Find the speed of an electron that has been accelerated
from rest by an electric field through a potential increase of

(a) 20.0 kV and

(b)
5.00 MV, typical of a high-voltage x-ray machine.

Give your answers in units of
the speed of light.

Solution:

- Concepts:

Relativistic energy and momentum - Reasoning:

The total energy E after acceleration is qV plus the rest energy, mc^{2}. - Details of the calculation:

E = qV + mc^{2}, p = (E^{2}- m^{2}c^{4})^{½}/c = ((qV)^{2}+ 2(qV)mc^{2})^{½}/c.

(a) E = (2*10^{4}eV)*(1.6*10^{-19}J /eV) + 9.1*10^{−31}*(3.00*10^{8})^{2}J = 8.51*10^{-15}J.

pc = (E^{2}- m^{2}c^{4})^{½}= 2.31*10^{-14}J.

v/c = pc/E = 0.227.

Non-relativistically, (since kinetic energy << rest energy) , we obtain v = (2K/m)^{½}= 0.28 c.

(b) E = 8.82*10^{-13}J.

pc = 8.78*10^{-13}J.

v/c = 0.996.

Spaceship A moving with speed v = 0.5 c with respect to earth is chased by
spaceship B, moving with speed v = 0.6 c with respect to earth. When ship A
passes earth at t_{earth} = t_{A} = 0, ship B is 10^{-3}
ly behind ship A, as measured by an observer on earth?

(a) When, as measured on earth, will ship B catch up with ship A.
What is the ships' distance from earth at that time?

(b) What is the speed with which ship B is approaching ship A, as measured
by ship A?

(c) When, as measured on ship A, will ship B catch up with ship A?

Solution:

- Concepts:

Relativistic kinematics, velocity addition - Reasoning:

The ships move with relativistic speeds. We are given the relative speed of the ships with respect to a reference frame and must add their velocities to find their speed with respect to each other.

Assume that a frame K' is moving with velocity v**i**with respect to a frame K. An object moves in K with velocity**u**= d**r**/dt. The particle's velocity in K',**u**' = d**r**'/dt', is given byu'

_{||}= (u_{||}- v)/(1 -**v∙u**/c^{2}),

u_{⊥}= u_{⊥}/(γ(1 -**v∙u**/c^{2})),where parallel and perpendicular refer to the direction of the relative velocity

**v**. - Details of the calculation:

Measure distances in ly and time in y. c = 1 ly/y.

(a) In the earth frame we have 0.5 c*t = 0.6 c*t - 10^{-3}ly. 0.1 t = 10^{-3}y. t = 10^{-2}y = 87.6 h.

The ships will be (0.5*1 ly/y)*(10^{-2}y) = 5*10^{-3}ly from earth.

(b) Let frame K refer to earth, frame K' refer to ship A, and let ship A travel in the +x direction.

The**v**= 0.5**i**,**u**= 0.6 c**i**,**u**' = (0.1 c**i**)/(1 - 0.3) = 0.143 c**i**.

Ship B is approaching ship A with speed 0.143 c, as measured by ship A?

(c) In frame K ship B catches up with ship A at ct = 10^{-2}ly and x = 5*10^{-3}ly.

Lorentz transformation: ct' = γ(ct - βx) = (10^{-2}- 0.5*5*10^{-3})ly/0.75^{½}= 8.66*10^{-3}ly.

In frame K' ship B catches up with ship A at t' = 8.66*10^{-3}y = 75.9 h.

Another approach: In frame K at t = 0 the position of ship B is x = -10^{-3}ly.

The Lorentz transformation then yields for frame K' that x' = -γx at t' = γβt.

At t' = 0, ship B was therefore at x'(t'=0) = -γx - u'γβt, since it approaches ship A with speed u'.

Ship B will reach ship A at t' = |x'(t'=0)|/u' = 8.66*10^{-3}y.

