Electromagnetic angular momentum, circularly polarized light, and Beth's experiment
Matt Reece
electrodynamics, and wondered if anyone here could shed some light on
this issue. Up to some constant factor which varies in every paper I
read because there are too many systems of units for electromagnetism,
the angular momentum is $\int_{all space} r x g d(volume)$, where $g =
E x H$ is the field momentum (in some system of units, anyway). Now,
notice that for a circularly polarized wave propagating in the
z-direction, the linear momentum is also in the z-direction, so $r x g
= 0$ everywhere. Hmm... never mind that there are lots of photons all
of the same spin propagating in that direction, carrying little
h-bars of spin with them. And that passing a circularly polarized wave
through a birefringent wave-plate gives you a torque on the plate - as
experimentally demonstrated a long time ago: Richard A. Beth, Physical
Review 50 (115).
Probably the best explanation I've found is, inconveniently for me, in
French... "Sur le moment d'impulsion d'une onde ŽlectromagnŽtique" by
J. Humblet, Physica 10 (585). Unfortunately I don't know French, so
I've been stumbling through this paper based on analogy to Spanish and
mostly unhelpful help from Babelfish. Anyhow, as far as I can tell,
they nicely show that for finite wave packets, the angular momentum is
transported at the border of the field... and then conclude "ˆ la
limite, pour le cas rigoureusement monochromatique, l'Žtendue spatiale
est nŽcessairement infinie, et les bords du paquet ainsi que son
contenu en moment d'impulsion sont rejetŽs ˆ l'infini."
Great. So now we have rejected the angular momentum to infinity, if I
read that right. The question then is, how do we get a torque on a
wave plate? The book _States, Waves, and Photons_ by Simmons and
Guttmann has some discussion of this, and seems to claim that some
nice little things happen at the boundary of the plate which magically
make everything turn out exactly as one would expect due to a naive
quantum computation counting h-bars, or to Poynting's little
mechanical analogy. However, they don't seem to give any sort of proof
of this, aside form their claim that a measurement in classical E&M
has to give the same result as a quantum one. But that explanation is
unsatisfactory, since it appears the edge effects would depend in
detail on the scattering at the boundary, but the answer doesn't.
I suppose we could just claim that infinite plane waves are
unphysical, but I don't care for that explanation much either, since
even for a finite wave packet much larger than the wave plate which
looks like a plane wave near the wave plate, we still have the issue
of the angular momentum density of the wave being concentrated at the
boundary which is outside the wave plate.
I wondered if anyone might know of any E&M books that treat this
(Jackson just relegates the discussion of angular momentum to a couple
of exercises that don't seem to address this issue).
Thanks for any insight or explanation,
Matt Reece
Matt Reece
which made me realize I had been a bit imprecise, stating (r x g) was
zero when I only meant that the z component was zero. Here is some
clarification, along with an attempt at showing that the torque must
be explained by field angular momentum rather than the angular
momentum tensor:
Jos Bergervoet <berg...@iaehv.iae.nl> wrote:
>In sci.physics.electromag Matt Reece <mre...@midway.uchicago.edu>
wrote:
>>...
>> notice that for a circularly polarized wave propagating in the
>> z-direction, the linear momentum is also in the z-direction, so $r
>>x g = 0$ everywhere.
>I lost you there. For r on the z-axis you'll indeed have r x g = 0,
>but I don't see how you could even think it would hold everywhere.
Sorry about the confusion; I meant that the z-component of r x g is
zero. The x and y components may be nonzero but they should vanish
upon integration, due to the symmetry of this problem. In Beth's
experiment, the torque was about the z-axis, so the vanishing of (r x
g)_z is problematic. I'm trying to understand where the necessary
z-component of angular momentum appears. The claim is that it somehow
originates at the boundary, but I have yet to encounter either an
analytical computation of this or a computer simulation (although I'm
attempting to simulate it now, the software I have at my disposal is
not really adequate for this problem). Clearly their must be a
z-component to the angular momentum density _somewhere_; after all,
conservation laws require it, since we know there is a change in
mechanical angular momentum.
