I Question about the reciprocity of time dilation

[Chenkel]

Hello everyone,

After studying time dilation I believe we pick a rest frame as a place we situate a clock and because of time dilation when we have a "moving" clock it will have a longer period when compared to the rest clock.

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.

If we have two clocks built exactly the same when each one counts to 10 does each clock observe the clock in the other reference frame as counting to 5 when gamma is 2 due to relative velocity?

Any insights will be appreciated! Thank you.

 

[PeterDonis]

After studying time dilation

Studying it how?

In your previous thread on time dilation [1], you were given quite a bit of feedback and were recommended a book that you said you would study. Have you done that?

In your latest relativity thread that just got closed, it seemed like your primary learning source was LLMs, which, as you were told in that thread, are not good sources for this. I'm not sure whether you fully grasped that.

[1] https://www.physicsforums.com/threa...ation-for-near-speed-of-light-travel.1049121/

 

[PeterDonis]

I've heard this is reciprocal

"I've heard" is not very encouraging. Studying an actual relativity textbook (in addition to the Morin one you were recommended in the previous thread, Taylor & Wheeler's Spacetime Physics is good) would give you a much better basis than "I've heard".

Taking the statement as you make it, yes, it is true that time dilation due to relative motion is reciprocal. However, to fully understand why, you have to also take into account length contraction and (most important) relativity of simultaneity. It is not possible to correctly understand any one of these phenomena without including the other two.

In fact, all three phenomena are effects of spacetime geometry, which is why modern relativity texts generally emphasize spacetime geometry, which is invariant, and only derive things like time dilation, length contraction, and relativity simultaneity afterwards.

 

[Chenkel]

"I've heard" is not very encouraging. Studying an actual relativity textbook (in addition to the Morin one you were recommended in the previous thread, Taylor & Wheeler's Spacetime Physics is good) would give you a much better basis than "I've heard".

Taking the statement as you make it, yes, it is true that time dilation due to relative motion is reciprocal. However, to fully understand why, you have to also take into account length contraction and (most important) relativity of simultaneity. It is not possible to correctly understand any one of these phenomena without including the other two.

In fact, all three phenomena are effects of spacetime geometry, which is why modern relativity texts generally emphasize spacetime geometry, which is invariant, and only derive things like time dilation, length contraction, and relativity simultaneity afterwards.

I've been reading Morin's book and it is good.

 

[PeterDonis]

If we have two clocks built exactly the same when each one counts to 10 does each clock observe the clock in the other reference frame as counting to 5 when gamma is 2 due to relative velocity?

"Observe" is a poorly chosen word in this context, although it is unfortunately widely used.

What each clock literally observes, as in "sees with eyes or a telescope or some similar device", will depend on the direction of relative motion as well as the relative speed. What is normally called "time dilation" has to be calculated from these direct observations by correcting for the travel time of light signals.

For example, for $\gamma = 2$, the relativistic Doppler factor is $2 + \sqrt{3}$. So, if the two clocks are moving apart, each will directly observe the other to be running slow by a factor $2 + \sqrt{3}$. To obtain the time dilation factor of $2$, each clock has to correct for the light travel time of the light signals arriving from the other clock--as the other clock gets farther away, it takes longer for the light signals from it to arrive, and some of the directly observed factor of $2 + \sqrt{3}$ is due to that; the time dilation factor of $2$ is what is left over after that is corrected for.

 

[PeterDonis]

I've been reading Morin's book and it is good.

Ok, so can you do any better than "I've heard" for the statements you make in your OP? Can you, for example, cite particular passages from the Morin textbook that are relevant, and tell us what you are having a problem understanding in those passages?

 

[Chenkel]

"Observe" is a poorly chosen word in this context, although it is unfortunately widely used.

What each clock literally observes, as in "sees with eyes or a telescope or some similar device", will depend on the direction of relative motion as well as the relative speed. What is normally called "time dilation" has to be calculated from these direct observations by correcting for the travel time of light signals.

For example, for $\gamma = 2$, the relativistic Doppler factor is $2 + \sqrt{3}$. So, if the two clocks are moving apart, each will directly observe the other to be running slow by a factor $2 + \sqrt{3}$. To obtain the time dilation factor of $2$, each clock has to correct for the light travel time of the light signals arriving from the other clock--as the other clock gets farther away, it takes longer for the light signals from it to arrive, and some of the directly observed factor of $2 + \sqrt{3}$ is due to that; the time dilation factor of $2$ is what is left over after that is corrected for.

