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PHYS 2100
Introduction to Methods of Theoretical Physics
Fall 1998
Midterm Exam #2
NAME:
Solution Key
You have two hours to complete this exam. There are a total of four problems and you are
to solve all of them. Not all the problems are worth the same number of points.
You may use your textbooks and class notes and handouts, or other books. You may not
share these resources with another student during the test.
Indicate any figures or tables you use in your calculations. Show all work!
GOOD LUCK!
Problem
Score
Worth
1.
30
2.
25
3.
15
4.
30
Total Score:
100
Problem 1 (5+7+8+7+3=30 Points):A magnetic field is given by B = Ax ĵ for some constant A .
a) Plot this field on the axes below, representing it by arrows of appropriate length. Use enough
arrows to accurately demonstrate the dependence over the xy plane:
y
x
b) Find the curl of B . Support your answer with an appropriate line integral over some closed
loop in the above figure.
∂B y
∇×B = ---------k̂ = Ak̂
∂x
This is a constant vector, so any closed line integral is nonzero.
c) Suppose this is a region of static fields. What current density gives rise to B ? Your answer must
include the direction of the current density, as well as its dependence on r and t .
There is no time dependence, so from Ampere’s law, ∇×B = Ak̂ = µ 0 J . Solving
for the current density gives J = ( A ⁄ µ 0 )k̂ . The current density is a constant.
d) Suppose instead that this region is free of all real currents. Find the electric field. Again, your
answer must include the direction, as well as its dependence on r and t .
1 ∂E
Now, the current density is zero, so Ampere’s law says ----- ------- = ∇×B = Ak̂ in
c 2 ∂t
which case we integrate to get E = Ac 2 tk̂ .
e) Prove that this region has zero charge density.
Gauss’ Law gives the charge density ρ = ε 0 ∇•E = 0 since the electric field does
not depend on position.
Problem 2 (5+10+10=25 points). The goal here is to demonstrate that Maxwell’s Equations
directly imply that electric charge must be conserved.
a) Briefly reproduce the argument given in class that conservation of mass in fluid flow leads to
∂ρ
∇ ⋅ ( ρv ) + ------ = 0 where ρ is the mass density and v is the velocity field of the fluid.
∂t
We did this in class in period #2 on Sept. 29. The point is that since mass is conserved, the rate of change of the mass within a closed volume must equal the negative of the total flux leaving that closed volume. Writing this down and using the
divergence theorem, along with bringing the time derivative inside the volume integral as a partial time derivative, gives you this equation.
∂E
b) Show from Maxwell’s Equations that ∇ ⋅  J + ε 0 ------- = 0 . Clearly state (but don’t bother to

∂t 
prove) any vector calculus identities you need.
∂E
Ampere’s Law says that ∇×B = µ 0 J + ε 0 µ 0 ------- . The divergence of a curl is always
∂t
zero (that’s a vector calculus identity) so if I take the divergence of both sides and
∂E
divide by µ 0 I get ∇ ⋅  J + ε 0 ------- = 0 which is what I was looking for.

∂t 
c) Prove that Maxwell’s Equations imply that electric charge is conserved.
∂(∇ ⋅ E )
Now just follow through the divergence to get ∇ ⋅ J + ε 0 -------------------- = 0 and then use
∂t
∂ρ
Gauss’ Law, i.e. ∇ ⋅ E = ρ ⁄ ε 0 , to get ∇ ⋅ J + ------ = 0 . Since the current density is
∂t
just J = ρv , this looks just like the equation in part (a) for the conservation of
mass. Therefore, Maxwell’s Equations directly imply that charge is conserved.
Problem 3 (15 points): A passenger, riding in an exceptionally fast train on an exceptionally long
track, takes two flash photographs while sitting still in his chair. Both events are observed by
8
someone on the ground, who finds that they happen 3 ×10 meters apart. She also records that the
second flash comes 5/3 seconds after the first. How long did the photographer wait between shots?
The invariant interval, as determined by the observer on the ground, is
2
2
8
2
8
2
8
( ∆t ) – ( ∆x ) = ( 5 ×10 m ) – ( 3 ×10 m ) = ( 4 ×10 m )
2
For the photographer, ∆x = 0 since he is standing still. Therefore, the time he measures between the “events” of his shots is just
2
2
2
8
( ∆t ) – ( 0 ) = ( ∆t ) = ( 4 ×10 m )
2
or, in conventional time units by dividing by c , ∆t = 4 ⁄ 3 seconds. This is also the
“proper time” between the events.
Problem 4 (15+10+5=30 points): A nucleus of mass M , sitting still in the laboratory, decays to
another nucleus of mass m ( m < M ) by emitting a single photon.
M 2 – m2
-.
a) Show that the daughter nucleus recoils with a velocity v = -------------------M 2 + m2
Energy conservation:
M = γm + E
where E is the photon energy.
Momentum conservation:
E = γmv
where v is the recoil velocity.
(The momentum conservation equation immediately assumes that E = p for the
photon, and that the photon and daughter move in opposite directions.) Therefore,
1+v
M = γm + γmv = γm ( 1 + v ) = m ---------------------------( 1 – v2 )1 ⁄ 2
1⁄2
( 1 + v )2
1+v 1⁄2
( 1 + v )2 1 ⁄ 2
= m ------------------= m --------------------------------= m -----------1–v
(1 – v)(1 + v)
1 – v2
Now, square both sides and rearrange the terms to solve for the velocity:
M 2( 1 – v ) = m2( 1 + v )
v( M 2 + m2 ) = M 2 – m2
v = ( M 2 – m2 ) ⁄ ( M 2 + m2 )
b) Find the energy of the photon emitted in the laboratory, in terms of M and m .
I think the easiest way to do this is to use the momentum equation, but first find
( M 2 – m 2 ) 2 –1 ⁄ 2
4M 2 m 2 – 1 ⁄ 2
M 2 + m2
γ = 1 – ----------------------------= – ----------------------------= --------------------2Mm
( M 2 + m2 )2
( M 2 + m2 )2
m2 + M 2 M 2 – m2
M 2 – m2
after which you plug in and get E = γmv = ---------------------m --------------------- = --------------------- .
2mM
2M
M 2 + m2
c) What answer do you expect if m = 0 ? Demonstrate that your answer to (b) gives this result.
If m = 0 then it should be the same as decaying to two photons, i.e. two massless
particles which would then have the same energies E and equal, but opposite,
momenta, also equal to E . They need to share the initial energy equally since nothing distinguishes one from the other. In other words, we expect E = M ⁄ 2 .
Indeed, if we substitute m = 0 in the equation for the energy, you get E = M ⁄ 2 .