Maxwell's Wave Equation and Degrees of Freedom?
Jay R. Yablon
Maxwell's equation is often written as J^u = d_v d^v A^u where A^u is
the photon potential and J^u is the current.
Because the current is conserved, d_u J^u = 0, we know that J^u has
three degrees of freedom (four independent components constrained by one
scalar condition).
A^u on the other hand, has two degrees of freedom.
In J^u = d_v d^v A^u, how does one balance a left side with three and a
right side which appears to have only two, degrees of freedom?
Thanks,
Jay.
_____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator, sci.physics,research
Igor
> Dear friends:
>
> Maxwell's equation is often written as J^u = d_v d^v A^u where A^u is
> the photon potential and J^u is the current.
>
> Because the current is conserved, d_u J^u = 0, we know that J^u has
> three degrees of freedom (four independent components constrained by one
> scalar condition).
>
> A^u on the other hand, has two degrees of freedom.
Please explain why you think it only has two degrees of freedom.
Puppet_Sock
> Maxwell's equation is often written as J^u = d_v d^v A^u where A^u is
> the photon potential and J^u is the current.
Um. That's not how I remember it. I recall an object usually
written as Fuv = Au,v - Av,u
Socks
Jay R. Yablon
news:1178907866....@h2g2000hsg.googlegroups.com...
states, one for +1 and the other for -1 helicity.
Jay R. Yablon
news:1178913682.8...@e65g2000hsc.googlegroups.com...
J^v = d_u F^uv with the covariant gauge condition d_u A_u = 0 then you
get the wave equation J^u = d_v d^v A^u. Do you not remember this wave
form of Maxwell's equation? Jay.
Timo A. Nieminen
> Maxwell's equation is often written as J^u = d_v d^v A^u where A^u is
> the photon potential and J^u is the current.
>
> Because the current is conserved, d_u J^u = 0, we know that J^u has
> three degrees of freedom (four independent components constrained by one
> scalar condition).
>
> A^u on the other hand, has two degrees of freedom.
>
> In J^u = d_v d^v A^u, how does one balance a left side with three and a
> right side which appears to have only two, degrees of freedom?
While a photon might have two degrees of freedom of polarisation *wrt to
a given axis*, what about momentum, energy, photon number (ie
probability of actually having the photon)?
What you have is a classical equation. The field quantities appearing
are a 4-current, constrained by charge conservation, and the
4-potential, constrained by the Lorenz condition.
But I wouldn't say 3 degrees of freedom on each side; I'd say three
independent field components on each side. Since J and A are functions
over all (t,r), they have infinite degrees of freedom.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
Igor
> "Igor" <thoov...@excite.com> wrote in message
That's only true if A is a wavefield. In general, if you choose the
Lorentz gauge, or any arbitrary gauge for that matter, that's only one
condition on four degrees of freedom, leaving you with three.
Jay R. Yablon
bosons of Yang-Mills theory are massless (two polarization states) until
they swallow a Goldstone degree of freedom (scalar) to become massive.
But to get to two polarization states, one must first employ the
covariant d_u A^v=0 to impose a first constraint going from 4 to 3,
then, e.g., the non-covariant Coulomb constraint A^0 = 0 from 3 down to
2. So, for the underlying question: why is it that the gauge bosons of
Yang-Mills are said to only have two polarization states until they
swallow a scalar? In different terms, if we impose d_u A^v=0 and A^0 =
0 then we are forcing them to start out with two polarizations, but what
is us that compels us to set A^0 = 0? Can't we just set d_u A^v=0 so
that each of J^u = d_v d^v A^u and A^v has three freedoms, where d_u J^u
= 0 and d_u A^v=0 are parallel constraints on J^u and A^u each taking
way the freedom of one component? I just don't quite see a "natural"
origin for there to be two polarizations at the outset, it seems like
something we force.
Actually, now I think I do see it. It is because when you pass over to
quantum theory, the wave equation is for a free on-shall photon is
invariant under a transformation eta^u --> eta^u + a q^u in the
polarization vector, where q^u is the four-momentum and a is a scalar
multiplier.
Jay
Hans de Vries
> Dear friends:
>
> Maxwell's equation is often written as J^u = d_v d^v A^u where A^u is
> the photon potential and J^u is the current.
>
> Because the current is conserved, d_u J^u = 0, we know that J^u has
> three degrees of freedom (four independent components constrained by one
> scalar condition).
>
> A^u on the other hand, has two degrees of freedom.
>
> In J^u = d_v d^v A^u, how does one balance a left side with three and a
> right side which appears to have only two, degrees of freedom?
