Time dilation and Einstein's theorem

In summary, Einstein's theorem states that if two clocks are moving towards each other in a circular orbit, the clock at the point of the orbit where they are moving towards each other will be slower than the clock at the point of the orbit where they are moving away from each other.
  • #1
worlov
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Hello!

Einstein's theorem is in the last sentence of the following quote (bold) [1]:

"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 1/2 tv2/c2 (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."

This theorem is better known as the twin paradox. Einstein did not stop at a preliminary remark and wrote
in more detail:

"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 traveled clock on its arrival at A will be 1/2tv2/c2 second slow. Thence we conclude that a balance-clock 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."

Each observer determines that all clocks in motion relative to that observer run slower than that observer’s own clock. If the moving observer returns to point A, he will notice the following: the clock at point A is slower than his own clock. but this clock shows that more time has passed since his departure than his own clock. For moving observers this conjucture makes no sense. If a clock is slower than its own clock, it should also show less time. Therefore, the clock in point A should be synchronized according to the rules of relativity ... but how?[1] ON THE ELECTRODYNAMICS OF MOVING BODIES By A. EINSTEIN June 30, 1905.
 
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  • #2
worlov said:
For moving observers this conjucture makes no sense. If a clock is slower than its own clock, it should also show less time. Therefore, the clock in point A should be synchronized according to the rules of relativity ... but how?
Time dilation and length contraction are well known. A third phenomenon, much more important but less cool-sounding, is the relativity of simultaneity. This resolves the problem. When the moving clock turns round it changes synchronisation convention, and the change in synchronisation exactly accounts for the "extra" time that the stationary clock shows compared to a naive time-dilation analysis done by the moving clock.

The simplest way to see this is to use the Lorentz transforms to work out what time the outbound frame says clock A will show at turnaround and what time the inbound frame will say it shows. The difference is the "extra" time you need.
 
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  • #3
So, the moving observer always watches the clock at point A. If he moves away from point A, in his view the clock at point A goes slower. If he moves to point A, the clock at point A goes faster. If he reaches point A, the clock will go renewed slower than his own clock.

Here I miss the actual function of a clock - even measurement of time.
 
  • #4
worlov said:
Each observer determines that all clocks in motion relative to that observer run slower than that observer’s own clock. If the moving observer returns to point A, he will notice the following: the clock at point A is slower than his own clock. but this clock shows that more time has passed since his departure than his own clock. For moving observers this conjucture makes no sense. If a clock is slower than its own clock, it should also show less time. Therefore, the clock in point A should be synchronized according to the rules of relativity ... but how?

In the case of a circular orbit, the "moving" clock is not moving inertially but continuously accelerating. This means that this clock is not an inertial observer and there is a critical and fundamental difference between that clock and a clock that is moving inertially.

Certainly, if the laws of SR could be applied directly and equally to both clocks, then a true contradiction must result. But, the moving observer (if he is accelerating in circular motion) must apply SR taking this constant acceleration into account. When he does that he agrees that his own clock runs continuously slower than clock A. There is an analysis of that scenario here:

https://www.physicsforums.com/threa...ther-in-a-circular-orbit.896607/#post-5660815

PS In the opening of part I of the Einstein paper, he says:

"Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good."

This excludes the case where an observer is moving in a circle, where the system of coordinates do not meet this criterion.
 
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  • #5
worlov said:
So, the moving observer always watches the clock at point A. If he moves away from point A, in his view the clock at point A goes slower. If he moves to point A, the clock at point A goes faster. If he reaches point A, the clock will go renewed slower than his own clock.
That's an accurate description of what the observer will see. But the speed of light is finite, so what she sees is not what the clock is currently showing. She needs to correct for the travel time of light in order to work out what the clock is showing now. Once she does that, she will find that the clock always runs slow - but that the time it is showing "now" changes at turnaround. The amount of that change accounts for the "extra" time the clock at A shows compared to a naive time-dilation-only analysis.
 
  • #6
PeroK said:
When he does that he agrees that his own clock runs continuously slower than clock A.

Exactly! One of the twins has to admit that his system is so to say subordinate. Then, why should not he believe his eyes, if he sees that the clock in point A is slower than his own clock?
 
