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Alright, so David Griffiths in his "Introduction to Electrodynamics" states that the Twin Paradox is not a paradox at all since the traveling twin returns to Earth. By returning to Earth, the twin had to reverse direction, thus undergoes acceleration, and therefore cannot claim to be a stationary observer.

However, what if the traveling twin simply Skypes the twin that is on Earth. The twin on earth will still appear older, which would make no sense since in that case the rocket can be seen as the stationary frame of reference while the Earth "travels" at a speed close to the speed of light. No acceleration is undergone, yet the paradox remains.

Is Griffiths just completely glossing over important nuance again?

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No. Griffiths is not glossing over this. Skype or any signal still travels (at best) at the speed of light. –  suresh May 11 '14 at 0:01
    
So the Skype signal is the deceleration needed for the paradoxical nature of the effect to disappear? –  Tyrion Lannister May 11 '14 at 0:05
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There has to be some acceleration, after all the twins were together when born, and now one of them is travelling at near the speed of light. How can there not be an acceleration for the travelling twin? –  LDC3 May 11 '14 at 0:19
    
You might want to read this –  PatronBernard Jan 12 at 15:29

5 Answers 5

The twin on earth will still appear older

No, that is not correct. If the twin on the rocket never reverses course and remains inertial, the twins never meet to compare ages at the some location.

Since the twins remain spatially separated, their ages must be compared by spatially separated clocks.

For example, when the twins are separated by 1 light-year, the observation of the age of the twin on the rocket, as observed by the twin on Earth must be made with a clock, synchronized with the clock on Earth but co-located with the rocket, i.e., located 1 light-year from Earth.

Similarly, the observation of the age of the twin on Earth, as observed by the twin on the rocket, must be made with a clock, synchronized with the clock on the rocket but co-located with Earth, i.e., located 1 light-year from the rocket.

But, as is well known, clocks synchronized in the Earth's frame of reference are not synchronized in the rocket's frame of reference and vice versa.

Thus, due to this relativity of simultaneity (synchronization), each twin observes the other to have aged less without contradiction.

See this answer for a useful diagram.

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So, when the twins meet-up again, they are the same age? Is this an observational effect or something to do with the travelers innate natures changes? –  Geremia May 14 '14 at 21:03
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@Geremia, the twins can only meet-up again if the twin on the rocket reverses course which will cause the twin on the rocket to age less. If the twin on the rocket doesn't reverse course, the twins never meet-up again. –  Alfred Centauri May 14 '14 at 21:13
    
@AlfredCentauri People have poked around the problem in the context of a closed universe small enough that a geodesic round trip can be made. I can't recall what conclusions were drawn. I assume the GR must be invoked. –  dmckee Jan 18 at 0:50

Another nuance that is sometimes skipped over is the doppler shift: that is, the number of wave-crests of light emitted by one twin and seen by the other is different. Imagine that each twin has an atomic clock that is counting the number of wave-crests emitted by an atomic clock held by the other twin.

When the travelling twin starts the journey, both twins see eachother's atomic clocks radiation red-shifted and so they both see the other moving more slowly though time.

When the travelling twin turns around some vast distance later, the frequency of the Earth bound atomic clock is blue-shifted immediately and Earth time is seen to speed up. The Earth twin has to wait some time for the travelling space clock becomes blue-shifted and so for this short period of time, the travelling twin sees Earth time running faster than normal and the Earth twin sees spaceship time running slower than normal.

For the remaining part of the journey, both twins see blue shifted light from the other and so see the other moving faster through time, but really they are just catching up on light already emitted.

When the travelling twin finally arrives back, the middle sequence is never made up: the Earth bound atomic clock has emitted more wavecrests than it has counted leaving the spaceship clock (and visa versa). This is why the Earth twin looks older and has gained more life experience.

Satellites are already in our future: gravity red-shifts our light as it reaches the satellites and blue-shifts satellite light when it falls back to Earth. Blue shifted light means more CPU clock cycles and so satellites are travelling through time at a faster rate than we on the Earth's surface are.

