# B Time dilation and Einstein's theorem

#### Dale

Mentor
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.

#### worlov

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.

#### jbriggs444

Science Advisor
Homework Helper
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?

#### Dale

Mentor
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.

#### Mister T

Science Advisor
Gold Member
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.

#### worlov

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.

#### Janus

Staff Emeritus
Science Advisor
Gold Member
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.

#### Mister T

Science Advisor
Gold Member
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.

#### Dale

Mentor
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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#### Ibix

Science Advisor
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 travelling clock (red worldline) in the frame of the stay-at-home: The turnaround happens at time $T$. But now let's think about the outbound frame, the frame in which the travelling 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: 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 travelling 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. Again, this is consistent with the time dilation formula. The travelling 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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• Mister T and Dale

#### Grimble

But the Lorentz transforms tell us that T′=γ(t−vx/c2)
but surely x - the distance the clock has travelled, OD, = vt
then
T'=γt(1-v2/c2) = t/γ

in the frame of the stay-at-home: . The distance OT' is the distance, if the travelling clock were a light clock, that light would have travelled from the initial event O to the red turnaround at time T' in the stay-at-home frame.
... and T' occurs at time T on the time axis on the stay-at-home clock's frame
One cannot get much more basic than time and distance axes, T and X at 90°.

I am sorry but I am just having difficulty following what is wrong with the most simple and basic mathematical principles - or maybe I am just being naïve... ?

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#### Janus

Staff Emeritus
Science Advisor
Gold Member
but surely x - the distance the clock has travelled, OD, = vt
then
T'=γt(1-v2/c2) = t/γ

. The distance OT' is the distance, if the travelling clock were a light clock, that light would have travelled from the initial event O to the red turnaround at time T' in the stay-at-home frame.
... and T' occurs at time T on the time axis on the stay-at-home clock's frame
One cannot get much more basic than time and distance axes, T and X at 90°.

I am sorry but I am just having difficulty following what is wrong with the most simple and basic mathematical principles - or maybe I am just being naïve... ?
To do a light clock analysis using the space-time diagram, you would have to add another axis, y (coming straight out of the screen) to represent the distance between the mirrors which is perpendicular to the relative motion of the clocks. As drawn the time-space diagram only deals with the time axis and the axis along the relative velocity of the clocks.

And while T and X axis at 90 degrees is basic, and both frames measure T and X at being 90 degrees, They don't measure T and X along the same axis.

Start with two clocks A and B, which start at the same point, both reading 0, and separate at 0.6c. When Clock A reads 1 it sends a light signal to B and B then sends it back.
In the frame of A (green world line), the space-time diagram looks likes this. The light (yellow line), arrives at B when it reads 2 and returns when A reads B. the light lines are drawn at 45 degree angles, as that is the time to space scale of the diagram. According to A, the light leaves some time before B reads 1 and returns a bit after B reads 3. B also ticks off it time at a slower rate.

From the frame of B (blue line), events unfold like this: The light still leaves A when it reads 1, arrives at B when it reads 2 and returns to A when A reads 4. To maintain a invariant speed for light the light lines must still be drawn at 45 degree angles. To keep this consistent with the departure and arrival readings, the light leaves A after clock B reads 1 and returns to A well after B reads 3 ( it looks like a bit after B reads 5)

The time and space axis are not fixed features of space-time, but are frame dependent.

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#### Ibix

Science Advisor
but surely x - the distance the clock has travelled, OD, = vt
then
T'=γt(1-v2/c2) = t/γ
But here you are telling me the time $t$ on a clock instantaneously colocated with the ship when the ship's clock reads $T'$. I'm telling you what a clock at any $x$ reads when it's instantaneously colocated with a moving clock (moving at the same speed as the ship, but not necessarily in the same place) reads $T'$.
The distance OT' is the distance, if the travelling clock were a light clock, that light would have travelled from the initial event O to the red turnaround at time T' in the stay-at-home frame.
This sentence is somewhat tortured, but I think it's correct.
One cannot get much more basic than time and distance axes, T and X at 90°.
As drawn on a Minkowski diagram, if the $x$ and $t$ axes are at right angles then $x'$ and $t'$ axes are not. They are orthogonal in Minkowski geometry, but the screen obeys Euclidean geometry so I can't draw things that obey Minkowski geometry accurately.

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#### sweet springs

Each observer determines that all clocks in motion relative to that observer run slower than that observer’s own clock.
Not so. Say a space pilot gets on a large turn table of radius R and turning speed R $\omega$ and make a cosmic round trip and get it off and return to the Earth.
The pilot is in round motion for the Earth. The pilot clock has ticked slower when they meet.
The Earth is in round motion for the pilot. However, the pilot clock has ticked slower when they meet.

#### jbriggs444

Science Advisor
Homework Helper
Not so. Say a space pilot gets on a large turn table of radius R and turning speed R $\omega$ and make a cosmic round trip and get it off and return to the Earth.
The pilot is in round motion for the Earth. The pilot clock has ticked slower when they meet.
The Earth is in round motion for the pilot. However, the pilot clock has ticked slower when they meet.
It is far from clear what measurements you are describing here. The unambiguous fact of the matter is that the clock held by the rotating pilot will be seen to have ticked fewer times when the clocks are reunited. This in no way contradicts the legitimacy of a series of tangent frames in which each clock ticked more slowly than the other throughout the scenario.

However, time dilation is indeed not the whole story. When hopping from tangent frame to tangent frame to build a global picture of the situation, relativity of simultaneity is an important concept to grasp.

