Twin Paradox

**ElectroPhysics** · Dec7-04, 11:45 PM

As I'm involved in electronics industry due to my job requirements, I can't manage myself for Physics study. That is why I'm putting my questions here and hope you guys will try to answer them. Similarly, I will also try to answer some of your questions.

Suppose A and B are two persons at rest on the Earth surface. Then suddenly B starts its journey in a spaceship towards space in a straight line. When they both meet again we see that B is younger than A. How is it possible?

**pervect** · Dec8-04, 02:16 AM

Look through this board for the words "twin paradox". You can also do a websearch for the same phrase, but there's a lot of misinformation on relativity on the web. You can rely on the sci.physics.faq entries, though, and there is definitely a discussion of the twin paradox there. I'm feeling a bit lazy tonight, so I'm not going to type the URL in, because this topic has been discusses a bazillion times already, and it should be easy to find.

**yogi** · Dec8-04, 03:22 AM

You will find literally hundreds of articles on the subject of time dilation - and you will find that the explanations fall into several different catagories. Those that strictly adhere to the formalism of SR usually explain the age difference as being due to one of two causes 1) The dynamic theory states that one of the two observers experiences acceleration in turning around and returning, and therefore he changes inertial frames whereas the stay at home observer remains in the same frame so either SR doesn't stricly apply, or what amounts to the same thing - the problem is resolved via resort to GR 2) The kinetic theory dismisses the acceleration explanation and presupposes that the totality of the path integral going away and returning will always involve less time as accumulated on a clock carried by the traveler. This line of inquiry is adopted by some well known and respected Relativity authors such as Robert Resnic. If you venture further into alternative theories like Lorentz Ether, the age difference is explained as a physical consequence of motion wrt space. The point of all this is that there is no universally accepted answer.

selfAdjoint · Dec8-04, 11:11 AM

Yogi, your point one is a confused double of point two. The curved worldline which accumulates less proper time is necessarily the world line of an accelerated body; the slope of a world line relative to the time axis gives its speed, and if the slope varies (curvature) so does the speed (acceleration). Then the lesser proper time comes directly out of the Minkowski version of Pythagoras' theorem (with a minus where P. has a plus). This is all entirely within SR, you don't need GR at all. The idea that you need GR to handle acceleration is a myth.

And your third explanation is an ether one that is not accepted by 99 and 44 one hundredths of physicists, so really shouldn't be included.

**yogi** · Dec9-04, 01:57 AM

SA - I would agree that you don't need GR to deal with time dilation in the turn around - but you will find it in many analysis of the subject which was the subject of my post (not my belief). Nor do I endorse the kinematic explanation proferred by Resnic.

99 and 44 one hundredths of the physics must be pure - like ivory, they float... but they cannot explain the triplet version of the twin paradox. Until the moral majority can provide answers that are verifiable, it is important to keep an open mind.

**RandallB** · Dec9-04, 01:11 PM

Originally Posted by yogi

..... but they cannot explain the triplet version of the twin paradox.

. Yogi
What is the "triplet" version of the twins 'paradox'.?
Once someone understands and can explain the twins COMPLETLY. I don't see how a triplet could be a problem.

RB

**yogi** · Dec10-04, 11:05 PM

RandallB - the triplets version has the two siblings clocks in sync as they depart (we avoid the issue of initial acceleration by having the outbound traveler (2) already up to speed as he flys by the stay at home sibling (1) at which point they sync their clocks. The traveling sibling (2) proceeds at a high uniform velocity to some destination where he would normally turn around and head home and incur the turn around acceleration forces that are normally invoked to explain why there will be an age difference. Instead of decelerating, travler (2) simply transfers his clock reading to an inbound traveler (3) (doesn't have to be a sibling - rather any inbound traveler will do) who had left some time previously and is now returning home to reunite with (1). Traveler (3) is inbound at the same velocity as traveler (2) outbound velocity - from the standpoint of the path integral approach, the total time accumulated by the clock (2) and clock (3) will be less that the time accumulated by clock (1).

