2
$\begingroup$

I have a problem understanding what proper time really means, well actually, it's the symmetry that confuses me... Suppose for convenience that particle A has a mean life-time of 10 seconds in its own rest frame, as in, when it was created in a laboratory, it decayed after 10 seconds (without it moving in a relativistic speed).

The scenario is: Let S be the laboratory's frame of reference, and S' be the frame of reference stuck to particle A. Suppose at t=0seconds, according to the lab's stationary clock (the clock in the lab), particle A is created and moves at a relativistic speed relative to the lab frame, which has a Gama factor equal to 5 (for convenience). (Particle A moves in X axis direction relative to the lab's frame)

As I understand, when 50 seconds pass according to the lab's clock, particle A will decay, cause relative to the lab's frame, particle A's clock has been moving slowly, and when 50 seconds pass according to the lab's clock, particle A's clock will look like that it had only ticked for 10 seconds relative to an observer (Bob) in the lab's frame. Thus according to the equation (Delta T = Gama * proper time---> 50 = gama* proper time---> proper time for particle A = 50/5 = 10seconds).

So now, we basically know that 10 seconds exactly passed in Particle A's frame of reference. As in, if an observer (Joe) was in S' frame of reference all along, he will say that according to the watch on his wrist, 10 seconds exactly passed before particle A decayed.

However, relative to observer Joe (in the particle's frame of reference), during these 10 seconds, the lab's frame of reference S has been moving with a gama factor of also 5, thus during these 10 seconds, he will see the lab's frame clock moving slowly, and he will conclude that the lab's frame clock (which is moving slowly relative to him) has ticked for 2 seconds only, cause delta T = gama * lab proper time ---> 10 = gama *proper time of lab ---> Proper time of lab = 2seconds.

How is this possible? We know that 50 seconds passed in the lab's frame and not 2 seconds... the situation is supposed to be symmetric, none of the observers could tell whether they were moving or not. I would be glad if someone could explain where I had gone wrong in my explanation. Thanks!

Improve this question
$\endgroup$

3 Answers 3

Reset to default
2
$\begingroup$

Both of the following statements are true:

(1) Joe observes the lab clocks to run slow

(2) According to the lab clocks, 50 seconds elapse between the creation and decay events of the muon.

Note that in the above, I have used the plural clocks. To determine the elapsed time in the lab, the readings of two spatially separated clocks are recorded, a clock co-located with the creation event of the muon and a clock co-located with the decay event.

Clearly, the lab clocks must by synchronized (in the lab frame of reference) for this to be valid.

But, according to Joe (or any relatively moving reference frame), these two lab clocks are not synchronized (relativity of simultaneity) and this is the resolution to the apparent contradiction.

According to Joe,

  1. the muon decays in 10 seconds
  2. the lab clocks run slow compared to his clock
  3. the two lab clocks show a difference of 50 seconds between the events since they aren't synchronized.

As always, drawing a spacetime diagram will make the above quite clear.

Improve this answer
$\endgroup$
2
  • $\begingroup$ A useful formula to know is that if two clocks are synchronized in their own rest frame, and a distance d apart in this frame, then in the frame of an observer who sees the clocks moving at speed v (parallel to the axis joining them), at a single moment in this frame the clocks will be out of sync by vd/c^2. In this example the speed of the particle is 0.9797959c in the lab frame so if two lab clocks were next to the emission and decay, their separation in the lab frame would be d = 50 * 0.9797959 light-seconds, so in the particle frame they're out of sync by that times v=0.9797959c, or 48 s. $\endgroup$
    – Hypnosifl
    Commented Jan 17, 2015 at 17:46
  • $\begingroup$ (Continued) So in the particle rest frame, at the same moment it's passing the first lab clock and that clock reads 0 seconds, the second lab clock already reads 48 seconds. Then in this frame it takes 10 seconds to reach the second lab clock, and that clock is slowed by a factor of 5 so it only ticks forward 2 seconds, so it reads 50 seconds as the particle reaches it and decays. $\endgroup$
    – Hypnosifl
    Commented Jan 17, 2015 at 18:11
1
$\begingroup$

However, relative to observer Joe (in the particle's frame of reference), during these 10 seconds, the lab's frame of reference S has been moving with a gama factor of also 5, thus during these 10 seconds, he will see the lab's frame clock moving slowly, and he will conclude that the lab's frame clock (which is moving slowly relative to him) has ticked for 2 seconds only, cause delta T = gama * lab proper time ---> 10 = gama *proper time of lab ---> Proper time of lab = 2seconds.

