Re: Have equation of motion, what is the Lagrangian?

On May 3, 11:12 am, andy everett <vze2q...@xxxxxxxxxxx> wrote:
The equation of motion for the anchored string is:

Psi,tt = c^2*Psi,zz - w^2*Psi

The equations can be written as (the KINEMATIC EQUATIONS)
phi = -Psi,t
A = Psi,z
From this also follow the identity (the "BIANCHI IDENTITIES")
-A,t - phi,z = 0.

Define (the CONSTITUTIVE EQUATIONS):
J = (1/m) A
rho = (e) phi
omega = s Psi,
such that
e m = (1/c)^2, and s = e w^2
with e and m constants.

Then, the equation above becomes (THE FIELD EQUATION)
rho,t + J,z = omega.

This is the Euler Lagrange equation of a Lagrangian theory if you can
devise a Lagrangian density L whose variational is given by:
delta(L) = -rho delta(phi) + J delta(A) + omega delta(Psi).
For, integrating by parts, you have
-rho delta(phi) = rho (delta Psi),t = (rho delta Psi),t - (rho,t)
delta Psi
J delta(A) = J (delta Psi),z = (J delta Psi),z - (J,z) delta Psi.
So, combining terms, you get:
delta(L) = (omega - rho,t - J,z) delta(Psi) + (J delta Psi),z +
(rho delta Psi),t.

Substituting the constitutive equations into the original expression
for delta(L), you get:
delta(L) = -e phi delta(phi) + 1/m A delta(A) + s Psi delta Psi.
This integrates to
delta(L) = -e delta(phi^2/2) + 1/m delta(A^2/2) + s delta(Psi^2/2).
Since we're assuming the constitutive coefficients e, m and s are
constant, then
delta(L) = delta(-e phi^2/2 + 1/m A^2/2 + s Psi^2/2).

Or, in terms of e and c; the Lagrangian density, up to an additive
constant, is:
L = -e/2 (phi^2 - (Ac)^2 - (w Psi)^2)
= -e/2 ((Psi,t)^2 - (c Psi,z)^2 - (w Psi)^2).

Note the term "density". The Lagrangian density is actually a factor
in a 2-form
L dt ^ dz.
So, I'll write this as L, instead, and refer to the density itself as
*L. Hence,
L = *L dt ^ dz.

The field Psi is a 0-form,
q = Psi.
The gradients form a 1-form
v = A dz - phi dt.

The conjugate fields also form a 1-form
p = rho dz - J dt
and a 2-form
f = omega dt ^ dz.
Thus, the variational of the Lagrangian can be expressed in
differential forms as:
delta(L) = delta(v) ^ p + delta(q) ^ f.
The variational is worked out in the language of forms, using the
exterior differential operator d. Hence, since
v = dq,
then
delta(v)^p = d(delta(q))^p = d (delta(q) ^ p) - delta(q) ^ dp.
So, combining, you get:
delta(L) = d(delta(q) ^ p) + delta(q) ^ (f - dp).
Hence, the field law is
dp = f.
The expression inside the differential is a "co-boundary term":
Theta = delta(q) ^ p.

In (sloppy) physicists applications you always hear the refrain "this
can be thrown away". This throwaway mentality is why the Buffalo
nearly went extinct. Everything is used, even the hides, in a more
ecological mentality.

The form
Theta = delta(q) ^ p = delta(Psi) ^ (rho dz - J dt)
is called the CANONICAL FORM. Out of it comes the definition of the
symplectic structure of the field theory. In particular, the "space-
like" part gives you the conjugacy relation for ordinary Hamiltonian
mechanics,
delta(Psi) ^ rho dz.
Thus, rho and Psi are conjugate.

Note that I defined rho with an extra coefficient, e. If this
formalism were to be subject to canonical quantization, e would be
extremely relevant. But at a classical level, e disappears from the
dynamics and the field law.

In a deDonder-Weyl Hamiltonian approach, ALL the parts of the
canonical form are used. The deDonder-Weyl Hamiltonian does a
Legrendre transfrom both the z and t part:
H_{DDW} = v ^ p - L -- expressed as a 2-form.
In contrast, the "ordinary" Hamiltonian is (as a 2-form)
H = (Psi,t dt) ^ rho - L.

The coefficients e, m and s bear mention. If you ONLY impose the
requirement that the Lagrangian be invariant under "Lorentz"
transforms (in the (t,z) coordinates, such that c is the invariant
speed), and leave the Lagrangian otherwise unspecified, then it
becomes a function of whatever Lorentz invariants can be defined from
the field components (Psi, phi and A). These are:
I = Psi^2/2 (or you could just take Psi); and J = (phi^2 - (cA)^2)/
2.
That is, the density *L becomes a function *L = *L(I, J) of just I and
J. So, then, DEFINE
s = dL/dI
e = -dL/dJ.
Then s, and e are functions of I and J and no longer just constant.
But the variational of the Lagrangian proceeds the same as before:
delta(L) = s delta(I) - e delta(J)
with
delta(I) = Psi delta(Psi)
delta(J) = phi delta(phi) - c^2 A delta(A).
Combining results, this yields,
delta(L) = s Psi delta(Psi) - e phi delta(phi) + (1/m) A delta(A)
where m is defined as 1/(ec^2). Thus, we derive the following
constitutive laws,
omega = s Psi, rho = e phi, J = A/m.

It's exactly the same form as before -- showing a degree of
independence of what the actual form of L is. But the coefficients s
and e (and m) are no longer constant. So the field equation now reads:
rho,t + J,z = omega
or
(e phi),t + c^2 (e A),z = s Psi
or
(e Psi,t),t = c^2 (e Psi,z),z - s Psi.

If s/e > 0, then you can define w^2 = s/e and write this as
1/e (e Psi,t),t = c^2 1/e (e Psi,z),z - w^2 Psi.

The difference from before is that now e is variable.

This is the classical analogue of renormalization. e is the classical
version of a renormalizaton coefficient. Renormalization refines a
field theory by "turning on" the constitutive coefficients (or turning
the BACK on, depending on your point of view of what ought to have
been the correct classical theory to start out with, in the first
place). Because e is variable, the field law is no longer invariant
under scale transformation (the original law, you expressed above, is
invariant under z -> kz, t -> kt, w -> w/k).

Most people who work in quantum field theory are now aware that what
they think of as the tools and tricks of quantum field theory (meaning
all the stuff related to renormalization theory and related concepts)
actually has nothing per se to do with quantum theory or quantum
anything at all. It's all grounded firmly in CLASSICAL field theory.

(So, let this solution be a lesson both to you AND the one presenting
the homework problem to you).
.

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