# Dingle's Dilemma again

mitesh9
Dingle's Dilemma again!

Dear all,
Following is a problem, I suppose, is related to some controversy called "Dingle's Dilemma", as I read somewhere on some relativity forum, and I wish to understand where is the catch.

Consider two inertial frames A and B, stationary with respect to each other, with two identical and synchronized clocks (one in each). Now, another inertial frame C (say a rocket) passes by A with a uniform velocity (say v). When the co-ordinates of A and C coincide while the process of passing by, C synchronizes its clock with A (and as A and B are still synchronized).

Now, by the time C reaches B (assuming that C's motion is in the line connecting A and B), for an observer in A, C's clock has been slowed down by some time (time dilation, say 1 sec), while, as there is no mechanism to establish wether A is in absolute motion or C, for an observer in C, the clock in A has been slowed down, and as clocks in A and B are already synchronized B has been slowed down as well.

What will happen, when the C reaches B? Will their clocks be still synchronized (I don't think so)! And if not, which one will be slower, B's clock or C's clock? (At the moment when B and C coincides, If the need arise, there should be some kind of accident, where, C doesn't move at all, abruptly stops, and merges with B.)

In case you wish to respond to this one, please note that, I'm not against SR (or Einstein for that matter, nor am I gifted enough to ever expect this), instead, being a chemist, the best fit for my role in PF can be as a hobbyist relativist.

paw
Since A and B are not moving wrt to each other they can be considered to be the same frame. It seems to me you can eliminate the second frame (A) and ask if the clocks at C and B were synchronised at some time t0 what is the state at some other time t1. This gives the standard answer C sees B's clock ticking slow while B sees C's clock ticking slow.

mitesh9
Do we do such generalizations? I mean, though A and B are stationary, their clocks are at different places (co-ordinates may be). I don't know if this has anything to do with the situation, however. By the way, is it the dingle's dilemma?

paw
Do we do such generalizations? I mean, though A and B are stationary, their clocks are at different places (co-ordinates may be).

Assuming a flat spacetime I don't see why not. A different position in a flat spacetime shouldn't affect the clocks rate in any way. It amounts to the same question as a spaceship travels from alpha centauri to procyon at constant speed 0.9c, what do observers on Earth and the ship 'see'. Obviously the ship observer sees Earths clock running slow while the Earth observer sees the ship clock running slow.

mitesh9
So is it the same thing that happens in muon lifetime case? Muons see our clocks slower, while we see muons' clock slower?

paw
So is it the same thing that happens in muon lifetime case? Muons see our clocks slower, while we see muons' clock slower?

Yes.

BTW Dingles Dilema seems to have been his inablity to accept experimental evidence. :grin

Look up Herbert Dingle on wiki and you'll see he had a number of objections to SR and ended up suggesting a conspiracy, more or less. I didn't see any reference to a specific 'dilema'.

mitesh9
Now, as we have concluded here, both observers on B and C frames think they are running faster than other in time (i.e. their clocks are faster then the other one), but when C coincides with B, he can actually check, wether what he thinks is true or not. What I wish to know is that, when comparing clocks from B and C, at the time of coinciding the two frames, keeping both clocks in two hands, what will one come to know?
That is to say, the rocket-man crashes (meets with an abrupt accident, though I don't think that's required) on B and being in B now, copares the clock in B with his own. What will he find then?

Edit: Or alternatively, when C coincides with B, someone from B throws the clock into C (which is rocket), now, the rocket-man can compare the clocks, keeping them in both hands. Or more simply, when the frames C and B coincides, the rocket-man can take a look at both the clocks (considering both of them to be big and observable )

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paw
Now, as we have concluded here, both observers on B and C frames think they are running faster than other in time (i.e. their clocks are faster then the other one

Ok so far...

but when C coincides with B, he can actually check, wether what he thinks is true or not.

Now we need to start being more careful. How and what is C checking?

What I wish to know is that, when comparing clocks from B and C, at the time of coinciding the two frames, keeping both clocks in two hands, what will one come to know?

Suppose C passes B close enough to actually see B's clock (remember it was previously synchronised at t0), without light travel delays, then C will observe B's clock has accumulated less time than C's. B, of course, can say the same about C's clock.

