Something in special relativity that is supposed to be easy

[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
I had two approaches:

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:
$$t'=\gamma (t-v \frac{x}{c^2})$$
$$t'_0=\gamma (\frac{L}{2\beta c}-\beta c \frac{L}{2} \frac{1}{c^2})<br /> =\gamma (\frac{L}{2 \beta c}-\frac{\beta^2 L}{2 \beta c})<br /> =\gamma L \frac{1-\beta^2}{2 \beta c}$$

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.

Can someone here please help me?
Thanks in advance.

 

[Doc Al]

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:
$$t'=\gamma (t-v \frac{x}{c^2})$$
$$t'_0=\gamma (\frac{L}{2\beta c}-\beta c \frac{L}{2} \frac{1}{c^2})<br /> =\gamma (\frac{L}{2 \beta c}-\frac{\beta^2 L}{2 \beta c})<br /> =\gamma L \frac{1-\beta^2}{2 \beta c}$$

It seems fine to me... but it's wrong.

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

 

[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...

I had L=40 meters, V=0.7c
My first answer was 0.101360544 microseconds and the second was 0.068 microseconds... both were rejected.

Thanks for the reply

 

[Doc Al]

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.)

 

[Doc Al]

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

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.

 

[Ahmes]

Thanks for your answer, but I don't think I got it...
You said:

you assume that the distance between the particles as seen by the left particle equals $L/\gamma$. Not so!

And then you said:

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.

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

Thanks again for your help.

 

[Doc Al]

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?

Ask yourself this: At what time (according to the left particle frame) does the right particle pass the point L/2? If we call the origin of the frame to be the point where the left particle crosses -L/2, then the spacetime coordinates of that event are x = L, t = 0 in the lab frame. Which means the time in the left particle frame is:
$$t' = \gamma(t - xv/c^2) = -\gamma(L/c^2)$$
Which means the right particle crosses L/2 before the left particle reaches -L/2. If you want, you can figure how far the right particle must have moved by the time the left particle is at -L/2 (which happens at t'=0).

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

It better get you the same answer!