>
> Thanks,
>
> Jay.
Jay,
You might want to look at Sakurai's treatment of the manifestly
covariant photon propagator in Advanced Quantum Mechanics, Chapter
4.6, eqn (4.329) and onwards. It comes down to the observation that a
virtual photon has in principle 4 states of polarization, time-like,
transversal (2x) and longitudinal, corresponding with the 4 components
of A_u. For real photons the fields E and B resulting from the time-
like A_t and longitudinal A_z states cancel each other exactly. This
cancellation goes beyond the E and B fields, all QED effects cancel.
The origin lays in the current J^u which: (1) as you say is conserved
and therefore removes one degree of freedom and (2) is such that the
effects caused by potentials A_t and A_z (resulting from J_t and J_z)
cancel each other in the case of massless propagation. Ultimately it's
how the motion of a charge density, (in the z direction) causes
related A_t and A_z components, which guarantees this.
If the propagation is massive as in the case of the vector bosons then
the cancellation fails and we're left with the 3 degrees of freedom
originating from the source current conservation.
In principle one should consider the manifestly covariant treatment
with four polarization states components as the physical correct one.
Photons restricted to only two transversal states are not Lorentz
invariant, they will have longitudinal and time-like components as
well when viewed from other reference frames. However, it's perfectly
OK to consider all external photon lines as having only transversal
states as long as you stay in one reference frame.
Regards, Hans.
dougsw...@gmail.com
Taking derivatives does not increase the degrees of freedom. Yet I
think of the E and B fields, which are contained in d^u A^v - d^v A^u,
as both being independent in 3 directions, or 6 degrees of freedom
overall. Oops.
So there is a right answer here, but I'm unsure which one it is.
Either, 1, E and B have 2 degrees of freedom because photons travel at
the speed of light, and the polarization constraint on A has only 2
degrees of freedom, or 2, E and B have 6 degrees of freedom because
there is no device that can measure the potential A^v directly - only
changes in A^v is physically measurable.
I'm leaning toward 2...
> In principle one should consider the manifestly covariant treatment
> with four polarization states components as the physical correct one.
> Photons restricted to only two transversal states are not Lorentz
> invariant, they will have longitudinal and time-like components as
> well when viewed from other reference frames. However, it's perfectly
> OK to consider all external photon lines as having only transversal
> states as long as you stay in one reference frame.
Most folks are probably aware that for a spin 1 field, the scalar
(time-like) mode implies negative probability densities, so must be
always set to be a virtual particle mode. There is a supplementary
condition that wipes out the scalar and longitudinal modes from the
books.
doug
Jay R. Yablon
"swee...@alum.mit.edu" <dougsw...@gmail.com> wrote in message
news:1179243099....@h2g2000hsg.googlegroups.com...
> Hello:
>
> Taking derivatives does not increase the degrees of freedom. Yet I
> think of the E and B fields, which are contained in d^u A^v - d^v A^u,
> as both being independent in 3 directions, or 6 degrees of freedom
> overall. Oops.
>
> So there is a right answer here, but I'm unsure which one it is.
> Either, 1, E and B have 2 degrees of freedom because photons travel at
> the speed of light, and the polarization constraint on A has only 2
> degrees of freedom, or 2, E and B have 6 degrees of freedom because
> there is no device that can measure the potential A^v directly - only
> changes in A^v is physically measurable.
>
> I'm leaning toward 2...
Actually, it seems to me that the A^u, which classically can be
represented as (phi, A_x, A_y, A_z), is equal to a velocity four-vector
times a scalar with dimensions of energy/momentum. One can change one's
Lorentz frame of observation, and thereby cause a covariant
transformation among the four components of A^u.
Turn then to F^uv, which has six components, E_xyz and B_xyz. Go into
the rest frame. All you have are three components of the electric
field, E_xyz. When we then transform into a frame of relative motion,
there is a covariant transformation of some components into the B_xyz,
but these only arise because of the relative motion and, it seems, ought
not be through of as three additional degrees of freedom on top of
E_xyz.
In more informal. colloquial terms, if magnetism is a relative
phenomenon, then the three components B_xyz of the magnetic field are
not separate degrees of freedom, but arise out of E_xyz simply from
relative motion.
So lining it all up, A^u with the covariant gauge condition d_u A^u = 0
has three degrees of freedom. J^u = d_s d^s A^u has three degrees of
freedom because of charge conservation d_u J^u =0 which directly
parallels the covariant gauge condition d_u A^u = 0. F_uv has three
degrees of freedom in E_xyz, which then transform over to B_xyz under
Lorentz motion, but that transformation does not add degrees of freedom.
