How long until two particle bunches catch up?

[scienceman2k9]

Problem:

An observer at rest in a laboratory sees particles emitted from a source. The first bunch of particles is emitted with speed v=0.92c; the second bunch emitted 10^-3 s later with v=0.99c. How long, according to the observer, will it be until the second bunch catches up with the first?

I'm not sure how straight forward this problem is. Using "classical" physics I get T=1.3*10^-2 s I am not sure what to account for in this problem...relativistic relative velocity maybe?

 

[HallsofIvy]

Yes, use the relativistic formula for the speed of the second particles relative to the first:
\frac{v_2-v_1}{1- \frac{v_2v_1}{c^2}}
\frac{.97- .92}{1- (.98)(.92)}c

(By the way, classically it is 1.3* 10^-3.)

 

[Doc Al]

I'm not sure how straight forward this problem is. Using "classical" physics I get T=1.3*10^-2 s I am not sure what to account for in this problem...relativistic relative velocity maybe?

The relativistic relative velocity would be needed if you were asked to solve the problem from the view of one of the particles. But here you only need to find the time according to the observer at rest, so "classical" physics is all you need. Yes, it's that simple.

 

[Andrew Mason]

Yes, use the relativistic formula for the speed of the second particles relative to the first:
\frac{v_2-v_1}{1- \frac{v_2v_1}{c^2}}
\frac{.97- .92}{1- (.98)(.92)}c

(By the way, classically it is 1.3* 10^-3.)

I am confused here. I don't see why the relativistic formula is used since all distance and time measurements take place in the observer's frame.

(First of all, shouldn't the .97 and .98 both be .99?)

If the observer sees the particles moving at .92c and .99c respectively, by definition he measures a speed difference of .07c. This means he measures their separation to decrease by 2.1x10^7 m (his metres) per second (his seconds). The first particle travels .92c x 10^-3 m = 2.8x10^5 m before the second particle starts. So it takes d/v = 1.3x10-2 seconds (his seconds) before the observer sees the two particles at the same position.

AM

 

[George Jones]

Yes, use the relativistic formula for the speed of the second particles relative to the first

If both speeds are with respect to the lab frame, then I don't think so.

Suppose that, in the lab frame, the first bunch of partcles propagates at speed v_1 from the spacetime origin, while the second bunch propagates at speed v_2 from x = 0 starting at t = T.

Then, it is straightforward to write down the equations of the two worldlines, and to find the lab time coordinate of the event at which the worldlines intersect.

 

[scienceman2k9]

thanks!...the problem was pretty straight forward afterall

 

[Meir Achuz]

Problem:

An observer at rest in a laboratory sees particles emitted from a source. The first bunch of particles is emitted with speed v=0.92c; the second bunch emitted 10^-3 s later with v=0.99c. How long, according to the observer, will it be until the second bunch catches up with the first?

I'm not sure how straight forward this problem is. Using "classical" physics I get T=1.3*10^-2 s I am not sure what to account for in this problem...relativistic relative velocity maybe?

I think the "classical" answr is correct here. Using relative velocity would give the time in the rest system of one of the particles.