Question about the reciprocity of time dilation

[Dale]

Are you saying that in the rest frame of clocks B and C the clocks B and C are synchronized (ticks of B and C happen at the same time) but relative to the rest frame of A (where B and C moving) the clocks B and C are not synchronized and tick at different times?

Yes. In A’s frame B and C both tick at the same rate (both are time dilated the same), but they show different values.

Sorry if I misunderstood some things, the example is a little difficult for my imagination.

Yes. This is the single most challenging concept of special relativity.

 

[Chenkel]

Yes. In A’s frame B and C both tick at the same rate (both are time dilated the same), but they show different values

Why should B and C show different values from the rest frame of A? In the rest from of A wouldn't the time on B be $\frac {T_{A}} {\gamma}$ and the time on C be $\frac {T_{A}} {\gamma}$?

I know I'm missing something but I'm not sure what.

I also have a strong feeling my math isn't good right now.

 

[Dale]

Why should B and C show different values from the rest frame of A?

This is the relativity of simultaneity. There are three main effects of relativity: time dilation, length contraction, and the relativity of simultaneity. While students tend to get length contraction and time dilation, the relativity of simultaneity is notoriously difficult to internalize. Things that are simultaneous in one frame are not simultaneous in other frames. Most relativistic paradoxes are based on incorrect intuitions about simultaneity.

In the rest from of A wouldn't the time on B be TAγ and the time on C be TAγ?

No. To account for the relativity of simultaneity you need to use the full Lorentz transforms: $$t’=\gamma \left(t-\frac{vx}{c^2}\right)$$$$x’=\gamma\left(x-vt\right)$$

Personally, I wish that the length contraction and time dilation formulas were never taught to beginners. They are very easy to misuse. It is better to use the Lorentz transforms instead. They will automatically simplify when appropriate, but will get the correct results when the simplified formulas are inappropriate

 

[Sagittarius A-Star]

Why should B and C show different values from the rest frame of A? In the rest from of A wouldn't the time on B be $\frac {T_{A}} {\gamma}$ and the time on C be $\frac {T_{A}} {\gamma}$?

I know I'm missing something but I'm not sure what.

Assume that clock A and clock B displayed 0 when the they met.
Assume that clock B and clock C are synchronous to the coordinate time of their common rest-frame $S'$ with reference to this frame.
Assume clock A is located at x=0, clock B is located at x'=0 and clock C is located at x'=2.
Assume $\gamma =2$

Inverse Lorentz transformation for time:
$t = \gamma (t' +vx'/c^2)$.

Reference frame A, time dilation of clock B (see posting #22):
$t_B = 2*(5+0)=10$
Reference frame A, time dilation of clock C:
$t_C = 2*(5+vx'/c^2) = 2*(5+2*v/c^2) \neq t_B$

 

[Chenkel]

Assume that clock A and clock B displayed 0 when the they met.
Assume that clock B and clock C are synchronous to the coordinate time of their common rest-frame $S'$ with reference to this frame.
Assume clock A is located at x=0, clock B is located at x'=0 and clock C is located at x'=2.
Assume $\gamma =2$

Inverse Lorentz transformation for time:
$t = \gamma (t' +vx'/c^2)$.

Reference frame A, time dilation of clock B (see posting #22):
$t_B = 2*(5+0)=10$
Reference frame A, time dilation of clock C:
$t_C = 2*(5+vx'/c^2) = 2*(5+2*v/c^2) \neq t_B$

I feel you know what you're talking about but I don't fully understand the full Lorentz transformation to say I can grock the example you gave.

I'm gonna study more, hopefully someday I understand.

 

[Herman Trivilino]

Are you saying that in the rest frame of clocks B and C the clocks B and C are synchronized (ticks of B and C happen at the same time) but relative to the rest frame of A (where B and C moving) the clocks B and C are not synchronized and tick at different times?

The clocks tick at the same rate, but they are not synchronized, that is, they are not set to the same clock reading. If you were at rest relative to A and you watched the process used by someone at rest relative to B and C synchronize the clocks, you would observe him making what would appear to you to be an error. It is this perceived error that leads you to the conclusion that he thinks your clocks are running slow when you know that it is his clocks that are running slow.

As I said before, any good textbook that serves as an introduction to special relativity will explain this in detail.

When developing special relativity, Einstein was struggling with the notion that time dilation had to be symmetrical to satisfy his first postulate, and that this realization came to him in the middle of the night, causing him to suddenly sit upright in his bed.

 

[Herman Trivilino]

Why should B and C show different values from the rest frame of A?

Because simultaneity is relative.