  • #7
worlov said:
Exactly! One of the twins has to admit that his system is so to say subordinate. Then, why should not he believe his eyes, if he sees that the clock in point A is slower than his own clock?

He doesn't measure clock A as running slower. He measures clock A as running faster than his own.
 
  • #8
worlov said:
One of the twins has to admit that his system is so to say subordinate.
No - one of the twins has a non-inertial coordinate system. It's not "subordinate", whatever you might mean by that. It's different, so naive application of time dilation formulae (which only apply between inertial coordinate systems) won't work.
 
  • #9
This "paradox" is not paradoxical at all - the predicted results are consistent from all viewpoints. But what it is, is confusing.

Some of you ask, why should an observer not believe what he sees with his own eyes? Because what he sees is carried to him by photons, traveling at "c", and his own velocity is of a similar magnitude. There's a Doppler Effect.
For example, suppose that you are traveling towards a clock which, objectively, according to Lorentz rules, is running slower than yours. You may see it running faster because you are racing into its stream of photons.

Let's make the Doppler effect vanishingly small. Place a "stationary" clock on a "ground" surface and place another "moving" clock one light-year above it. We will move this clock horizontally for only an hour. It will remain almost exactly above the "stationary" clock, not moving toward or away from it by any significant amount.

At a certain time, the "moving" clock sees the "stationary" clock displaying Noon. (of course, this display happened a year before). At this moment we give the "moving" clock an almighty whack with a cricket bat, accelerating it to relativistic speed. After the hour has passed, a friend whacks it back toward us, and an hour later we stop it where it originally was, with a third whack. Thanks to the cricket bats, the clock has spent very little time accelerating.

In our "stationary" reference frame, this "moving" clock was running at half speed. It now reads 1 o'clock but we can see the "stationary" clock, far below us, displaying 2 o'clock.

Because the photons rising up from the "stationary" clock are effectively parallel at both ends of the journey, and orthogonal to the path of the journey, we can be assured that when the "moving" clock read 12:30 o'clock, it was being hit by photons from the "stationary" clock that carried an image of 1 o'clock.

Surely an observer on the "moving" clock would think that the "stationary" clock was the faster one? And isn't that contrary to Relativity? How could an observer on the moving clock possibly conclude that the "stationary" clock was running slower than his own?

Because when he looks out his window while moving, he does not see the "stationary" clock below him. He sees it close to being ahead of him. This is a consequence of light always traveling at "c" relative to inertial observers; the angle of the light incoming to the "moving" clock depends on what reference frame you use.

We tried to elminate the Doppler effect, but we eliminated it only from the "stationary" frame. It's present in the "moving" frame.

Now, suppose that he measures the angle at which he sees this clock ahead of him, and calculates the Doppler effect, and compensates for it, what will he discover? He will discover that the "stationary" clock is actually running slow. It simply looks like it's running fast.

When he gets whacked back in the opposite direction, the clock (and the whole universe) appears to swing around so that it's ahead of him again. He continues to see a clock running at double speed, while his calculations tell him it's really at half speed.

That is how the paradox resolves itself.

David
 
  • #10
Ibix said:
so naive application of time dilation formulae (which only apply between inertial coordinate systems) won't work.
What does "naive" mean? It is a basis of the theory of relativity: the principle of relativity! Every observer can consider himself at rest. Therefore he uses the Lorentz transformations. In this way every observer gets the same result. This also affects time dilation.

By the way, the movement in the circle is accelerated, but this has no effect on the formula for time dilation:
https://www.researchgate.net/publication/30398795_Measurements_of_relativistic_time_dilatation_for_positive_and_negative_muons_in_a_circular_orbit
 
  • #11
worlov said:
What does "naive" mean? It is a basis of the theory of relativity: the principle of relativity! Every observer can consider himself at rest. Therefore he uses the Lorentz transformations. In this way every observer gets the same result. This also affects time dilation.

By the way, the movement in the circle is accelerated, but this has no effect on the formula for time dilation:
https://www.researchgate.net/publication/30398795_Measurements_of_relativistic_time_dilatation_for_positive_and_negative_muons_in_a_circular_orbit

What Einstein wrote, essentially:

"Consider a system of coordinates in which the laws of Newtonian mechanics hold good ... we assume (as a postulate) that the speed of light is independent of the source ... we show that time dilation applies to a clock moving in this system of coordinates".

