Which Clock Shows Less Time When They Collide?

[AntigenX]

Simple question (derived from some unanswered posts from various posters, mostly mitesh9)...

Two identical clocks A (stationary) and B (moving towards A) with a relative velocity v, so that it will collide with A after some time. (the notion of stationary/moving is for the sake of simplicity, otherwise, its impossible to judge it as per SR)

The clocks are set to read "zero" by a flesh of light equidistant from both. That means, they are set to the origin of their time axis. their spetial co-ordinates are different but their time co-ordinates at the start (flash of light) are same, which can be used as a reference to calculate their respective accumulated time, to which both the clocks will agree.

The clocks are of special kind, and have some mechanism that is infinitely sensitive to touch. so the clocks will stop the moment they are touched by the other clock. Note that, acceleration has no part to play, because, the clocks stop before the acceleration starts.

Now the questions.

1. When they collide, the clocks will stop. Which clock has accumulated less time? Presumably B, but how, because, we can not tell which clock is moving/stationary.

2. For clocks to collide, they should have same spacetime co-ordinates. Though their space co-ordinates are same at the time of collision, their time co-ordinates will be different (due to time dilation sufferred by B), in which case, they should not collide at all? How can they collide then? And if they do, then where has gone the time dilation?

3. More questions If the discussion proceeds ...

 

[Dale]

Please draw the spacetime diagrams for your scenario. That will answer all your questions.

 

[AntigenX]

Please draw the spacetime diagrams for your scenario. That will answer all your questions.

Thank you for your suggestion, but If I draw spacetime diagram considering A stationary, it will say B is slow and vice versa. Further, the solutions must be possible even without spacetime diagrams, No?

 

[Dale]

Thank you for your suggestion, but If I draw spacetime diagram considering A stationary, it will say B is slow and vice versa.

Draw them and see. Don't to forget to include the spacetime diagrams for the flash of light that you are using to start the clocks.

 

[DocZaius]

The method of starting both clocks at the same time is problematic to me. Be very careful about the beam of light "equidistant" from each clock sending out the signal to start clocks.

 

[AntigenX]

Draw them and see. Don't to forget to include the spacetime diagrams for the flash of light that you are using to start the clocks.

Well, space time diagrams for A and B are two right angle triangles, mirror images of each other, having three vertices...

1. The origin
2. Clock Start Event (flash of light)
3. Collision Event

I don't see any point in drawing spacetime diagram for flash of light's point of view (or for that matter any point of view). The thing is, two clocks started at the same time, stopped an can be analyzed by anybody now. They are no longer dynamic things but static.

The method of starting both clocks at the same time is problematic to me. Be very careful about the beam of light "equidistant" from each clock sending out the signal to start clocks.

The method can be any arbitrary method. This is not the question at all, except, if it is not possible to make both clocks to read zero at same instant by any means. If at all it is possible by any way, we may include that as the standard way of doing this.

As to your point, the isotropy of light speed makes it possible to set both clocks zero at any arbitrary instant by an equidistant light flash.

 

[Dale]

Well, space time diagrams for A and B are two right angle triangles, mirror images of each other, having three vertices...

1. The origin
2. Clock Start Event (flash of light)
3. Collision Event

OK, here is your first problem. The spacetime diagrams are not triangles, they are quadrilaterals. There are four events of interest.

1. the flash of light
2. clock A receives the flash (and is set to 0)
3. clock B receives the flash (and is set to 0)
4. clock A and B collide (and read out their times)

I don't see any point in drawing spacetime diagram for flash of light's point of view (or for that matter any point of view). The thing is, two clocks started at the same time, stopped an can be analyzed by anybody now. They are no longer dynamic things but static.

In SR light doesn't have a "point of view", so just draw the diagrams for the rest frames of A and B, but include the light on each diagram.

You may not see the point, but that is just because you haven't done it. Try it and see. Honestly, if you are unwilling to even attempt it then there is really no point in the rest of us even bothering to respond to your posts. You cannot expect to overcome your confusion without some minimal effort on your part.

PS You will undoubtedly make some mistakes at first, but I can help you correct your diagram and the process will teach you more than anything else I can think of. I really consider it the most valuable exercise someone can do to learn SR.

