Relativity (Homework 7, Physics 323)

2. (simple) A 20 year old captain of a spaceship moving at constant velocity β=1-0.002 passes near Earth. After one year of spaceship time, the captain sends a message back to Earth. By how many

years have the people of Earth aged when the message arrives?

This is pretty simple: moving clocks indicate a smaller time.  If $t_0=1$ year (spaceship) then

$t=γt_0=[1/\sqrt{(1-β^2)}] t_0=[1/\sqrt{(1-0.998^2)}] t_0=1.582*t_0=1.582$ years

So when the message is released from spaceship the people on earth have aged $t=1.582$ years

Distance from spaceship (in Earth reference) is

$d=t*(βc)=1.582*0.998 c=1.5788*c   (l y)$

Time to travel back for message is

$t_1=1.5788 years$

So in total the people from Earth have aged (when the message arrives)

$t=1.582+1.5788=3.1608 years$

4. An initially slowly-moving charged particle (charge q, mass m) enters the uniform-electric-field (E) of a Van de Graff particle accelerator. The particle is in the field region for time t measured in

the laboratory.

a. Find the velocity of the charge on leaving the accelerator.

b. After acceleration, suppose the particle disintegrates into two fragments after a time t0 (as measured in the particle’s rest frame). How long did the particle take to disintegrate as viewed by the


Suppose that the particle is accelerated at high energies (relativistic). Total energy is

$E=m_0 c^2+q E$ and $E^2=p^2 c^2+m_0^2 c^4$

$p^2=((m_0 c^2+q E)^2-m_0^2 c^4)/c^2 =(q^2 E^2+2m_0 c^2 q E)/c^2$

$γ^2 m_0^2 v^2=(q^2 E^2+2m_0 c^2 q E)/c^2$ or

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

$1/v^2 =1/c^2 +(m_0^2)/((qE/c)^2+2m_0 qE)$

If the particle is moving non relativistic the speed after acceleration is

$a=q E/m$ and $v=at=q E/m*t$


Moving clocks are slower ($t_0$) so that in the laboratory frame ($t$)

$t=γt_0 with γ=1/√(1-v^2/c^2 )$