If you have C stop at (or crash into) B you introduce an acceleration and that's going to complicate the situation. Acceleration will break the symmetry. This would best be solved with a spacetime diagram. I'm reasonably sure that C will have the longer world line and thus the shortest 'proper' time in this case. Or to put it another way, when the situation is no longer symmetric C will have aged less than B (or A for that matter since A and B are equivalent).

mitesh9
Suppose C passes B close enough to actually see B's clock (remember it was previously synchronised at t0), without light travel delays, then C will observe B's clock has accumulated less time than C's. B, of course, can say the same about C's clock.

Let's remove crash, or anything related to acceleration. And let's go a little slow as well. What I gather is that, when C and B coincide (that is they are sharing the same coordinates of space), and removing light travel delays, both will see each other's clocks slow? When they are face to face (for a fraction of time may be) each having his clock in his hand so that he and other one, both can see the clock? This seems improbable, because, they are not calculating time, but actually seeing the clocks, and clocks can give a single time to all observers sharing same spatial coordinates!

paw
This seems improbable, because, they are not calculating time, but actually seeing the clocks, and clocks can give a single time to all observers sharing same spatial coordinates!

That's the problem right there. They may share the same spatial coordinates momentarily but it's the spacetime coordinates that matter. B and C can only share the same spacetime coordinates if C stops (or crashes) and that has to involve an acceleration.

mitesh9
Well, I get that now, but that means that they are at same place, but different time, because, they are sharing same spatial coordinates (x,y z), but their time is different. I think that's what you wish to convey (or is there any other definition of space time coordinates, that I am not aware of).

So in such a case, will they collide (being at same place but different times)? This is like, we are both in same school and same home, and you were rushing from school to home following same path in the evening, which I followed in opposite direction to reach to school in the same morning. Should we collide?

paw
Well, I get that now, but that means that they are at same place, but different time, because, they are sharing same spatial coordinates (x,y z), but their time is different. I think that's what you wish to convey (or is there any other definition of space time coordinates, that I am not aware of).

So in such a case, will they collide (being at same place but different times)? This is like, we are both in same school and same home, and you were rushing from school to home following same path in the evening, which I followed in opposite direction to reach to school in the same morning. Should we collide?

Try to think of a body that's in relative motion to an observer as getting further and further away from that observer on the time axis. Try not to think of the time coordinate as an actual time like a clock on the wall. Remember, there is no absolute time.

This way you can see that, even though C might miss B by the smallest distance in space, C will be at a completely different point in spacetime. The only way C and B can share the same point in space time is for one of them to accelerate to match speeds as well as position.

They can collide of course but they won't agree on what time the collision occurred at. They can only work with their clocks that were synchronised at some earlier time t0.

mitesh9
Sure, but my point is, as there is no absolute time (irrelative to any other time, but truly absolute), there is no absolute place as well (we arbitrarily select them as per our convenience).
Ideally, they must not collide. Two observers can only collide if their all spacetime coordinates (that is x, y, z & t) are same for both of them (as I tried to explain in my school-home example), other than that they can not collide.
And If they collide, they must have "observed" same space time coordinates. That is to say, even if they don't try to observe the other's time (or clock), the collision itself is the proof for both of them, that their spacetime coordinates were same at the time of collision (i.e. before acceleration/de-acceleration, as that will happen after collision). If they remember their own time of collision, as the collision it self is evidence of them being at sametime-sameplace, they may confirm afterwords, without looking their clocks again, that they were same.

Edit: To make it more clear, though there is no absolute time, they defined their own relative time by synchronization of their clocks, so at the time of synchronization, they were both at the origin of the time axis, yet having different spatial coordinates (x, y, z) and thus they did not collide. Similarly, if one of the observer is moving away from the other observer on time axis, even if their spatial coordinates coincide, they must not collide. This further clarifies why should their all four spacetime coordinates be same for a collision to occur.

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paw
the collision itself is the proof for both of them, that their spacetime coordinates were same at the time of collision (i.e. before acceleration/de-acceleration, as that will happen after collision). If they remember their own time of collision, as the collision it self is evidence of them being at sametime-sameplace, they may confirm afterwords, without looking their clocks again, that they were same.

Sure the collision is proof they share the same spacetime coords. But my point is you can't have a collision without an acceleration. If in the split instant before C hits B they compare clocks the situation will be as before, C sees B's clock as ticking slow and vise versa. When the collision is complete, and assuming both observers and their clocks aren't just puffs of vapor, both will agree that C has accumulated less proper time.