 

[AntigenX]

OK, here is your first problem. The spacetime diagrams are not triangles, they are quadrilaterals. There are four events of interest.

1. the flash of light
2. clock A receives the flash (and is set to 0)
3. clock B receives the flash (and is set to 0)
4. clock A and B collide (and read out their times)

This is equivalent to my system of two spacetime diagrams drawn using flash of the light to be the start point, and putting the stationary observer at origin.

In SR light doesn't have a "point of view", so just draw the diagrams for the rest frames of A and B, but include the light on each diagram.

My sentence meant the spacetime diagram you just proposed. I do not imply any point of view of light, but the point of view of the observer stationed at the spacetime co-ordinates of light flash.

You may not see the point, but that is just because you haven't done it. Try it and see. Honestly, if you are unwilling to even attempt it then there is really no point in the rest of us even bothering to respond to your posts. You cannot expect to overcome your confusion without some minimal effort on your part.

PS You will undoubtedly make some mistakes at first, but I can help you correct your diagram and the process will teach you more than anything else I can think of. I really consider it the most valuable exercise someone can do to learn SR.

As I have recently said earlier in some other post, I have never disregarded spacetime diagrams, but It seems that the current trend in SR is "Draw spacetime diagrams else you won't learn SR". My efforts are no less than anyone else trying to learn SR. It's just that I have chosen the other way. I do not want to let spacetime diagrams dictate my thinking at this stage.

As to your PS, I don't have any doubt that spacetime diagrams are helpful, but it's not the only way, No? I have many reasons for not drawing spacetime diagrams, the chief being, it's perception and interpretations are highly personal and can create more conflicts, driving us far from the original question (as is the case now). Further, I do not doubt the intentions of others who may wish to help me, and as you asked me, I figured out the spacetime diagram from my point of view (I hope you are not telling me to really draw it! though it won't make any difference).

Apart from these, Is the question so non-trivial that can not be solved without spacetime diagram? We are deviating from the original questions.

 

[Dale]

As to your PS, I don't have any doubt that spacetime diagrams are helpful, but it's not the only way, No?

Of course they are not the only way. However, you obviously are not learning using your way, so why not try my way?

When I was learning SR I was very much in your situation. I had also failed to grasp it using the other ways, and I struggled with the basics for years. It wasn't until I sat down and went through the personal effort to draw a few spacetime diagrams that SR finally clicked. I really wish that someone had done for me what I am trying to do for you.

Apart from these, Is the question so non-trivial that can not be solved without spacetime diagram?

No, it is so trivial that the fact that you even asked it indicates that you need to spend some effort learning the basics. I am deliberately not answering your question, instead I am trying to help you learn how to answer it yourself.

 

[AntigenX]

Of course they are not the only way. However, you obviously are not learning using your way, so why not try my way?

Well, then see the attached gif.

 

[Janus]

Well, then see the attached gif.

You haven't shown the light in these diagrams, as asked.

 

[AntigenX]

You haven't shown the light in these diagrams, as asked.

May be I am too bad at drawing diagrams, however, the midway point between A and B, which is labeled "Flash" is light flash.
The time for light signals to travel to the clocks is ignored (which does not make any difference to the situation I suppose), because the clocks starts after the signal reaching them, so anything before that is irrelevant.

 

[yuiop]

... so anything before that is irrelevant.

It does make a difference. If the flash occurs at a point a point that is truly midway between A and B (in A's frame) then A and B will not be equidistant from the flash when the signal arrives at A and B. If you would like to draw a more accurate version of your initial diagram that includes the light signals just draw the light signals as lines that are 45 degrees from the vertical t axis while all physicals objects travel on lines that are less than 45 degrees from the t axis. This is true in any frame.

For every event with coordinates (x,t) in A's frame there should be a corresponding event (x',t') in B's frame where x' and t' are defined by the Lorentz Transformation

... so anything before that is irrelevant.

What happens before is relevant. You are seem to be assuming your flah will reach A and B simultaneously in A's frame but B is moving away from the flash so B's clock is "zeroed" later than A's clock in A's frame.

 

[AntigenX]

It does make a difference. If the flash occurs at a point a point that is truly midway between A and B (in A's frame) then A and B will not be equidistant from the flash when the signal arrives at A and B. If you would like to draw a more accurate version of your initial diagram that includes the light signals just draw the light signals as lines that are 45 degrees from the vertical t axis while all physicals objects travel on lines that are less than 45 degrees from the t axis. This is true in any frame.

