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I have just begun with special relativity so pardon me if my question seems too obvious.

In the books I am following, there is an example of time dilation which says:

The half life of muons is $\tau$(in the proper frame of muon). We have muon beam moving with a speed of $0.999 c$ and so the time taken for the beam intensity to reduce to half ,in the lab frame, would be $\tau \gamma$ where $\gamma$ is the Lorentz factor for this beam.

However we can also say, that the people in the lab frame would have aged only $\frac{\tau}{\gamma}$ with respect to the muon frame, because the muons feel that the people in the lab frame are going backwards at the same speed.

Correct me if I am wrong, but I feel that it is a contradiction, that on seeing one way the observers in the lab have aged $\tau \gamma$, and in the other $\frac{\tau}{\gamma}$.

Where I am going wrong?


marked as duplicate by WillO, ZeroTheHero, Jon Custer, Kyle Kanos, David Z Sep 18 '18 at 6:24

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  • $\begingroup$ You can't be in two different reference frames at once, just like you can't be in two places at once, so there is no contradiction. $\endgroup$ – probably_someone Sep 14 '18 at 13:13
  • 4
    $\begingroup$ Welcome to Special Relativity. $\endgroup$ – JEB Sep 14 '18 at 15:51
  • $\begingroup$ Welcome to Physics! Please note that this site is not a place to obtain solutions to worked problems. Please see this Meta post on asking homework-like questions and this Meta post for "check my work problems". $\endgroup$ – Kyle Kanos Sep 17 '18 at 10:15
  • $\begingroup$ I strongly disagree that this violates our homework policy. Like, it's probably a duplicate but just because it has some actual numbers in it doesn't make it a homework problem. $\endgroup$ – CR Drost Sep 17 '18 at 13:17
  • $\begingroup$ Yeah, I also disagree that this violates the homework policy. It does seem pretty familiar, though, so I think it's a duplicate of something. $\endgroup$ – David Z Sep 18 '18 at 6:24

Yes, you're right that something is not OK here. How come time in one frame seems to slow down, while in the other one it seems to speed up?

Short answer: it's not what happens.

Actually, from both frames, the time in the other frame appears to slow down. While this sounds impossible at first, it turns out special relativity comes with a whole bunch of strange phenomena. This is one of them.

Have you heard of the Minkowski diagram? It's an intuitive, visual representation of the relation of different reference frames.

Minkowski diagram

For any reference frame, events (=points in spacetime) appear to happen at the same time only if the line that connects them is paralell with the x (space) axis of the frame.

From the black frame, A and B appear to happen at the same time, and if you look closely, OB is shorter than OA. Time seems to run slower in the blue frame.

But from the blue frame, B and C appear to happen at the same time. OC is shorter than OB, so time seems to run slower in the black frame.


You need to be careful to be clear who observes a time. The lab people observe the muons to take $\tau \gamma$ to decay from their point of view. They infer that, in the muon's frame, it experienced $\tau$. From the muon's frame it experiences $\tau$ before decaying. It sees the lab folk experiencing $\frac{\tau}{\gamma}$ during that time.

Different observers will not agree on times that separate events. Note that this is entirely symmetrical. The lab folk see the muon slowed by a factor of $\tau$ and the muon sees the lab folk slowed by the same factor.

  • $\begingroup$ But if the age of the lab guy is x units. Then according to one frame his age after muon half life would be $x+\tau \gamma$ and according to the other frame the age would be x+ $\frac{\tau}{\gamma}$. Can this be so? $\endgroup$ – K.A. Siddiqui Sep 14 '18 at 15:15
  • $\begingroup$ Yes, because there is spatial (space-like) separation between the to events (the event of deciding how much the other aged). If you turn the muon beam around and bring it back to the same spatial position as the grad student in the lab: boom, the muon has aged less. This is the twin paradox. The key (imho) to resolving the paradox is realizing that "now" (aka the hyper slice of simultaneity) is frame dependent, and the difference between 2 slices of "now" for 2 observers is position dependent. $\endgroup$ – JEB Sep 14 '18 at 15:54

Your question is far from trivial. You can't imagine how many people are confused as you are. I'll do my best to explain the reason behind that.

My answer to a related question, a few days ago, began:

First of all, I think you should learn the basic concepts for this kind of problems. There are three:

  1. objects

  2. events

  3. reference frames.

Let's apply my prescription to your problem.

  • Objects. The problems refers to a beam of muons, but it is better to think of a single muon. This is the object.

  • Events. There are two: the muon's birth $B$ and its decay $D$.

  • Reference frames. Two frames are involved: the muon's rest frame $K$ (problem's text says "proper frame") and the laboratory frame $K'$.

Now for time measurements. No problem in frame $K$, where the muon is at rest. One clock $C$ is enough, and $\tau$ is the time elapsed from event $B$ to event $D$, as recorded by $C$. Note that we can safely assume that $C$ is placed just near the resting muon.

Measuring time in $K'$ is another story: since the muon is moving in this frame, one clock is not enough. This was Einstein's main intuition: unless you assume there is an absolute time, when you are interested in events happening in different places of your reference frame you must read times via two clocks, each located in the place where an event happens. (Of course, clocks must have been previously synchronized. I cannot dwell explaining how synchronization may be effected).

Actually clocks are not all, as you have to detect emission $B$ and decay $D$ of the muon. So in two places you will set two more complex instruments, $I_1$ and $I_2$, able to detect the relevant events and record their times. Then you have only to read these times and calculate the difference $\tau'$. You expect this difference equates $\gamma\,\tau$.

After that, you begin another story, which talks of aging people. But I urge you to apply my prescription again:

  • Which is the object? It is the aging person $P$, no longer the muon.
  • Which are the frames? (I exchanged the order of questions for reasons you will presently understand.) They are $K$ and $K'$, as before, but the person is resting in $K'$, not in $K$.
  • Which are the events? Here you have to decide: Where do you want to put the person? Suppose you choose to put him near $I_1$. Then the first event can coincide with $B$ (muon's emission) but the second cannot coincide with $D$, which in $K'$ happens elsewhere. Take whatever event you like in $P$'s life, say $F$, and call $\theta'$ the time interval (measured by $I_1$) from $B$ to $F$.

Surely you may take measurements on $B$ and on $F$ in frame $K$, but in this frame these events do not happen in the same place, so two clocks are needed. If $\theta$ is the time interval as measured by these clocks, which relation do you expect between $\theta$ and $\theta\,'$?

I am sure you see that now the situation is exactly reversed wrt the muon case: events $B$ and $F$ (relating to $P$) happen in the same place of $K'$ but in different places of $K$. Therefore time dilation acts in reverse: $\theta=\gamma\,\theta'$. There's no contradiction, since the $\tau$'s referred to events $B$ and $D$, whereas the $\theta\,$'s refer to $B$ and $F$.


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