Poynting Vector for static fields
Mike Freeman
sources claim that momentum density and angular momentum density are
proportional to E x B (or E x H). While this makes sense for EM waves, what
meaning can it have for static, crossed fields? (Yes, this is related to
Feynman's spinning disc.)
- Mike Freeman -
t...@rosencrantz.stcloudstate.edu
Mike Freeman <mfre...@capcollege.bc.ca> wrote:
It means just what it says it means: in a region of space where there
are electromagnetic fields, there is momentum, and the momentum
density is proportional to E x B! "What," you say, "even if
everything is static?" Yes!
One way to think about this is to imagine starting with a region of
space in which there was initially no electromagnetic field and then
moving charges and currents into place to produce the desired final
field. If you work out in detail how to do that, and if you calculate
the forces that have to be exerted on the charges and currents to get
them into place, you'll find that the momentum lost by the charges and
currents exactly equals the momentum gained by the electromagnetic
fields. If you then remove the charges and currents and bring the field
back to zero, that momentum that was stored in the fields gets transferred
back to mechanical momentum of the charges and currents.
In other words, if you want to hang onto conservation of momentum, you
have to assume that the electromagnetic fields carry momentum
proportional to E x B.
[There's a bit of fine print here. First, in doing your thought
experiments about all this stuff, you should be careful to consider
only cases where the fields die away to zero at infinity fast enough
that things like the total momentum are finite. Otherwise, it's easy
to get confused. (For instance, the angular momentum density of a
circularly polarized plane wave is zero everywhere, but that of a
wave packet that has a finite extent is not.)
Second, E x B is not the only expression you can use for the momentum
density. You can add to it any other vector that's automatically
conserved. We had a big discussion about this a year or two ago on
this newsgroup. E x B is the standard expression to use, and I think
it's the only one that has all of the "nice" properties one would
want: gauge invariance, zero for regions of zero field, proper
behavior on change of coordinate systems, etc. (I don't know how
many of those you need to assume for E x B to be the unique answer.
Presumably one could work that out.)]
All of that is probably more detail than you need right now, although
since I don't know exactly what your background is, maybe it's not.
It's very possible that all of this doesn't answer your question. If
not, try asking a more specific question, and I'll be glad to try
again!
-Ted
John Baez
<t...@rosencrantz.stcloudstate.edu> wrote:
>[...] in a region of space where there
>are electromagnetic fields, there is momentum, and the momentum
>density is proportional to E x B! "What," you say, "even if
>everything is static?" Yes!
Just to throw an extra worm into the can:
In the Hamiltonian approach to electromagnetism, the variables analogous
to "position" and "momentum" are the vector potential A and the electric
field E. If we work in temporal gauge - which brings out this analogy
the best - E is just the time derivative of A. So whenever there's an
E field, it's a bit odd to say "everything is static". The A field is
changing!
WARNING:
Of course this remark is bound to rile the masses, because in many
contexts the A field is regarded as "unphysical", except up to gauge
transformations, and I'm picking a gauge in order to conclude A is
changing when E is nonzero.
So, to all of you about to fire off an angry post in my direction,
I should say:
I'M JUST JOKING! HOLD YOUR FIRE!
But of course, I'm also *not* joking. If you are trying to derive
the formula for the momentum of the electromagnetic field by
seeking a quantity whose Poisson brackets generates translations
in space, you really need the Hamiltonian formalism. And for this,
it's very natural to work with the canonically conjugate fields A and
E. And from this point of view, the formula above is not so surprising!
Also, in situations where there's a Bohm-Aharonov effect, you
can have an A field without any B field. Then the above remarks
become *especially* relevant. My student Miguel Carrion-Alvarez
has worked out some fascinating things about how you have to deal
with these Bohm-Aharonov modes when quantizing the electromagnetic
field. Turns out people have been being a bit sloppy about this.
> We had a big discussion about this a year or two ago on
> this newsgroup.
Yeah - I don't know why I'm contributing to the flogging of this
dead horse! I guess I just saw it twitch slightly.
Pieter Kuiper
Mike Freeman <mfre...@capcollege.bc.ca> wrote:
> Forgive me if this has been dealt with previously, but I'm perplexed: many
> sources claim that momentum density and angular momentum density are
> proportional to E x B (or E x H). While this makes sense for EM waves, what
> meaning can it have for static, crossed fields?
I did this once for a coaxial cable. The Poynting vecor is non-zero only
between the concentric conductors, and it can easily be shown that its
surface integral is equal to the power P = IV of the current in the
cable.
Mike Freeman
momentum is physically real:
First, let's consider the simpler (?) case where we just look at
energy. Let's have a (finite) solenoid, Y-axis, and a pair of charged
plates, field in X direction. Since there is no power dissipation, no
change in stored energy, and no flux of the Poynting vector out of a
closed surface, Poynting's theorem says that 0 = 0. Poynting's vector
is nonzero, but I don't see its physical significance.
Now, for momentum. We can create the magnetic field first, no
Poynting vector yet. Now, I move some charges from one plate to the
other to create the electric field. Yes, there will be a force in the
Z-direction on the charges, due to the magnetic field, which I will
have to balance. Still, I don't see where the alleged momentum comes
from.
Gerard Westendorp
E x B also works for static fields.
There are a couple of things that make this counter-intuitive. For
example, a static E-field and a static B-field both contain zero
momentum on their own, but contain non-zero momentum when superimposed.
Intuition can tempt us into thinking the energies and momenta of
superimposed fields simply add up. But when you think about it, fields
from a positive and a negative charge actually cancel. In other words,
you can make electrostatic energy disappear by adding more of it.