For every event with coordinates (x,t) in A's frame there should be a corresponding event (x',t') in B's frame where x' and t' are defined by the Lorentz Transformation

Two questions,

1. Is my diagram wrong?
2. If yes, can you please correct it (though I shouldn't be asking this)?

Again, I think we are deviating from the original questions. The method of starting the clocks is not important, but what happens after the clocks are matched (not synchronized)!

 

[Janus]

May be I am too bad at drawing diagrams, however, the midway point between A and B, which is labeled "Flash" is light flash.
The time for light signals to travel to the clocks is ignored (which does not make any difference to the situation I suppose), because the clocks starts after the signal reaching them, so anything before that is irrelevant.

And herein is your mistake. Because the time it takes for the light signals to travel cannot be ignored because It doesn't take an equal amount of time for the light to travel from emission point to A and B, according to A or B. Each determines that the flash reached the other before them.

Heres an modified attachment of your diagram showing the travel of the light flash(in blue). Light paths in space time diagrams follow 45° angles. Maybe it will clear things up for you.

 

[Mentz114]

Space time diagrams

Here is my attempt to draw the space time diagram for the observer midway between A and B. Where did I go wrong ?

 

[AntigenX]

Thank you kev for your suggestion and Janus for pointing out my mistake and revising my diagram...

All said and done, the original question is still unanswered. Which will be the slower clock? We are stressing upon the method of setting both clocks at zero, which may be any arbitrary method. But what after that? As is now clear from revised diagrams, the equidistant light flash can't work. Let's say then that there was some other non-equidistant flash that set them zero. Which clock will be slower then? Let the flesh be from such a distance so that the light travels same distance to reach clocks A and B. It's settled now I suppose.

Or is it impossible to set both clocks read zero? It doesn't seem impossible!

And even in the revised diagrams, from both point of views, the other clock is slower. How to decide which clock is slower? Or should we say the clock which is "really" moving is slower? But we have no way to decide which clock is moving!

Further, If at all any clock is slower, the collision should not happen, because, due to different time co-ordinates, clocks A and B will reach the collision point at different instants.

 

[Mentz114]

AntigenX,

Which will be the slower clock?

I depends who is asking. If A and B merely pass very close and observe one and others clocks, each will see the other clock slower. If they pass an observer located at the midpoint between them, she will see their clocks as the same. They can all disagree happily and there is no problem.

However if A and B are brought to rest at the midpoint, then they will compare clocks and find the same elapsed time provided they used identical procedures for stopping.

Further, If at all any clock is slower, the collision should not happen, because, due to different time co-ordinates, clocks A and B will reach the collision point at different instants.

Think again. If two objects travel inertially (no forces) on a collision course, they must collide, regardless of what local clocks do.

Regarding my earlier post, I see now that it correct, because the midpoint observer sees the same time on both clocks as they pass.

M

 

[kahoomann]

Thank you kev for your suggestion and Janus for pointing out my mistake and revising my diagram...

All said and done, the original question is still unanswered. Which will be the slower clock?
Or is it impossible to set both clocks read zero? It doesn't seem impossible!

Further, If at all any clock is slower, the collision should not happen, because, due to different time co-ordinates, clocks A and B will reach the collision point at different instants.

Do you realize how many clocks you need to compare time with a moving clock?
That's the key question

 

[AntigenX]

Hi Mentz114,

AntigenX,
Which will be the slower clock?

I depends who is asking. If A and B merely pass very close and observe one and others clocks, each will see the other clock slower. If they pass an observer located at the midpoint between them, she will see their clocks as the same. They can all disagree happily and there is no problem.

I understood that.

However if A and B are brought to rest at the midpoint, then they will compare clocks and find the same elapsed time provided they used identical procedures for stopping.

M

Now see, where is the time dilation then?
Because I was told in other thread that time dilation is real and not apparent. In our scenario, at least one of the two clocks is moving, and thus should slow down and accumulate less time. If both clocks are brought together after stopping them in previously described method, they are bound to differ in their accumulated time. But which clock that be?


Edit:

Think again. If two objects travel inertially (no forces) on a collision course, they must collide, regardless of what local clocks do.