Perhaps I wasn't clear about the situation I was envisioning. Let me
try to set up the problem more precisely. To avoid the problem of
waves extending to infinity, envision a finite wave packet, large in
extent, which looks like a plane wave in the center and only tapers
off at a large distance. Clearly near the edges this wave has a
z-component, but I'm imagining that we are concerned with it
interacting with a much smaller object placed in its center, so that
in the region of interest it looks like a plane wave. Also, we require
it to be circularly polarized. In that situation, we have the linear
momentum g pointing in the z direction (since it is proportional to
the Poynting vector). In other words, Ex, Ey, Hx, and Hy may be
nonzero but Ez and Hz are zero since we are imagining a wave that
locally looks like a plane wave. Then Sx and Sy are zero, but Sz is
nonzero, and g is proportional to S and so points toward z. But then,
for any vector r, regardless of choice of origin, (r x g)_z = 0, since
r x g is perpendicular to g.
In order for this to work out as it should, something interesting must
be happening at the boundary, and I want to try to get a better
understanding of how this happens (i.e., something of the mechanism of
the scattering or diffraction process that gives rise to the z
component of the field, rather than an appeal to Noether).
I'm fairly sure that the source of the torque must appear in the form
of angular momentum, not in the form of some tensor current, as I will
now attempt to show (warning: lots of TeX notation). The argument
should apply so long as integration by parts is acceptable, based on
the symmetry of the Maxwell tensor and the antisymmetry of the cross
product.
The conservation law generally takes a form:
$L_{em} + L_{mech} = \int M \cdot n dS$
where $M = r \times T$ is the angular-momentum tensor. $T$ is the
Maxwell stress tensor $T_{jk} = \epsilon (E_{j}E_{k} - (E_{l}E_{l})/2)
+ (magnetic terms)$, so it is symmetric. As a result, if we assume the
field vanishes sufficiently rapidly at infinity so that integration by
parts is allowed, we have:
$\int M \cdot n dS = \int div(M) dv = \int x \times div(T) dv =
\int \epsilon_{ijk} x_{j} (div T)_{k} dv =
\int \epsilon_{ijk} x_{j} \partial_{l} T_{lk} dv$
which we can integrate by parts to get:
$\int \epsilon_{ijk} T_{lk} \partial_{l}x_{j} dv =
\int \epsilon_{ijk} T_{lk} \delta_{lj} dv = \int \epsilon_{ijk} T_{jk}
dv$
Since \epsilon is antisymmetric in j and k, and T is symmetric, this
integral vanishes.
In short, I think we can state definitely that the field angular
momentum must have a z component in this situation. The question is:
where does it come from?
Thanks again for any insight. I apologize for my lack of precision
before.
Matt Reece
t...@rosencrantz.stcloudstate.edu
Matt Reece wrote:
>notice that for a circularly polarized wave propagating in the
>z-direction, the linear momentum is also in the z-direction, so $r x g
>= 0$ everywhere. Hmm... never mind that there are lots of photons all
>of the same spin propagating in that direction, carrying little
>h-bars of spin with them. And that passing a circularly polarized wave
>through a birefringent wave-plate gives you a torque on the plate - as
>experimentally demonstrated a long time ago: Richard A. Beth, Physical
>Review 50 (115).
This is a very nice puzzle. It's well worth thinking about. We
discussed it at some length on this newsgroup some time within the
last couple of years.
Here are some things I can say that may be helpful. (I think that you
already know these first few points, but I can't resist repeating them
out of some combination of a desire for completeness and a tendency to
long-windedness.)
1. The proof that the field angular momentum is the integral over all
space of r x (E x H), strictly speaking, only works if the
electromagnetic fields go to zero fast enough at infinity. To be a
bit more precise, the theorem I'm talking about is the one that says
that the above integral, plus the mechanical angular momentum, gives a
conserved quantity. To prove this theorem, you integrate by parts and
throw away boundary terms. So in particular the result doesn't
apply to plane waves, which extend out to infinity.
2. If you want, you can work out the field angular momentum contained
in a finite-sized volume of space by going through the same argument
as the one I hinted at above, but not dropping boundary terms. If you
do, you'll find that the expression for the angular momentum within a
volume is the volume integral of r x (E x H) *plus* a surface integral
over the boundary. For the case of a circularly polarized plane wave,
any compact volume does indeed have nonzero angular momentum, but it
comes entirely from the surface term.