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But, the number of wavecrests measured to be emitted from each side is the same as received at the other side. This is just a fancy clock. –  Blackbody Blacklight May 13 '14 at 1:20

Tyrion Lannister,

All answers to the problem claiming that there is no such thing as the twins paradox I saw were based on acceleration. Acceleration is said to be actually responsible for the difference in time-flow between the two twins.

However, before heralding all over the world to have found the solution to the twins paradox, one should simply take a look at the equation that lead to the conclusion that the paradox must exist:

$$ \Delta t' = \frac {\Delta t} {\sqrt{1 - \frac{v^2}{c^2}}} $$

Immediately a question arises: Where is the acceleration in this equation? If acceleration is the culprit, it should appear somewhere in the equation that shows time dilatation. And yet, all we can see is just constant $v$. So how come acceleration is responsible for time dilatation, if there is no acceleration at all ...?

We can also approach the problem from another angle. Claiming that acceleration solves the problem of twin paradox is equivalent to claiming that all time dilatation in Special Relativity is due to acceleration, isn't it? If acceleration solves the paradox, than there must be no other source of time dilatation in this theory than acceleration, right?

But this would mean that Einstein simply didn't notice the fact that all the unusual phenomena in his own theory are all due to acceleration. This would mean that Einstein must have been plain wrong saying that the Special Relativity pertains to inertial frames only. Claiming that SR (time dilatation) simply boils down to acceleration must mean that SR is actually equivalent to General Relativity - that, in fact, there is no SR at all.

So, the dismissal of the twins paradox based on acceleration means simply the dismissal of the very Special Relativity Theory. But for this to be a legitimate claim, someone needs to prove it. Is anyone able to provide such a proof? I haven't seen any so far ...

On the other hand, one obviously cannot take off Earth and travel in space without accelerating. No doubt about it. But then, it is not acceleration that lead somebody to ponder what happens to twins after one of them went on a space travel. Therefore, any acceleration contributing to time dilatation in the twins paradox is only a part of the problem. It must be so, unless we see the proof that SR = GR.

EDIT: Any comment as to what in my answer is incorrect? I just love to learn more and more about SR.

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The point you make is legitimate. See Herbert Dingle's Science at the Crossroads ch. 9. –  Geremia May 14 '14 at 21:00
    
I'm aware of Dingle's battle and his criticism of contemporary physics, but I never read the book. As to the clock paradox. The key thing is that any experimental proof of SR time dilatation in favor of any inertial reference frame would at the same time disprove the axiom that there is no preferred inertial frame of reference. That's it; end of story. –  bright magus May 15 '14 at 13:05
    
Dingle ultimately concluded that something must be wrong with the principles of SR. He admitted that given those principles, the twin paradox is not contradictory per se. –  Geremia May 15 '14 at 17:01
    
Something must be wrong? How about this: time dilatation and length (distance) contraction means that the two variables change in inverse proportions, i.e. if $t$ is larger than $x$ is smaller, right? Now, the constancy of light expressed by $c=x/t$ and $c=x'/t'$, and therefore $x/t=x'/t'$ requires that they should change in direct proportions, i.e. if $t$ is larger than $x$ should also be larger. Definitely, something is wrong. –  bright magus May 15 '14 at 17:18
    
Read Science at the Crossroads ch. 9 and let me know what you think. –  Geremia May 15 '14 at 19:13

No absolute relativity as long as a constant reference exists this reference is speed of light (this reference will be responsible for defining the moving body

Earth twin will see that other twin's clock is slow because it is really slow But space twin will not see that other twin's clock is slow because his time Perception is also slow

Space twin will see earth moves far from him but by shorter distance than the distance seen by earth twin, so if we supposed that space twin is moving far from earth with the exact speed of light he will never see that earth is moving because he is moving with the light that is moving from earth so he will not receive image updates

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Space twin will see Earth twin's clock as running slow. That's the basis of the twin paradox –  Jims Bond Jan 12 at 14:53
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Relativity isn't as simple as you make it out. And I'm not quite sure what you meant by 'time Perception' - it seems to be more philosophy than physics. –  Jon Custer Jan 12 at 15:02