Regarding my earlier post, I see now that it correct, because the midpoint observer sees the same time on both clocks as they pass.

M

You are right of course, but not answering my questions. The collision is bound to happen, but can not happen without all four spacetime co-ordinates being same. The clocks can not collide if they are at same place at different times (viz. 2:00 and 4:00 O'clocks). And who's spacetime co-ordinates should be same? For both clocks spacetime co-ordinates should be same (It indicates their local time and not absolute time). However, their local times once matched, can not be same, as at least one of them is moving.

 

[AntigenX]

Do you realize how many clocks you need to compare time with a moving clock?
That's the key question

No, I don't. Please throw some light on this, and explain how is it related to our scenario.

Yet, I would like to point out that we are comparing the time of stopped clocks after the collision, when they are neither running nor moving. While comparing the time, they are in rest frame, but they were stopped before changing the frame(s).

Also, the method of setting the clocks to zero may be arbitrary and is irrelevant for our discussion.

 

[matheinste]

Hello AntigenX.

Just a quick thought withought any real analysis I suspect the problem may lie in the fact that given an event ( flash ) only one of the relatively moving observers can be at rest relative to it and depending on which one it is will deecide who's clock appears to run slower. So either can be said to run slower depending on which frame you choose to be stationary relative to the event ( flash ).

Matheinste.

 

[Janus]

Thank you kev for your suggestion and Janus for pointing out my mistake and revising my diagram...

All said and done, the original question is still unanswered. Which will be the slower clock? We are stressing upon the method of setting both clocks at zero, which may be any arbitrary method. But what after that? As is now clear from revised diagrams, the equidistant light flash can't work. Let's say then that there was some other non-equidistant flash that set them zero. Which clock will be slower then? Let the flesh be from such a distance so that the light travels same distance to reach clocks A and B. It's settled now I suppose.

Or is it impossible to set both clocks read zero? It doesn't seem impossible!

It is quite possible to simultaneously set both clocks to zero in any given frame. Mentz114 shows in his diagram the clocks simultaneously set to zero in the frame of the origin of the light flash. However, it is impossible to set both clocks simultaneously to zero in all frames. If the clocks are set to zero in according to A, they will not be according to B or the origin of the flash. The same goes for setting them to zero according to B.

And even in the revised diagrams, from both point of views, the other clock is slower. How to decide which clock is slower? Or should we say the clock which is "really" moving is slower? But we have no way to decide which clock is moving!

Time is relative. Which clock runs slows depends on which frame you are in. There is no absolute method of saying which clock "really" runs slower in the way you mean. It is a meaningless question
Let's use an analogy: You have two people facing each other in a room with a chair against one wall. One of them says the chair is against the right wall and the other says it is against the left wall. Who is "really" correct? Time dilation is like this, you can only says which clock runs slow according to which frame, not which clock really slows down.

Further, If at all any clock is slower, the collision should not happen, because, due to different time co-ordinates, clocks A and B will reach the collision point at different instants.

When the clocks meet they will read the same time according to all frames. They collision point at the same instant in all frames, and thus collide in all frames. What will differ form frame to frame is how fast each clock ran and when it started. In a frame where one clock ran slower, it will have also started sooner.

 

[Mentz114]

Hi Mentz114,

Now see, where is the time dilation then?
Because I was told in other thread that time dilation is real and not apparent. In our scenario, at least one of the two clocks is moving, and thus should slow down and accumulate less time. If both clocks are brought together after stopping them in previously described method, they are bound to differ in their accumulated time. But which clock that be?

A and B's clocks would be slower compared with one left at the midpoint and synchronised with the light flash.

Don't confuse the two stuations

1. observing a moving clock ( it runs slower) depends on relative velocity.
2. round trip setup, where two clocks are together, travel separately and then meet. Depends on trip proper length, and is real ( verified by experiment).

I have to go out, so no more until tomorrow.
M

 

[AntigenX]

Hello AntigenX.

Just a quick thought withought any real analysis I suspect the problem may lie in the fact that given an event ( flash ) only one of the relatively moving observers can be at rest relative to it and depending on which one it is will deecide who's clock appears to run slower. So either can be said to run slower depending on which frame you choose to be stationary relative to the event ( flash ).

Matheinste.