 

[Chenkel]

When developing special relativity, Einstein was struggling with the notion that time dilation had to be symmetrical to satisfy his first postulate, and that this realization came to him in the middle of the night, causing him to suddenly sit upright in his bed.

That's a interesting story, maybe I'll have that eureka moment too.

 

[Chenkel]

So I've been playing around with these ideas and came up with the following thought experiment.

(for the example I have clocks that can visibly clap and cover themselves in paint)

If you have two light clocks moving relative to each other (A and B) (let's also say they are synchronized when they meet) where each one counts to 4 ticks and claps on each tick and covers itself in paint on the 4th tick lets also say that gamma equals 2.

The rest frame of B will observe A to be clapping half as fast as B and observe A to cover A in paint after B covers B in paint.

The rest frame of A will observe B to be clapping half as fast as A and B will cover itself in paint after A covers A in paint.

Both clocks tick a total of 4 times and cover themselves in paint but the order of paint dumping is reversed based on which reference frame you use.

Hopefully my intuitions getting a little better.

 

[Sagittarius A-Star]

So I've been playing around with these ideas and came up with the following thought experiment.

I propose that you calculate around with Lorentz transformation and apply it to this thought experiment.

 

[Chenkel]

I propose that you calculate around with Lorentz transformation and apply it to this thought experiment.

I'll try that once I understand the full Lorentz transform. Thanks for the suggestion.

 

[Sagittarius A-Star]

I'll try that once I understand the full Lorentz transform. Thanks for the suggestion.

You can find a link to a LaTeX Guide on the left bottom side under the input field. Important is to read the chapter "Delimiting your LaTeX code".

 

[Dale]

Both clocks tick a total of 4 times and cover themselves in paint but the order of paint dumping is reversed based on which reference frame you use.

Yes.

One way to see this visually is to draw a spacetime diagram. I recommend $\gamma=5/4$ instead of 2. Draw light pulses from each “dump” event, one that is emitted and one that was received at the dumping. Each frame will claim that the other frame’s dump is simultaneous with the event on their own worldline which is halfway between the emitted and received light pulses.

 

[FactChecker]

Maybe this will help to clear up the difficulty of the symmetry.

 

[Chenkel]

Yes.

One way to see this visually is to draw a spacetime diagram. I recommend $\gamma=5/4$ instead of 2. Draw light pulses from each “dump” event, one that is emitted and one that was received at the dumping. Each frame will claim that the other frame’s dump is simultaneous with the event on their own worldline which is halfway between the emitted and received light pulses.

I'm not sure how to do that, but I'll try if I get good enough at the language of SR.

 

[Chenkel]

Maybe this will help to clear up the difficulty of the symmetry. View attachment 336519

Thanks for diagram, I didn't understand it the first time I looked at it, maybe in time I will understand.

 

[FactChecker]

Thanks for diagram, I didn't understand it the first time I looked at it, maybe in time I will understand.

I think it is fundamental to understanding SR. If you have questions about any particular step, I will try to clarify it. I just put it together today and it might be a little rough.

 

[Chenkel]

Maybe this will help to clear up the difficulty of the symmetry. View attachment 336519

In my attempt to understand this is the red arrow showing the movement of R relative to a rest frame B?.

Does the blue arrow show the movement of B relative to a rest frame R?

The first slide puzzles me a little, it says "Blue clocks is ahead of red clocks" does this mean blue's clock is ticking faster than the moving red clock?

Thanks for making the diagram.

 

[FactChecker]

In my attempt to understand this is the red arrow showing the movement of R relative to a rest frame B?.

Yes, but it is just relative to B without indicating which frame is a "rest" frame. Both frames have equal rights to call themselves a rest frame.

Does the blue arrow show the movement of B relative to a rest frame R?

Yes.

The first slide puzzles me a little, it says "Blue clocks is ahead of red clocks" does this mean blue's clock is ticking faster than the moving red clock?

No, that is a step too far. You should not be immediately thinking about ticking speed, but just think about how the clocks are set to be synchronized so that the speed of light is always c. Experiments show that is true. We are initially considering how clocks are set up and down the line rather than their ticking speed. The only clocks that we will consider the ticking speed of are the ones labeled B and R that are initially at the center. We will consider their speed by comparing them to the string of opposite-colored clocks along their path. But that will come in the later steps.

Looking at the left side from Red's point of view. Red can assume that he is stationary and everything is normal. Suppose Red has measured one light-second to the left of his center. He sees that Blue on the left side is rushing toward the center and will hit a light pulse (from the center) too early. Yet Blue still measures the speed of a left-directed pulse of light as c. How can that be? Blue clocks on that side must be set so that one second of Blue time elapses too early. Blue clocks are set ahead of his Red clocks. The farther to the left you go, the more they are set ahead.
Looking at the right side from Blue's point of view says that Blue will think, similarly, that the Red clocks are set ahead of his.

