Classical Mechanics and Field Theory
QFT Seminar, Fall 2001
We start with Newton’s law:
F = ma
which we think about as follows: once an initial velocity x0 (t0 ) and position
x(t0 ) of a particle of mass m are given, as well as the force F exerted on the
particle at all times (as a function of position and/or velocity), then all future
positions and velocities are obtained by solving the differential equation:
F (x(t), x0 (t)) = mx00 (t)
Examples: (a) No force exerted (F = 0). Then:
x0 (t) = x0 (t0 ) and x(t) = x(t0 ) + (t − t0 )x0 (t0 )
i.e. the motion is straight-line motion with constant velocity.
(b) The (simple) harmonic oscillator (F (x) = −kx). Then:
x(t) = A cos(ωt) + B cos(ωt)
where ω =
pm
k
and A, B are determined by the initial conditions.
The Lagrangian formulation of mechanics characterizes the “classical”
paths x(t), or more precisely their lifts (x(t), x0 (t)) to paths in the tangent space
(or “configuration space”) T Rn . Start with a Lagrangian:
L(q, q̇) : T Rn → R
and its action integrals:
Z
S[a,b] [x] :=
b
L(x(t), x0 (t))dt
a
and declare that paths (x(t), x0 (t)) are classical if they satisfy the variational
principle:
d
S[a,b] [x + y]|=0 = 0
d
for all intervals [a, b] and all paths y(t) satisfying y(a) = y(b) = 0.
This principle gives us the Euler-Lagrange equations of motion:
∂L
d ∂L
(x(t), x0 (t)) =
(x(t), x0 (t)) ∀i = 1, · · · n
∂qi
dt ∂ q̇i
which also characterize the classical paths (x(t), x0 (t)).
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We see this by integrating by parts:
Z
0=
a
b
d
L(x + y, x0 + y 0 )|=0 =
d
Z
b
a
d ∂L
∂L
−
∂qi
dt ∂ q̇i
Z
b
a
∂L 0
∂L
yi +
y =
∂qi
∂ q̇i i
∂L
yi +
yi
∂ q̇i
b
a
and the last term is zero because yi (a) = yi (b) = 0. Of course this holds for all
the paths y(t) if and only if the Euler-Lagrange equations are satisfied.
Remark: The Einstein summation convention is in effect! In this context, that
will mean that we sum over the index i without writing the summation symbol.
Examples: (a) The Euler-Lagrange equations for L(q, q̇) = 21 mq̇i2 are:
0 = mx00 (t)
which is Newton’s law (with no force).
When we introduce a potential energy term: L(q, q̇) = 12 mq̇i2 − V (q) we
obtain the Euler-Lagrange equations:
−∇V (x(t)) = mx00 (t)
which is Newton’s law with force F = −∇V (x).
(b) When n = 1 and V = 12 kq 2 , we get the simple harmonic oscillator.
(c) We can “couple” simple harmonic oscillators via a Lagrangian:
L(q, q̇) =
1
1
mi q̇i2 + ki qi2 + mij (qi − qj )2
2
2
Here our T Rn is the configuration space of n particles in R1 , and it is
interesting to note that one can always “uncouple” such harmonic oscillators
via an orthogonal change of coordinates u = Aq giving:
L(u, u̇) =
1
1
mi u̇2i + Ki u2i
2
2
in the new “normal modes” coordinates.
Some More Vocabulary: A term in the Lagrangian which is quadratic in
the q̇i (as in example (a)) is called a kinetic energy term. A term which is a
function of position only is (as above) called a potential energy term, and in
addition the quantities:
∂L
pi :=
∂ q̇i
are called the momenta.
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In the Hamiltonian formulation, we define:
H(q, q̇, p) := q̇i pi − L(q, q̇)
as a function of T Rn and the momenta pi (this is the “Legendre transform”).
When we compute:
dH = q̇i dpi + pi dq̇i −
∂L
∂L
∂L
dqi −
dq̇i = q̇i dpi −
dqi
∂qi
∂ q̇i
∂qi
we find that H is (locally) a function of the qi and pi only. Moreover, when
L(q, q̇) is of the form “kinetic” − “potential” then the pi are naturally sections
of the cotangent bundle T ∗ Rn and they frame the cotangent bundle when the
kinetic term is non-degenerate. Thus in this situation, we can think of:
H(q, p) : T ∗ Rn → R
as a function on the cotangent bundle (also called the phase space).