When inbound 3 reaches (1) back on earth ---If the total time logged by clocks (2) and (3) taken together is less than the time logged by (1) there is an intrinsic difference in the frames of the moving clocks. But this is contrary to postulate that neither (1) nor (2) nor (3) can measure velocity wrt absolute space, and therefore there is no reason to presume that clocks (2) and (3) run slower than clock (1). Since the transference of the reading from outbound (2) to the inbound traveler (3) does not involve acceleration - the paradox continues

**Garth** · Dec11-04, 03:21 AM

ElectroPhysics - The fact that the two observers age at different rates is not a paradox. What is a paradox is that as each observer is in an inertial frame of reference each twin thinks the other is 'younger'. When they meet again one must be actually be younger than the other, but which one? It is the one who has had to accelerate to turn round, and therefore not continue in an inertial frame of reference. The SR twin paradox is resolved by considering the two observers different surfaces of simultaneity.

However in the cosmological twin paradox in a closed universe two observers could meet and one circumnavigate the universe and they meet again much later. Now each one has been in an inertial frame of reference, each one thinks the other has circumnavigated the universe and is therefore the 'younger'. So when they meet again which one will actually be the younger?

This has been discussed here.

Garth

**jdstokes** · Dec11-04, 09:54 AM

Originally Posted by yogi

RandallB - the triplets version has the two siblings clocks in sync as they depart (we avoid the issue of initial acceleration by having the outbound traveler (2) already up to speed as he flys by the stay at home sibling (1) at which point they sync their clocks. The traveling sibling (2) proceeds at a high uniform velocity to some destination where he would normally turn around and head home and incur the turn around acceleration forces that are normally invoked to explain why there will be an age difference. Instead of decelerating, travler (2) simply transfers his clock reading to an inbound traveler (3) (doesn't have to be a sibling - rather any inbound traveler will do) who had left some time previously and is now returning home to reunite with (1). Traveler (3) is inbound at the same velocity as traveler (2) outbound velocity - from the standpoint of the path integral approach, the total time accumulated by the clock (2) and clock (3) will be less that the time accumulated by clock (1).

When inbound 3 reaches (1) back on earth ---If the total time logged by clocks (2) and (3) taken together is less than the time logged by (1) there is an intrinsic difference in the frames of the moving clocks. But this is contrary to postulate that neither (1) nor (2) nor (3) can measure velocity wrt absolute space, and therefore there is no reason to presume that clocks (2) and (3) run slower than clock (1). Since the transference of the reading from outbound (2) to the inbound traveler (3) does not involve acceleration - the paradox continues

The resolution is trivial. Proper time is defined as the time as measured by a single clock. It is obtained by integrating the proper time differential along an observer's worldline. The clocks in (2) and (3) belong to different inertial frames of reference. It makes no sense to add the proper times of two different clocks in the same integral.

selfAdjoint · Dec11-04, 11:12 AM

Exactly, it is the proper time of the clock, not the observer's, and when you separate the two, and accelerate the clock, by passing it between two frames with a speed difference between them, you invoke the Minkowskian rule, and the clock shows less. The triplets all in their own frames remain symmetric.

**yogi** · Dec11-04, 12:57 PM

If The total time logged by clock (2) going out is the same as the total time logged by clock (3) in returning to earth and the total time (2) + (3) is less than (1) then clocks (2) and (3) will be running at a different rate than clock (1). There is no acceleration at any point in the experiment -reading a passing clock cannot have a physical affect.

**jdstokes** · Dec12-04, 08:32 AM

Originally Posted by yogi

If The total time logged by clock (2) going out is the same as the total time logged by clock (3) in returning to earth and the total time (2) + (3) is less than (1) then clocks (2) and (3) will be running at a different rate than clock (1). There is no acceleration at any point in the experiment -reading a passing clock cannot have a physical affect.