How @worlov interpreted this:

"Consider any system of coordinates (whether Newton's laws apply or not) ... assume time dilation and the Lorentz Transformation apply with this coordinate system ... state that any observer can consider himself at rest and consider himself an inertial observer - reach a contradiction."

You are ignoring the critical criterion of Einstein's paper that his result apply to observations made in an inertial reference frame. Not in any reference frame.

Movement in a circle has no effect on the formula for time dilation for an inertial obsever, observing the object in circular motion. But, the object in circular motion is not an inertial observer and cannot apply the time dilation formula to objects moving in his system of coordiates.

That is, incidentally, the same in Newtonian mechanics. For example, if you are an inertial observer and an object is moving in a circle, you can conclude and calculate that the object is subject to a centripetal force. However, for the object moving in a circle, it sees you moving in a circle, but if it concludes that you must be subject to a force, then it is wrong. There is no force on you. Your circular motion is due to its acceleration, not your own.

Or, think of someone jumping up and down on a trampoline. You can study the trampolinist's motion using Netwon's laws (or SR if your prefer). But, the person on the trampoline cannot directly use Newton's laws (or SR) to explain why the rest of the world is bouncing up and down!

If you do not accept this, then you can go ahead and find contradictions all over physics. No physics will hold together if you take results derived in inertial reference frames and apply them in non-inertial reference frames. Time dilation is one example. Newton's laws of motion are another example.

You must understand the difference between observing accelerated motion (from an inertial reference frame) and an observer in a state of acceleration trying to use the laws of physics that do not apply in an accelerated coordinate system.
 
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  • #12
worlov said:
What does "naive" mean? It is a basis of the theory of relativity: the principle of relativity! Every observer can consider himself at rest.
Yes, every observer can consider themselves at rest, but that is not what the principle of relativity says. The principle of relativity says that all inertial frames are equivalent. For some observers, the frame in which they are at rest is not inertial. And...
Therefore he uses the Lorentz transformations. In this way every observer gets the same result. This also affects time dilation.
The derivation of the Lorentz transformations assumes that the frames in question are inertial; thus they cannot be correctly applied to transform between non-inertial frames. The researchgate paper you've cited doesn't suggest otherwise.
 
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  • #13
worlov said:
What does "naive" mean? It is a basis of the theory of relativity: the principle of relativity! Every observer can consider himself at rest.
Yes, everybody can regard themself as at rest. Assuming that this means that all observers can regard themself as inertial is naive.
worlov said:
Therefore he uses the Lorentz transformations.
No. Not unless he remains in an inertial frame.
worlov said:
In this way every observer gets the same result. This also affects time dilation.
Clearly not - or the twin paradox would be genuinely paradoxical.
worlov said:
By the way, the movement in the circle is accelerated, but this has no effect on the formula for time dilation:
Yes it does. Between inertial frames, both frames regard the other's clock as running slow by a factor of ##1/\gamma##. In the case of circular motion the orbiting observer sees the inertial clock running fast by a factor of ##\gamma## while the inertial observer sees the orbiting clock running slow by a factor of ##1/\gamma##. The situation is not symmetric, unlike two inertial observers.

Edit: I can't seem to get the ResearchGate paper you cite to download (on the move - dodgy internet connection). I rather suspect from the abstract that it does not consider the perspective of the muons, since there are no physicists co-moving with them. Thus I presume it won't do the bit of maths that would show the asymmetry. Itxs easy enough to do if you want to do it.
 