3. You can of course use the usual expression for the angular momentum
as long as you consider, not a true plane wave, but a wave packet of
finite (but possibly very large) extent. In this case, the angular
momentum density r x (E x H) does integrate up to give the right
value, but the contributions come mostly from the fringes of the wave
packet. Near the center, where the wave packet looks most like a
plane wave, the angular momentum density is zero.
This seems to be what your French source is saying:
>"Ã la
>limite, pour le cas rigoureusement monochromatique, l'étendue spatiale
>est nécessairement infinie, et les bords du paquet ainsi que son
>contenu en moment d'impulsion sont rejetés à l'infini."
If my school French is up to the task, this says "In the limit, for
the perfectly monochromatic case, the spatial extent is necessarily
infinite, and the edges of the wave packet as well as its angular
momentum content are rejected to infinity."
In other words, as your wave packet gets bigger and bigger, looking
more and more like a monochromatic plane wave, the angular momentum
density r x (E x H) marches off to spatial infinity.
>Great. So now we have rejected the angular momentum to infinity, if I
>read that right. The question then is, how do we get a torque on a
>wave plate?
Great question! Since I'm a theorist, let me consider a simple case
of a perfect absorber rather than a birefringent material.
You have a circularly-polarized plane wave incident on a circular disk of
perfectly absorbing material. To make sure everything's nice
and convergent, say that the plane wave is really a great big
wave packet, and to keep things simple, say that
wavelength << radius of disk << extent of wave packet.
After the wave packet has passed by the disk, it looks essentially
like a plane wave with a cylindrical hole cut out of it. Right in the
neighborhood of edges of the hole, the E and B fields are quite
complicated (diffraction and all that). But if you tried hard enough,
you could work out the fields and integrate r x (E x H) over all
space. If you did that, you'd find a large contribution from the
edges of the wave packet, and another large contribution, of the
opposite sign, from the region right around the edges of that
cylindrical hole. The first contribution is of course unaffected by
the fact that some of the light has been absorbed (since it's very far
away). The second contribution gives you exactly the right amount of
absorbed angular momentum to account for the torque on the disk.
In the limit of short wavelengths, you can do this calculation in a
much simpler way: neglect diffraction and say that the field drops
sharply to zero at the edge of the shadowed cylinder. Then calculate
the field angular momentum by integrating r x (E x H) only over the
non-shadowed region *and* including the boundary term at the surface
of the shadowed cylinder. The boundary term is what'll give you the
absorbed angular momentum.
In the spirit of full disclosure, let me acknowledge that I've never
actually done this calculation! But I have worked through the proof
of the theorem that angular momentum is conserved in situations like
this, so I don't have to! I know that the answer must come out this
way.
>The book _States, Waves, and Photons_ by Simmons and
>Guttmann has some discussion of this, and seems to claim that some
>nice little things happen at the boundary of the plate which magically
>make everything turn out exactly as one would expect due to a naive
>quantum computation counting h-bars, or to Poynting's little
>mechanical analogy. However, they don't seem to give any sort of proof
>of this, aside form their claim that a measurement in classical E&M
>has to give the same result as a quantum one. But that explanation is
>unsatisfactory, since it appears the edge effects would depend in
>detail on the scattering at the boundary, but the answer doesn't.
I don't really understand this objection. Angular momentum is conserved,
no matter what in detail happens with the scattering at the boundary.
-Ted
Matt Reece
> To prove this theorem, you integrate by parts and
> throw away boundary terms. So in particular the result doesn't
> apply to plane waves, which extend out to infinity.
That's correct. Interestingly, in the case that we can throw away
boundary terms, we also have a decomposition into a term (E x A) and
another, more messy term, but it seems that the (E x A) term is
actually in the right direction for Beth's experiment. I wonder if I
should be looking into calculating the torque this way, even though
the thought of using equations that aren't gauge-invariant bothers me
a bit.
> [...] is the volume integral of r x (E x H) *plus* a surface integral
> over the boundary. For the case of a circularly polarized plane wave,
> any compact volume does indeed have nonzero angular momentum, but it
> comes entirely from the surface term.
