# Something in special relativity that is supposed to be easy

1. Jan 19, 2005

### Ahmes

and it probably is...
A spectator in the lab sees at t=0 two particles that are L meters apart, moving towards each other.
The speed of each particle in the system of the lab (S) is $$v=\beta c$$.
Calculate the time of their collision, in the system of one of the particles.
O---------> v v <------------O

1) In the system S it is obvious that the collision will take place at $$t_0=\frac{L}{2 \beta c}$$
It is also clear that the collision's x coordinate will be $$x_0=\frac{L}{2}$$ also is the system of the lab.
So I used the Lorentz transformation:
$$t'=\gamma (t-v \frac{x}{c^2})$$
$$t'_0=\gamma (\frac{L}{2\beta c}-\beta c \frac{L}{2} \frac{1}{c^2}) =\gamma (\frac{L}{2 \beta c}-\frac{\beta^2 L}{2 \beta c}) =\gamma L \frac{1-\beta^2}{2 \beta c}$$

It seems fine to me... but it's wrong. So I tried another thing:
I calculated the relative speed of the right particle in the system of the left particle, according to the transformation of velocities:
$$u'_x=\frac{u_x-v}{1-v \frac{u_x}{c^2}}$$
$$u_x=-\beta c \mbox{(the speed of the right particle in the system of the lab (S))}$$
$$v=\beta c$$
$$u'_x=\frac{-2\beta}{1+\beta^2} c$$
Now, the left particle sees shorter distance, i.e. $$L'=L/\gamma$$
and $$t'_0=L'/u'_x \Rightarrow t'_0=\frac{L (1+\beta^2)}{2 \beta \gamma c}$$
So i have another, different, seemingly right answer, which is also wrong.

2. Jan 19, 2005

### Staff: Mentor

Looks fine to me. What makes you think it's wrong? Note: You can simplify your answer if you realize that
$$1-\beta^2 = 1/\gamma^2$$

3. Jan 19, 2005

### Ahmes

Well, I think it is wrong because the automated system that checks the answers didn't accept this...
And the fact that I have two different answers (both look right) also throws me off balance...

My first answer was 0.101360544 microseconds and the second was 0.068 microseconds... both were rejected.

4. Jan 19, 2005

### Staff: Mentor

Automated systems?
My answer agrees with 0.068 microseconds. Note that you are giving the time as measured by the "left" particle assuming that its clock read t' = 0 just as it passed the observer.

Try redoing it assuming that the observer is at the collision point. Let t' = 0 just as the left particle passes x = - L/2.

(I'll take another look at the problem when I get a chance; I'm running out the door.)

5. Jan 19, 2005

### Staff: Mentor

As soon as I hit enter I realized that this will give you the same answer. (As it better! )

I looked at your second method, where you found the relative velocity of the particles. Everything is OK except the last step where you assume that the distance between the particles as seen by the left particle equals $L/\gamma$. Not so! While the lab observer sees the left particle pass x = -L/2 at the same time that the right particle passes x = +L/2, the particle frame observers disagree. According to the left particle frame, when the left particle reaches x = -L/2, the right particle has already passed x = +L/2. Stick with the first method.

6. Jan 20, 2005

### Ahmes

You said:
And then you said:
So where is the right particle in the left particle's frame (when the left one passes -L/2) if not $L/\gamma$ meters from it?
If I'm not mistaken (and I might be and probably am mistaken) the calculation you suggest will give the same expression I got before

$$t' = \gamma(t - xv/c^2) = -\gamma(L/c^2)$$