# Time Dilation and travel

1. Feb 6, 2014

### MostlyHarmless

The proper mean lifetime of $\pi$ mesons(pion) is $2.6x10^{-8}s$. Suppose a beam of such particles has a speed of .9c.

a) What would their mean life be as measured in the lab?

b) How far would they travel(on average)before they decay?

c)What would your answer be to part (b) if you neglected time dilation?

d) What is the interval in spacetime between the creation of a typical pion and its decay?

For part a) I just found the dilated time interval.

$Δt={\frac{Δt'}{{\sqrt{1-v^2/c^2}}}}$ Where Δt' is the proper time given, and V is the speed at which the pion's reference frame, S', is moving wrt to the lab, S.

Simply plugging in I got $Δt=6.65x10^{-8}s$

For part b) I'm assuming it means with respect to the lab, so would I just multiply $Δt$ by V?

(This is where the my confusion from the my thread in the physics section comes from: https://www.physicsforums.com/showthread.php?t=736893)

For this step should I be able to find the same distance if I applied the length contraction formula or used Δt and V?

Edit: After messing with some formulas I found that multiplying v by $Δt$ wouldn't give the same as length contraction.

I found that, V, the velocity relative to the lab, S, could be expressed as proper length/ proper time. (Right?)

So $Δt • V$ would give $L_p • \gamma$
While Length Contraction, L, is given by: $L={\frac{L_p}{\gamma}}$

I also found an example in my book that is essentially part b) and they used $Δx=vΔt$. So then what is the difference between L and Δx?

Last edited: Feb 6, 2014
2. Feb 6, 2014

### Rococo

I am getting a different answer for part (a):

$Δt={\frac{2.6\times10^{-8}}{{\sqrt{1-(0.9c)^2/c^2}}}}$

$Δt=5.96\times10^{-8}s$

For part (b), you are correct. The distance traveled by the pion with respect to lab, $Δx$, will be equal to the velocity of the pion with respect to the lab, $v$, multiplied by the lifetime of the pion as measured in the lab, $Δt$.

The symbol $L$ means the length of something, such as a rod, as measured in the lab. When you measure the length of an object, it must be the case that both ends of the rod are measured at the same time. If you found the position of one end of the rod, and then found the position of the other end of the rod some time later, you would not know the length of the rod because it may have moved in this time.

The symbol $Δx$ means the distance between two points in space, as measured in the lab. If Event 1 occurs at a point $x_1$ and a time $t_1$, and if Event 2 occurs at a point $x_2$ and at a different time $t_2$, then the spatial separation between the two events, as measured in the lab, will be $x_2-x_1=Δx$.

So, you cannot use the length contraction formula to find the distance traveled by the pion in the lab reference frame. You would use the length contraction formula to determine, say, the physical length of a moving rod in the lab frame, which has a length of 1 meter in its rest frame. But, it cannot be applied in this question, because we are not talking about a moving object with a particular length. The question is about a particle travelling a particular distance through space. The distance is found as you said: $Δx=vΔt$

3. Feb 6, 2014

### MostlyHarmless

Apologies, I typed 2.6, rather than 2.9.

And that's makes sense, I was thinking that length and interval were interchangeable.

Now, the formula I used for time dilation, I was reading and I realized it was derived(at least in my book) from a triangle. Basically a ship traveling at speed v wrt S, has a source or light and a mirror a distance L apart in the y' direction. In S', the ships rest frame, the light simply travels a distance 2L. But in S, it travels some distance based on the triangle it makes by moving. The specifics aren't really relevant at the moment. Is this a general formula for time dilation or would it only pertain to that specific example? I ask because the next question asks about the ship except the light source and mirror are oriented in the x' direction. At its face it doesn't seem like the same problem, and then trying to draw a diagram didn't help much.

See this post for a diagram of the ship, with light source and mirror oriented in y'.

And the light and mirror rotated to be wrt to x'. Same problem? Or completely different?

Last edited: Feb 6, 2014
4. Feb 7, 2014

### Rococo

I'm getting an error when I try to follow the link, maybe you can post the same picture here.

Yes, it is a general formula for time dilation. Even when the light and mirror are arranged horizontally, it will still apply. That is, in the S' frame, the time for the light to reach one mirror and return back again $Δt'=\frac{2L_0}{c}$, where $L_0$ is the length between the two mirrors as measured in the S' frame. But in the S frame, this time will be measured to be $Δt=γΔt'=γ\frac{2L_0}{c}$.

The horizontal arrangement of the mirror and the light source in the S' frame is usually the way the length contraction formula is derived. If we let $Δt_{out}$ be the time, as measured in the S frame, for light to go out the far mirror, and $Δt_{in}$ be the time, as measured in the S frame, for the light to back in again, then it can be shown:

$cΔt_{out}=L+vΔt_{out} → Δt_{out}=\frac{L}{c-v}$

$cΔt_{in}=L-vΔt_{in} → Δt_{in}=\frac{L}{c+v}$

Can you see where these formula have come from? $L$ is the length between the mirrors in the S frame. In the first case, the light must travel this length, plus the extra distance the far mirror moves away from it in the time $Δt_{out}$, $vΔt_{out}$. In the second case, the light travels a shorter distance on its return journey because the first mirror is moving towards it, i.e. it must travel a distance $L-vΔt_{in}$.

So, that means that in the S frame, the total time for the light to travel to one mirror and back again in given by $Δt=Δt_{out} + Δt_{in}=\frac{L}{c-v}+\frac{L}{c+v}$.

But, now we must use the time dilation result I mentioned earlier, $Δt=γΔt'=γ\frac{2L_0}{c}$.

$\frac{L}{c-v}+\frac{L}{c+v}=γ\frac{2L_0}{c}$

$\frac{L}{c-v}+\frac{L}{c+v}={\frac{2L_0}{{c\sqrt{1-v^2/c^2}}}}$

$\frac{2Lc}{c^2-v^2} = \frac{2L_0}{\sqrt{c^2-v^2}}$

And through algebraic rearrangement:

$L=\sqrt{1-\frac{V^2}{c^2}}L_0$

$L=\frac{L_0}{γ}$

Which is the length contraction formula.

Last edited: Feb 7, 2014