Classical Mechanics, Fourier Analysis and Optics
Physics And Electronics
Field: Physics – Electromagnetism
Physics and Electronics
PHYC20014 Physical Systems
Classical Mechanics, Fourier Analysis and Optics
Due Friday, 22 September, 2017 at 5 pm
1. Willy Wien’s selection machine. Suppose a positron (mass m and charge e) moves through an electromagnetic field with scalar potential �(x, t) and vector potential A(x, t). As you know from your EM class, this means the electric and magnetic fields can be written
E = �@A @t
�r�, B = r⇥A.
The Lagrangian for the positron is
L(t,x, ẋ) = 1
2 mẋ
j
ẋ j
� e (�� ẋ j
A j
) , (1)
where x i
is the ith Cartesian coordinate and we sum over j (Einstein summation convention).
(a) Show that the conjugate momentum associated with coordinate x i
is
p i
= mẋ i
+ eA i
. (2)
(b) When is p i
conserved?
(c) Convert L into the Hamiltonian for the system. You should find
H(t,x,p) = 1
2m (p
j
� eA j
)2 + e�. (3)
(d) Express Hamilton’s equations for (3) in the following form:
ẋ i
= 1
m (p
i
� eA i
) (4)
ṗ i
= e
m (p
j
� eA j
) @A
j
@x i
� e @� @x
i
. (5)
(e) In a Wien filter, the potentials are
� = Ex, A = (�By, 0, 0)
where E and B are constants. Without calculating anything, explain why p z
is con- served.
(f) Specialise Hamilton’s equations to the filter. You should find
ẋ = 1
m (p
x
+Bey) , ṗ x
= �Ee, ṗ y
= �Beẋ, ẏ = py m .
(g) This arrangement acts as a velocity selector: a positron initially travelling in the positive y direction will be deflected unless it has a specific velocity. Show that the “magic” velocity is
v = E
B .
(h) What happens when E > cB?
[1 + 1 + 2 + 3 + 1 + 3 + 3 + 1 = 15 marks]
2. Fishy sums and hot donuts. Poisson summation is a deep relationship between Fourier series and Fourier transforms. We can exploit this to learn about hot donuts!
(a) Let f : R ! R be a real function, and define a new function
h(x) ⌘ 1X
n=�1 f(x+ n). (6)
Check that h(x) has period T = 1.
(b) Assuming h(x) satisfies the Dirichlet conditions, calculate the coe�cients of the expo- nential Fourier series
h(x) = 1X
n=�1 c n
e2⇡inx.
You should find c n
= F (n) , (7)
where F ⌘ F̂ [f ] is the Fourier transform of f . Deduce the Poisson summation formula, 1X
n=�1 f(x+ n) =
1X
n=�1 F (n)e2⇡inx. (8)
This relates the periodic sums of a function and its Fourier transform.
(c) Consider heat flow on an infinite, 1D wire. The temperature T (x, t) obeys the di↵usion equation,
@T
@t = D
@2T
@x2 . (9)
Suppose we start with a point-like spike, T (x, 0) = �(x). Show that the function
T (x, t) ⌘ 1p 4⇡Dt
e�x 2 /4Dt
solves (9) with initial condition T (x, 0) = �(x).1 This is called the heat kernel. It is just the Green’s function with the sign flipped.
1 For checking the delta property, you are encouraged to use results from lectures.
(d) Now wrap the wire into a circle of unit circumference C. Without doing any calcula- tions, argue that
S(✓, t) = 1p 4⇡Dt
1X
n=�1 e�(✓+n)
2 /4Dt (10)
is the heat kernel on the circle.2 Hint. Remember that the periodic version of �(x) is the Dirac comb X
T
(✓) from Tutorial 1.
(e) Use Poisson summation to rewrite (10) as
S(✓, t) = 1X
n=�1 e�(2⇡n)
2 Dte2⇡in✓. (11)
(f) We can modify the initial temperature distribution to S g
(✓, 0) = g(✓), where g has the exponential Fourier series
g(✓) = 1X
n=�1 d n
e2⇡in✓.
Using (11) and the method of Green’s functions (or otherwise), derive the identity
S g
(✓, t) = 1X
n=�1 d n
e�(2⇡n) 2 Dte2⇡in✓.
(g) Mathematically, you can define a donut as the product of two circles, C ⇥ C. The di↵usion equation (9) becomes
@K
@t = D
✓ @2K
@✓21 +
@2K
@✓22
◆ . (12)
Verify that the heat kernel on the donut is
K(✓1, ✓2, t) = 1X
m,n=�1 e�4⇡
2(m2+n2)Dte2⇡i(✓1m+✓2n).
(h) Bonus. Heat flow on higher dimensional donuts is considered a trade secret by liti- gious, higher-dimensional donut vendors. Suppose we have an N -dimensional donut
N timesz }| { C ⇥ · · ·⇥ C
with heat equation
@K
@t = D
NX
i=1
@2K
@✓2 i
.
Generalise the answer from (f) to find the heat kernel. Don’t tell the vendors!
[2 + 4 + 4 + 2 + 3 + 2 + 3 + (2) = 20 marks]
2 The heat equation on the circle is identical to (9), with ✓ replacing x.
Question 3 [20 marks]
A particle undergoes motion in one dimension governed by the Hamiltonian
)ln(),( 0 qpHpqH = , (1) where )(tq and )(tp are the generalized coordinate and conjugate momentum respectively as functions of time t , and 0H is a constant. The motion is restricted to the top right quadrant of the phase space, i.e. we have 0)( >tq and 0)( >tp for all t .
(a) What are the SI units of 0H ?
(b) Write down Hamilton’s equations of motion for )(tq! and )(tp! .
(c) Without solving the equations in part (b), prove that ),( pqH is a constant of the motion.
(d) Apply the result from part (c) to sketch the trajectory corresponding to the initial conditions 1)0( =q and )1exp()0( == ep .
(e) Now solve Hamilton’s equations of motion explicitly for the trajectory in part (d). You should find
)/exp()( 0 etHtq = (2) and
)/exp()( 0 etHetp -= . (3)
(f) A canonical transformation is a transformation ),( pqQQ = , ),( pqPP = that leaves the Hamiltonian unchanged. Starting from the Principle of Least Action, one can show (you don’t need to!) that the transformation satisfies
dt dGPQKQPpqHqp +-=- ),(),( !! (4)
in one dimension, where we write )],(),,([),( PQpPQqHPQK = , and G is an arbitrary function called the generating function. Consider the specific generating function qeQqG Q ln),( = . By substituting into (4) and thinking carefully about what quantities are linearly independent, prove that the transformation generated by G is given by
)ln(),( qppqQ = and
qqppqP ln),( -= .
(g) Write down the transformed Hamiltonian ),( PQK in terms of Q and P . You should find that it is simpler than ),( pqH !
(h) Write down Hamilton’s equations of motion for )(tQ! and )(tP! .
(i) Solve the equations in part (i) for the trajectory in part (d). You should find 1)( =tQ and tHtP 0)( -= . Does this solution agree with equations (2) and (3), when you invert the transformation?
[1 + 4 + 3 + 2 + 2 + 3 + 1 + 2 + 2 = 20 marks]