Example: When L(q, q̇) = 12 mq̇i2 − V (q), then:
H(q, p) =
1 2
p2
mq̇i + V (q) = i + V (q)
2
2m
The Hamiltonian has some exceedingly useful properties:
d
H(x(t), p(t)) = 0
dt
on all classical paths (where p(t) are the p-coordinates of the path). That is,
the Hamiltonian is a conserved quantity along paths. This is an immediate
consequence of the Euler-Lagrange equations:
∂L
d ∂L
∂L 0 ∂L 00
d
H(x(t), p(t)) = x00i
+ x0i
−
x −
x =0
dt
∂ q̇i
dt ∂ q̇i
∂qi i ∂ q̇i i
And from the expression for dH, we obtain Hamilton’s equations:
∂H
= x0i (t) and
∂pi
∂H
∂L
=−
= −p0i (t)
∂qi
∂qi
of motion along classical paths. Thus the Hamiltonian vector field
H=
∂H ∂
∂H ∂
−
∂pi ∂qi
∂qi ∂pi
is precisely the velocity vector field along the classical paths in phase space!
A “quantity” A : T ∗ Rn → R is conserved (on classical paths) if and only if:
{A, H} := H(A) =
∂A ∂H
∂A ∂H
−
=0
∂qi ∂pi
∂pi ∂qi
(and one double-checks that H itself is a conserved quantity)
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Of course, the momenta pi and position qi are not (in general) conserved:
{pi , H} =
∂L
=: ṗi and {qi , H} = q̇i
∂qi
The Poisson bracket is the pairing {, } on “quantities” defined by:
{A, B} =
∂A ∂B
∂A ∂B
−
∂qi ∂pi
∂pi ∂qi
which satisfies the following identities (check!):
(1) (Skew symmetry) {A, B} = −{B, A}
(2) (Jacobi identity) {A, {B, C}} + {B, {C, A}} + {C, {A, B}} = 0
(3) (Leibniz rule) {AB, C} = A{B, C} + {A, C}B
Examples: {qi , qj } = 0 = {pi , pj } and {qi , pj } = δij
Let’s consider the role of symmetry before moving on to field theory:
Example: Consider the Lagrangian (in polar coordinates, why not?):
L(r, θ, ṙ, θ̇) =
1 2 1 2 2
mṙ + mr θ̇ − V (r)
2
2
(which is of the form “kinetic” − “potential”). This has an obvious rotational
symmetry:
L(r, θ, ṙ, θ̇) = L(r, θ + s, ṙ, θ̇)
When we apply the Euler-Lagrange equations, we therefore get:
0=
∂L
d ∂L
d
=
= mr2 θ̇ = 0
∂θ
dt ∂ θ̇
dt
on classical paths, so the “angular momentum” pθ = ∂L
= mr2 θ̇ is a conserved
∂ θ̇
quantity. This generalizes to an easy version of Noether’s theorem: any oneparameter group Φs : Rn → Rn of symmetries of the Lagrangian yields a
conserved quantity. (Choose a coordinate system so that ∂q∂ 1 is the derivative
of Φs at s = 0. In that coordinate system, p1 will be conserved.)
There is a time translation symmetry which will also lead to a conserved
quantity, though it a little is harder to see (but crucial for field theory). Let Φs
be one-parameter group acting on paths: Φs (x(t)) = x(t + s). While it is not
true that L(x(t), x0 (t)) is invariant under Φs , the action satisfies:
Z b
S[a,b] [Φs (x)] =
L(x(t + s), x0 (t + s))dt = S[a+s,b+s] [x]
a
We get:
Z
b
0
0
Z
L(x(t+s), x (t+s))−L(x(t), x (t)) =
a
b
4
b+s
Z
L(x(t), x (t))−
0
a
a+s
L(x(t), x0 (t))
and when we take
Z
a
b
d
ds
and set s = 0, we get:
d ∂L
∂L
−
∂qi
dt ∂ q̇i
x0i
∂L 0
+
x
∂ q̇i i
b
= L(x(b), x0 (b)) − L(x(a), x0 (a))
a
with x0 here playing the role that y played previously, except that x0 doesn’t
vanish at the endpoints. But now assume that x(t) is a path satisfying the
Euler-Lagrange equations. Then the first term above vanishes, and we get:
pi q̇i − L(q, q̇)
as a conserved quantity. This is the Hamiltonian! Thus the Hamiltonian is the
conserved quantity corresponding to time translation.
(I think that this is the more serious version of Noether’s theorem.)
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