I still think you are looking at the time axis as a clock rather than a coordinate axis. In order to share the same spacetime coords both observers must be at rest wrt to each other and in the same place. As long as one is moving wrt to the other they can't share the same spacetime coords without an acceleration.

mitesh9
I exactly see what you want to show me (see the Edit, sorry for the delay), however, the acceleration (or de-acceleration) requires collision. And without (or before) collision, there won't be any acceleration (or de-acceleration). In turn collision requires same spacetime coordinates, which as you correctly pointed out, can not happen without acceleration (or de-acceleration) within the domain of SR.

I think, we both know now, that there is some catch, which we are not able to figure-out.

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mitesh9
Dear pav,

Thank you for your extensive support and help. See you tomorrow again...
In the meantime, is not there anybody willing to help or join the discussion?

phyti
The time interval from event AC to BC is the same for all, (in terms of the invariant interval).
The clocks do not indicate time, only the rate of time. Time dilation is a function of v/c. The clock that moves faster has a slower rate. When C is coincident with B (event BC), they could transmit a signal containing the time to each other. Coincident meaning close enough to observe without colliding. Objects cannot truly be 'in the same place', therefore there is always a distance involved, but not significant enough to affect measurements. The clock with the smaller value would be the faster.

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MeJennifer
Do we do such generalizations? I mean, though A and B are stationary, their clocks are at different places (co-ordinates may be). I don't know if this has anything to do with the situation, however. By the way, is it the dingle's dilemma?
Clocks at different places do generally not matter except in the case of accelerating frames.

For instance a frame that is accelerated cannot maintain a constant time for each coordinate value in the direction of acceleration.

When C meets B, C will read less time than B.
B and C could still claim that the other clock experienced less time during the experiment. Because, from C's perspective, A and B are not synchronized, there is no problem with the above result.

mitesh9
The time interval from event AC to BC is the same for all, (in terms of the invariant interval).
Which of the observer will observe invariant interval? If none, than we must not bring in any invariant time interval into discussion.
The clocks do not indicate time, only the rate of time. Time dilation is a function of v/c. The clock that moves faster has a slower rate. When C is coincident with B (event BC), they could transmit a signal containing the time to each other.
Well, yes, clocks do not indicate the time, but the rate of time, however, our problem also involves rate of time only, and not absolute time. And surely, While coincident, B and C can both transmit the signal, but what we are interested in is, will their signals show same passage of time interval or different intervals for both clocks?
Coincident meaning close enough to observe without colliding. Objects cannot truly be 'in the same place', therefore there is always a distance involved, but not significant enough to affect measurements.
Sure, but instead of observers, if the entities traveling in the frames C and B (as there is not absolute motion, but only relative one, both are traveling with respect to each other) are two particles, and if they collide, is not the point of impact where both particles are at same place same time (may be their centers are separated, yet at the point where their surfaces meet for impact, they are there same place same time), yet as you pointed out, this are not significant details.
The clock with the smaller value would be the faster.
But which clock will that be?

I think I should repeat my question.
Three inertial frames A, B and C. A and B stationary while C moving with a relative velocity v with respect to A and B, and traveling in the line connecting A and B, so that it will first coincide with A and eventually B (coincide means their origins share same spatial co-ordinates x, y and z). A's clock is slightly (say 1m) away from A's spatial Origin, while, B and C's clocks are perfectly at their spatial origins. Further assume that, clocks have an intuitive mechanism, by virtue of which, they stop as soon as they are touched by anything. Now when C coincides with A, all three clocks are synchronized. Considering that A's clock was off-origin 1m, and clocks were smaller than that, the clocks of A and C won't collide, even when their origins coincide.
Eventually, C reaches B and their origins coincide, However, there are clocks at their origins, so they will collide. At the same time of collision (when the clocks have touched each other) the clocks will stop. If the clocks are intact, will there be any time difference in the time value (or rate or interval) shown by the clocks?
We should note that, As has already been posted earlier in this thread, The clocks should not collide, if their spacetime coordinates (of at least one point on their surface, where the point of impact is) are same. But, if we apply Lorentz transform on any of the clock, we will come to know that it is off-sync with the other, and hence, as it's spacetime co-ordinates are different then the other clock (time coordinate to be precise), it must not collide, yet we know it will collide.

To make it more clear, though there is no absolute time, they defined their own relative time by synchronization of their clocks, so at the time of synchronization, they were both at the origin of the time axis, yet having different spatial coordinates (x, y, z) and thus they did not collide. Similarly, if one of the observer is moving away from the other observer on time axis, even if their spatial coordinates coincide, they must not collide. This further clarifies why should their all four spacetime coordinates be same for a collision to occur. (this has also been pointed out in earlier posts)

I'm pretty sure, there is some catch in the situation, but can't see what it is, may you people help me out.