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  • #14
worlov said:
What does "naive" mean? It is a basis of the theory of relativity: the principle of relativity! Every observer can consider himself at rest. Therefore he uses the Lorentz transformations. In this way every observer gets the same result. This also affects time dilation.
But How he uses the Lorentz transformations are different when he is accelerating than if he is in an inertial frame. To know what another clock is doing, an observer in inertial motion only has to know the relative velocity of the other clock with respect to himself. If he is accelerating there is an additional factor he has to account for. This factor depends on the magnitude of the acceleration, the direction of the other clock with respect to the acceleration, and the distance between them along the line of acceleration. If you are accelerating towards a the clock, then this factor has the clock running fast, the further away, the faster it runs. Accelerating away slows the clock according to the observer*
By the way, the movement in the circle is accelerated, but this has no effect on the formula for time dilation:
https://www.researchgate.net/publication/30398795_Measurements_of_relativistic_time_dilatation_for_positive_and_negative_muons_in_a_circular_orbit
But it does. Imagine three clocks: one at the center of the circle, one circling it at some speed, and one passing it in a straight line at the same speed relative to the center clock. Assume that the third clock travels at a tangent to the second clock's path, and at the moment of interest they are right next to each other. Both clocks will be getting the same information in terms of light information from the center clock at that moment. However, the circling clock will determine that the center clock is running fast compared to itself, while the inertially moving clock will determine that the center clock is running slow compared to itself.
However both the circling clock and the inertial moving clock will agree that they are running at the same rate at the moment they pass each other.

* this is true even if the clocks are not moving relative to each other. Take two clocks, put them both in the same accelerating frame, with the clocks separated along the line of the acceleration, and the clock in the direction of the acceleration will run fast compared to the other, even though they are at rest with respect to each other and under the same acceleration.

This does no
 
  • #15
PeroK said:
How @worlov interpreted this
It should not be forgotten that Einstein formulated his theorem in the context of the special theory of relativity. He begins with a polygonal route:

"...It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line..."

and only after passes to the circle:

"If we assume that the result proved for a polygonal line is also valid for a continuously curved line... Thence we conclude that a balance-clock 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."

Well, Einstein ignored the acceleration phases.
 
  • #16
Janus said:
But How he uses the Lorentz transformations are different when he is accelerating than if he is in an inertial frame.
Let me try to expand on that a bit.

The Lorentz transforms let you take the coordinates (x,y,z,t) of an event in one frame and obtain the coordinates (x', y', z', t') of the same event in some other frame. There is no particular problem in using the Lorentz transforms to do this for an accelerating frame. It is a simple process.

Suppose that we start with the coordinates (x,y,z,t) in the accelerating frame. We translate to coordinates in the inertial frame that is momentarily co-moving with the accelerating frame at time t. This is dead easy since it is the identity translation. We still have coordinates (x,y,z,t). Then we apply the Lorentz transform to shift to (x',y',z',t') coordinates in the target inertial frame. For the "v" term in the transform, we, of course, use the relative velocity between the two inertial frames.

Easy, peazy. But there is a catch. This works for translating coordinates of events. If we want to do something fancier -- like computing a time dilation factor, a velocity or an acceleration things are not so simple. Let's us do a time dilation.

In principle, computing time dilation is fairly simple. We pick out two events that are separated by ##\Delta t## in our accelerated coordinate system. Then we find the coordinates for the same two events in the inertial coordinate system and look for the ##\Delta t'##.

So let us do that. We have two events ##(x_1, y_1, z_1, t_1)## and ##(x_2, y_2, z_2, t_1 + \Delta t)##. We do the Lorentz transform on the first one and get ##(x'_1, y'_1, z'_1, t'_1)##. We do the Lorentz transform on the second one and get ##(x'_2, y'_2, z'_2, t'_1 + something)##.

We might naively expect to find a time dilation factor of ##\gamma##. But there is a gotcha.

The gotcha is that in the interval between the first event and the second, our starting frame accelerated. The relative velocity between the frames changed. The second Lorentz transform does not use the same v as the first. That means that the time dilation factor is not as simple as ##\gamma##. Nor is it something simple-minded like ##\frac{\gamma_1 + \gamma_2}{2}##. The Lorentz transform for time looks like:
$$t' = \gamma (t-\frac{vx}{c^2})$$
To a first order approximation, if you change v between the two transforms, that gives rise to a change of ##\frac{\gamma \ x \ \Delta v}{c^2}##. The presence of the ##\Delta v## is a clue that acceleration factors in. The presence of the x means that separation in the x direction also factors in.
 
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  • #17
worlov said:
Well, Einstein ignored the acceleration phases.

Yes, and by doing this an observer moving with the clock, in a polygon or a circle, is not an inertial observer. So, that observer cannot directly and simply employ the theory of SR as set out in the previous sections of the paper. In the polygon approximation, which is how I analysed the problem in the link I posted earlier, the "moving" clock is instantaneously jumping from one inertial frame to another. You must, therefore, apply the Lorentz Transformation separately for each side of the polygon and add up all those phases. When you do that, you see that the clock at A is advancing more quickly than the "moving" clock.

