In the twin paradox, the twin who has gone into space and back at relativistic speeds supposedly comes back younger than the one who stayed on earth, the reason being that from the earth twin's reference frame, time has gone more slowly for space twin. However, what has puzzled me is that from space twin's reference frame, time has gone slower for earth twin, so presumably when space twin returns, it will be space twin who is the younger. Who is younger upon space twin's return? Could somebody explain why there is this paradox - in fact, is this the twin paradox, in which case reply with the answer 'yes' and ignore this thread thereafter. But if not, what is wrong with my reasoning?

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I have seen it argued that the event that breaks the paradox is event T; the turnaround of twin B who's on a relativistic journey. Twin A who remained home continues happily in his own reference frame on earth (call it frame S), but twin B who began his journey in reference frame S' switches reference frames at event T to reference frame S''. So twin A travels from event D (the departure) to event R (the return) in a single reference frame, while twin B changes reference frames.

The longest straight worldline is the method to choose the preferred frame of reference. Worldline length depends on staying in ONE frame of reference, too, so to choose the preferred frame of reference you choose the frame with the longest continuous worldline. In the twin paradox twin A's worldline is longest.

Hurkyl
Staff Emeritus
Gold Member
SR can handle acceleration and accelerated frames of reference perfectly well.

Janus
Staff Emeritus
Gold Member
Okay, let's take this step by step.

You have two twins, Stacy and Travis.

Stacy stays home and Travis gets in the ship.

The following is what each experiences:

Stacy:
He sees Travis accelerate off to some high fraction of c, say ,866c, travel for some distance, say 10ly, turn around and come back. During this time, he will expect to see Travis' time as running slow due to time dilation and expects Travis' to be younger when he returns. In fact, with these numbers he would expect to have aged 23 yrs, while Travis only ages 11.5 yrs.

Travis:
He accelerates to .866c travels for a distance and returns. The trick here is that due to length contraction, Travis only measures that he has traveled 5ly away from Stacy rather than 10ly, It takes 11.5 years to travel 5ly back and forth at .866c, so he will expect to be 11.5 yrs older when he returns; the exact same difference in age that Stacy expects.

Both Twins agree as to how old Travis is when he returns, so no paradox here.

But what does Travis see as happening to Stacy's clock?

Well, during the time he is coasting, he see's Stacy as aging slower as SR predicts. It is during the acceleration phases of the trip to which we must look.

Relativity has rules as to what an accelerating object will measure with respect to clocks.

If the clock is in the direction in which you are accelerating, you will see it speed up. (even if it is accelerating at the same rate as you are)

If the Clock is in the opposite direction of your acceleration, you will see it slow down. (even if it is accelerating at the same rate as you are)

The rate at which you see the clock speed up or slow down depends on the degree of your acceleration and the distance separating you along the direction of acceleration.

Thus Travis see this:

As he accelerates away from Stacy, he sees Stacy's clock run slow (Due both to velocity difference and the Acceleration Travis is undergoing) But the effect due to acceleration will be small because at this point of the trip, the distance between Stacy and Travis is still fairly small.

As he coasts away from Stacy he see's Stacy's clock running slow. (relative velocity alone.)

As he nears 5 ly distance he begins to slow down, and now see Stacy's clock run fast. (By a great deal since they are separated by 5 ly)

After stopping, he continues to aplly power in order to accelerate back up to .866c for the return trip. He sees Stacy's clock continue to run fast.

As he coasts back towards Stacy, he see's Stacy's clock run slow.

While braking to a stop he sees Stacy's clock run slow. (But again since they are once again close together, this effect will be small.)

Once Travis adds up all the elasped time on Stacy's clock, including all the periods of running slow and running fast, He will discover that it will be equal to 23 yrs, the exact time that Stacy measured on his own clock.

Thus both Travis and Stacy will both agree on which is older and by how much, they just won't agree as how this came to be.

EL
Mentors, you are of course correct...

David
Originally posted by speeding electron
In the twin paradox, the twin who has gone into space and back
The original “clock paradox” of the 1905 SR theory had nothing to do with “twins” or traveling into space. The “paradox” was caused by Einstein making both of two clocks “synchronous” (running at the same rate) before he began the “relative motion” between the two clocks, so no one could tell which clock “should” be the one that slows down. There was no “earth” and no “travel into space” and no “blast off” in the 1905 paper, and acceleration effects were not considered, and there were no “turn-arounds” in his original paradox thought experiment in Section 4 of the paper.

You can read his own “resolution” of the paradox in his 1918 paper on the subject.

Thank you all, especially Janus, for the clear explanation. I follow your reasoning. Would there be the same sort of cancelling out of time differences if somone were to say travel at close to c away from somebody else and then stop, without coming back, and to look back to his origin? Is this acceleration effect more SR or GR? Yes, Pound & Rebka and all that, the same effect as clocks running slower higher up in G fields, the same thing because of the principle of equivalence, right? Thanks again for your time.