Regards,

Mitesh

mitesh9
Clocks at different places do generally not matter except in the case of accelerating frames.
For instance a frame that is accelerated cannot maintain a constant time for each coordinate value in the direction of acceleration.
Well, paw told me similar thing. Though I'm not sure will I be able to translate the situation in two frame situation, because there may be some complications.

When C meets B, C will read less time than B.
B and C could still claim that the other clock experienced less time during the experiment. Because, from C's perspective, A and B are not synchronized, there is no problem with the above result.
Dear Sir, as MeJennifer and pawhas pointed out, A and B being stationary with respect to each other, and no acceleration being involved, how can C see them not synchronized? And yet, A's purpose was to synchronize the clocks of C and B, so that there can be a fair competition between them. Now when B and C are synchronized, C moves towards B, their clocks collides and stops, which would have accumulated more time? In fact, the first question is should they collide at all?

Mentor
Dear Sir, as MeJennifer and pawhas pointed out, A and B being stationary with respect to each other, and no acceleration being involved, how can C see them not synchronized?
Dear Sir, the moving frame in which C is at rest will not see clocks A and B as synchronized. This is the relativity of simultaneity: Look it up!
And yet, A's purpose was to synchronize the clocks of C and B, so that there can be a fair competition between them.
Since the clocks of C and B are in relative motion, they cannot be synchronized. Of course, you could arrange that when the two clocks pass each other that they read the same at that instant. But that's hardly the same as being synchronized.

Dear Sir, as MeJennifer and pawhas pointed out, A and B being stationary with respect to each other, and no acceleration being involved, how can C see them not synchronized?
Dear Sir, I am in the fortunate position that I don't have to rely on others to work this kind of problem.
C and B meet.
C reads less time than B when they meet.
A and B are not synchronized in C's frame.

mitesh9
I think I should stop this discussion from my side.

It was good to learn that this is how "spirit of the game" is defined on PF. In last three days of my participation in PF discussions on relativity, I have collected more infraction points (rather than information) than anybody else who keeps on insulting others. I would suggest to review my posts, if you do not believe me. And hang me till death, If you find any insulting word in any of my posts. PF is not a learning forum, but a rigid community I must say, who do not want to answer questions, but instead prefers insulting people for their ignorance.

Anyways, learning physics may be less difficult task, than learning it from any man, I suppose.

Bye bye, PF.

Regards,

Mitesh Patel

If you find any insulting word in any of my posts. PF is not a learning forum, but a rigid community I must say, who do not want to answer questions, but instead prefers insulting people for their ignorance.
Maybe you want to read my statement in the appropriate context: "Dear Sir, as MeJennifer and pawhas pointed out, A and B being stationary with respect to each other, and no acceleration being involved, how can C see them not synchronized?".
I don't have to rely on other user's statements to work the problem, meaning that I'm confident enough to contradict them if necessary.
I never meant to ridicule you for asking here. Asking and giving answers is the intention of this forum.

paw
Dear Sir, as MeJennifer and pawhas pointed out, A and B being stationary with respect to each other, and no acceleration being involved, how can C see them not synchronized?

I'm sorry but you have misunderstood. I (and MeJennifer) said A and B are equivalent frames. As such you don't need A at all. In your scenario A serves only to define the time t0 that B and C's clocks have the same reading. That could be done just as easily with a light flash equidistant between B and C. B and C set their clocks when the light pulse arrives.

Any time after t0, C will observe B's clock as running slow and B will observe C's clock as running slow. This will remain true for all time unless B or C undergo an acceleration.

As for whether B and C will collide if they are at some time coincident in space, the answer is yes they will. However B and C will not agree on the time the collision occured. When the collision has occurred C will have accumulated less proper time than B because C has undergone an (rather severe) acceleration.

paw
I don't have to rely on other user's statements to work the problem, meaning that I'm confident enough to contradict them if necessary.

No need to contradict either me (or MeJennifer). : ) Mitesh9 misunderstood the statements. See above.

paw
Dear Sir, I am in the fortunate position that I don't have to rely on others to work this kind of problem.
C and B meet.
C reads less time than B when they meet.
A and B are not synchronized in C's frame.

Now there's an interesting point. You agree A and B are equivalent frames right? That is, there's no acceleration and no relative motion between them.

