A question about Relativity of Time using Time dilation experiment

  • #1
rahaverhma
69
1
The time dilation experiment involves two frames in relative motion, let one at ground and other at train with velocity V. The light clock runs faster in rest frame, as seen by an observer A at rest in train ( just beside clock ) than that observed by an observer B in ground frame which observes moving clock. But here the discussion in both conditions are just about clocks in Train's frame observed by both. So, how do I conclude that the time goes faster in ground frame while slower in moving frame, that is the man at ground ages faster than the man in train.
One thing that I think that I can do is that the light clock, which is at rest in A's frame, can also be kept in ground frame and I also think that the time of rest-clock in ground frame will be same as that of rest-clock in train frame as the light covers same distance D with same speed of light, c. Then, the time can be compared for the two light-clock set-ups after looking at both : one kept on the ground and the other at train, both observed from ground by observer B. And, phenomenon of Ages of observers A and B can be explained. But here is another problem that then the equation

∆t'(observer:B) = (gamma)*∆t(proper/observer:A)

will not even be concluded because as I mentioned earlier that this one is for the same event placed at train's frame as seen by both observers A and B. Please tell me if my reasoning is wrong or right 😬 and help to sort out this issue i.e. Explain me the relativity of Time with this time dilation experiment.
 
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  • #2
rahaverhma said:
The time dilation experiment involves two frames in relative motion, let one at ground and other at train with velocity V. The light clock runs faster in rest frame, as seen by an observer A at rest in train ( just beside clock ) than that observed by an observer B in ground frame which observes moving clock. But here the discussion in both conditions are just about clocks in Train's frame observed by both. So, how do I conclude that the time goes faster in ground frame while slower in moving frame, that is the man at ground ages faster than the man in train.
Velocity-based time dilation is symmetric and the measurements are reciprocal. The ground clock runs slow as measured from the train. And the train clock runs slow as measured from the ground.
rahaverhma said:
One thing that I think that I can do is that the light clock, which is at rest in A's frame, can also be kept in ground frame and I also think that the time of rest-clock in ground frame will be same as that of rest-clock in train frame as the light covers same distance D with same speed of light, c. Then, the time can be compared for the two light-clock set-ups after looking at both : one kept on the ground and the other at train, both observed from ground by observer B. And, phenomenon of Ages of observers A and B can be explained. But here is another problem that then the equation

∆t'(observer:B) = (gamma)*∆t(proper/observer:A)

will not even be concluded because as I mentioned earlier that this one is for the same event placed at train's frame as seen by both observers A and B. Please tell me if my reasoning is wrong or right 😬 and help to sort out this issue i.e. Explain me the relativity of Time with this time dilation experiment.
I don't understand what you are saying here.
 
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  • #3
rahaverhma said:
, how do I conclude that the time goes faster in ground frame while slower in moving frame, that is the man at ground ages faster than the man in train.
You don't, because he doesn't.

The problem is that there is no assumption-free way to define "at the same time" for things that are not at the same place. The two clocks are only in the same place once, so they can be unambiguously zeroed, but they cannot be compared again without making some kind of assumption about what "at the same time" means in the sentence "at the same time as my clock reads 1 minute, the other clock reads X minutes". If you use the natural assumption for the train observer the platform clock will have ticked less. If you use the natural assumption for the platform observer then the train clock will have ticked less.

This is time dilation. The effect is symmetric and it is an over-simplification (and wrong) to say that one observer ages faster than the other.

A related phenomenon is differential aging. If the clocks do meet again then you can compare their time readings unambiguously at the second meeting. Which one shows more time depends on the choices you make about how to move the clocks so they meet
 
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  • #4
Another thing to note is that there is also no such thing as ”a moving” inertial frame or ”a stationary” imertial frame. They are moving/stationary relative to something. Neither frame can be universally acclaimed to be ”the” stationary frame. This is a basic cornerstone of relativity and something many people have difficulty grasping.
 