Hmm... are you sure about this? It's the component of angular momentum
in the direction of propagation we're concerned with here, and I think
I worked out that the surface term gives us zero for this as well.
I'll look back over that.
> [...] say that the plane wave is really a great big
> wave packet, and to keep things simple, say that
>
> wavelength << radius of disk << extent of wave packet.
OK, this is exactly the sort of regime that I was looking at.
> After the wave packet has passed by the disk, it looks essentially
> like a plane wave with a cylindrical hole cut out of it. Right in the
> neighborhood of edges of the hole, the E and B fields are quite
> complicated (diffraction and all that).
That's right, and in this case it's pretty easy to see how we get a
nonzero angular momentum density in the propagation direction. The
book I mentioned by Simmons and Guttmann even has a nice plot of what
happens. The angular momentum density has a term that's related to the
intensity gradient, and we obviously have to have a gradient near the
edges.
> The second contribution gives you exactly the right amount of
> absorbed angular momentum to account for the torque on the disk.
OK, that makes sense. But what I still want to understand is how we
get a torque in the case of a birefringent material. The problem here
is that we don't have a large intensity gradient, so you don't see
this kind of "shadow" effect.
> In the spirit of full disclosure, let me acknowledge that I've never
> actually done this calculation! But I have worked through the proof
> of the theorem that angular momentum is conserved in situations like
> this, so I don't have to! I know that the answer must come out this
> way.
Right, that's why conservation theorems are so great. The problem
comes about when we're trying to find a way of calculating forces and
torques based on computer simulations of the fields. Forces seem to be
reasonably straightforward, but in order to make sure torque
calculations are working out as they should, and that we've set up the
simulation correctly, I want to understand in a bit more detail how
the fields come out in such a way that we get the angular momentum
components that we need.
> >[...] has to give the same result as a quantum one. But that explanation
> >is unsatisfactory, since it appears the edge effects would depend in
> >detail on the scattering at the boundary, but the answer doesn't.
> I don't really understand this objection. Angular momentum is conserved,
> no matter what in detail happens with the scattering at the boundary.
I'm sorry, I didn't mean to imply that there was anything wrong with
that explanation. Of course the conservation theorem is correct. The
problem is that we want to be able to compute these torques based on
the field values, so I want to understand in more detail what happens
to the fields that ensures that we get exactly the right value. For
instance, why does the ratio L_z / U = 2/omega hold when we integrate
the field over the entire half-wave plate, since we don't have a nice
approximation like the fields going to zero at the boundary to use? We
know we have to get a result like this (or at least close to it;
reflection and some other things might mess it up a bit) since the
photons each change from +hbar to -hbar spin and the energy of each
one is hbar*omega.
Thanks for your reply; I think you've summarized some of the issues
much better than I did, and your explanation for an absorbing material
makes sense. Unfortunately, I don't see that it's so easy for a
birefringent material, and I need a bit more insight into what the
fields are doing.
Matt Reece
Matt Reece
> mre...@midway.uchicago.edu (Matt Reece) wrote:
[cut]
> >[...] passing a circularly polarized wave
> >through a birefringent wave-plate gives you a torque on the plate - as
> >experimentally demonstrated a long time ago: Richard A. Beth, Physical
> >Review 50 (115).
> In my opinion, the easy way to do this is to invoke photons. Not exactly
> classical, but as you have noticed, a classical calculation of a circularly
> polarised infinite plane wave is problematic. See Nieminen et al, Journal of
> Modern Optics 48, 405-413 (2001) for our semi-classical treatment of Beth's
> experiment.
I just read the paper, and I agree that it's a nice, simple approach
to the problem. However, I'm not sure it will work for my purpose. I'm
working for the summer at the Electro-Optics lab at the University of
Louisville, and what we're trying to do here is find a way to use some
software in the lab to compute forces and torques due to
electromagnetic fields. The goal is to be able to compute these
quantities based on simulations that give us the fields E and H. This
is why I'm concerned about trying to see how we can compute the
correct answer for torque in Beth's experiment based on a classical
description of the field. Unfortunately, the software we have here
isn't especially well suited for the purpose, and I'm going back to
school in Chicago soon so I don't know what is going to be done with
this project after this, but I was hoping to find some sort of
classical description so we can see if the information we're trying to
get out of our software is anything close to what we should be
getting.
