# Calculating time/velocity for relativistic cosmic ray

1. Dec 12, 2008

### vaizard

1. The problem statement, all variables and given/known data
A primary cosmic ray (which comes from elsewhere in our Galaxy) hits the nucleus of an
atom in our atmosphere 38 km above the ground and creates an unstable secondary particle,
which then travels straight down toward the ground at a speed of 0.99400c. (The altitude and the speed given in the previous sentence are measured by a person at rest on the ground.)
a) As measured by a person on the ground, how long does it take this particle to reach the
ground?
b) Using time dilation, calculate the time it takes the particle to reach the ground, as
measured in the frame of the particle.
c) Using length contraction, calculate the distance from the point of the particle’s creation to
the ground as measured in the frame of the particle.
d) Using your answer to part (c), calculate the time to reach the ground, measured in the
particle’s frame.

2. Relevant equations
$$\Delta t = \frac{\Delta t_0}{\sqrt{1-{u^2/c^2}}}$$
$$l = l_0 \sqrt{1-u^2/c^2}$$
$$\gamma = \frac{1}{\sqrt{1-u^2/c^2}}$$

3. The attempt at a solution
Part (a): $$\Delta t = \frac{l}{u} = \frac{38*10^3}{0.994*(3*10^8)} = 1.274*10^{-4} \, \mbox{seconds}$$

Part (b): $$\Delta t_0 = \Delta t \sqrt{1-u^2/c^2} = 1.274*10^{-4} \sqrt{1-0.994^2} = 1.394*10^{-5} \, \mbox{seconds}$$

Here is where I'm running into a problem, on part (c):
I think that for this part, you solve for $$l_0$$, because that would give the proper length:
$$l_0 = \frac{l}{\sqrt{1-u^2/c^2}} = \frac{38*10^3}{\sqrt{1-0.994^2}} = 3.474*10^5 \, \mbox{meters}$$

The solution to this problem, however, says:
$$l = \frac{l_0}{\gamma} = \frac{38*10^3}{9.142} = 4.156*10^3 \, \mbox{meters}$$

This doesn't make any sense according to length contraction either, because this answer should be the longest possible distance, since this answer is the length in the frame of the object being measured. But this value is less than that from the observer on Earth.

Is there something that I'm missing here?

2. Dec 12, 2008

### cepheid

Staff Emeritus
In the rest frame of the particle, the particle is at rest (by definition) and the Earth is rushing up towards it at 0.994c. For this reason, the Earth's dimensions (in this case distance through atmosphere between particle and ground) appear length contracted to the particle (because the Earth is in motion relative to it). As a result, the 38 km (proper length) as measured by the Earth-bound observer appears length contracted to the "stationary" observer to something much smaller than 38 km.

By the way, if you're still questioning the solution given in the book even after my explanation, I should also point out that it has the virtue of being in exact agreement (with YOUR answer) for delta t0, the time elapsed before the surface of the Earth hits the particle (as measured in the particle reference frame). So you know the book's answer is correct, it's just a matter of convincing yourself of that.

3. Dec 13, 2008

### vaizard

Does this mean that 38 km is the proper length?

4. Dec 14, 2008

### cepheid

Staff Emeritus
Sure. Proper length is the length as measured in the frame of reference in which the thing you're measuring is stationary. If you're on a racecar, and you somehow manage to get out a ruler and measure its length, (while moving) to be 3 m, you'll get the same value as what you did when you measured it when it was sitting in the garage. However, some guy at the side of the track who measures the length of the car while you're moving will get a value that is smaller than 3 m, because you're moving relative to him, therefore, from his perspective, the car is length contracted. This is not the proper length, because it is not the length as measured in the rest frame of the car.

The only thing that (I suspect) you're having trouble wrapping your head around is that for part c, we're using the following perspective:

guy in racecar = earth bound observer

guy at the side of the track = particle

5. Dec 14, 2008

### vaizard

Yes, that's the problem, but I think I'm starting to get it. What's being measured is the Earth, right? And therefore, the Earth bound observer is the one that measures the proper length.

6. Dec 14, 2008

### cepheid

Staff Emeritus
Exactly. To reiterate, part a was done in the rest frame of the earth, whereas in parts b and c, we switched to the rest frame of the particle. Therefore, our definition of who was "moving" and who was "stationary" changed.