# Isotropic emitter (Homework 7-323)

1. (simple problem) An isotropic emitter of 2 GeV total-energy protons (“isotropic” and “2 GeV” in the rest frame of the emitter) moves with velocity β=0.6 in a trajectory radially away from the Earth.

a. Find the speed of a proton emitted towards Earth in the Earth’s frame and in the emitter’s frame.

b. Find the time it takes for a proton to reach the Earth in the Earth’s frame and in the emitter’s frame when the emitter is at a distance 1015 km from Earth.

We know that

$E^2=p ⃗^2 c^2+m_0^2 c^4$    so that $p ⃗ ^2=(E^2-m_0^2 c^4)/c^2$
The relativistic 3D momentum is defined as

$p ⃗=mv ⃗=γm_0 v ⃗$

Thus in the emitter frame:

$(m_0^2 v^2)/(1-v^2/c^2 )=(E^2-m_0^2 c^4)/c^2$      or
$(m_0^2)/(1/v^2 -1/c^2 )=(E^2-E_0^2)/c^2$    or
$1/v^2 =1/c^2 +(m_0^2)/((E^2-E_0^2)/c^2)$

We have:

$E_0=938 MeV,m_0=1.67*10^(-27) kg ,E=2 GeV$

$1/v^2 =1/c^2 +(1.67*10^{-27} )^2/(([(2*10^9 )^2-(938*10^6 )^2]*(1.6*10^{-19} )^2)/(3*10^8)^2 )=$

$=1/(3*10^8 )^2 +3.14*10^{-18}$

$v=2.649*10^8 m/s$      if $c=3*10^8 m/s$
In the Earth frame

$u_{emitter}=-βc=-0.6 c=-1.8*10^8 m/s$
$s=(v+u)/(1+vu/c^2 )=(2.649*10^8-1.8*10^8)/(1-(2.649*10^8*1.8*10^8)/(3*10^8 )^2 )=$

$=(8.49*10^7)/(1-1.5894)=-1.44*10^8 m/s$
b)

The time is (for emitter)

$t_0=d/v=(10^15*10^3)/(2.649*10^8 )=3.775*10^9 s$

Time is (in Earth frame)

$t=d/s=10^18/(1.44*10^8 )=6.94*10^9$