The time difference is not caused by acceleration. The transfer of the clock reading from (2) to (3) changes the inertial frame of reference, instantaneously swinging the hyperplane of simultaneity through time so that the Earth time jumps many years into the future, accounting for the time difference when (3) passes by (1).

**yogi** · Dec13-04, 07:28 AM

This is the absurdity of trying to rationalize age difference using the "apparent times" observed by clocks in motion relative to the observer as "physically real times." Reading a local clock (2) cannot influence the reading on the clock, nor can it influence the proper time of the earth clock. The total proper time logged on clock (2) at the time it is read when added to the proper time logged by clock (3) after being set to the clock (2) reading at the meeting point of (2) and (3) will determine the age difference when compared to the proper time accumulated by (1) when (3) returns.

Changing the inertial frame by reading clock (2) and setting clock (3) in reference thereto is no different that the pass-by sync of clock (2) with (1) at the outset - where is the physics that requires the hands on clock (1) to instantly spin forward at the time of the transfer. Apparent times cannot be magically invoked to bring about physical changes.

**gonzo** · Dec13-04, 08:42 AM

I'm curious about something in general here. I've seen a lot of references to thought experiments involving the reading or syncing of two clocks in different reference frames as they "pass by" each other in an attempt to avoid dealing with acceleration.

I really must be missing something here, because it seems to me that this is invoking some magic faster than light method of conveying information. If a ship passes an oberserver at high eough speed that we are dealing with noticeable relativistic effects, aren't there problems with defining how you read each other's clock when they pass, and what that even means?

How do you even theoretically attempt to sync two clocks rapidly moving with respect to each other?

**Garth** · Dec13-04, 11:18 AM

The observers are passing close by each other, they can measure their relative velocity, predict the point of closest approach, pass a light signal and make any necessary adjustments as they read their respective clocks. In due time they can send each other the time recorded on their own clock for comparison.

Garth

**RandallB** · Dec13-04, 05:32 PM

ElectroPhysics
When they both meet again we see that B is younger than A. How is it possible? Well first you will need to define when and where B is going to turn around. Because that's the only way they will meet again and unless they meet nobody gets older than anybody - at least not that they can tell.

Originally Posted by jdstokes

The time difference is not caused by acceleration.

jd is letting you know that the Triplets thing and using Sync data Xfer infromation to create 'new clocks' in other referance frames all to avoid "Acceleration" is not needed at all.
Acceleration has nothing to do with what your looking at - if needed just assume you make your speed changes in a very short time like one minute - or even less.
To account for acceleration when when make speed changes you may just assume time clock going with B simply stops for the duration of the acceleration. That missing one minute ( or nano second) makes no differace to the rest of your problem.

The important part is that during the transfer of the Twin from frame (2) to the returning refrance frame (3) -- or as jd puts it:

Originally Posted by jdstokes

The transfer of the clock reading from (2) to (3) changes the inertial frame of reference, instantaneously swinging the hyperplane of simultaneity through time so that the Earth time jumps many years into the future, accounting for the time difference when (3) passes by (1).

.... that you know when and where you are.

Now as jd put it - it's a little hard to figure -- here is the EASY way to understand it.
Figure the returning speed change in two steps. from Ref frame (2) to Ref Frame (1). Stay for just a minute (Or just a nano second if you like).. Then go from Frame (1) to Frame (3) for the return trip to Earth.

Now the key thing is to take the time to figure out just exactly WHERE and WHEN you are while in sync with Earth in frame (1) as your making this turn in direction. Remember only the clocks in Frame one are the same as back on Earth the clock with the twin, having been 'slow' will be behind.
Now that you know for sure what time it already is back on Earth, and how far away it is in Earth's frame, then start your trip back.

You won't understanding the "swinging the hyperplane of simultaneity" (not that you need to, & I'm not sure there is such a thing a "Hypeplane") untill you understand the numbers you get this way, And you should be starting to understand simultaneity whitch is jd's point.

Randall B