Originally posted by speeding electron
Would there be the same sort of cancelling out of time differences if somone were to say travel at close to c away from somebody else and then stop, without coming back, and to look back to his origin?
When you look back, do you see what happens right now or in the past?

It's similar to the following paradox:
Two spaceships are loaded with bombs that are both set to explode in 1 year. The timers are set off at the same moment, and the two ships start travelling away from each other at relativistic speeds. Which bomb will explode first?

The answer to that paradox relies on the fact that it takes time from the moment the event occured until light from it gets to you.

Janus gave the conventional rationale to the twin exercise - but it fails in the context of the triplets scenereo - here there is no turn around and therefore no acceleration or deceleration - the outbound sibling simply transfers his reading to an inbound triplet at the so called turn around pylon - but the path integral for the outbound triplet plus the inbound triplet leads to the same time discrepency in relation to the stay at home triplet as though there was a turn around. If however, time dilation is regarded as actual, the twin paradox dissolves - take for instance the high speed muon that has a laboratory lifetime of 2 usec. A clock attached and traveling with this muon will always record a time interval of 2 usec no matter how fast the muon travels - even if travels at nearly the speed of light so it can reach a star 5 light years from earth before it decays. In other words, the proper time for the muon from creation to death is 2usec - now we in this exercise, we have specified the proper length for the journey of this muon to the distance to the star which is a fixed proper distance of 5 light years in the earth star frame, the reading of a clock in our earth star system will be 5 years when the muon reaches the star - but the muon's clock (which reads proper time in the traveling frame of the muon) will read only 2 usec. These are actual times as measured by the muon clock (proper time measured in the frame of the traveling muon) and 5 years as measured in the earth star frame for the time of the trip. The muon is analogous to the traveling twin - it has a different age upon arrival at the star as measured by its own clock which measures its lifetime as 2usec. There is no turn around acceleration - all you consider is that the spacetime interval is the same in both reference frames. The root of the problem lies in the fact that we are specifing a fixed proper time in one frame and a fixed proper distance in another frame.

Originally posted by Hurkyl
SR can handle acceleration and accelerated frames of reference perfectly well.
SR can handle acceleration, but not an accelerated reference frame. SR requires (unaccelerated) inertial reference frames.

speeding electron wrote

However, what has puzzled me is that from space twin's reference frame, time has gone slower for earth twin, so presumably when space twin returns, it will be space twin who is the younger. Who is younger upon space twin's return?
The traveling twin does not always reckon the stay-at-home twin to a age more slowly. During the turn around point the stay-at-home twin's clock runs fast by the amount gd/c2 where g is the acceleration that the traveling twin undergoes. This speeding up of the stay-at-home twin's clock as rekonded by the traveling twin is the key to the differences in times. You can think of this in terms of the equivalence principle: During the acceleration of the traveling twin towards the earth the Earth can be thought of as being high up in a gravitational field. An observer low in a gravitational field will observe identical clocks higher up in his gravitational field to be running faster.

Cosmological Physics, John A. Peacock pages 7-8 at
http://assets.cambridge.org/0521422701/sample/0521422701WS.pdf

Originally posted by Hurkyl
SR can handle acceleration and accelerated frames of reference perfectly well.
SR is defined according to the postulates of SR which are

(1) Principle of Relativity - The laws of physics are the same in all inertial frames of reference

(2) Light Postulate - The speed of light in an inertial frame of reference is independent of the source

It is as follows

Principle of covariance - all observers, inertial or not, observer the same laws of physics.

That means that the laws can be expressed in terms of tensors, since tensors are geometric objects defined independant of any coordinate system. This principle belongs, not to SR, but to GR. And an accelerating frame of reference is considered a different coordinate system. To transform to it one must use the assumption/postulate that the laws of physics are valid under such a transfromation. This was why Einstein specifically defined "special relativity" to be relativity in inertial frames of reference.

Hurkyl
Staff Emeritus
Gold Member
Principle of covariance - all observers, inertial or not, observer the same laws of physics.
This is true in GR. However, I'm not talking about treating accelerated frames of reference in GR! In SR, when you transform into an accelerated reference frame, the laws of physics also change. The principle is similar to generalized coordinates in classical mechanics.

turin
Homework Helper
Chen said:
Two spaceships are loaded with bombs ...
...
The timers are set off at the same moment, ...
...
Which bomb will explode first?

The answer to that paradox relies on the fact that it takes time from the moment the event occured until light from it gets to you.
No. The answer to the paradox is that it remains an ostensible paradox by virtue of the phraseology. Simultaneity is not absolute in relativity; causality is. This issue is one of the most prominent sources of confusion for beginners/dabblers in the subject.

To apply the idea to this "paradox":
Each ship observes its own bomb to explode first causally.
The bombs explode simultaneously in the frame that observes both ships moving at the same speed.
The bomb explodes first temporally in frame that observes the ship containing that bomb to move more slowly.