A problem on time taken by relativistic particles

[visharad]

Homework Statement

a) What is the speed of proton that has relativistic γ factor of 10^10? Write as (1-ε)c

b) A proton with γ=10^10 is chased by a photon across a galaxy. Observer A at one end of the galaxy (xA=0) sees the proton at t1=0 and the photon at t2 = 1.33 x 10^-8 s. According to A, how far apart are the two particles at t1?

c) Another observer B is at another end of the galaxy (xB=10^21 m). The two observers are at rest relative to each other. At what times will B see the particles go past?

d) How many years does it take the proton to cross the galaxy according to A and B?

Homework Equations

L = Lo $\sqrt{1 - v^2/c^2}$
Δt = Δto/$\sqrt{1 - v^2/c^2}$

The Attempt at a Solution

a) Y = 1/$\sqrt{1 - v^2/c^2}$
1 - v^2/c^2 = 1/Y^2
v^2/c^2 = 1 - 1/Y^2
v^2/c^2 = 1 - 1/(10^10)^2
v^2/c^2 = 1- 1/10^20
v/c = sqrt(1 - 1/10^20)
v/c = (1 - 5 * 10^-21)
v = (1 - 5*10^-21) c
Is this correct?

b) The photon covers the distance in t2 - t1 = 1.33 x 10^-8 s
The distance = c t^2
= 3 x 10^8 x 1.33 x 10^-8
= 3.99 m
Is this correct?

c) Can we say that the observer B will see the proton pass by at t3 = t1 + xB/v and the photon pass by at t4 = t2 + xB/c ?
If not, then what is the correct method?

d) Can we simply divide distance (which is 10^21 m) by speed and convert into years? If not, then what is the correct method?

For c and d, I am not sure if we need to use Lorenz transformation equations. The observers are at rest relative to each other. So I think we do not need.

 

[visharad]

For part c, is it possible to express the answers as combination of a large number and a small number? I want to express t3 and t4 in such a way that it is easy to see the difference between them.

 

[visharad]

Please see if this is correct
c) t3 = time at which observer B sees the proton fly by
t4 = time at which observer B sees the photon fly by

t3 = t1 + 10^21/(1-ϵ)c
t3 = 3.33 * 10^12/(1-ϵ)
t3 = 3.33 * 10^12[1 + 1/(1-ϵ) - 1]
t3 = 3.33 * 10^12[1 + ϵ/(1-ϵ)]
Since ϵ << 1, therefore using the approximation 1-ϵ = 1, we get
t3 = 3.33 * 10^12(1 + ϵ)
t3 = 3.33 * 10^12(1 + 5*10^-21)
t3 = 3.33 * 10^12 + 1.67 * 10^-8 s
t4 = t2 + 10^21/c
t4 = 1.33 * 10^-8 + 3.33 * 10^12
t4 = 3.33 * 10^12 + 1.33 * 10^-8 s

d) For both observers, the time taken by the cosmic ray is approximately 3.33 * 10^12 s
= 10^5 years