So, while C will always say both A's clock and B's clock are running slow, A's clock and B's clock will always read the same time according to C. Right?

Isn't this the same as saying A and B's clocks are synchronized according to C? Or is the word 'synchronized' reserved for more specialised use?

Mentor
simultaneity

Now there's an interesting point. You agree A and B are equivalent frames right? That is, there's no acceleration and no relative motion between them.

So, while C will always say both A's clock and B's clock are running slow, A's clock and B's clock will always read the same time according to C. Right?
Wrong.
Isn't this the same as saying A and B's clocks are synchronized according to C? Or is the word 'synchronized' reserved for more specialised use?
A and B are two clocks in the same frame and are presumeably synchronized within their own frame. They are not synchronized according to frame C, since they are separated along the direction of motion. According to frame C, clock B is ahead of clock A.

When clock C passes clock A let's assume that they both read the same time (by special arrangement). According to frame A-B, at the moment that C passes A all three clocks read the same time. Frame C agrees that at the moment he passes A, clocks C and A both read the same (again, by arrangement), but does not agree that clock B reads the same time.

paw
A and B are two clocks in the same frame and are presumeably synchronized within their own frame.

Right. That's what I meant.

They are not synchronized according to frame C, since they are separated along the direction of motion. According to frame C, clock B is ahead of clock A.

Arrrrr, I missed that. They do tick at the same rate though, correct?

When clock C passes clock A let's assume that they both read the same time (by special arrangement). According to frame A-B, at the moment that C passes A all three clocks read the same time. Frame C agrees that at the moment he passes A, clocks C and A both read the same (again, by arrangement), but does not agree that clock B reads the same time.

Thanks. Luckily I don't think any of that changes the results I predicted though.

1. Any time after C passes A, C will see both A and B ticking slower while A and B will see C ticking slower.

2. A collision will occur.

3. B and C will not agree on the time the collision occured.

4. After collision C will have accumulated more proper time than B.

5. A was not required for any of this analysis.

Does this all seem right to you?

Mentor
Arrrrr, I missed that. They do tick at the same rate though, correct?
Sure.

Thanks. Luckily I don't think any of that changes the results I predicted though.

1. Any time after C passes A, C will see both A and B ticking slower while A and B will see C ticking slower.
Sounds good.

2. A collision will occur.

3. B and C will not agree on the time the collision occured.
Sounds good. Assume that clocks A & B are a distance L apart (along the direction of motion) and are synchronized and that C passes A when both read t = 0. Then when C collides with B, B will read t = L/v and C will read $L/(\gamma v)$.

(If I misunderstood the scenario, please correct me.)

4. After collision C will have accumulated more proper time than B.
Not sure what you mean. Proper time between what two events?

paw
Sounds good. Assume that clocks A & B are a distance L apart (along the direction of motion) and are synchronized and that C passes A when both read t = 0. Then when C collides with B, B will read t = L/v and C will read $L/(\gamma v)$.

(If I misunderstood the scenario, please correct me.)

No, you didn't misunderstand. That's exactly how I saw the scenario.

Not sure what you mean. Proper time between what two events?

Between t=0, that is when C passed A, and when C collides with B. But I see you answered in the affirmative anyway in the previous quote.

AntigenX
the acceleration (or de-acceleration) requires collision. And without (or before) collision, there won't be any acceleration (or de-acceleration). In turn collision requires same spacetime coordinates, which as you correctly pointed out, can not happen without acceleration (or de-acceleration) within the domain of SR.

This is like running in circles...

Three inertial frames A, B and C...
...though there is no absolute time, they defined their own relative time by synchronization of their clocks, so at the time of synchronization, they were both at the origin of the time axis, yet having different spatial coordinates (x, y, z) and thus they did not collide. Similarly, if one of the observer is moving away from the other observer on time axis, even if their spatial coordinates coincide, they must not collide. This further clarifies why should their all four spacetime coordinates be same for a collision to occur. (this has also been pointed out in earlier posts)...

Is this correct? Do we require all four spacetime coordinates equal for a collision?

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Mentor
Do we require all four spacetime coordinates equal for a collision?
By definition, a collision means that two things are at the same place at the same time.

AntigenX
By definition, a collision means that two things are at the same place at the same time.

But if collision occurs, whose coordinates we should consider? For event collision, the coordinates of space will naturally be same for two frames colliding, but the time coordinates of both will be different, because one is moving and other is stationary. And if both are moving, then also their time coordinates will be different. This means that two things can never collide!