Homework 12 Answers, 95.658 Spring 2010, Electromagnetic Theory II
Dr. Christopher S. Baird, UMass Lowell
Problem 1
For the purposes of helping to write a practice final exam, please create and write very neatly on a
separate sheet of paper: 2 multiple choice problems and 1 short work problem pertaining to any subject
covered in class this semester. Do not solve the problems or mark the correct answers.
SOLUTION:
Any reasonable problem receives credit
Problem 2
Suppose we have already applied the law of energy conservation and derived the relativistic expression
for a particle's total energy, E = γu E0, but do not yet know the rest energy E0. Use this expression to
find the relativistic kinetic energy of the particle, take the low-speed limit, and force this to equal the
classical expression in order to derive the rest energy.
SOLUTION:
The relativistic kinetic energy is the total energy minus the rest energy:
T = E0


1
−E 0
1−u 2 /c 2
In the low-speed limit this should reduce to the classical result:
1
T = m u2
2
Make a Taylor series expansion of the term in parentheses in the relativistic expression
−1 / 2
1−x 


1
3 2
=1 x x ...
2
8
T = E 0 1
T = E0

1 u2 3 u4

... − E 0
2 c2 8 c4
1 u2 3 u4

...
2 c2 8 c4

This is the relativistic kinetic energy.
For low speeds, u << c we have u/c << 1 and we can drop all terms except the first non-vanishing one:
T = E0
 
1 u2
2 c2
We force this to equal the classical result and solve for E0
 
1
1 u2
2
mu = E 0
2
2 c2
E 0=mc 2
Problem 3
Prove that the 4-velocity U is indeed a 4-vector, where U = (γu c, γu u). Do this by applying a Lorentz
transformation to U and reducing the expressions down to the velocity addition formulas (Eq. 11.31 in
Jackson)
SOLUTION:
The Lorentz transformation for the coordinate variables is:
x 0 '=  x 0− x par 
x par '= x par − x 0 
x perp '=x perp
The Lorentz transformation of the four velocity is then:
U 0 '=v  U 0 − U par 
U par '=v U par −U 0 
U perp '=U perp
Write out each element in terms of the classical velocity:
u ' c=v  u c− u u par 
u ' u par '=v u u par − u c
u ' u perp '=u u perp
Solve the first equation for γu' and substitute into the last two equations:
u par '=c
v u u par − u c
v  u c− u u par 
u perp '=c
u u perp
v  u c− u u par 
Simplify
u par '=
u par −v
vu
1− 2par
c
u perp
u perp '=

v 1−
v u par
c2

Exchange labels to match the book (being careful to exchange “-v” for “+v”):
u par =
u par 'v
vu '
1 par
c2
u perp '
u perp =

v 1
v u par '
c2

These are the velocity addition formulas derived explicitly previously.
We have therefore proved that U, as defined above, is indeed a 4-vector and obeys Lorentz
transformations if the classical velocity u obeys the velocity addition formulas, which we had proven
previously to be true.
Problem 4
A free particle of rest mass m with a total energy E that is four times its rest mass, collides with an
identical particle at rest. They stick together after the collision. What is the rest mass, velocity, and
relativistic kinetic energy of the composite particle? What is the loss of kinetic energy in this collision
and where did it go? Where does this energy go according to classical mechanics?
SOLUTION:
Let us call the first particle A,the particle at rest B, and the composite particle C. The initial particles are
identical so that they have the same rest mass and therefore the same rest energy:
E 0=mc 2
Particle A is initially traveling such that its total energy E is four times its rest energy:
E A=4 m c
2
We can use this result and the equation derived in the class notes to find the momentum of particle A:
E 2= p 2 c 2m2 c 4
p A=

E 2A
2
c
−m2 c 2
p A= 15 m c
Particle B is initially at rest so that its total energy is just its rest energy:
E B=mc 2
and its momentum is zero:
p B =0
Particle C has a total energy EC which depends on its rest mass mC and velocity uC, both unknown at
this point:
E C=
1
2
mC c
2
2
 1−uC /c
as well as a momentum that depends on these unknowns:
p=
1
mC uC
 1−u 2C /c 2
Apply the conservation of energy:
E AE B=E C
2
2
4 mc m c =
2
1
2
mC c
2
2
1−u C /c
2
2
mC =251−uC /c  m
2
Apply the conservation of momentum:
p A p B = p C
The conservation of momentum is a vector equation but this problem is simple enough that everything
happens in the same direction, so that it reduces to a scalar equation.
 15 m c0=
1
mC u C
2
2
1−u
/c
 C
15 m2 c 2=
1
mC2 u2C
2
2
1−u C /c
The two equations in boxes now form a system of equations that let us solve uniquely for the two
unknowns. Substitute the first into the second and solve for the velocity.
u C=

3
c
5
Now plug this back in to solve for the rest mass of particle C.
mC = 10 m
The total rest mass of the system before the collision is 2m and the rest mass of the composite particle
after the collision is approximately 3.2m. We can therefore conclude that there is a total system mass
gain of 1.2m :
mC − mAmB =  10−2 m
mC − mAm B ≈1.2 m
The relativistic kinetic energy of the composite particle is:
T C =mC c 2


1
−mC c 2
2
2
 1−u c /c
2
2
T C =5 mc −  10 m c
T C =5−  10 m c 2
T C ≈1.8 m c
2
The total initial kinetic energy of the system is
T AT B= E A −m c 2E B−mc 2
T AT B=3 mc
2
The total kinetic energy loss of this system due to the collision is:
T AT B −T C =  10−2 mc 2
T AT B −T C ≈1.2 m c
2
We can now see that the total system kinetic energy loss equals total system rest mass-energy gained.
Kinetic energy has been transformed into mass. According to classical mechanics, the kinetic energy is
converted to potential energy.