri to 0.866c, then T would have to wait 5.2 years, but that isn't what happened. In the S/T experiment, Alpha Centauri seems to start out closer. We know it got this way because it was in fact S and T that were both accelerated to 0.866c, but S and T do not know this. The point of my thought experiment is that we should not need to know the history of S and T (or U, V, and W) in order to correctly predict what their clocks will read when the experiment is finished.

mrmux · Sep 1, 2005

BP, you aren't the only one gaining understanding from this thread. I'm enjoying it a lot. Kudos to you all, and drag on! (Dragon?)

Relativity and the "Twin Paradox"

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