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Nonimaging Optics Tutor: Jewel Kumar Ghosh Problem Set 1
In the lecture we have discussed about the Lagrangian and Hamiltonian mechanics and its application to optics. In the Lagrangian mechanics the starting point is the action which is written as:
S =
∫ (^) t 2
t 1
dtL [xi(t), x˙i(t)] (1)
where L [xi(t), x˙i(t)] is called the Lagrangian which is a functional of the positions xi(t), their derivatives ˙xi(t). The Lagrangian can be explicitly dependent on time t but we will assume it is not. The equations of motion emerge by varying this action and setting it to zero. These are called the Euler-Lagrange equations of motion:
d dt
∂L
∂ x˙i
∂L
∂xi
In the Hamiltonian formulation we define the canonical momenta by:
pi =
∂L
∂ x˙i
The Hamiltonian is defined as the Legendre transform of the Lagrangian:
H =
i
xipi − L (4)
The Hamiltonian is a functional of positions and momenta although this is a bit obscure in the formula written above. This is because we can invert Eq. (3) and write ˙xi = x˙i(pi) and then insert it into the definition of Hamiltonian. In this notation, the Euler-Lagrange equations are called the Hamilton equations and they are:
x˙i(t) =
∂H
∂pi
, p˙i(t) = −
∂H
∂xi
In this problem set, we will solve problems related to the Lagrangian and Hamil- tonian mechanics.
Problem 1 First we will derive what we have learnt from our familiar Newtonian mechanics. Consider the following Lagrangian:
L =
m x˙^2 − V (x). (6)
a) Find the Euler-Lagrange equations of motion. Does it give Newton’s law? b) Construct the Hamiltonian. Can you derive Newton’s law from Hamilton’s equation?
Problem 2 From Fermat’s principle derive the optical Lagrangian. Show that for constant refractive index, the canonical momenta is given by the direction cosines.
Problem 3 From your mechanics course you know that a particle at height z gains a velocity v =
2 gz. Now consider the following:
T =
dt =
ds v
dx^2 + dz^2 √ 2 gz
dx
( (^) dz dx
2 gz
From this “Lagrangian” obtain the Euler-Lagrange equations and find the solution. You can observe that this is the same as optical Lagrangian when the refractive index is n(z) ∼ √^1 z.
Problem 4 Consider the following Lagrangian:
L =
mv^2 − e
[
φ(t, x, y, z) − A~ · ~v
]
where
A^ ~(t, x, y, z) = Ax(t, x, y, z)ˆx + Ay(t, x, y, z)ˆy + Az (t, x, y, z)ˆz, (9)
~v =
dx dt
ˆx +
dy dt
yˆ +
dz dt
z, vˆ^2 = ~v · ~v. (10)
Using the Euler-Lagrange equations show that it gives the Lorentz force
F^ ~ = e
E~ + ~v × B~
where
E^ ~ = −∇~φ − ∂
A~
∂t
, B~ = ∇ ×~ A~ (12)