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  • #5
rahaverhma said:
The time dilation experiment involves two frames in relative motion, let one at ground and other at train with velocity V. The light clock runs faster in rest frame, as seen by an observer A at rest in train ( just beside clock ) than that observed by an observer B in ground frame which observes moving clock. But here the discussion in both conditions are just about clocks in Train's frame observed by both. So, how do I conclude that the time goes faster in ground frame while slower in moving frame, that is the man at ground ages faster than the man in train.
Suppose we start the experiment at one location, ##g_0## in the ground frame and follow a moving clock on the train, ##C_t##, for one of its seconds. Then the moving clock is at another ground location, ##g_1## at the end of its (the train's) second. How much time passed in the ground frame depends on how the two ground frame clocks, ##C_{g0}## and ##C_{g1}##, at ##g_0## and ##g_1##, respectively, are synchronized. Suppose the ground people think that their two clocks, ##C_{g0}## and ##C_{g1}##, are perfectly synchronized. Then an observer on the train who rode with the train clock will think that the ground clocks are NOT perfectly synchronized. That is called the "relativity of simultaneity".
rahaverhma said:
One thing that I think that I can do is that the light clock, which is at rest in A's frame, can also be kept in ground frame and I also think that the time of rest-clock in ground frame will be same as that of rest-clock in train frame as the light covers same distance D with same speed of light, c.
In that case, there will be two clocks at different train locations to compare to the one ground clock.

In both these cases, you are not considering enough clocks and how the two clocks in the "stationary" frame are synchronized.
 
  • #6
The time dilation result derived is a bit strange, no doubt, but there doesn’t
seem to be anything downright incorrect about it until we look at the situation from A’s
point of view. A sees B flying by at a speed v in the other direction. The ground is no
more fundamental than a train, so the same reasoning applies. The time dilation factor, γ ,
doesn’t depend on the sign of v, so A sees the same time dilation factor that B sees. That
is, A sees B’s clock running slow. But how can this be? Are we claiming that A’s clock is
slower than B’s, and also that B’s clock is slower than A’s? Well ... yes and no.

Remember that the above time-dilation reasoning applies only to a situation where
something is motionless in the appropriate frame. In the second situation (where A sees B
flying by), the statement tA = γ tB holds only when the two events (say, two ticks on
B’s
clock) happen at the same place in
B’s frame. But for two such events, they are certainly
not in the same place in A’s frame, so the tB = γ tA result in Eq. (11.9) does not hold. The
conditions of being motionless in each frame never both hold for a given setup (unless
v = 0, in which case γ = 1 and tA = tB). So, the answer to the question at the end of the
previous paragraph is “yes” if you ask the questions in the appropriate frames, and “no”
if you think the answer should be frame independent.


A passage on Time Dilation from David Morin's book on Classical Mechanics. Help me understand what the 2nd para in the passage means to say.
 
  • #7
rahaverhma said:
A passage on Time Dilation from David Morin's book on Classical Mechanics. Help me understand what the 2nd para in the passage means to say.
What don't you understand about it? Time dilation is symmetric, as it must be.
 
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  • #8
rahaverhma said:
A passage on Time Dilation from David Morin's book on Classical Mechanics. Help me understand what the 2nd para in the passage means to say.
It says that A determines that B's clock is ticking slowly and B determines that A's clock is ticking slowly. And that this is not a contradiction since A derives the result based on an analysis of two events that are at the same place in B's rest frame, which are therefore not in the same place in A's rest frame. So B cannot be using the same pair of events as A, since B needs a pair of events that are in the same location in A's frame.

Note that this doesn't guarantee that you don't have a contradiction, just that you don't have a contradiction yet. It will turn out that you don't have a contradiction at all once you derive the Lorentz transforms and come to understand relativity of simultaneity, but at this point all Morin is doing is explaining that A and B are following the same process, but cannot be applying it to the same events, so one shouldn't look at the symmetry of time dilation and immediately conclude it's crazy.
 
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  • #9
Ibix said:
It says that A determines that B's clock is ticking slowly and B determines that A's clock is ticking slowly. And that this is not a contradiction since A derives the result based on an analysis of two events that are at the same place in B's rest frame, which are therefore not in the same place in A's rest frame. So B cannot be using the same pair of events as A, since B needs a pair of events that are in the same location in A's frame.

Note that this doesn't guarantee that you don't have a contradiction, just that you don't have a contradiction yet. It will turn out that you don't have a contradiction at all once you derive the Lorentz transforms and come to understand relativity of simultaneity, but at this point all Morin is doing is explaining that A and B are following the same process, but cannot be applying it to the same events, so one shouldn't look at the symmetry of time dilation and immediately conclude it's crazy.
👍I understood it. One thing more though, what is meaning of an "appropriate frame" as mentioned in the passage ?
 
  • #10
rahaverhma said:
👍I understood it. One thing more though, what is meaning of an "appropriate frame" as mentioned in the passage ?
It means the relevant frame in the context he's talking about. So where he says "In the second situation (where A sees B flying by), the statement tA = γ tB holds only when the two events (say, two ticks on B’s clock) happen at the same place in B’s frame" he means the "appropriate frame" is B's frame. He could have swapped A and B in that sentence and the "appropriate frame" would be A's.
 
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