> IIRC, Jackson's angular momentum exercises are specifically for non-plane wave
> beams for a very good reason. The infinite plane wave is indeed unphysical
> (how does one generate a circularly polarised infinite plane wave). Consider
> Jackson's finite width beam angular momentum problem for a very wide beam.
Right, he has a problem which involves showing that the ration
(Angular momentum in direction of propagation)/(Energy) = (+/-)
1/omega for a finite wave. The field gradient at the edges of the wave
lead to this. What I want to understand, though, is why (in terms of
the classical field values) we get the same result if we have a
situation in which the finite wave is still much larger than the
object it interacts with. Why (from a classical standpoint) does the
interaction process always lead to the same (1/omega) result, and has
this been accurately modeled, either analytically or numerically?
Thanks,
Matt Reece
Matt Reece
>You earlier mentioned M = r x g, but now you write $M = r \times T$
>which seems incorrect.
No, I meant that (r x g) is the electromagnetic angular momentum
density (a vector) while (r x T) is the angular momentum density
tensor.
The differential form of the conservation law (in terms of 3-vectors
and 3 x 3 tensors... or at least pseudovectors and pseudotensors)
looks like:
(d/dt) (l_mech + l_elec) = div(M).
Here l_mech is mechanical angular momentum density, l_elec = r x g,
and M = r x T.
> The full relation using 4-vector indices is:
>
>$$ M^{\alpha\beta\gamma} =
> \Theta^{\alpha\beta} x^\gamma - \Theta^{\alpha\gamma} x^\beta $$
>
>(see Jackson, 2nd ed., Eq. 12.117) but I don't think you can use the
>simple cross-product notation to denote this tensor contraction!
That expression for M is correct, but we can use cross-product
notation because $\Theta$ is a symmetric tensor. This means that we
can define
$$ M^{\alpha\beta} = \epsilon^{\beta \gamma \delta}
\Theta^{\alpha}_{\delta} x_{\gamma} $$,
where I may have mixed up a few of the indices but the general idea is
that we can antisymmetrize over two indices and still keep all of the
information that was in your M^{\alpha\beta\gamma}.
>In addition, the ordinary three-dimensional angular momentum is
>contained in the components $M^{0ij}$. With the equation above, they
>depend only on the time-space components $\Theta^{0k}$ (with k=1,2,3)
>of the stress tensor. But you seem to use the space-space components
>$\Theta^{ij}$, which actually don't contribute to the three-dimensional
>angular momentum (which just comes out as r x g, as it should!)
Everything you say here seems to be true, but I think that you
misunderstood my point. The angular momentum does in fact come out to
be (r x g), but the conservation law includes a sort of "angular
momentum current" in the form of the tensor M.
Jackson, deals with this in problem 6.10 (3rd edition). It's problem
6.11 in the second edition. Let me quote part of it:
"Show that the differential and integral forms of the conservation law
are
$ \frac{\partial}{\partial t} (l_{mech} + l_{field}) + \nabla \cdot M
= 0 $
and $ \frac{d}{dt} \int_{V} ((l_{mech} + l_{field}) d^{3}x + \int_{S}
n \cdot M da = 0 $ where the field angular-momentum density is $
L_{field} = x \times g = \mu epsilon x \times (E \times H)$ and the
flux of angular momentum is described by the tensor $ M = T \times x$"
He goes on to describe that M can be written as a third-rank tensor
(as you did above) but may "be written as a pseudo-tensor of the
second rank."
I hope this clears up the confusion over what I mean by "angular
momentum" and by the "angular momentum tensor."
Matt Reece
t...@rosencrantz.stcloudstate.edu
Matt Reece <mre...@midway.uchicago.edu> wrote:
>t...@rosencrantz.stcloudstate.edu wrote in message
>news:<9mk1rs$mqq$1...@news.state.mn.us>...
>> [...] is the volume integral of r x (E x H) *plus* a surface integral
>> over the boundary. For the case of a circularly polarized plane wave,
>> any compact volume does indeed have nonzero angular momentum, but it
>> comes entirely from the surface term.
>Hmm... are you sure about this?
