Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Is this true or false: If A and B have clocks and are traveling at relative velocity to each other, then to B it APPEARS that A's clock moving slower, but A sees his own clock moving at normal speed. Similarly, to A it APPEARS that B's clock is moving slower, but B sees his own clock moving at normal speed.

If the above is true, then both A seeing his own clock moving at normal speed and B seeing his own clock moving at normal speed means that in reality both clocks are moving at normal speed, and neither has slowed down, whereas the other person's clock APPEARING to move slowly is merely an illusion.

Now is this true or false: If A and B have clocks and are traveling at relative velocity to each other, then A sees his own clock move slowly (compared to the speed of the clock when A was at rest with respect to B) and B sees his own clock move slowly, so that in reality both clocks are moving slowly, but they still remain synchronized (since both are slow by the same amount)

If the above statement is not true, then why do muons decay slowly when moving fast? [it could only be possible if the muon saw its own clock as moving slowly. If we saw the muon's clock moving slowly, but the muon saw its own clock moving at the normal rate, then the muon would decay at the normal rate, and not slowly]

Can anyone please explain where I went wrong?

share|improve this question
1  
"in reality both clocks are moving at normal speed"? This does not follow. –  santa claus Feb 6 '13 at 18:29
    
Furthermore, the clocks are no longer synchronized with each other. –  santa claus Feb 6 '13 at 18:30
add comment

4 Answers

Take the two events when the muon is created, and when it decays. We'll choose the origins of the muon rest frame and the lab rest frame so their origins coincide at the point and time the muon is created i.e. the muon is created at (0, 0) in both frames. Both frames will agree about their relative velocity $v$.

In the muon's rest frame it is stationary, so it is created at $(0, 0)$ and decays at $(t_\mu, 0)$, where $t_\mu$ is the muon lifetime (about 2.2$\mu$s).

Now work out what happens in the lab. We've agreed the muon is created at $(0, 0)$ in both frames, so we just have to work out where the spacetime point $(t_\mu, 0)$ is in lab co-ordinates. Once we do that we'll have the lifetime of the muon as seen in the lab.

The only safe way to do this (especially for beginners in SR) is to use the Lorentz transformations, which are:

$$ t' = \gamma \left( t - \frac{vx}{c^2}\right) $$

$$ x' = \gamma \left( x- vt \right) $$

So the point $(t, x) = (t_\mu, 0)$ transforms to $(t', x') = (\gamma t_\mu, -\gamma vt_\mu)$. The lifetime of the muon in the lab frame is $\gamma t_\mu$, and in that time it manages to travel a distance $\gamma v t_\mu$: a result that is reassuring because it's just the velocity in the lab frame times the lifetime in the lab frame.

The trouble is that you're trying to use hand waving arguments, and these are usually a minefield for the beginner because there are so many conceptual issues with SR. If you're trying to work through an apparent paradox in SR the only safe way to procede is to identify the key spacetime points and work out how they transform between frames.

Response to comment:

I didn't address your question about the relative rate clocks run because it isn't a helpful concept. Let me try an illustrate this by addressing your question about the muon. You are correct that the lab sees the muon clock run slow and the muon sees the lab clock run slow. I'm guessing (comment if I'm wrong) that you are puzzled because the situation is apparently symmetrical but the muon lifetime is different in the two frames. How can the apparent symmetry in clock rates produce an asymmetrical result?

The reason for the asymmetry gets at the heart of SR, so actually you've asked an excellent question. The reason for the asymmetry is that in the muon rest frame the creation and decay take place at the same place, $x = 0$, but in the lab frame they take place in different places: creation at $x = 0$ and decay at $x = \gamma vt_\mu$. It's the asymmetry in the position that is related to the asymmetry in time.

To see how this works you need to understand that the fundamental basis of SR is an invariance called the line element (also known as the proper time). Suppose you have two spacetime points $(t, x, y, z)$ and $(t+dt, x+dx, y+dy, z+dz)$ then the line element is defined by:

$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$

This should remind you of Pythagorus' theorem for the distance between two points in space, and indeed that's exactly the role it plays in SR. It is the spacetime distance between the two spacetime points. However you should note that unlike Pythagorus' theorem the $dt^2$ has a minus sign, and it's this minus sign that is responsible for all the weird effects in SR.

The key point in SR (and GR in fact) is that the quantity $ds^2$ is an invariant i.e. all observers in all frames will agree it has the same value.