Then, the question arise how one goes from three down to two for a
massless vector boson e.g., photon. I think Hans DeVries has given the
most satisfactory answer to this which I have seen to date in this
thread. The other question is how two goes up to three for a massive
vector boson. That takes us into spontaneous symmetry breaking, Higgs
method, Goldstone scalars, and the like.
Jay.
maxwell
wrote:
> You might want to look at Sakurai's treatment of the manifestly
> covariant photon propagator in Advanced Quantum Mechanics, Chapter
> 4.6, eqn (4.329) and onwards. It comes down to the observation that a
> virtual photon has in principle 4 states of polarization, time-like,
> transversal (2x) and longitudinal, corresponding with the 4 components
> of A_u. For real photons the fields E and B resulting from the time-
> like A_t and longitudinal A_z states cancel each other exactly. This
> cancellation goes beyond the E and B fields, all QED effects cancel.
>
> The origin lays in the current J^u which: (1) as you say is conserved
> and therefore removes one degree of freedom and (2) is such that the
> effects caused by potentials A_t and A_z (resulting from J_t and J_z)
> cancel each other in the case of massless propagation. Ultimately it's
> how the motion of a charge density, (in the z direction) causes
> related A_t and A_z components, which guarantees this.
>
> If the propagation is massive as in the case of the vector bosons then
> the cancellation fails and we're left with the 3 degrees of freedom
> originating from the source current conservation.
>
> In principle one should consider the manifestly covariant treatment
> with four polarization states components as the physical correct one.
> Photons restricted to only two transversal states are not Lorentz
> invariant, they will have longitudinal and time-like components as
> well when viewed from other reference frames. However, it's perfectly
> OK to consider all external photon lines as having only transversal
> states as long as you stay in one reference frame.
>
(A submu) in EM & not its derivatives (the E & B fields) - the point
that John Carpenter has been making now for over 20 years following
the pioneering work of LV Lorenz in 1867.
Arnold Neumaier
>
> Taking derivatives does not increase the degrees of freedom. Yet I
> think of the E and B fields, which are contained in d^u A^v - d^v A^u,
> as both being independent in 3 directions, or 6 degrees of freedom
> overall. Oops.
>
> So there is a right answer here, but I'm unsure which one it is.
> Either, 1, E and B have 2 degrees of freedom because photons travel at
> the speed of light, and the polarization constraint on A has only 2
> degrees of freedom, or 2, E and B have 6 degrees of freedom because
> there is no device that can measure the potential A^v directly - only
> changes in A^v is physically measurable.
Classically, pure radiation is a solution of the free Maxwell equation,
and has two polarization degrees of freedom. A general electromagnetic
field, however, has 3 degrees of freedom, because the field generated by
an external source has no reason to be transversal, since it does not
satisfy the free Maxwell equation.
Quantum mechanically, the free photon field has 2 degrees of freedom;
but the QED field has 3 degrees of freedom, since the interaction
produces off-shell contributions from the Fermions.
Arnold Neumaier
Hans de Vries
wrote:
> Classically, pure radiation is a solution of the free Maxwell equation,
> and has two polarization degrees of freedom. A general electromagnetic
> field, however, has 3 degrees of freedom, because the field generated by
> an external source has no reason to be transversal, since it does not
> satisfy the free Maxwell equation.
One can construct longitudinal polarized solutions of the Maxwell
equations for the free field but these can arise only from NON-
conserved currents.
To calculate the EM field of an arbitrary non-conserved current one
has to apply the operator 1/q^2 (the inverse d'Alembertian in
configuration space) to the charge current and use F = d^u A^v - d^v
A^u to get the EM fields. Now, a simple but interesting example:
Say we have a non conserved charge at rest, and varying in time like
q=sin(wt). Since it is at rest it will have only an A_0 field
propagating away on the light cone, while decreasing with 1/r.
V = (1/r) sin(wr/c + phase)
The E field is longitudinal polarized. Being the derivative of V, it
has a component which decreases with only 1/r, as is typical for a
radiation field. The spherical wave approaches a plane wave far enough
away from the center.
Now this is a solution of the Maxwell equations of the free field. A
changing electric flux through a surface is normally accompanied by a
B field along the contour. On the spherical surface however all the B
fields around the contours of the sub-surfaces cancel thanks to Stokes
law and no net B field remains.
This longitudinal polarized radiation field can be linearly added to
the classical transverse polarized radiation field to obtain a
radiation field with three degrees of freedom. However, since charge
currents are conserved we will never observe this in nature.
Regards, Hans.