Well, I've never done the calculation, but I won't let that stop me
from confidently asserting that I'm sure! After all, I'm sure that
1. There is field angular momentum in that volume (since an absorber
would feel a torque if it absorbed all the radiation)
2. The volume integral is zero.
3. The conservation law (volume integral + surface integral + mechanical
angular momentum is conserved) is correct.
The only way out is for the surface integral to give the correct
contribution.
>OK, that makes sense. But what I still want to understand is how we
>get a torque in the case of a birefringent material. The problem here
>is that we don't have a large intensity gradient, so you don't see
>this kind of "shadow" effect.
Well, the fields aren't zero inside the "shadowed" region, but they're
very different from what they are outside of it. (I forget whether
we're imagining a quarter-wave or a half-wave plate. In one case, the
field inside the shadowed region is linearly polarized, and in the
other case it's circularly polarized in the opposite sense to the
incident wave.) So near the boundary of that cylinder you have strong
field gradients, and the exact solution to Maxwell's equations is a
big hairy mess. If you could work it out that exact solution, you'd
find that the integral of r x (E x H) in the region near the surface
of that cylinder would give the correct contribution to the angular
momentum.
I have no intention of doing an explicit calculation to prove this,
but once again I assert with utter confidence that it must be true,
because it's the only way for the conservation law to be satisfied!
-Ted
Gerard Westendorp
>
> I am becoming very confused about angular momentum in classical
> electrodynamics, and wondered if anyone here could shed some light on
> this issue.
We had some long and partially unresolved threads on photon
A.M. (see "the problematic nature of photon spin".)
It is quite tricky, because (as you also say) the circulating
momentum density in a wave packet is located on the fringes of the
wave packet, but a plane wave actually does not seem to have any
circulating momentum density. However, even a plane wave is able to
torque a conductor that is in its way.
[..]
> Great. So now we have rejected the angular momentum to infinity, if I
> read that right. The question then is, how do we get a torque on a
> wave plate?
I think when a big circularly polarized wave hits a small conductor,
the conductor is torqued, and a "shadow" of opposite A.M. is cast
in the wake of the conductor. This takes care of the locality
problem: The wave's original AM remains far away on the fringes of the
wave, while the new opposite AM is created on the shadow fringes.
Apparently, we must distinguish
between "ability to exert torque" , which is a carried locally
by the plane wave, and "angular momentum density" which is carried
far away on the fringes of the wave.
However, it should be noted that A.M. is a tricky quantity, which
can be manipulated by a choice of reference frame. One may also
for example add a constant A.M. density field to the universe without
changing any physics. I had suggested using ExA as a local EM. AM.
density. This had some problems of its own. Maybe this time I will
figure it out.
Gerard
Timo Nieminen
>
>what we're trying to do here is find a way to use some
>software in the lab to compute forces and torques due to
>electromagnetic fields. The goal is to be able to compute these
>quantities based on simulations that give us the fields E and H.
Will these be plane waves? If not, then you won't have the infinite plane wave
problem.
>This
>is why I'm concerned about trying to see how we can compute the
>correct answer for torque in Beth's experiment based on a classical
>description of the field.
There are two classical methods for finding the torque on the birefringent
material due to the field. From above, it's easy to see why you want to
calculate the force using the change in angular momentum of the field. (Just an
idle thought - why not divide the infinite plane wave into an infinite number
of finite beams?) However, if your simulations also give you the fields inside
the birefringent disk, you will know (1) E at all points within the disk, and
(2) the induced dielectric polarisation per unit volume, which means you can
find the force and torque per unit volume for all points within the disk.
Perhaps of use?
>Right, he has a problem which involves showing that the ratio
>(Angular momentum in direction of propagation)/(Energy) = (+/-)
>1/omega for a finite wave. The field gradient at the edges of the wave
>lead to this. What I want to understand, though, is why (in terms of
>the classical field values) we get the same result if we have a
>situation in which the finite wave is still much larger than the
>object it interacts with. Why (from a classical standpoint) does the
>interaction process always lead to the same (1/omega) result, and has
>this been accurately modeled, either analytically or numerically?
If a finite, but wide, beam interacts with a birefringent disk, this will
change the polarisation of the portion of the beam that passes through the
disk. Note that the central region of the beam will now be non-uniform. The
region where the polarisation-changed and original-polarisation portions of the
beam meet should produce the required mathematical result.