Let's see how this applies to the muon. We can ignore $dy$ and $dz$ because we'll take the muon to be travelling along the $x$ axis, so $ds^2 = -c^2dt^2 + dx^2$. First calculate $ds^2$ between the muon creation and decay in the muon rest frame. Because in its rest frame the muon is stationary $ds^2 = -c^2dt^2$, so for the muon in its rest frame:

$$ ds^2 = -c^2t_\mu^2 $$

So if we calculate $ds^2$ in the lab frame we should also find it's $-c^2t_\mu^2$, and I'll show this in a moment, but first I want to point out the underlying principles.

I've said $ds^2$ has to be invariant, and that means I can add zero to it because adding zero doesn't change it's value. This may seem a silly thing to say, but suppose we take a change in time and $x$ such that:

$$ ds_1^2 = -c^2dt_1^2 + dx_1^2 = 0 $$

i.e. we choose $dt_1$ and $dx_1$ so that when you calculate the line element $ds_1^2$ comes out zero. If $ds_1^2$ is zero I can add it to my line element $ds^2$ that I calculated above without changing its value. And this is the key point: I can add some $dx_1$ to the spacing between the spacetime points provided I add a corresponding $dt_1$ that ensures the net change in $ds^2$ is zero. This is exactly what happens in the lab frame. The $x$ spacing has changed because the creation and decay no longer happen in the same place, and to balance this the $t$ spacing has to change to keep $ds^2$ constant. This is the origin of time dilation. It's not really clocks running at different rates, it's that different observers will disagree about the $x$ and $t$ spacing of the events.

It just remains to prove that $ds^2$ really is constant in the muon experiment. In the lab frame the two events are $(0, 0)$ and $(\gamma t_\mu, -\gamma vt_\mu)$ so $ds^2$ is given by:

$$ \begin{split} ds^2 &= -c^2\gamma^2t_\mu^2 + \gamma^2v^2t_\mu^2 \\ &= \gamma^2 t_\mu^2 (v^2 - c^2) \\ &= \frac{v^2 - c^2}{1 - v^2/c^2} t_\mu^2 \\ &= \frac{v^2 - c^2}{c^2 - v^2} c^2 t_\mu^2 \\ &= -c^2t_\mu^2 \end{split} $$

and this is the same value we got in the muon rest frame. QED!

share|improve this answer
2  
John you are an absolute physics.SE beast. –  joshphysics Feb 7 '13 at 2:34
    
I was not trying to solve the muon problem using Lorentz transform. I was trying to conceptually understand the concept behind time dilation. If A with a clock is moving relative to B, who sees that the clock is slow? Does A see that his own clock is slow, or does B in another frame see that A's clock is slow, and A sees his clock running at normal speed? –  khushro Feb 7 '13 at 6:36
    
@khushro: I've attempted to address your question about the clocks, though I can't answer it directly because comparing the clock rates isn't a helpful thing to do. Anyhow have a read through what I've added and see if it helps. –  John Rennie Feb 7 '13 at 7:44
    
@joshphysics: Well, Get in line. We all have told that already..! :D –  Waffle's Crazy Peanut Feb 7 '13 at 13:37
add comment

Is this true or false: If A and B have clocks and are traveling at relative velocity to each other, then to B it APPEARS that A's clock moving slower, but A sees his own clock moving at normal speed. Similarly, to A it APPEARS that B's clock is moving slower, but B sees his own clock moving at normal speed.

This is true.

If the above is true, then both A seeing his own clock moving at normal speed and B seeing his own clock moving at normal speed means that in reality both clocks are moving at normal speed, and neither has slowed down, whereas the other person's clock APPEARING to move slowly is merely an illusion.

That isn't the right way to think about it. $A$'s clock really is running slowly in $B$'s reference frame, and $B$'s clock really is running slowly in $A$'s reference frame. This clashes with our Galileian intuitions and will probably seem like a paradox until you learn how to reason about Minkowski space-time. I'm not going to explain how to do that in this answer, because any good text book will do that, but the point is that you went wrong at the point where you concluded that time dilation is an illusion.

Now is this true or false: If A and B have clocks and are traveling at relative velocity to each other, then A sees his own clock move slowly (compared to the speed of the clock when A was at rest with respect to B) and B sees his own clock move slowly, so that in reality both clocks are moving slowly, but they still remain synchronized (since both are slow by the same amount)

This is false. (I think you know this.)