Check out the link I posted, where as others have mentioned, the relativity of simultaneity is critical to the calculations.
 
  • #18
worlov said:
Well, Einstein ignored the acceleration phases
But not the relativity of simultaneity. Which is the source of the "extra time" I've been banging on about since #2. You can either treat the polygonal lines using inertial frames and correct for the relativity of simultaneity at the corners, or regard the whole thing as a non-inertial frame. Either way, simply applying the inertial frame time dilation formula isn't correct for describing the inertial clock from the traveller's point of view.
 
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  • #19
PeroK said:
Check out the link I posted, where as others have mentioned, the relativity of simultaneity is critical to the calculations.
We can mark all clocks. The moving observer will then enter in a table which clock which time has indicated (compared with his own time). When the moving observer sees the first clock again, he can check his table. He will notice that this clock shows too much time, although it is just as slow as before.
 
  • #20
worlov said:
We can mark all clocks. The moving observer will then enter in a table which clock which time has indicated (compared with his own time). When the moving observer sees the first clock again, he can check his table. He will notice that this clock shows too much time, although it is just as slow as before.
Let's try to clarify just what is meant by how the Relativity of Simultaneity is critical.
You have three clocks, Red, Green, and Blue.
Red and Green remain at rest with respect to each other, and it their rest frame are synchronized. Blue travels from Red to Green and then back to Red.
Thus according to the Rest frame of Red and Green the sequence of events unfolds like this assuming that blue has a relative velocity of 0.8c, the distance between Red and Green is 1 light hour, as measured in the rest frame of Red and Green.
exam1.png

Blue leaves Red while All three clocks read zero.
Blue arrives at green when both Red and Green reads 1.25 hr, and it reads 0.75 hr due to time dilation.
Blue reverses direction. All the clock readings remain the same.
Blue arrives back at Red, reading 1.5 hrs, while Red and Green read 2.5 hrs.

Now the same events according to Blue.
exam2.png

Blue is at rest while Red and Green are moving. Since Red and Green are moving, they and the distance between them is length contracted to 0.6 light hr. Red and Blue still both read zero when they start off next to each other. However, due to the Relativity of Simultaneity, Green already reads 0.8 hr
It takes 0.75 hr for Green to reach Blue (0.6 light hr/ 0.8 c) During which time both Red and Green run 0.6 as fast as Blue and advance by 0.45 hr. Green now reads 1.25 and Blue reads 0.75, just like when we looked at things from Red and Green's frame. Red however only reads 0.45 hr. (0.8 hr behind Green)
Red and Green reverse direction. The reading on Red and Green do not change, But now that the relative velocity of Red and Green with respect to Blue have changed, Relativity of Simultaneity requires that Red is the clock that is ahead of Green by 0.8 hr. Thus now Red read 2.05 hr.
Again, it take 0.75 for Red to reach Blue, during which time both Red and Green advance 0.45 hr. Upon rejoining, Blue reads 1.5 hr and Red reads 2.5 hr. In perfect agreement with the end result when we assumed that Red and Green were at rest the whole time.

Red and Blue are in perfect agreement as to the start and end of the scenario in terms of what each of them reads, but they disagree as to what happened between start and finish.
 

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  • #21
worlov said:
Here I miss the actual function of a clock - even measurement of time.
The type of time measured by a clock is called proper time. It is different from coordinate time. It is a relativistic invariant that can be calculated by the integral of ##\sqrt{dt^2-dx^2-dy^2-dz^2}##. All inertial frames agree on this formula and even non inertial frames agree on its value but not the formula.

worlov said:
One of the twins has to admit that his system is so to say subordinate.
Not subordinate. The term is “non inertial”. Everybody agrees that the traveling twin’s frame is non inertial, including the traveling twin himself.

worlov said:
Well, Einstein ignored the acceleration phases.
True. He also performed all of his calculations in the inertial frame only.
 
