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}.

As a rocket ship passes Earth at speed (3/5)c, clocks on Earth and on the
ship are synchronized at t = 0. The rocket ship sends a light signal back to
Earth when its clock reads one hour.

(a) According to Earth's clock, when was the signal sent?

(b) According to Earth's clock, how long after the rocket passed did the signal
arrive back on Earth?

(c) According to the ship clock, when did the signal arrive back on Earth?

Solution:

- Concepts:

Proper time - Reasoning:

We are given the proper time interval in one frame and the relative speed of the frames. - Details of the calculation:

(a) The ship's time interval t is the proper interval between passing earth and ship emitting the signal.

Earth time interval t' = γ(1 h) = (1 h)/(1 - 9/25)^{½}= 1.25 h = 75 min = 4500 s.

(b) The signal was emitted at t' = 4500 s. It had to travel a distance d = (3/5) c*t' with speed c, which takes Δt' = d/c = 2700 s. It arrived t'_{a}= 7200 s = 120 min = 2 h.

(c) The earth time interval between ship passing and signal arriving is the proper time interval.

Ship time interval t_{a}= γ(2 h) = 2.5 h.

We can also use kinematics.

When earth is at a distance d = (3/5)c*(1 h) the light signal is emitted. It travels a time t' before reaching Earth.

d + (3/5)c*t' = ct', t' = 1.5 h. The signal is emitted at t = 1 h, so the arrival time is t_{a}= 2.5 h.

An entrepreneur decides to operate a spacewash, which is normally at rest
relative to its home planet. Although the device is only 100 meters long, he
advertises the spacewash as "the only one that simultaneously washes the front
and back of a WizzFlizz 200TM." According to the catalog, the WizzFlizz 200TM
is 200 meters long.

(a) An inspector arrived at spacewash to check how it operates. What is the
minimum speed v_{0} (in units of c) at which the WizzFlizz 200TM must go
through the spacewash for the advertisement to be true? (Assume that the wash
is instantaneous.)

(b) The pilot of the WizzFlizz 200TM observes the wash process from the
spaceship. At the minimum speed v_{0}, how accurate does the pilot's
clock need to be to see that the simultaneity claim is false? (Assuming that
it's true if observed from the spacewash.)

(c) Suppose a spaceship is approaching the home planet with insufficient
speed (4/5)v_{0}. The spacewash has its own engine that allows it to
accelerate toward the ship. To what speed should it accelerate (relative to its
home planet) to perform the wash as promised in the advertisement?

Solution:

- Concepts:

Length contraction, the invariant space-time interval, velocity addition - Reasoning:

Simultaneity is not absolute, but relative.

(a) Length contraction: γ = L_{0}/L = 2, 1 - β^{2}= ¼, β = √3/2.

(b) c^{2}∆t^{2}- ∆x^{2}= constant. In the spacewash frame ∆t = 0, ∆x = 100 m.

In the spaceship frame ∆x^{ }= 200 m. c^{2}∆t^{2}- 4*10^{4}= -10^{4}. ∆t = (1/√3)*10^{-6}s.

The clock must be able to measure time intervals smaller than that (1/√3)*10^{-6}s.

(c)

Velocity addition: - Let frame K be the ship and frame K' be the planet. Let v = 4v
_{0}/5 be the planet with respect to the ship and u_{||}= v_{0}be the speed of the spacewhash with respect to the ship. Then u'_{||}= (u_{||}- v)/(1 -**v∙u**/c^{2}) = (v_{0}- 4v_{0}/5)/(1 - 4v_{0}^{2}/(5c^{2}) is the speed of the spacewash with respect to the planet.

u'_{||}= v_{0}/(5 - 4v_{0}^{2}/c^{2}).

u'_{||}/c = √3/4. The speed of the spacewash with respect to the planet must be v = c√3/4, or ½v_{0}.

In inertial frame O a rod of length L is
oriented along the x-axis and moving with velocity **u** in the positive y
direction. This rod is then viewed from an inertial reference frame O' moving
with velocity **v** in the positive x direction.

(a) What is the
velocity of the rod in O'?

(b) What is the length of the rod in O'?

(c)
What angle does the rod make with respect to the x' axis?

Solution:

- Concepts:

Relativistic kinematics - Reasoning:

We are asked to relate observations in an inertial frame O to observations in an inertial frame O'. - Details of the calculation:

(a) The stick moves with velocity u**j**in O. It is oriented parallel to the x-axis, so its length in O is L.

O' moves with velocity v**i**with respect to O. The velocity of**u**' the stick in O' can be found from the velocity addition formulas.

u'_{||}= (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**.Here

**v**= v**i**, u_{||}= 0, u_{⊥}= u**j**.

Therefore v_{x}' = -v, v_{y}' = u/γ = (1 - u^{2}/c^{2})^{½}u.We can also proceed by transforming the coordinates of events. Let us look at the positions of the left and the right end of the stick at t = 0 and at t = t

_{1}in O and O'.

To transform the coordinates of an event from O to O' we use the matrix equation below.

