Question

SPECIAL RELATIVITY QUESTION
A physicist determines that a sub-atomic particle created at one end of the laboratory moved at 0.94c and survived for 0.32 us, decaying just as it reached the other end.
a) How far did the particle move in the frame of the laboratory?
b) How long did the particle survive as measured in its rest frame?
c) In the rest frame of the particle, how long is the laboratory?

Answer 1

Are you familiar with the consequences of the Lorentz Transformations that allow different reference frames to represent the same events?

Answer 2

a.k.a. Length Contraction and Time Dilation?

Answer 3

Yes.

Answer 4

Just confused about the mathematics.

Answer 5

Okay. So the information given is from the observer in the lab.
v = 0.94 c
t = 0.32 us

So, the time measured is the dilated time. This is what the lab saw.
To calculate distance, that is simply velocity multiplied by time.
x = vt

In it's own rest frame, the particle will survive a shorter time. This is because of time dilation. What we see in the lab would be larger than what the particle would experience.
Relationship: t = (gamma) to, where t is 0.32 us. gamma is the expression 1/sqrt(1-v^2)

To the particle, the length of the lab will be larger. This is because what we observe in the lab is shorter due to length contraction.
Relationship: L = (gamma) / Lo, where L is x from above. gamma is the same as above.

Steps:
(1) x = vt
(2) solve for to
(3) solve for Lo

Answer 6

Thanks! I'll solve and see if everything makes sense according to my spacetime diagram.

Answer 7

In the particle frame of reference the lab will be shorter, not longer, than it is in the lab rest frame.

Answer 8

Yep, I just caught that. The largest length is always in the object's rest frame, which in this case is the frame of the laboratory.