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  • #22
Janus said:
In perfect agreement with the end result when we assumed that Red and Green were at rest the whole time.
You presented one of the possible calculation for the twin paradox. But that is not the point. Einstein assumes the existence of such a clock, which indicates more time than would have passed according to its speed. In nature there is no such clock. Therefore his assumptions should be reconsidered anew.
 
  • #23
@worlov - can you explain to me what is the relativity of simultaneity? Because it seems to me that you are totally failing to grasp this and its implications for twin paradox-like scenarios.
 
  • #24
To determine the simultaneity at different locations, the running time of the light must be considered... I do not have against the twin paradox - the calculation is correct. But ... it's about natural processes. Physically, no clock can exist that displays more time than it counts with "ticks".

Suppose the moving observer before departure has hidden the clock at point A in a safe. No one can see or even manipulate this clock during his travel. If the moving observer returns and opens the safe, which displayed time he will expect? He sees that the clock is just as slow as when he left. Of course, he assumes that this clock was always uniformly in his absence. And he knows how long he was gone.
 
  • #25
worlov said:
Physically, no clock can exist that displays more time than it counts with "ticks".
And no clock is doing so anywhere in this setup. The point about the relativity of simultaneity is that if you naively apply the time dilation formula to a polygonal path (edit: or, indeed, a curved path) from the perspective of the person following that path, then there are ticks of the inertial clock that you forgot to account for.
worlov said:
Suppose the moving observer before departure has hidden the clock at point A in a safe. No one can see or even manipulate this clock during his travel. If the moving observer returns and opens the safe, which displayed time he will expect? He sees that the clock is just as slow as when he left. Of course, he assumes that this clock was always uniformly in his absence. And he knows how long he was gone.
You are asking what time will the traveling twin expect to see on the inertial clock? If he sees ##\Delta t## on his clock then he'll expect to see ##\gamma\Delta t## on the inertial clock. Edit 2: Or, at least, if he accounts correctly for ALL relativistic effects he will.
 
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  • #26
worlov said:
Physically, no clock can exist that displays more time than it counts with "ticks".
Please look at my previous post. All correctly functioning clocks "tick" proper time, which is the integral of ##\sqrt{dt^2-dx^2-dy^2-dz^2}## where ##(t,x,y,z)## are the coordinates in an inertial frame (in units where c=1). That quantity is what defines the clock's ticks, and the integral of that is what the clock displays.
 
  • #27
Dale said:
Please look at my previous post. All correctly functioning clocks "tick" proper time, which is the integral of √dt2−dx2−dy2−dz2
Well, from that the time dilation can be deduced. Otherwise I see no contradiction to my consideration.
 
  • #28
worlov said:
Suppose the moving observer before departure has hidden the clock at point A in a safe. No one can see or even manipulate this clock during his travel. If the moving observer returns and opens the safe, which displayed time he will expect? He sees that the clock is just as slow as when he left. Of course, he assumes that this clock was always uniformly in his absence. And he knows how long he was gone.
I do not understand what the existence of the safe adds to the scenario.

The clock in the safe ticks at the same rate as a clock sitting on top of the safe. When the safe is opened, the readings of the clock inside and the clock on top will be identical, of course.

Are you trying to reason that they are different? Are you trying to say that someone would expect them to be different?
 
  • #29
worlov said:
Well, from that the time dilation can be deduced. Otherwise I see no contradiction to my consideration.
It shows how to calculate both the ticks and the display. If you actually work out the numbers then you get a clear statement about the reading of any clock after any arbitrary motion.

The issue is that the proper time ##d\tau## is only given by ##d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2## in an inertial frame. In non inertial frames ##d\tau## will have a different formula. The contradiction to your consideration is that in the traveling clock's frame a different expression is necessary, in contradiction to your assumption that the same expression would be used.
 
  • #30
worlov said:
Einstein assumes the existence of such a clock, which indicates more time than would have passed according to its speed.

He assumes the opposite. The speed of a clock's motion has no effect on the clock's behavior. If it did it would constitute a way to distinguish between rest and uniform motion.
 
  • #31
jbriggs444 said:
I do not understand what the existence of the safe adds to the scenario.
I just wanted to emphasize that a clock works autonomously. Regardless of whether it is watched or not, a clock goes uniformly.