 

[Chenkel]

Yes, but it is just relative to B without indicating which frame is a "rest" frame. Both frames have equal rights to call themselves a rest frame.

Yes.

No, that is a step too far. You should not be immediately thinking about ticking speed, but just think about how the clocks are set to be synchronized so that the speed of light is always c. Experiments show that is true. We are initially considering how clocks are set up and down the line rather than their ticking speed. The only clocks that we will consider the ticking speed of are the ones labeled B and R that are initially at the center. We will consider their speed by comparing them to the string of opposite-colored clocks along their path. But that will come in the later steps.

Looking at the left side from Red's point of view. Red can assume that he is stationary and everything is normal. Suppose Red has measured one light-second to the left of his center. He sees that Blue on the left side is rushing toward the center and will hit a light pulse (from the center) too early. Yet Blue still measures the speed of a left-directed pulse of light as c. How can that be? Blue clocks on that side must be set so that one second of Blue time elapses too early. Blue clocks are set ahead of his Red clocks. The farther to the left you go, the more they are set ahead.
Looking at the right side from Blue's point of view says that Blue will think, similarly, that the Red clocks are set ahead of his.

I just read this and didn't really grock the last two paragraphs, I'll read it again later, hopefully I can understand.

In your example are you using light clocks where the pulses travel vertically on the graph?

Usually it doesn't matter how the light clock is oriented I'm pretty sure so I'm not sure if my question gets to the meat and potatoes.

 

[FactChecker]

I just read this and didn't really grock the last two paragraphs, I'll read it again later, hopefully I can understand.

In your example are you using light clocks where the pulses travel vertically on the graph?

No. The light pulse is horizontal and is just to verify that the Red and Blue clocks are initially synchronized (versus their respective central clocks, R and B) so that the speed of light is c in their respective inertial reference frame.

 

[Chenkel]

No. The light pulse is horizontal and is just to verify that the Red and Blue clocks are initially synchronized (versus their respective central clocks, R and B) so that the speed of light is c in their respective inertial reference frame.

I'm a little confused, the light pulses are horizontal?

Maybe my imagination is lacking a little bit here, but I can assume you most likely know what you are talking about, I'll study it.

 

[FactChecker]

I'm a little confused, the light pulses are horizontal?

Yes. You are thinking about the vertical light used to illustrate the mathematics of the time dilation. I am talking about a (possible too casual) verification that the clocks along the way are initially synchronized so that the speed of light is c in either direction along the relative path. Einstein synchronization is a more formal way of synchronizing clocks that reflects a horizontal pulse so that it goes in a round trip.

Maybe my imagination is lacking a little bit here, but I can assume you most likely know what you are talking about, I'll study it.

Keep in mind that my goal is not to formally derive an equation for time dilation, but only to give intuition into why each inertial reference frame can measure the other's clock (R or B) to be running slow. The quick and dirty explanation is that one reference frame (say Blue) would measure the ticking clock of the other reference frame (R) by comparing it to his own blue clocks as it moves TO THE LEFT of the figure. On the other hand, the Red reference frame would measure the ticking clock B by comparing it to his own red clocks as it moves TO THE RIGHT of the figure. So we are talking about two different sets of clocks in opposite directions from the initial center clock positions.

 

[PeroK]

Looking at the left side from Red's point of view. Red can assume that he is stationary and everything is normal. Suppose Red has measured one light-second to the left of his center. He sees that Blue on the left side is rushing toward the center and will hit a light pulse (from the center) too early. Yet Blue still measures the speed of a left-directed pulse of light as c. How can that be? Blue clocks on that side must be set so that one second of Blue time elapses too early. Blue clocks are set ahead of his Red clocks. The farther to the left you go, the more they are set ahead.

This can't be right. Time dilation depends only on speed, not on direction of motion. In this case, the Blue clocks would show the same time dilation in the Red frame if they are moving away from a light source.

Time dilation is not directly derived from the time delay in the transmission of light signals, which seems to be the implication of your post.

 

[Histspec]

I've heard this is reciprocal and if we situate a clock at either reference frame we will measure the clock in the other reference frame to have a longer period than the clock at "rest."