Event: coordinates in O _{ }coordinates in O' _{ }(1) left edge at t = 0 (0, 0, 0, 0) (0, 0, 0, 0) (2) right edge at t = 0 (0, L, 0, 0) (-γβL, γL, 0, 0) (3) left edge at t = t (ct, 0, ut, 0) (γct, -γβct, ut, 0) (4) right edge at t = t (ct, L, ut, 0) (γct - γβL, -γβct + γL, ut, 0) From events 1 and 3 we find u

_{x}' = ∆x'/∆t' = -γβct/γt = -βc = -v.

From events 1 and 3 we find u_{y}' = ∆y'/∆t' = u/γ.

(b) In O' at t' = 0 the left edge of the stick is at x' = 0, y' = 0.

At t' = -γβL/c the right edge of the stick is at x' = γL, y' = 0.

The stick travels with velocity u' = -v**i**+ (u/γ)**j**. Therefore at t' = 0 the right edge of the stick is at

x' = γL - vγβL/c, y' = (u/γ)γβL/c, or x' = γL(1 - β^{2}) = L/γ, y' = (u/c)βL.

The length of the stick in O' is (x'^{2}+ y'^{2})^{½}= L(1 - β^{2}+ ((u/c)β)^{2})^{½}.

(c) tanα' = y'/x' = (u/c)βγ.

A relativistic particle is launched at the origin (0,0) with initial momentum
**p**(0) = (p_{x}(0), p_{y}(0)), with p_{x}(0)
> 0 and p_{y}(0) > 0, and is subject to a
constant force pointing in the negative y direction.

(a) Solve the equations of motion for x(t) and y(t).

(b) Determine the time T at which the particle reaches the x-axis
again (i.e. y(T) = 0).

(c) Find the trajectory of the particle, i.e. y = y(x).

NOTE: Give all answers for the laboratory frame.

Solution:

- Concepts:

Motion in two dimensions:**F**= d**p**/dt, E = (m^{2}c^{4}+ p^{2}c^{2})^{½},**v**/c =**p**c/E - Reasoning:

We are asked to solve the equation of motion for a relativistic particle. We have motion in two dimensions. - Details of the calculation:

(a) d**p**/dt =**F**, dp_{x}/dt = 0, dp_{y}/dt = -F.

p_{x}(t) = p_{x0}, p_{y}(t) = p_{y0}- Ft.

**v**=**p**c^{2}/E, E = (m^{2}c^{4}+ p_{x0}^{2}c^{2}+ (p_{y0}- Ft)^{2}c^{2})^{½}.

x(t) = ∫_{0}^{t}dt cp_{x0}/(m^{2}c^{2}+ p_{x0}^{2}+ (p_{y0}- Ft)^{2})^{½}.

Let m^{2}c^{2}+ p_{x0}^{2}= E_{0}^{2}/c^{2}and t' = p_{y0}- Ft, dt' = -Fdt.

x(t) = -(cp_{x0}/F)∫_{py0}^{py0-Ft}dt'/(E_{0}^{2}/c^{2}+ t'^{2})^{½ }= -(cp_{x0}/F)sinh^{-1}(t'/(E_{0}/c))|_{py0}^{py0-Ft}

= -(cp_{x0}/F)[sinh^{-1}((p_{y0}- Ft)c/E_{0}) - sinh^{-1}((p_{y0}c/E_{0})].

x(t) = (cp_{x0}/F)sinh^{-1}((Ft - p_{y0})c/E_{0}) + c_{1}, c_{1}= (cp_{x0}/F)sinh^{-1}((p_{y0}c/E_{0}) = constant.

y(t) = ∫_{0}^{t}dt c^{2}(p_{y0}- Ft)/(m^{2}c^{4}+ p_{x0}^{2}c^{2}+ (p_{y0}- Ft)^{2}c^{2})^{½}

= -(c/F)∫_{0}^{t}dt' t'/(E_{0}^{2}/c^{2}+ t'^{2})^{½ }= -(c/F)(E_{0}^{2}/c^{2}+ t'^{2})^{½}|_{py0}^{py0-Ft}

= -(c/F)(E_{0}^{2}/c^{2}+ (p_{y0}- Ft)^{2})^{½}+ c_{2}, c_{2}= (c/F)(E_{0}^{2}/c^{2}+ p_{y0}^{2})^{½}= constant.

(b) y(T) = -(c/F)(E_{0}^{2}/c^{2}+ (p_{y0}- FT)^{2})^{½}+ (c/F)(E_{0}^{2}/c^{2}+ p_{y0}^{2})^{½}= 0.

p_{y0}= ±(p_{y0}- FT), T = 0 or T = 2p_{y0}/F.

This is also the non-relativistic result.

(c) (F/(cp_{x0}))(x(t) - c_{1}) = sinh^{-1}((Ft - p_{y0})c/E_{0}).

Ft - p_{y0}= (E_{0}/c)sinh(F/(cp_{x0}))(x(t) - c_{1}).

y(t) - c_{2}= -(c/F)(E_{0}^{2}/c^{2}+ (E_{0}^{2}/c^{2})sinh^{2}(F/(cp_{x0}))(x(t) - c_{1})^{2})^{½ }= -(1/F)(E_{0}^{2}+ E_{0}^{2}sinh^{2}(F/(cp_{x0}))(x(t) - c_{1})^{2})^{½}

= -(E_{0}/F)(cosh^{2}(F/(cp_{x0}))(x(t) - c_{1})^{2})^{½}.