peter

Dear all,

The enduring discussion on some of my recently posted questions encourages me
to post the following question. (Thank you all!)

1) How the field variables are to be treated in a canonical theory of the
classical elm field?

Some books treat the vector potential, A(r,t), like the position
variable, so that the conjugated momentum reads

P_A = &L/&(&A/&t)

Is this appropriate?


2) What are the correct Hamiltonian eqs. of motion,

&P/&t = - &H/&Q,

or

&P/&t = - &H/&Q + &/&x &H/&(&Q/&x) ?


3) Which is the correct field Lagrangian? In many books, I see

L = E^2 - B^2

Heisenberg & Pauli (1929) and Fock & Podolsky (1932) used

2L = (E^2 - B^2) - (div A + &Phi/&t)


Thank you very much in advance!

Peter

igor

Sure. All that is required in general is that the Lagrangian be a
function of some set of variables and their ordinary time
derivatives. The generalized momenta are defined as the partial
derivatives of the Lagrangian wrt the ordinary time derivatives of
this set of variables. The generalized forces are then defined as the
partial derivatives wrt to the original set of variables. The latter
follows automatically from the Euler-Lagrange equations.


The Hamiltonian is always derived from the Lagrangian by way of a
Legendre transform: H = L - p_i v^i, where p_i are the generalized
momentum components derived from the Lagrangian and v^i are the
corresponing generalized velocity components. This transform takes
L(x, v) into H(x, p) by knowing p(v) derived from the Lagrangian. The
Hamilton equations are then easily derived by differentiating H wrt to
p and x (however they have been defined). A good check is that the
derivative of H wrt v should always vanish.

There's really no such thing as a correct Lagrangian for a given
system. Many different Lagrangians can define the same system. This
is because we can always add to any particular Lagrangian the ordinary
time derivative of a given scalar function and still have the same
underlying Euler-Lagrange equations. For a given system, the
Hamiltonian will be fixed, but this is not necessarily true for the
Lagrangian.

peter

Igor <thoovler@excite.com> writes:


Why "sure"? The time variable is singled out ...


Is (a) or (b) correct?


How that?

&H(x,p(v))/&v = &H(x,p(v))/&p &p/&v =/= 0

The 2 Lagrangians above make not the impression that they differ by merely a total time derivative


How would you fix the Hamiltonian w.r.t. gauge freedom of A and Phi?

Thank you,
Peter

igor

I'm not sure what you are asking here. Can you be a little more specific?

The answer depends on the form of the Lagrangian that you start with.
What does it look like?

The variable v is a generalized velocity, so whatever you choose as
your generalized coordinates, the derivatives of the Hamiltonian wrt
their time derivatives should always vanish.

They didn't make that impression with me right away either. There's a
way to make it work though, but it requires a bit of calculation.
Besides, that second parenthesis vanishes under the Lorentz gauge
condition which is usually assumed in most modern texts.

The Hamiltonian corresponds to an observable and is automatically
gauge invariant. Same for inertial momentum. Canonical momentum is
not gauge invariant, however. Neither will be the Lagrangian in
general. The Legendre tranform removes gauge freedom, leaving the
Hamiltonian invariant.

peter

Igor <thoovler@excite.com> writes:

Point mechanics:

&L/&r = d/dt &L/&(dr/dt) = dp_can/dt

Continuum mechanics:

&L/&f(r,t) = &/&t &L/&(&f/&t) + &/&r &L/&(&f/&r)

Here, t is not singled out


2L = (E^2 - B^2) - (div A + &Phi/&t)


What is wrong with this:

Sorry,

2L = E^2 - B^2

All 'L' are Lagrange densities, of course

The difference equals

1/2 (div A + &Phi/&t)^2

and seems not to give a contribution to the action

Yes, but this is not my goal, I wish to remove the gauge freedom


H = (p-qA)^2/2m + q.Phi

is not gauge invariant, or did I overlook something?