For some reason some of your latex isn't completing.

 

[Chenkel]

"Observe" is a poorly chosen word in this context, although it is unfortunately widely used.

What each clock literally observes, as in "sees with eyes or a telescope or some similar device", will depend on the direction of relative motion as well as the relative speed. What is normally called "time dilation" has to be calculated from these direct observations by correcting for the travel time of light signals.

For example, for $\gamma = 2$, the relativistic Doppler factor is $2 + \sqrt{3}$. So, if the two clocks are moving apart, each will directly observe the other to be running slow by a factor $2 + \sqrt{3}$. To obtain the time dilation factor of $2$, each clock has to correct for the light travel time of the light signals arriving from the other clock--as the other clock gets farther away, it takes longer for the light signals from it to arrive, and some of the directly observed factor of $2 + \sqrt{3}$ is due to that; the time dilation factor of $2$ is what is left over after that is corrected for.

I'm not well versed in doppler effects, is doppler referring to a frequency change of a light ray due to relative motion?

 

[Ibix]

I'm not well versed in doppler effects, is doppler referring to a frequency change of a light ray due to relative motion?

Frequency change of anything due to motion. If you have a clock moving away from you you will see a longer than one second gap between the tocks for two reasons - one is time dilation, two is that there is a longer light speed delay for light from each subsequent tick to reach you. The combination of these two effects is the relativistic Doppler effect. To "see" time dilation you need to observe the relativistic Doppler effect (capital D, it's a man's name) and subtract out the changing light speed delay.

 

[PeterDonis]

For some reason some of your latex isn't completing.

Try refreshing the page. Sometimes the Javascript that does the images needs to be prodded.

 

[PeterDonis]

I'm not well versed in doppler effects

Is the relativistic Doppler effect covered in Morin's textbook?

 

[Chenkel]

Is the relativistic Doppler effect covered in Morin's textbook?

I'll check it out. I have his book but I misplaced it so I'm reading an online copy, hopefully it includes the relevant chapters.

 

[Dale]

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

It has nothing to do with the location.

Time dilation is the rate of change of coordinate time with respect to proper time. So suppose you have a clock A at rest in an inertial frame A. The proper time on the clock is $\tau_A$ and the coordinate time is $t_A$. Similarly with a clock B. So the time dilation is $$\frac{dt_B}{d\tau_A}=\frac{dt_A}{d\tau_B}=\frac{1}{\sqrt{1-v^2/c^2}}$$

The proper time is always slower than the coordinate time. It is not a comparison between two clocks. It is a comparison between one clock and a coordinate system.

 

[FactChecker]

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.

If we have two clocks built exactly the same when each one counts to 10 does each clock observe the clock in the other reference frame as counting to 5 when gamma is 2 due to relative velocity?

When a "stationary" IRF observes a relatively moving clock as it moves forward, it is not using the same "stationary" clock all the time. It is using the synchronized "stationary" clocks down the line of motion of the moving clock. If you switch the IRF that you call "stationary" then the "stationary" clocks that are used are in the opposite direction. So completely different sets of clocks are being used. That is how each will think that the other's clock runs slow.

 

[Chenkel]

When a "stationary" IRF observes a relatively moving clock as it moves forward, it is not using the same "stationary" clock all the time. It is using the synchronized "stationary" clocks down the line of motion of the moving clock. If you switch the IRF that you call "stationary" then the "stationary" clocks that are used are in the opposite direction. So completely different sets of clocks are being used. That is how each will think that the other's clock runs slow.

Does IRF stand for inertial reference frame?

 

[Nugatory]

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.

You are confused because you have not allowed for the relativity of simultaneity.

Say both clocks are set to 12:00 noon at the moment that they pass each other. One hour later A looks at their clock and sees that it reads 1:00 PM, as we expect. A uses the Lorentz transformation to calculate what B's clock reads AT THE SAME TIME that their clock reads 1:00, or equivalently they are watching B's clock through a telescope and by allowing for light travel time they can see what B's clock read when light left B AT THE SAME TIME that A's clock reads 1:00 PM. Either way, they find that B's clock reads 12:30 AT THE SAME TIME that A's clocks reads 1:00. In other words, the events "A's clock reads 1:00" and "B's clock reads 12:30" are simultaneous when we use the frame in which A is at rest.
Clearly B's clock is running slow, by a factor of two.

