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A spectator in the lab sees att=0two particles that areLmeters apart, moving towards each other.

The speed of each particle in the system of the lab (S) is [tex]v=\beta c[/tex].

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 [tex]t_0=\frac{L}{2 \beta c}[/tex]

It is also clear that the collision's x coordinate will be [tex]x_0=\frac{L}{2}[/tex] also is the system of the lab.

So I used the Lorentz transformation:

[tex]t'=\gamma (t-v \frac{x}{c^2}) [/tex]

[tex]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}

[/tex]

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:

[tex] u'_x=\frac{u_x-v}{1-v \frac{u_x}{c^2}} [/tex]

[tex]u_x=-\beta c \mbox{(the speed of the right particle in the system of the lab (S))}[/tex]

[tex]v=\beta c[/tex]

[tex]u'_x=\frac{-2\beta}{1+\beta^2} c[/tex]

Now, the left particle sees shorter distance, i.e. [tex]L'=L/\gamma[/tex]

and [tex] t'_0=L'/u'_x \Rightarrow t'_0=\frac{L (1+\beta^2)}{2 \beta \gamma c}[/tex]

So i have another, different, seemingly right answer, which is also wrong.

Can someone here please help me?

Thanks in advance.

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# Homework Help: Something in special relativity that is supposed to be easy

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