- #1

Pushoam

- 962

- 52

## Homework Statement

## Homework Equations

1) k.e. = (1/2)mv

^{2}, non- relatvistic

2)K.E. = m

_{0}c

^{2}(ϒ -1), relativistic

3) L=L

_{0}/ϒ , Length contraction

## The Attempt at a Solution

(A) Taking 10

^{19}eV to be the kinetic energy of the proton,

Non- relativistic calculation

(1/2)mv

^{2}= 10

^{19}eV = 1.6

v = √[ (3.2/1.67)* 10

^{27}] = 4.3 * 10

^{13}m/s

How is kinetic energy defined relativistically?

K.E. = m

_{0}c

^{2}(ϒ -1)

We can't take 10

^{19}eV as total energy of the proton as if we take so then we won't be able to calculate speed of the proton.

So, for calculating speed of the proton, I have to take 10

^{19}eV as the kinetic energy of proton.

So,

ϒ

^{2}= [ {10

^{19}eV/ m

_{0}c

^{2}}+1]

^{2}

m

_{0}c

^{2}= 939 MeV

Solving it gives,

1 - v

^{2}/c

^{2}= [939 * 10

^{-13}]

^{2}

v ≈ c

Hence, w.r.t. galaxy frame the proton will take 10

^{5}years to cross the galaxy.

B)

W.r.t. the proton frame,

the length of the galaxy is shortened to 10

^{5}[939 * 10

^{-13}] light years.Hence, the time taken = 10

^{5}[939 * 10

^{-13}] yearsIs this correct?