When you physically do this, you have to bring together charges from a
long distance, while you extract the electromagnetic energy by letting
the attractive forces do work. Similarly, when you pull together
an arrangement of charges and magnets from a long distance, you generate
all kinds of Lorentz forces. These are proportional to the velocity at
which you pull the arrangement together. So you could minimize the
forces by keeping the velocity low. But this
would proportionally stretch the time you need to pull the arrangement
together, keeping the product F*t constant. But F*t = momentum.
Another counterintuitive thing is momentum without motion. Actually, if
you want you can still envisage electromagnetic energy being transported
in static fields. Because div(ExB) = d/dt(E^2+B^2)/2 = 0, the Poynting
vector field is divergencyless for static fields. In other words,
momentum is being pumped round in closed loops.
Gerard
t...@rosencrantz.stcloudstate.edu
Mike Freeman <mfre...@capcollege.bc.ca> wrote:
>First, let's consider the simpler (?) case where we just look at
>energy. Let's have a (finite) solenoid, Y-axis, and a pair of charged
>plates, field in X direction. Since there is no power dissipation, no
>change in stored energy, and no flux of the Poynting vector out of a
>closed surface, Poynting's theorem says that 0 = 0. Poynting's vector
>is nonzero, but I don't see its physical significance.
The Poynting vector is the momentum density. This system has momentum
stored in it. To see this, imagine turning off the current in the
solenoid. As B drops to zero, there's an induced electric field
(Faraday's law). That induced electric field pushes on the charged
plates. By the time the B field has dropped to zero, the plates have
picked up exactly the same amount of momentum as the amount that was
initially stored in the fields (the amount you would get by
integrating the Poynting vector).
>Now, for momentum. We can create the magnetic field first, no
>Poynting vector yet. Now, I move some charges from one plate to the
>other to create the electric field. Yes, there will be a force in the
>Z-direction on the charges, due to the magnetic field, which I will
>have to balance. Still, I don't see where the alleged momentum comes
>from.
Your description sounds perfectly correct. If you hadn't included
the last sentence, I'd conclude that you understand the situation
perfectly. Since you say you don't, I'll take your word for it,
but I don't understand what you don't see!
You move a charge from one plate to another. As you do so, there's a
magnetic force on the charge, so there's a change in the charge's
momentum. But momentum is conserved, so that means something else
must have changed momentum in the opposite direction. Sure enough,
there was a change in the electric field, and hence a change in the
Poynting vector. If you work out the numbers, the change in
(mechanical) momentum of the charge equals minus the change in field
momentum.
In other words, the answer to your last statement ("Still, I don't see
where the alleged momentum comes from") is this: momentum was
transferred from the charge to the electromagnetic fields. Total
momentum is conserved, and everyone's happy.
Except, I gather, you. I'm not sure what to say about this, since I'm
not sure exactly what's making you unhappy. The only thing I can
think of to do is to state precisely what's true about momentum
conservation in electrodynamics. What I'm about to say may all be
completely obvious to you, in which case please feel free to ignore it
and sneer at me for misunderstanding your point.
When you first learn about momentum in freshman physics, you learn
that the sum of mv for an isolated system of objects is conserved.
(If you then learn special relativity, you learn to replace that with
gamma mv, but let's ignore that.) When you learn electrodynamics, you
find that this is not true. But it is still true that
Mechanical momentum + field momentum
is conserved. Here "mechanical momentum" means the momentum that
you knew and loved before you learned electrodynamics, and "field
momentum" is the integral of the Poynting vector.
Since momentum's entire purpose in life is to be conserved, it makes
sense to expand our notion of momentum and to say that field momentum
really is momentum. If you don't say this, then you have to give up
the idea that momentum is conserved, which would make Isaac Newton
cry.
It's a bit like potential energy. Imagine that you had figured out
conservation of energy by studying perfectly elastic collisions of
superballs on a level surface. At first, you'd only know about
kinetic energy: when you said "energy" you'd really mean "kinetic
energy."
Then (as I'll be doing in a couple of weeks) you move away
from Minnesota to a place where there are hills, and you notice that
energy is not conserved when one of the superballs rolls up a hill.
Eventually, you'd realize that you could rescue conservation of energy
by introducing a new form of energy, potential energy. At first, you
might resist the notion that energy could mean more than just kinetic
energy, but if it was a choice between accepting that and giving up
conservation of energy, you'd suck it up and decide that energy could
mean more than just kinetic energy. Similarly, when you learn
electrodynamics, you have to decide that momentum means more than just
mechanical momentum.
(Of course, if you then want to be even more highbrow and impress your
friends, you learn to say things like "momentum is the generator of
spatial translations.")
-Ted
observer
Put some conductor in that field (sea water or semiconductor) and
there would be some movement.
Mechanist
Firstly, E x H is a flux of the energy.
A stationary magnetic field is generated by a moving line of charges.
In this case:
E = const, H = const ,
E x H /= 0 ,
div(E x H) = 0 .
The next particular example illustrates what may seem counterintuitive
in a general case.
I'll superimpose on the above configuration an immobile line of
opposite electric charges that creates the static electric field E0:
E + E0 = 0 .
The total energy flux is nullified:
(E + E0) x H = 0 .
However, it consists of two unvanishing components.
The flux of the energy due to the moving line of charges
E x H /= 0 ,
and the flux of the energy due to the interaction of the moving line
of charges with the static line of charges
E0 x H /= 0 .
Altering E0 we change the flux of the interaction energy thus
changing the value and the direction of the total energy flux.