Question

4-4 Our galaxy is about 10^5 light years across, and the most energetic particles known have an energy of about 10^19eV. How long would it take a proton with this energy to traverse the galaxy as measured in the rest frame of
The galaxy ? (10^5 yrs)
The particle ? (5 mins)

Answer 1

It's the 10^19eV that SEEMS to be causing me problems here. The numbers I get seem to go into a lot of decimal places just shy of c. and my calculator and brain explode ...

Answer 2

Using rest energy of a proton \( m_p = 938.272 MeV/c^2\), we can see

\(\gamma = \dfrac{E}{E_0} = \dfrac{10^{19}}{938.272 \cdot 10^6} \approx 1.07 \cdot 10^{10}\)


With events:

- \(E_1\): proton starts journey,

- \(E_2\) proton finishes journey,

...... we have space-time co-ordinates:

S: rest frame of Universe

\(E_1: (0,0)\)
\(E_2: (x,t)\)

S': rest frame of proton

\(E'_1: (0,0)\)
\(E'_2: (0,t')\)

In frame S, it is simply:

\(t = \dfrac{ x}{v}\)


And given \(\beta = \sqrt { 1- \dfrac{1}{\gamma^2} } \approx 1\) then

\(t = \dfrac{ 10^5 c }{c} = 10^5\)

In frame S', firrstly, the Lorentx transforms have simplified because \(x' = 0\) to these:


\(x = \gamma v t'\)

and

\(t = \gamma t'\)

from the second

\(t' = \dfrac{t}{\gamma } = \dfrac{10^5 * 365 * 24 * 60}{1.07 * 10^{10} } \approx 4.96 \) min


in terms of calculator......i used the net :-))
https://www.wolframalpha.com/input/?i=(10%5E5+*+365+*+24+*+60)%2F(1.06+*+10%5E(10)+)

https://www.wolframalpha.com/input/?i=((10%5E19)%2F(938.272+*+10%5E6)+)