Thank you,
Peter

hendrik van hees

I'd like to point out a little subtlety. It's of course right what you
say, the Hamiltonian, i.e., the energy density integrated over all
space is gauge invariant. The energy density, however, in general is
not. If you take the energy-momentum tensor you get via Noether's
theorem from time-translation invariance of the action (or better you
obtain the whole energy-momentum tensor from space-time-translation
invariance to make everyting manifestly Lorentz-covariant),
the "canonical energy density" is not gauge invariant, at least not if
you start from the usual Lagrangian

\Lag=-1/4 F_{mu nu} F^{mu nu}

This is not of big concern since the energy density is not an
observable, but only the total energy of the field, as long as you
don't try to treat gravity. In GTR the energy-momentum tensor, coupling
as sources to the gravitational field must be gauge invariant since the
gravitational field is observable. Indeed, within GRT, you get the
right energy-momentum tensor of the em. field by variation of the
metric tensor g_{mu nu}. Then you obtain the Belinfante energ-momentum
tensor which is gauge invariant and differs from the canonical
energy-momentum tensor only by a divergence, i.e., it leads to the same
total field energy and momentum as it must be since energy and momentum
are defined via Noether's theorem as the generators of time and space
translations respectively.

For more explanations, see

http://theory.gsi.de/~vanhees/publ/lect.pdf

Sect. 3.3.

--
Hendrik van Hees Institut für Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
Fax: +49 641 99-33309 D-35392 Gießen
http://theory.gsi.de/~vanhees/faq/

peter

Hendrik van Hees <Hendrik.vanHees@theo.physik.uni-giessen.de> writes:

Why, then, Goenner, Spezielle Relativitätstheorie, 5.2.6, makes so artificial
tricks to remove the gauge freedom from the Hamiltonian equations of motion?

My idea consists in - following Boltzmann - to go back to Maxwell's 'Dynamical
theory' (1864/65). His original set of "20 eqs. for 20 variables" presented
there is underestimated and redundant (but, due to the latter property, not
inconsistent). Removing both deficiencies,
- gauge freedom vanishes like the grin of Cheshire's cat,
- the propagating elm field is transverse from the very beginning.


How severe is this limitation/constraint?

Thank you for this link! Having read it, I would like to make the following
comment.

Obviously, the common canonical energy-momentum tensor exhibit severe
deficiencies such, that various modifications/corrections have been proposed.
Isn't this an indication of deficiencies in the common canonical in general?

Is there a simple example for a gauge-invariant H (3.63)?

Looking forward,
Peter

igor

Now I see what you are saying. It depends on the context. And there
are many variations of the Euler-Lagrange equations. A good mechanics
text or field theory text can usually sort them out for you. Goldstein
deals with both type of Lagrangian extensively.


The problem is that for purposes of the Legendre transform, p and v
are taken as independent variables. Any relation between p and v is
derived and not assumed and thus will be different in different
systems. So the only thing we can say in general is &H(x,p(v))/&v
= - &L(x,v)/&v + p = 0.

Probably what I should have said is that the form of the Hamiltonian
is gauge invariant. Certainly, the quantity H - q phi, as well as p -
q A, is gauge invariant, leaving the form invariant.

peter

Igor <thoovler@excite.com> writes:

...

Will study it, thanks for the hint

...

What about the term

v &^2L/&v^2 ?

If p=p(v) in H, p=p(v) in

H = p.v - L

...

Why not?

Maxwell's original equations (1864) suggest to use the Helmholtz decomposed
field components and corresponding potentials.

E_L = -&A_L/&t - grad Phi == - grad phi_E

E_T = -&A_T/&t == curl a_E

In Maxwell's theory, there is only one equation for the two variables A_L and
Phi. This is the kernel and root of the gauge freedom in this theory. It is
circumvented when working with

phi_E == -& phi_A/&t + Phi

and A_T, where A is Helmholtz decomposed, too.

A = A_L + A_T = -grad phi_A + curl a_A

What do you think?


What is the physical meaning of

H - q Phi ?

Thank you,
Peter

igor

It took me a while, but I think understand what you're saying.
However, since p and v are independent in the context of the Legendre
transform, there will be no such term. The Legendre transform by
itself makes no assumption about a relation between p and v. It's the
Euler-Lagrange equations that provide such a relationship.

I'm not really sure. Why not just use the modern forms with a single
vector and scalar potential?


Kinetic energy.