Einstein later tried to prove his „twin“ theorem with the general relativity:
https://en.wikisource.org/wiki/Translation:Dialog_about_Objections_against_the_Theory_of_Relativity

He modeled the acceleration phases with the imaginary gravitational field. Obviously, he relied on the principle of equivalence. However, the CERN researchers claim that the accelerations have no effect on the formula for time dilation of the special theory of relativity. So, this exchange – acceleration => gravitational field – is not justified. Moreover, the correctness of the principle of equivalence is also questionable. In this respect, Einstein's proof with the polygonal lines deserves more respect.
 
  • #32
worlov said:
I just wanted to emphasize that a clock works autonomously. Regardless of whether it is watched or not, a clock goes uniformly.
Relativity does not require one to be watching the clocks. You are making the mistake that Relativity relies on the exchange of light signals, it doesn't. We use light in examples because it is convenient to illustrate the effects of Relativity, not because it is crucial to it. The only real reason that would require watching the clock would be to unsure that nothing other than it running at its normal rate happened during the course of the test ( it didn't stop running and start running again, a fault didn't cause it to start running fast, etc.)
Einstein later tried to prove his „twin“ theorem with the general relativity:
https://en.wikisource.org/wiki/Translation:Dialog_about_Objections_against_the_Theory_of_Relativity

He modeled the acceleration phases with the imaginary gravitational field. Obviously, he relied on the principle of equivalence. However, the CERN researchers claim that the accelerations have no effect on the formula for time dilation of the special theory of relativity. So, this exchange – acceleration => gravitational field – is not justified. Moreover, the correctness of the principle of equivalence is also questionable. In this respect, Einstein's proof with the polygonal lines deserves more respect.
The CERN test just verified the "clock postulate"; That when measured from an inertial frame, the rate at which a clock will tick is only effected by it velocity with respect to that frame and no additional effect is caused by any acceleration it may be undergoing.
But the equivalence principle deals with how the acceleration effects measurements made by the accelerating observer. CERN verified that particles accelerated do not show any additional time dilation as measured from the lab, but it did not measure how the accelerated particles measured the ticking of clocks in the lab.
 
  • #33
worlov said:
He modeled the acceleration phases with the imaginary gravitational field. Obviously, he relied on the principle of equivalence. However, the CERN researchers claim that the accelerations have no effect on the formula for time dilation of the special theory of relativity. So, this exchange – acceleration => gravitational field – is not justified. Moreover, the correctness of the principle of equivalence is also questionable. In this respect, Einstein's proof with the polygonal lines deserves more respect.

Many of the finest minds of the time used arguments far better than that to refute what Einstein had done, but since that time experiment and observation show that Einstein got it right. It is now an every day fact of life for thousands of scientists and engineers working all over the world.
 
  • #34
worlov said:
CERN researchers claim that the accelerations have no effect on the formula for time dilation of the special theory of relativity. So, this exchange – acceleration => gravitational field – is not justified.
The equivalence principle is fully justified in many experiments and the results I believe you are referring to are fully consistent with relativity.

The problem is that people who have not worked through the math mistakenly believe that gravitational time dilation is related to gravitational acceleration, but it is actually related to gravitational potential. So the fact that accelerations have no effect is predicted by both, and what has the effect is the potential whether it is the gravitational potential or the inertial potential that arises in an equivalent accelerating reference frame.

The mathematical concept that Einstein used to model the artificial gravitational field is called Christoffel symbols. The relationship is completely rigorous, even though it is not a large focus of relativity in practice. But he is fully justified in his description in that paper.
 
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  • #35
Here are some spacetime diagrams to explain why the time dilation formulae don't work here, and why the relativity of simultaneity is important. I've done this for a simple out-and-back twin paradox with instantaneous turnaround and a constant velocity of 0.6c. The extension to more complicated cases is messy, but fundamentally the same.