A rocket with a proper length of 700 m is moving in the positive x-direction at a speed of 0.9c. It has two clocks, one in the nose and one in the
tail, that have been synchronized in the frame of the rocket. A clock on the
ground and the nose clock on the rocket both read t = 0 as they pass, i.e. have
the same x-coordinate.

(a) At t = 0, what does the tail clock on the rocket read in the frame of
an
observer on the ground?

(b) When the tail clock on the rocket passes the ground clock,

i. what does the tail clock read in the frame of an
observer on the ground?

ii. what does the nose clock read in the frame of an
observer on the ground?

iii. what does the nose clock read in the frame of an
observer on the rocket?

(c) At t = 1 h, as measured on the rocket, a light signal is sent from
the nose of the rocket to an observer standing by the ground clock. What does
the ground clock read when the observer receives this signal? (Assume
distances perpendicular to the x-axis are negligibly small compared to distances
along the x-axis.)

(d) When the observer on the ground receives the signal, he sends a return
signal to the nose of the rocket. When is this signal received at the nose of
the rocket as seen on the rocket?

Solution:

- Concepts:

The Lorentz transformation - Reasoning:

We are asked to transform events from one frame to another reference frame. - Details of the calculation:

(a) β = 0.9, γ = 2.294. In the ground frame the length of the rocket is 700 m/γ.

Let K be the ground frame and K' be the rocket frame.

In K': event 1: tail clock reads t' = 0, since in K' the tail clock and the nose clocks are synchronized.

ct' = 0, x' = -700 m.

In K: event 1: ct = -γβ 700 m. x = -γ700 m.

In K, when the nose clock reads zero the tail clock reads t = -(γβ/c) 700 m. = -4.82 μs.

An observer on the ground concludes that the ship clocks are not synchronized, but that the tail clock is 4.82 μs behind the nose clock.

(b) i. In K': event 2: ground clock passes tail clock

t' = 700 m/(βc), x' = -700 m.

In K: event 2: ct = γct' + γβx' = γ 700 m(1/β - β). t = 1.13 μs.

In K, when the tail clock on the rocket passes the ground clock, it reads t = 1.13 μs.

ii. In K': event 3: nose clock reads t' = 700 m/(βc), since in K' the tail clock and the nose clocks are synchronized.

t' = 700 m/(βc), x' = 0.

In K: event 3: t = γt' = γ700 m/(βc) = 5.95 μs.

When the tail clock on the rocket passes the ground clock, the nose clock in K reads t' = 5.95 μs.

This is expected, since the tail clock is 4.82 μs behind of the nose clock.

iii. In K': event 3: t' = 700 m/(βc), x' = 0.

When the tail clock on the rocket passes the ground clock, the nose clock in K' reads t' = 700 m/(βc) = 2.59 μs.

(c) In the rocket frame the signal is fired at x' = 0, T' = 3600 s.

In the ground frame the signal is fired at t = γT' at x = cγβT'.

It reaches x = 0 at T = γT' + γβT' = γ(1 + β)T' = 15692 s.

The ground clock reads T when the signal arrives.

(d) In the rocket frame the signal arrives at T'' = γT = γ^{2}(1 + β)T' = T'/(1 - β) = 3.6*10^{4}s at the observer.

The travel time for the signal is T'' - 3600 s = 3.24*10^{4}s in the rocket frame.

It takes the same time to travel back to x' = 0.

The return signal is received at the nose of the rocket at t' = (2*3.24*10^{4}+ 3600)s = 6.84*10^{4}s in the rocket frame.

The space ship leaving Earth is constructed to make the occupants feel
comfortable as it accelerates. It has an acceleration g = 10 m/s^{2}
in its own rest frame.

(a) How long does it take the ship to reach a speed v = 0.1 c in its own
rest frame and in the Earth frame?

(b) How long does it take the ship to reach a speed v = 0.5 c in its own
rest frame and in the Earth frame?

Solution:

- Concepts:

The proper time interval - Reasoning:

We are asked to find the proper time interval in an accelerating frame and the corresponding time interval in an inertial frame. - Details of the calculation:

(a) For speeds up to ~0.1c, we can use the non-relativistic approximation.

v = at, t = (3*10^{6}s)/(86400 s/day) = 35 days.