If they skype each other, then because the skype signals travel at the speed of light, as long as they are moving apart they will each symmetrically see the other one as slowed down by the amount predicted by the relativistic Doppler effect equation. For example, suppose they are moving apart inertially, with a relative velocity of 0.6c. Then according to the equation, visually each will see the frequency of the other's clock slowed down by a factor of:

$\sqrt{(1 - 0.6)/(1 + 0.6)} = 0.5$

(Note that this is different from the time dilation factor, which is designed to factor out the effects of signal delays)

You can see how this works if you actually imagine following the path of successive signals sent by each one. Let's analyze things in the frame of the Earth twin, after the traveling twin has started moving away from the Earth. Let's say each starts their stopwatch at the moment of departure, so at that moment the Earth twin's watch reads $T_e = 0$ seconds and the traveling twin's watch reads $T_t = 0$ s. Then when each one's clock reads 10 seconds, they send a light signal showing their clock reading 10 second to the other, and when each one's clock reads 20 seconds, they send a light signal showing their clock reading 20 seconds to the other.

First we can figure out when the signals from the Earth twin will reach the traveling twin. This signal is sent at coordinate time t = 10 s in the inertial rest frame of the Earth, and if Earth is defined to be at position x = 0 light-seconds in this frame and the traveling twin is moving in the +x direction, then at this moment the traveling twin is at x = 6 ls in this frame. Then at 15 seconds later at t = 25 s in this frame, the light ray will be at x = 15 ls, and the traveling twin moving at 0.6c will be at x = 6 + (0.6 * 15) = 15 ls as well, so t = 25 s will be when the first signal catches up to the traveling twin in this frame. But in this frame the traveling twin's clock is running slow by a factor of $\sqrt{1 - 0.6^2} = 0.8$ due to time dilation, so the traveling twin's clock reads $T_t = 25 * 0.8 = 20$ seconds when the first signal reaches him.

The Earth twin sends the second signal at t = 20 s in this frame, when the traveling twin is at position x = 12 ls. 30 seconds later at t = 50 s, the second signal from the Earth twin will have reached x = 30 ls, and the traveling twin will have reached x = 12 * (0.6 * 30) = 30 ls as well, so this is when the second signal catches up to the traveling twin in this frame. Since the traveling twin's clock is running slow by a factor of 0.8 in this frame, the traveling twin's clock reads $T_t = 50 * 0.8 = 40$ seconds when the second signal reaches him. So the traveling twin sees signals sent 10 seconds apart by the Earth twin's clock reach him 20 seconds apart according to his own clock, meaning he that visually he sees the Earth twin's clock running slow by a factor of 0.5, just as predicted by the Doppler formula.

Now we can also analyze when the signals sent by the traveling twin will reach the Earth twin, again using the Earth rest frame. The traveling twin sends signals when his clock reads $T_t = 10$ and $T_t = 20$ seconds, but because his clock is running slow by a factor of 0.8 in this frame, the coordinate times of his sending the signals are t = 10/0.8 = 12.5 s and t = 20/0.8 = 25 s. Since he is traveling at 0.6c, at t = 12.5 s he is at position x = 0.6 * 12.5 = 7.5 ls when he sends the first signal, so the signal traveling at the speed of light will take 7.5 s to reach the Earth twin, reaching him at t = 12.5 + 7.5 = 20 s in this frame, when the Earth twin's clock reads $T_e = 20$ seconds. And at t = 25 s when the traveling twin sends the second signal, he is at position x = 0.6 * 25 = 15 ls when sending the second signal, so the signal takes another 15 s to reach the Earth twin, arriving at t = 25 + 15 = 40 s, when the Earth twin's clock reads $T_e = 40$ seconds. So you can see that everything is totally symmetrical--each one sends signals at 10 seconds and 20 signals on their clocks, and receives the other one's signals at 20 seconds and 40 seconds on their clock, meaning each sees the other one slowed down visually by a factor of 0.5. Only if one of them accelerates is this symmetry broken, just as Griffiths suggests.

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