But this is where the relativity of simultaneity comes in. Using the frame in which B is at rest, these two events do not happen at the same time. Instead, the event "A's clock reads 12:15" is simultaneous with the event "B's clock reads 12:15 12:30" so just as clearly A's clock is running slow by a factor of two.

As an aside, this setup is different from the twin paradox because the two clocks are not at the same place when we read them. That's even part of why we bother with the twin paradox as a teaching tool.

 

[Nugatory]

Does IRF stand for inertial reference frame?

Yes

 

[Chenkel]

You are confused because you have not allowed for the relativity of simultaneity.

Say both clocks are set to 12:00 noon at the moment that they pass each other. One hour later A looks at their clock and sees that it reads 1:00 PM, as we expect. A uses the Lorentz transformation to calculate what B's clock reads AT THE SAME TIME that their clock reads 1:00, or equivalently they are watching B's clock through a telescope and by allowing for light travel time they can see what B's clock read when light left B AT THE SAME TIME that A's clock reads 1:00 PM. Either way, they find that B's clock reads 12:30 AT THE SAME TIME that A's clocks reads 1:00. In other words, the events "A's clock reads 1:00" and "B's clock reads 12:30" are simultaneous when we use the frame in which A is at rest.
Clearly B's clock is running slow, by a factor of two.

But this is where the relativity of simultaneity comes in. Using the frame in which B is at rest, these two events do not happen at the same time. Instead, the event "A's clock reads 12:15" is simultaneous with the event "B's clock reads 12:15" so just as clearly A's clock is running slow by a factor of two.

As an aside, this setup is different from the twin paradox because the two clocks are not at the same place when we read them. That's even part of why we bother with the twin paradox as a teaching tool.

The A clock reads 12:15 at the same time B's clock reads 12:15?

 

[Dale]

The A clock reads 12:15 at the same time B's clock reads 12:15?

In some reference frame, yes. Whenever you say “at the same time” you need to specify the frame.

 

[Nugatory]

The A clock reads 12:15 at the same time B's clock reads 12:15?

Sorry, that was a typo. In the frame in which B is at rest, the "B reads 12:30" and the "A reads 12:15" events are simultaneous.

 

[Chenkel]

If we have two clocks built exactly the same when each one counts to 10 does each clock observe the clock in the other reference frame as counting to 5 when gamma is 2 due to relative velocity?

Let's label the two clocks A and B.

This must mean the event that one clock counts to 10 and one clock counts to 5 are simultaneous depending on which reference frame we choose as rest.

If we choose A as the rest frame then when clock A counts to 10 clock B counts to 5. (two simultaneous events from A's perspective)

If we choose B as the rest frame then when clock B counts to 10 clock A counts to 5. (Two simultaneous events from Bs perspective)

Just playing with this idea and trying to get my language correct, apologies if I'm being pedantic.

 

[Sagittarius A-Star]

This must mean the event that one clock counts to 10 and one clock counts to 5 are simultaneous depending on which reference frame we choose as rest.

That is correct if both clocks displayed 0 when the they met.

Assume clock A is located at x=0 and clock B is located at x'=0.

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

Reference frame A, time dilation of clock B:
$t = 2*(5+0)=10$

Reference frame B, time dilation of clock A:
$t'= 2*(5-0)=10$

 

[Nugatory]

Just playing with this idea and trying to get my language correct, apologies if I'm being pedantic.

That's pretty much it, although thinking in terms of "the rest frame" is a bad habit - it still tempts us into thinking of rest frames as something special. Better to think in terms of "using the frame in which A is at rest", "using the frame in which B is at rest", reminds us that choosing a frame is like choosing the direction we're facing before we use words like "left" and "right".

 

[Dale]

This must mean the event that one clock counts to 10 and one clock counts to 5 are simultaneous depending on which reference frame we choose as rest.

Not necessarily. Remember that time dilation is a comparison between $t$ and $\tau$. For example comparing $\tau_A$ vs $t_B$. However when one clock reads 10 is $\tau_A=10$ and when the other clock reads 5 is $\tau_B=5$. So the mere fact that you know that $d\tau_A/dt_B=d\tau_B/dt_A=1/\gamma$ doesn't really tell you anything directly about the relationship between the events at $\tau_A=10$ and at $\tau_B=5$.

It is possible that there is a reference frame where those events are simultaneous. It is also possible that there is no frame where they are simultaneous. @Sagittarius A-Star above outlined one example where there is such a frame.

