Question

A cosmic ray (an electron or nucleus moving ar speeds close to the speed of light) travels across the Milky Way at a speed of 0.98 times the spee of light. If the Milky way is 30,000 pc across, how long does this journey take in our (essentially stationary) frame of reference and the cosmic ray's frame of reference, in years?

Answer 1

Answer:

Cosmic ray's frame of reference: 99,875 years

Stationary frame of reference: 501,891 years

Explanation:

First of all, we convert the distance from parsec into metres:

$d=30,000 pc =9.26\cdot 10^{20} m$

The speed of the cosmic ray is

$v=0.98 c$

where

$c=3.0 \cdot 10^8 m/s$ is the speed of light. Substituting,

$v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s$

And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:

$T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s$

Converting into years,

$T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years$

Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:

$T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}$

And substituting v = 0.98c, we find:

$T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years$