If the above statement is not true, then why do muons decay slowly when moving fast? [it could only be possible if the muon saw its own clock as moving slowly. If we saw the muon's clock moving slowly, but the muon saw its own clock moving at the normal rate, then the muon would decay at the normal rate, and not slowly]

In its own reference frame the muon does indeed see its own clock moving at the normal rate, and therefore decays at the normal rate, as seen from its own reference frame. However, in our reference frame their clocks run slowly, and therefore they decay more slowly.

share|improve this answer
add comment

If A and B have clocks and are traveling at relative velocity to each other, then to B it APPEARS that A's clock moving slower, but A sees his own clock moving at normal speed. Similarly, to A it APPEARS that B's clock is moving slower, but B sees his own clock moving at normal speed.

[...] Is this true or false

Foremost: it is improper.
(In a specific technical sense; but this kind of setup description is plainly prone to lead to misunderstandings or misinterpretations, and it is therefore depreciated -- it is deservedly called "improper". It can be considered as some sort of more or less sloppy jargon that might be of some benefit to people who have and rely on an unambiguous "proper" understanding of how to express geometric-kinematic relations in RT in the first place.)

A proper way to express the situation described above would be (for instance):
The motion(s) of each clock separately (e.g. whether it had been a member of an intertial system throughout the trial, or whether and how it had been accelerating) and
the geometric-kinematic relations of the two clocks between each other (e.g. whether they had jointly been members of the same inertial system; or whether and how they (also together with suitable auxiliary participants under consideration) had been only rigid to each other, or whether and how they had even been moving wrt. each other)
must be determined and taken into account when trying to compare "rates" (i.e. "proper rates", of course) between clocks;
and when trying to determine whether either of the clocks had been "good" in the first place (e.g. whether either of them had been "ticking at regular/constant rate" at all).

why do muons decay slowly when moving fast?

The experimentally determined "mean life $\tau$" of muons (i.e. the mean duration for muons of a sample, for each from having been generated until having decayed; of sufficiently free muons) is listed by the Particle Data Group (PDG) as one specific value (under pretty much any circumstances experimentally accessible so far).

Of course, the duration of some race track, for instance, from the starting block releasing a (newly generated) muon until the finish line registering the muon decay is another quantity that may be considered (as far as it can be defined/determined at all; i.e. if the starting block and the finish line of the race track were and remained jointly members of the same inertial system).
Indeed, the comparison between this described duration of a (suitable) racetrack and the corresponding "life" duration of a muon under consideration is closely related to how to compare "rates" of separate clocks. From the detailed definition of how to measure and compare these quantities it follows that the described duration of a (suitable) racetrack is necessarily at least equal, or larger, than the corresponding "life" duration of a muon. (The expression for this ratio, in terms of measured geometric-kinematic relations, often involves the "dilation factor $\gamma \ge 1$".)

But foremost: it is proper and important to rigorously distinguish the described duration of a (suitable) racetrack from the corresponding "life" duration of a muon under consideration.

share|improve this answer
add comment

Time dilation actually occurs with objects that are moving at some portion the speed of light relative to any observer "that is to say" relatively stationary to that object compared with light speed; However to the "OBJECT" in relative motion, it would of course see this effect on the stationary observer as the one's clock moving slow. but that is only because of the protraction of signaling it receives, MEANING MEASUREMENTS HAVE NO MEANING AT HYPERLUMINUS SPEEDS THAT GO BEYOND THE TIME PENTICLE DISTANCE/VELOCITY-but in reality; the closer to light speed, the more decimated the time for any object (even if both objects are traveling in opposite directions from one another "then this effect would indeed be mutual in that case" for both objects) and thus there clocks would be in sync, but also consider that if for an object that was moving half the speed of light "in the direction of that light" that relative to that object the speed of light would be moving twice as slow, (or so you would think) but this is how time dilation corrects the in balance buy slowing the objects time causing velocities of outside objects plus there rate of time dilation to speed up, so that the speed of light "relative to the object" be once again the speed of light buy comparison. but as for light travelling opposite and away from you, it would be travelling faster than light added your speed moving in the opposite direction; but you could only observe how fast its going only as fast as the light from it is travelling toward you-Hence the speed of light: (then everything from that direction would be therefore, slow) but then if this light moving opposite of you where coming from the light moving ahead of you (the one moving in the same direction relatively light speed to your time frame) you would think it move faster but you can only observe the effects as fast as the light moving ahead of you-again, light speed].

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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