It will take ~35 days to reach a speed v = 0.1 c in both frames.

(b) The accelerating space ship is not an inertial frame. It accelerates with respect to an imagined second ship, which is at any instant happening to move alongside with identical instantaneous speed v, but is not accelerating. The imagined ship provides a momentary inertial frame of reference, relative to which the acceleration is g.

After a time interval dτ measured in the imagined ship, the rocket ship will have speed gdτ with respect to the imagined ship.

The speed with respect to earth will be v' = (v + gdτ)/(1 + vgdτ/c^{2}).

β' = (β + gdτ/c)/(1 + βgdτ/c).

dβ = β' - β = (1 - ββ')(g/c)dτ.

Keeping only terms to first order in the small quantities,

dβ = (1 - β^{2})(g/c)dτ, (g/c)dτ = dβ(1 - β^{2}).

(g/c)τ = ∫_{0}^{β(τ)}dβ'/(1 - β'^{2}) = tanh^{-1}(β(τ)), β(τ) = tanh((g/c)τ).

If β(τ) = 0.5, then τ = (c/g)tanh^{-1}(0.5) = 1.65*10^{7}s = 191 days.

It will take ~191 days to reach a speed v = 0.5 c in the ships frame.

Consider the decay Λ^{0} --> n + π^{0}, followed by π^{0}
--> 2γ.

(a) Given the masses M_{Λ}, M_{n} and M_{π}, find the
energy of the decay products n and π^{0} in the rest frame of the Λ^{0}.

(b) The two gamma rays from the decay of the π^{0} are observed to have
equal energies in the rest frame of the Λ^{0}. Find the angle between
the two gamma rays in this frame, in terms of the particle masses.

Solution:

- Concepts:

Particle decay, energy and momentum conservation - Reasoning:

The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved. - Details of the calculation:

(a) M_{Λ}c^{2}= E_{n}+ E_{π},**p**_{n}+**p**_{π}= 0. (conservation of energy and momentum)

E_{n}^{2}= p_{n}^{2}c^{2}+ M_{n}^{2}c^{4}, E_{π}^{2}= p_{π}^{2}c^{2}+ M_{π}^{2}c^{4}, E_{n}^{2}- E_{π}^{2}= (M_{n}^{2}- M_{π}^{2})c^{4}.

(E_{n}- E_{π})(E_{n}+ E_{π}) = (M_{n}^{2}- M_{π}^{2})c^{4}, (E_{n}- E_{π}) = (M_{n}^{2}- M_{π}^{2})c^{2}/M_{Λ}._{ }E_{n}= (M_{n}^{2}- M_{π}^{2}+ M_{Λ}^{2})c^{2}/(2M_{Λ}), E_{π}= (M_{π}^{2}- M_{n}^{2}+ M_{Λ}^{2})c^{2}/(2M_{Λ}).

(b) E_{π}= E_{1}+ E_{2},**p**_{π}=**p**_{1}+**p**_{2}. (conservation of energy and momentum)

Here E_{1}= E_{2}= E = E_{π}/2, |**p**_{1}| = |**p**_{2}| = E/c = E_{π}/(2c).

**|p**_{π}|^{2}= |**p**_{1}|^{2}+ |**p**_{2}|^{2}+ 2|**p**_{1}| |**p**_{2}|cosθ, where θ is the angle between the two gamma rays.

E_{π}^{2}- M_{π}^{2}c^{4}= (E_{π}^{2}/2)(1 + cosθ), cosθ = 2(E_{π}^{2}- M_{π}^{2}c^{4})/ E_{π}^{2}- 1.

cosθ = 1 - 2M_{π}^{2}c^{4}/ E_{π}^{2}= 1 - 8M_{Λ}^{2}M_{π}^{2}/(M_{π}^{2}- M_{n}^{2}+ M_{Λ}^{2})^{2}.