Well, I think light flash from a source at rest or moving state does not make any difference. And we are analyzing the clocks after stopping them. So what we are really interested in is accumulated time of both clocks and their comparison.

It is quite possible to simultaneously set both clocks to zero in any given frame. Mentz114 shows in his diagram the clocks simultaneously set to zero in the frame of the origin of the light flash. However, it is impossible to set both clocks simultaneously to zero in all frames. If the clocks are set to zero in according to A, they will not be according to B or the origin of the flash. The same goes for setting them to zero according to B.Time is relative. Which clock runs slows depends on which frame you are in. There is no absolute method of saying which clock "really" runs slower in the way you mean. It is a meaningless question

Again, what we are really interested in is accumulated time of both (stopped) clocks and their comparison, not their time when they are moving or running.

Let's use an analogy: You have two people facing each other in a room with a chair against one wall. One of them says the chair is against the right wall and the other says it is against the left wall. Who is "really" correct? Time dilation is like this, you can only says which clock runs slow according to which frame, not which clock really slows down.

I think I'm saying the same thing.
Edit: Sorry, I missed the last line. What is the meaning of saying that time dilation is real then?

When the clocks meet they will read the same time according to all frames. They collision point at the same instant in all frames, and thus collide in all frames. What will differ form frame to frame is how fast each clock ran and when it started. In a frame where one clock ran slower, it will have also started sooner.

How is it possible for both clocks to accumulate same time, when at least one of them was moving before it stopped?

A and B's clocks would be slower compared with one left at the midpoint and synchronised with the light flash.

Don't confuse the two stuations

1. observing a moving clock ( it runs slower) depends on relative velocity.
2. round trip setup, where two clocks are together, travel separately and then meet. Depends on trip proper length, and is real ( verified by experiment).

I have to go out, so no more until tomorrow.
M

Once again, what we are really interested in is accumulated time of both (stopped) clocks and their comparison, not their time when they are moving or running. So no situation "1. observing a moving clock ( it runs slower) depends on relative velocity". As is clear from the first post, the situation is close to point 2, but no round trip, and no acceleration. Further, I know "...is real ( verified by experiment)", and wish to figure out how?

 

[matheinste]

Hello AntigenX.

Having just read the post from Janus i think the question has been answered. Read it carefully.

Mstheinste

 

[Janus]

Again, what we are really interested in is accumulated time of both (stopped) clocks and their comparison, not their time when they are moving or running.

How is it possible for both clocks to accumulate same time, when at least one of them was moving before it stopped?

To determine how much time each clock accumulates, you have to take into account:
1) When did it start running?
2) At what rate did it tick while running.
3) When did it stop running?

Since we are comparing two clocks, we can figure that above as seen by each clock.

Thus for A and B considered from their own reference they start at a certain time reading 0, run at a certain rate and stop at a certain time reading time "t"

But what of B as seen from A, and A as seen from B.

A sees B start at some time before A itself starts(IOW, by the time A itself starts ticking clock B has already built up a "head start"), then tick at a slower rate than A, and then stop at the same time as A (when they collide), at which time it will read the same (t) as Clock A(by ticking faster, A closes the gap of B's head start). So even though clock B ticked slower than A while running, it started ticking before A, and thus accumulates the same time as A between starting and stopping.

B sees the same thing with A and B switched.

 

[Dale]

Well, then see the attached gif.

Hi Antigen, Sorry about the delay, had a picnic to attend . I see that the conversation has run away a little. I will try to rewind a bit and go step-by-step.

Your first spacetime diagram was pretty close. I corrected it so that the speed of light from the flash was not infinite! The other thing that I did was to add the ct' and x' axes representing coordinates in the other frame.

Before we go on to the diagram in the other frame, do you understand my additions/changes to this diagram? Do you disagree with any of the changes?

 

[AntigenX]

Hello AntigenX.

Having just read the post from Janus i think the question has been answered. Read it carefully.

Mstheinste

I had read and reread all the posts (Janus's posts more carefully), but can't see how is my question answered. Can you point out precisely?

Again situation restated, from the frame of the origin of light flash both clocks started at zero time "simultaneously". The observer in this frame obviously knows that A is stationary and B is moving wrt him. Now, after collision, he brings the stopped clocks and compares times. According to Janus's post and Mentz114's spacetime diagram, he finds that both stopped clocks have accumulated same time.