#### sweet springs

When hopping from tangent frame to tangent frame to build a global picture of the situation, relativity of simultaneity is an important concept to grasp.
Go-return rocket experiences only one hopping when turning around. Turning round trip pilot does continuous and homogeneous hoppings in tangent IFRs. Accumulated time dilation results when meeting again are same ratio of $\sqrt{1-v^2/c^2}$.　It applies for any closed trajectory in the Earth's IFR with tangent speed v.

Rotating frame of reference which is interpreted as integrated tangent IFRs gives up synchronization of clocks as well as simultaneity. Absolute past and future are important but simultaneity depends on frames of reference and not defined in some frames. Should we rely on concept of simultaneity to the end ?

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#### sweet springs

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 scenario of go-return rocket, we have three IFRs of A, say the Earth, go rocket and return rocket.

Say as for coordinate settings, the Earth IFR and go rocket IFR have coordinate time zero and space zero for the departure event, and go rocket IFR and return rocket IFR have coordinate of same time and same space for the turning event.

As for meeting again event, the three IFRs' networked clocks superposed on the Earth show different ticks. Readings of three superposed IFR clocks on the meeting again event are

return rocket < the Earth < go rocket

Each of three clock is synchronized in each IFR's clock network. Though each criticizes other two clocks for wrong synchronization, you do not have to worry about it. Synchronization is relative.

PS
For another setting, say all three IFRs have coordinate of time zero and space zero for the departure event, the readings of three superposed clocks at the event of meeting again are

the Earth < go rocket = return rocket

Say T the difference of go rocket IFR clock and return rocket IFR clock at the event of turning. Go rocket IFR clock is delayed much with relative speed $\frac{2v}{1+v^2/c^2}\approx 2v$ for return rocket IFR clock. By draw back of T in the return rocket clock

return rocket - T < the Earth < go rocket

This is same as the original setting above.

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#### Grimble

But here you are telling me the time tt on a clock instantaneously colocated with the ship when the ship's clock reads T′T'. I'm telling you what a clock at any xx reads when it's instantaneously colocated with a moving clock (moving at the same speed as the ship, but not necessarily in the same place) reads T′T'.
You have lost me here... - what I am saying is that the distance the traveller (on his spaceship) has traversed is the velocity he is travelling at x the time he has travelled - x=vt - (pretty basic mechanics surely) and that that substituted into the Lorentz time equation ...

#### Ibix

Science Advisor
You have lost me here... - what I am saying is that the distance the traveller (on his spaceship) has traversed is the velocity he is travelling at x the time he has travelled - x=vt - (pretty basic mechanics surely) and that that substituted into the Lorentz time equation ...
$x$ is where something is, not the distance it's travelled. However, the ship started at $x=0$, so in the specific case of the ship $x$ is equal to the distance travelled. For other clocks moving at the same speed as the ship but not at the same place, $x\neq vt$.

#### Grimble

But what 'other clocks' ?
The travelling clock is on board the spaceship. That is what this is all about - what have other clocks to do with it?

#### jbriggs444

Science Advisor
Homework Helper
But what 'other clocks' ?
The travelling clock is on board the spaceship. That is what this is all about - what have other clocks to do with it?
If there is a "travelling" clock, there is a frame in which it is travelling. In that frame, it makes sense to speak of x=vt. However, the clock that measures that t is not the travelling clock.

There is also a frame in which the travelling clock is at rest. The travelling clock measures time too, of course. But in its rest frame, v=0 and x=0.

#### Ibix

Science Advisor
But what 'other clocks' ?
The travelling clock is on board the spaceship. That is what this is all about - what have other clocks to do with it?
The point is that "at the same time as the ship turns around, back on Earth" means different things to different frames. So the interesting question is: when a clock at rest in the outbound ship frame passes the Earth showing the same time as the ship clock shows at turnaround, what time is it according to the Earth's clocks? And similarly for the inbound frame.

#### Grimble

Yes, of course, but in the frame of that clock it is at rest, not travelling.
It is, and can only be (in this thought experiment) ... only be travelling in the resting frame of clock T.
So how can v=0, x=0 in the Lorentz equation for a travelling clock? - that just seems to make no sense ...

#### Grimble

I'm sorry but we seem to be missing something here - Einstein's first postulate.

Let us look at this in its simplest view.
We have two frames, two inertial frames, in each of which there is a stationary clock. To an observer in frame A clock A is at rest and clock B, at rest in frame B , moving at v relative to A.
Two identical clocks in inertial frames.
Let us suppose they are light clocks.
When the light in clock A has travelled 1 light second to its mirror, it will measure 1 second to have passed and clock A will display (read) 1 second.
When the light in clock B has travelled 1 light second to its mirror, it will measure 1 second to have passed and will display (read) 1 second.
Frame A and Frame B are inertial frames and all we are measuring is how long it takes for light to travel 1 light second in each frame. the same time - 1 second measured from O when the clocks were co-located.
I am not saying that this is simultaneous because that judgement depends on the convention for simultaneity chosen; but I will say that they both happen at the end of equal intervals measured from event O.
Both clocks will read the same at turnround.
However, I am not saying that the travelling clock is not slow - it certainly is - but compared to what? Surely oit must be compared to A's measurement of the turnround time of T' that is the turnround time of B, in A's frame in which the light in clock B has travelled 1 light second to its mirror AND the distance that clock has travelled from clock A.
That is it will have travelledthe distance OT' or γT.

#### Dale

Mentor
I'm sorry but we seem to be missing something here - Einstein's first postulate.
If you are using the Lorentz transform then you are not missing either of Einstein’s postulates. Both postulates are built into the Lorentz transform.

• Ibix

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