First, here's a spacetime diagram of the stay-at-home clock (blue worldline) and the traveling clock (red worldline) in the frame of the stay-at-home:
upload_2019-1-13_21-52-58.png

The turnaround happens at time ##T##. But now let's think about the outbound frame, the frame in which the traveling clock is at rest as it travels to the right. The turnaround happens at time ##T'=T/\gamma## in this frame. But the Lorentz transforms tell us that ##T'=\gamma(t-vx/c^2)##, which is to say that "at the same time as the turnaround according to the outbound frame" is a sloped line on this graph, satisfying ##t=T'/\gamma+vx/c^2=T/\gamma^2+vx/c^2##. That's the dashed line in the diagram below:
upload_2019-1-13_22-0-22.png

The dashed line intercepts the stay-at-home clock's (blue) worldline at ##t=T/\gamma^2##. This is exactly consistent with time dilation - the traveling clock experiences ##T/\gamma## of elapsed time, and measures ##1/\gamma## less time on the stay-at-home clock.

Now let's turn our attention to the inbound frame. The maths here is a bit messier. The turnaround in this frame happens at ##T''=\gamma(1+v^2/c^2)T##. Again, the Lorentz transforms tell us that ##T''=\gamma(t+vx/c^2)##, or ##t=(2T-T/\gamma^2)-vx/c^2##. Again, this is a sloped line, but it is not the same sloped line as above - it slopes in the opposite direction.
upload_2019-1-13_22-11-50.png

Again, this is consistent with the time dilation formula. The traveling clock experiences time ##T/\gamma## on the inbound leg, and measures the stay-at-home clock to experience less time by another factor of ##1/\gamma##.

But now we can see what's gone wrong. The time up to ##T/\gamma^2## on the stay-at-home clock is "during the outbound leg" according to the outbound frame, and the time after ##2T-T/\gamma^2## is "during the inbound leg" according to the inbound frame. But because we changed our definition of "at the same time as the turnaround", to naively apply the time dilation formula is to forget about the middle portion of the worldline. This is why there is no problem with the behaviour of the stay-at-home clock - it ticks 2T times. But if you don't account for the relativity of simultaneity you have forgotten about ##2Tv^2/c^2## of them.
 

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<h2>1. What is time dilation?</h2><p>Time dilation is a concept in physics that states time passes at different rates for objects moving at different speeds. This is a fundamental principle of Einstein's theory of relativity.</p><h2>2. How does time dilation occur?</h2><p>Time dilation occurs due to the fact that the speed of light is constant in all frames of reference. This means that as an object approaches the speed of light, time slows down for that object relative to an observer who is not moving at the same speed.</p><h2>3. How is time dilation related to Einstein's theory of relativity?</h2><p>Einstein's theory of relativity is based on the idea that the laws of physics are the same for all observers, regardless of their relative motion. Time dilation is a direct consequence of this principle, as it shows that time is relative and can appear to pass at different rates for different observers.</p><h2>4. Is time dilation a proven phenomenon?</h2><p>Yes, time dilation has been proven through numerous experiments and observations. One of the most famous examples is the Hafele-Keating experiment, which showed that atomic clocks on airplanes moving at high speeds ran slightly slower than clocks on the ground.</p><h2>5. Can time dilation be experienced in everyday life?</h2><p>Yes, time dilation is a real phenomenon that occurs in everyday life. However, the effects are only noticeable at extremely high speeds, such as those reached by objects traveling near the speed of light. For most people, the effects of time dilation are too small to be perceived in their daily lives.</p>

1. What is time dilation?

Time dilation is a concept in physics that states time passes at different rates for objects moving at different speeds. This is a fundamental principle of Einstein's theory of relativity.

2. How does time dilation occur?

Time dilation occurs due to the fact that the speed of light is constant in all frames of reference. This means that as an object approaches the speed of light, time slows down for that object relative to an observer who is not moving at the same speed.

3. How is time dilation related to Einstein's theory of relativity?

Einstein's theory of relativity is based on the idea that the laws of physics are the same for all observers, regardless of their relative motion. Time dilation is a direct consequence of this principle, as it shows that time is relative and can appear to pass at different rates for different observers.

4. Is time dilation a proven phenomenon?

Yes, time dilation has been proven through numerous experiments and observations. One of the most famous examples is the Hafele-Keating experiment, which showed that atomic clocks on airplanes moving at high speeds ran slightly slower than clocks on the ground.

5. Can time dilation be experienced in everyday life?

Yes, time dilation is a real phenomenon that occurs in everyday life. However, the effects are only noticeable at extremely high speeds, such as those reached by objects traveling near the speed of light. For most people, the effects of time dilation are too small to be perceived in their daily lives.

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