Now, as the observer in light-flash frame knows that B was moving while A was stationary, thus B should have accumulated less time, for the same reason we say time dilation is real and not apparent. clocks actually slow when moving which is experimentally tested as has been pointed out earlier. Then how come we say that the light-flash frame observer (or any other observer because the clocks had been stopped earlier) would see the same accumulated time? where is the time dilation?

 

[AntigenX]

Greetings!

Hi Antigen, Sorry about the delay, had a picnic to attend . I see that the conversation has run away a little. I will try to rewind a bit and go step-by-step.

Your first spacetime diagram was pretty close. I corrected it so that the speed of light from the flash was not infinite! The other thing that I did was to add the ct' and x' axes representing coordinates in the other frame.

Before we go on to the diagram in the other frame, do you understand my additions/changes to this diagram? Do you disagree with any of the changes?

I Do agree.

 

[matheinste]

Hello all.

Question for Janus.

---When the clocks meet they will read the same time according to all frames. ----

Just to clear up a semantic point does this mean that what an observer in any/all frames will see the clocks reading the same times, not necessarily the same as each other. That is they can show different times but all will agree what these times are.

Thanks Matheinste.

 

[Dale]

I Do agree.

OK, now, note carefully that the worldline of B from B0 to AB is about two ct' units while the worldline of A form A0 to AB is about 5 ct units. Note also that the x-axis is parallel to the line from A0 to B0 (they are simultaneous in A's frame) while the x' axis is not parallel to that line (they are not simultaneous in B's frame). Finally, note that the worldline from Flash to B0 is about 3 (ct+x) or 20 (ct'+x') and the worldline from Flash to A0 is about 3 (ct-x) or 0.5 (ct'-x')

Does that give you enough information to draw the other graph?

 

[Janus]

I had read and reread all the posts (Janus's posts more carefully), but can't see how is my question answered. Can you point out precisely?

Again situation restated, from the frame of the origin of light flash both clocks started at zero time "simultaneously". The observer in this frame obviously knows that A is stationary and B is moving wrt him. Now, after collision, he brings the stopped clocks and compares times. According to Janus's post and Mentz114's spacetime diagram, he finds that both stopped clocks have accumulated same time.

Now, as the observer in light-flash frame knows that B was moving while A was stationary, thus B should have accumulated less time, for the same reason we say time dilation is real and not apparent. clocks actually slow when moving which is experimentally tested as has been pointed out earlier. Then how come we say that the light-flash frame observer (or any other observer because the clocks had been stopped earlier) would see the same accumulated time? where is the time dilation?

I think I see a problem here when referring to the frame of the origin of the light flash. In Mentz114's diagrams in my posts, it is treated as a frame in which the origin of the flash remains equidistant from both A and B at all times. In this frame, neither A or B is stationary but both have a equal velocity towards the origin of the flash. OTOH, you seem to be treating it as a frame in which A is stationary. I didn't catch this earlier because this is just the same as A's frame, when it comes to dealing with the problem and so I wouldn't treat it as a separate frame of its own.

So when you treat this like we did, then A and B start together, tick at the same rate and stop together, accumulating the same time.

When you treat it like you did, then things behave just as they would in A's frame where B starts ticking before A.

 

[Dale]

Hi Janus,

Yes, I noticed that also after I posted my diagram and then looked back at yours and Mentz's in detail. You and I interpereted the original scenario significantly differently from each other. The original phrasing is ambiguous since the word "equidistant" is frame-variant and no reference frame is specified. It is also ambiguous if the distance is equal at the time of emission or reception of the flash (again, frame variant).

IMO, this kind of ambiguity is yet another reason to use diagrams rather than words. A diagram can be wrong, but it cannot be ambiguous.

 

[Janus]

Hello all.

Question for Janus.

---When the clocks meet they will read the same time according to all frames. ----

Just to clear up a semantic point does this mean that what an observer in any/all frames will see the clocks reading the same times, not necessarily the same as each other. That is they can show different times but all will agree what these times are.

Thanks Matheinste.

In this particular situation as posed by the OP, the clocks real read identical times when they meet. In a different situation they could read different times (such as in the standard twin paradox), but all observers in all frame would agree as to what each clock read.