# Symmetry in special relativity

## Main Question or Discussion Point

I have read Einstein's 1905 paper and a number of other explanations and have a question I can't resolve.

The basic problem was that the mathematics differed between the case of a conductor moving relative to a magnet vs. a magnet moving relative to a conductor. Einstein used the Lorentz transform to transform the experiment into a frame where the electrons are not moving, then transforms back to get the result. In this way he can dispense with the concept of magnetism. So far so good.

He also mentions this. You have one clock mounted on the earth and another on a moving railway car. According to the Lorentz transform the clock on the moving car moves more slowly as the car moves away. If the car turns around and returns at the same velocity, then the clock on the moving car also moves more slowly, so when the two clocks meet again at the same place then the railway clock will be shown to be slower.

All right, but the situation seems symmetrical to me. If we are seated in the railway car then the same argument applies, and the stationary clock should be the slower one. So each clock is slower than the other, which can't be. I've also been exposed to that argument about the parallel mirrors used as a clock, but that has the same symmetry so the same difficulty is there. If I am on the railway car looking at mirrors mounted to solid ground then as far as I'm concerned the light travels farther between those mirrors and the earthbound clock will be slower than my clock on the car.

By example it is implied that the clock in the frame that is always inertial will be the faster, but I don't see why this should be short of appealing to experiments.

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Matterwave
Gold Member

The symmetry is broken because one of the clocks (the one on the train) is not in an inertial reference frame the whole time. You can't "stitch together" 2 different inertial frames (one leaving and one coming back) and hope to get sensible results.

The symmetry is broken because one of the clocks (the one on the train) is not in an inertial reference frame the whole time. You can't "stitch together" 2 different inertial frames (one leaving and one coming back) and hope to get sensible results.
Yes, I read one of those long threads and it did not help me.

I can follow the procedure and get the answer that agrees with experiment, I see that the situation is asymmetric in a certain way, but I don't see where he uses the symmetry breaking in his derivation. If he doesn't use it, then the symmetry argument applies.

That is my question. Where does the symmetry breaking enter into his derivation?

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by $\frac{1}{2}tv^2/c^2$(up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.

It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide.

If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be $\frac{1}{2}tv^2/c^2$ second slow. Thence we conclude that a balance-clock7 at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions.

--Albert Einstein, on the Electrodynamics of Moving Bodies

Matterwave
Gold Member
He broke the symmetry by choosing to analyze the situation in the frame of B rather than A because A's frame is non-inertial.

ghwellsjr
Gold Member
The first paragraph which you quoted from Einstein's 1905 paper is not describing the Twin Paradox. There he is talking about two previously synchronized clocks that are separated by a distance, one at A and the other at B, and the one at A is moved to location B. He shows that it will no longer be synchronized and gives the equation to show the difference in time based on the total time for the trip and the speed.

The last paragraph is where he introduces the Twin Paradox where both clocks start out at A and one of them takes a circular path and returns to A. But he only analyzes the situation from the inertial Frame of Reference in which A is stationary. While the one clock is taking the round-trip, it is never inertial (its speed is constant but its direction is changing so it is always accelerating) and so there is no inertial Frame of Reference in which the traveling clock has a speed of zero for more than an instant in time.

So in Einstein's scenario, there is no symmetry and therefore no argument based on breaking symmetry.

But usually, when the Twin Paradox is discussed now-a-days, it is like your first post where the trip for the traveling clock is divided into two parts, one where it is traveling away from the stationary clock and one where it is traveling back toward the stationary clock. So in either half of the trip, there is a symmetry between the two clocks so that they each see the other as being time dilated by the same amount but if you use either of the traveling clock's Frame of Reference for the entire trip, you would see that during the other half of the trip, the traveling clock was time dilated even more than the stationary clock and that is why it ends up with less time on it when they reunite.

I found my answer in the writing of Daniel F. Styer of Oberlin University. If looked at from the point of view of General Relativity then it is all explained. The answer is that all of the time dilation that is observed when the two clocks meet again is due to the acceleration they experience.

http://muj.optol.cz/richterek/data/media/ref_str/styer2007.pdf

How do two moving clocks fall out of sync?

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Don’t take my post too serious, just a reflection I’ve had for some time.

It’s known that the travelling and therefore accelerating twin comes back younger, but why doesn’t s/he not come back shorter? Apparently it’s possible to catch up in length but not in time.
Mind you, has a travelling clock ever actually been checked for loss of size on return? (or checked for increase in weight?)

In the reference frame of A, B is moving at a certain speed, its clock is moving slower, then it turns around and is still moving at a certain speed. So in the reference frame of A, B is constantly moving very fast and therefore its clock will be slower.

Now let's start with the reference frame of B. He sees A moving away very fast and the clock on A moving slower. Then he reaches the point where he turns around. Keep in mind, you cannot change reference frames and mash the results together. The reference frame keeps on moving forward. Now from that reference frame, A is still moving away and its clock is moving slower, but B is flying towards A and in this reference frame it is moving even faster than A, therefore it is experiencing more time dilation than A. In the end, in either reference frame the result should be the same.

You just have to keep in mind that you cannot change frames half way. You cannot take a speeding bullet, observe from the earth frame that the earth is stationary, then observe from the bullet's frame and determine the bullet is stationary and then mash the results together to conclude that both the bullet and earth are stationary compared to each other.

Nabeshin
Don’t take my post too serious, just a reflection I’ve had for some time.

It’s known that the travelling and therefore accelerating twin comes back younger, but why doesn’t s/he not come back shorter? Apparently it’s possible to catch up in length but not in time.
Mind you, has a travelling clock ever actually been checked for loss of size on return? (or checked for increase in weight?)
Because time is a cumulative effect. When the two clocks are running at different rates, they rack up a time difference proportional to how long they are out of sync for. Thus, once the two twins return and are in the same inertial reference frame, there is still a lack of synchronization due to the fact that the clock ran slower in the past.

Now look at length contraction. Clearly there is no memory here. No height difference is accumulated at any point along the trajectory, so when the twins return and are again in the same reference frame, by definition they have the same heights.

PAllen
2019 Award
Because time is a cumulative effect. When the two clocks are running at different rates, they rack up a time difference proportional to how long they are out of sync for. Thus, once the two twins return and are in the same inertial reference frame, there is still a lack of synchronization due to the fact that the clock ran slower in the past.

Now look at length contraction. Clearly there is no memory here. No height difference is accumulated at any point along the trajectory, so when the twins return and are again in the same reference frame, by definition they have the same heights.
To amplify this, ghwellsjr has noted that if you define the concept of an odometer to accumulate length, then you would have a length version of the twin paradox. Of course, while clocks are easy to build and conceptualize, odometers that work in space are a bit problematic. But the conceptual analogy is nice:

odometer <---> clock
ruler <---> metronome

The latter have no memory, the former do.