 

[Chenkel]

That is correct if both clocks displayed 0 when the they met.

Assume clock A is located at x=0 and clock B is located at x'=0.

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

Reference frame A, time dilation of clock B:
$t = 2*(5+0)=10$

Reference frame B, time dilation of clock A:
$t'= 2*(5-0)=10$

Where in Morin's book are those formulations of Lorentz transforms so I can understand them better?

 

[Sagittarius A-Star]

Where in Morin's book are those formulations of Lorentz transforms so I can understand them better?

They are in chapter 2.1. Unfortunately, only chapter 1 is freely available online.

You can find online a discussion (without a derivation) of the Lorentz transformation in a shortened version of Wolfgang Rindler's SR book, see equations (4) and (6) to (8):
http://www.scholarpedia.org/article...nematics#Galilean_and_Lorentz_transformations

 

[Chenkel]

They are in chapter 2.1. Unfortunately, only chapter 1 is freely available online.

You can find online a discussion (without a derivation) of the Lorentz transformation in a shortened version of Wolfgang Rindler's SR book, see equations (4) and (6) to (8):
http://www.scholarpedia.org/article...nematics#Galilean_and_Lorentz_transformations

Thank you.

 

[Herman Trivilino]

If we have two clocks built exactly the same

You need more than two clocks moving relative to each other. Let's call them A an B. Say you have a third clock, C, at rest relative to B. They are synchronized in their rest frame and lie along the line of relative motion. When A passes B we note the readings on each clock. Then later, when A passes C we again note the readings on these clocks. By examining the readings we conclude that A is running slow compared to B and C.

But in the rest frame of A the clocks B and C are not synchronized.

Now introduce a fourth clock, D, at rest relative to A and repeat the above procedure as B moves past them. Now we conclude, by the same reasoning, that B is running slow compared to A and D.

Any good textbook will work out the details and show that it's the relativity of simultaneity that accounts for the symmetry of time dilation.

I thought I already asked you to work this out in another thread and instead of responding with an attempt you simply hit the like button.

I'm not well versed in doppler effects, is doppler referring to a frequency change of a light ray due to relative motion?

Go to YouTube and do a search for Paul Hewitt Trip. Watch that old cartoon.

Edit: Paul Hewitt Twin Trip.

 

[Chenkel]

You need more than two clocks moving relative to each other. Let's call them A an B. Say you have a third clock, C, at rest relative to B. They are synchronized in their rest frame and lie along the line of relative motion. When A passes B we note the readings on each clock. Then later, when A passes C we again note the readings on these clocks. By examining the readings we conclude that A is running slow compared to B and C.

But in the rest frame of A the clocks B and C are not synchronized.

Now introduce a fourth clock, D, at rest relative to A and repeat the above procedure as B moves past them. Now we conclude, by the same reasoning, that B is running slow compared to A and D.

Any good textbook will work out the details and show that it's the relativity of simultaneity that accounts for the symmetry of time dilation.

I thought I already asked you to work this out in another thread and instead of responding with an attempt you simply hit the like button.

Go to YouTube and do a search for Paul Hewitt Trip. Watch that old cartoon.

Edit: Paul Hewitt Twin Trip.

Thanks for the Paul Hewitt reference, I just watched one of his videos.

The exercise you gave me was a little unclear because I'm inexperienced with some of the concepts but I am trying to learn. I'll make an effort to work out the thought experiment again a little later.

 

[Chenkel]

You need more than two clocks moving relative to each other. Let's call them A an B. Say you have a third clock, C, at rest relative to B. They are synchronized in their rest frame and lie along the line of relative motion. When A passes B we note the readings on each clock. Then later, when A passes C we again note the readings on these clocks. By examining the readings we conclude that A is running slow compared to B and C.

But in the rest frame of A the clocks B and C are not synchronized.

Now introduce a fourth clock, D, at rest relative to A and repeat the above procedure as B moves past them. Now we conclude, by the same reasoning, that B is running slow compared to A and D.

Any good textbook will work out the details and show that it's the relativity of simultaneity that accounts for the symmetry of time dilation.

I thought I already asked you to work this out in another thread and instead of responding with an attempt you simply hit the like button.

Go to YouTube and do a search for Paul Hewitt Trip. Watch that old cartoon.

Edit: Paul Hewitt Twin Trip.

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?

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

I wish I still knew where my Morin book was, reading it might help me understand the symmetry 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.