In S, the particle is moving from A to B. So the times of birth and death of the particle were read at two space points, A and B. Imagine that there are two clocks sticked to frame S at A and B to measure these time points. The point is that: looking from S', although the two clocks are both slow by a factor of gamma, their time pointers are not pointing the same time point. One of the clock is lagged in comparison to the other. You can verify this easily with the Lorentz transformation. In summary, Joe sees the two clocks moving to the particle at O at time points separated by 10s according to his clock, but the pointers of the clocks that are sticked to S (moving clocks at different space points) are not the same, and the difference they say is 50s.

Improve this answer
$\endgroup$
0
$\begingroup$

first of all, you said "when 50 seconds pass according to the lab's clock, particle A's clock will look like that it had only ticked for 10 seconds relative to an observer (Bob) in the lab's frame", which is not correct. Because the 10s is not relative to any observer in the lab's frame, it's the time that the moving observer reads from his watch.

now come back to your question, suppose we have two inertial frames $S$ and $S'$, one is moving relative to the other (it's meaningless to say which one is moving since it depends on which frame you are in). In order to illustrate the symmetry you mentioned, we further assume there are two events $E$ and $E'$, where $E$ has $S$ as its rest frame and $E'$ has $S'$ as its rest frame.

Then the proper time for the two events are:

  1. $\tau$=the time for $E$ observed by someone $P$ at rest in $S$

  2. $\tau'$=the time for $E'$ observed by someone $P'$ at rest in $S'$

Now the symmetry you mentioned should be expressed as the following:

  1. if $P$ observe $E'$, the observed time is given by $\gamma \tau'$

  2. if $P'$ observe $E$, the observed time is given by $\gamma \tau$ (Note that the $\gamma$ is the same with the one in case 1 since $S$ is moving away from $S'$ in exactly the same way that $S'$ is moving away from $S$)

which means when we talk about symmetry between two inertial frames, we should have two different events happening in the two frames and the two events are observed by two different observers in the two frames. furthermore, it's the ratio $\gamma$ that matters, not the value for the observed time. The exact values for the time depends on the different events.

Improve this answer
$\endgroup$
9
  • $\begingroup$ What would happen in this scenario: Two particles A,B are created at t=0 in the lab frame. A, same as in the first scenario, moves with gamma factor equal to 5, and B stays stationary. You said, when 50 seconds pass in the lab's clock, 10 seconds will pass in particle's A frame of reference, since the creation and decay of particle A in its own frame of reference is its proper time and is equal to 10 seconds. Yet, particle B will decay after 10 seconds also in the lab's frame of reference, and since it was measured at the same place in the lab's frame, it's also its proper time. $\endgroup$
    – Dylan132
    Commented Jan 17, 2015 at 18:06
  • $\begingroup$ Are you saying that, relative to an observer in particle A's frame of reference, particle A will decay after 10 seconds according to his watch, and particle B will decay after 50 seconds too? As in, when he sees particle B decay, how much time has already passed in the lab's clock? $\endgroup$
    – Dylan132
    Commented Jan 17, 2015 at 18:08
  • $\begingroup$ @Dylan132 for your first comment, yes you are right. $\endgroup$
    – M. Zeng
    Commented Jan 18, 2015 at 5:37
  • $\begingroup$ @Dylan132 for your second comment, yes for an observer in A's frame, particle B will decay after 50s according to his watch since from his point of view it is particle B that is moving away from him. "As in, when he sees particle B decay, how much time has already passed in the lab's clock?"---when particle B decays, the the lab observer will read 10s from his watch since B is at rest in the lab, while the moving observer will think that 50s has passed if he doesn't know relativity. So now if you think about this, this is a manifestation of the symmetry of the two frames that you mentioned. $\endgroup$
    – M. Zeng
    Commented Jan 18, 2015 at 5:46
  • $\begingroup$ Ok sorry but this is pretty much confusing! This is what I know or what I think I know. When 10 seconds pass in the lab's frame particle B decays and 50 seconds have already passed in particle A's frame. Then when extra 40 seconds pass in the lab's frame (thus 50 seconds have passed in the lab's frame), particle A decays (and 10 seconds pass in particle A's frame). It's like time is moving back in particle's A frame! Anyways, is there a book you know of that explains time dilation and proper time thoroughly? $\endgroup$
    – Dylan132
    Commented Jan 18, 2015 at 20:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.