As explained by others before, one can understand the symmetry of time dilation only in connection with relativity of simultaneity.
So let's say we have two clocks A and B at the same position, both indicating 0 at the start.
In the rest frame S1 of A we have another clock C synchronous with A.
In the rest frame S2 of B we have another clock D synchronous with B.
In order to simplify matters, we chose another frame S3 in which S1 (together with A and C) is moving with +v, while S2 (together with B and D) is moving with -v. This has the advantage that all four clocks have the same time dilation as seen in S3, and we furthermore choose v so that the Lorentz factor becomes 2 in S3.
We also observe that when A and C are synchronous in S1, they are not synchronous in S3 due to relativity of simultaneity, but C is ahead of A by 5 seconds. Likewise, when B and D are synchronous in S2 they are not synchronous in S3 due to relativity of simultaneity, but D is ahead of B by 5 seconds. This is described in the first part of the following image.
In the second part of the image we see the result when S1 and S2 continue their motion for 10 seconds (S3 time) until the clocks coincide. Since the time dilation factor is 2 and all clocks have same speed, all of the clocks will advance by 5 seconds.
We read off from the second image part: B is retarded with respect to C. However, since C is synchronous with A in S1, the observer in S1 concludes that B is retarded with respect to A.
We also read off from the second image part: A is retarded with respect to D. However, since D is synchronous with B in S2, the observer in S2 concludes that A is retarded with respect to B.

Summarizing: B is retarded with respect to A in S1, while A is retarded with respect to B in S2. Time dilation is symmetric.

 

[jartsa]

I've heard this is reciprocal and if we situate a clock at either reference frame we will measure the clock in the other reference frame to have a longer period than the clock at "rest."

I'm a little confused by the reciprocity and I wonder how the period of either clock can be smaller depending on where we set up our rest clock.

Instead of being confused about reciprocity, you should imagine a scenario where observer A is confused about reciprocity.

A says: B moves and B's clock is slow. My clock is normal. Why the heck does B think that my clock is slower than his own clock?

The task is now to explain the thoughts of B in the frame of A.

 

[Sagittarius A-Star]

If you have two light clocks moving relative to each other (A and B) (let's also say they are synchronized when they meet) where each one counts to 4 ticks and claps on each tick and covers itself in paint on the 4th tick lets also say that gamma equals 2.

Please be aware, that only every second light-pulse refection can be a "tick".

Caution, fake light clock!
Occasionally one sees animations with light clocks ticking twice as fast as the ones shown here. For them the counter counts when the pulse reaches the upper mirror and also when it reaches the lower mirror. Such light clocks lead the simple functional principle ad absurdum, because how does the counter know when the pulse arrives at the lower mirror? This information would first have to be laboriously transferred from the lower mirror to the counter. But this transmission cannot be done faster than the speed of light. In particular, the information would not reach the counter before the light pulse itself already arrives again at the upper mirror.

Source:
https://www.einstein-online.info/en/spotlight/light-clocks-time-dilation/

 

[Dale]

@Chenkel I found this good guide on spacetime diagrams

https://cs.westminstercollege.edu/~ccline/courses/phys301/spacetime_diagrams.pdf

 

[Chenkel]

@Chenkel I found this good guide on spacetime diagrams

https://cs.westminstercollege.edu/~ccline/courses/phys301/spacetime_diagrams.pdf

Thanks, I'll check it out.

 

[robphy]

A spacetime diagram is essentially a position-vs-time diagram (as opposed to a spatial diagram of moving boxcars). By convention, the time axis runs upwards instead of to the right, as in PHY 101. The usual stumbling block is how to motivate and determine the ticks along lines that are not parallel to the coordinate axes.

Here's my introduction to spacetime diagrams, motivating a graphical method using light clocks.
https://www.physicsforums.com/insights/spacetime-diagrams-light-clocks/
and more recently
https://www.physicsforums.com/insights/relativity-on-rotated-graph-paper-a-graphical-motivation/
which uses the Doppler effect and what you call "reciprocity" to develop special relativity.

My answer to https://physics.stackexchange.com/questions/383248/how-can-time-dilation-be-symmetric
uses diagrams like this to display the reciprocity

which is analogous to what happens in ordinary Euclidean geometry

(using https://www.desmos.com/calculator/wm9jmrqnw2 with E=-1 ).

Yes.

One way to see this visually is to draw a spacetime diagram. I recommend $\gamma=5/4$ instead of 2. Draw light pulses from each “dump” event, one that is emitted and one that was received at the dumping. Each frame will claim that the other frame’s dump is simultaneous with the event on their own worldline which is halfway between the emitted and received light pulses.

The essence of the first diagram is essentially what @Dale is suggesting, with \gamma=5/4 (i.e. v/c =3/5 and k=2), which is arithmetically nicer than either v/c =1/2 or \gamma=2.