Anyway, when faced with the same problem, I just went semi-classical. Photons
made my life easier. Agreed, a purely classical treatment would be more
satisfactory, but starting with a purely classical description of a field doesn't mean
you _need_ to stay purely classical - just find your photon flux and
local polarisation at any point of interest, and you immediately have the
momentum and angular momentum fluxes.
--
Timo Nieminen - http://www.physics.uq.edu.au/people/nieminen/nieminen.html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
t...@rosencrantz.stcloudstate.edu
Gerard Westendorp <wes...@xs4all.nl> wrote:
>I think when a big circularly polarized wave hits a small conductor,
>the conductor is torqued, and a "shadow" of opposite A.M. is cast
>in the wake of the conductor. This takes care of the locality
>problem: The wave's original AM remains far away on the fringes of the
>wave, while the new opposite AM is created on the shadow fringes.
>Apparently, we must distinguish
>between "ability to exert torque" , which is a carried locally
>by the plane wave, and "angular momentum density" which is carried
>far away on the fringes of the wave.
Well, of course we must! Those are utterly different things.
Anything can exert torque, regardless of whether or not it has angular
momentum. I don't have any angular momentum right now, but I can
exert torque on things!
Take an example closer to the situation at hand. Suppose we have a
*linearly* polarized plane wave and we send it through a quarter-wave
plate. It'll come out circularly polarized. Presumably everyone
agrees that the linearly polarized wave had no angular momentum. But
when it passes through the quarter-wave plate, it exerts a torque on
it.
(I do mean a quarter-wave plate, right? That's the one that turns
linear to circular polarization, isn't it?)
-Ted
Jos Bergervoet
>> Great. So now we have rejected the angular momentum to infinity, if I
>> read that right. The question then is, how do we get a torque on a
>> wave plate?
> I think when a big circularly polarized wave hits a small conductor,
> the conductor is torqued, and a "shadow" of opposite A.M. is cast
> in the wake of the conductor. This takes care of the locality
> problem: The wave's original AM remains far away on the fringes of the
> wave, while the new opposite AM is created on the shadow fringes.
> Apparently, we must distinguish
> between "ability to exert torque" , which is a carried locally
> by the plane wave, and "angular momentum density" which is carried
> far away on the fringes of the wave.
In the conservation law for the angular mopmentum tensor M,
$$
\partial_\alpha M^{\alpha\beta\gamma} = 0,
$$
we have the density of angular momentum, $M^{0ij}$ (where
the pair $ij$ selects an angular momentum component) and we
have a kind of "flux" in direction $k$ of this same angular
momentum component, given by $M^{kij}$.
You could simply compute the flux of angular momentum
entering some closed volume around an object, to find the
torque on this object. For instance, to find the z-component
of torque ($ij=12$ in the above), integrate the in-flux given
by the vector $f^k = M^{k12}$ into the boundary of the volume.
-- Jos
Jos R Bergervoet
> Jos Bergervoet <berg...@iaehv.iae.nl> wrote:
>>In addition, the ordinary three-dimensional angular momentum is
>>contained in the components $M^{0ij}$. With the equation above, they
>>depend only on the time-space components $\Theta^{0k}$ (with k=1,2,3)
>>of the stress tensor. But you seem to use the space-space components
>>$\Theta^{ij}$, which actually don't contribute to the three-dimensional
>>angular momentum (which just comes out as r x g, as it should!)
> Everything you say here seems to be true, but I think that you
> misunderstood my point. The angular momentum does in fact come out to
> be (r x g), but the conservation law includes a sort of "angular
> momentum current" in the form of the tensor M.
Yes. In the conservation law for M,
$$
\partial_\alpha M^{\alpha\beta\gamma} = 0,
$$
we have the density of angular momentum in $M^{0ij}$ (where
the pair $ij$ selects an angular momentum component) and we
have a kind of "flux" in direction $k$ of this same angular
momentum component, given by $M^{kij}$. You also wrote:
> ... to be able to compute
> forces and torques due to electromagnetic fields. The goal
> is to be able to compute these quantities based on
> simulations that give us the fields E and H.
Therefore, it seems that you should simply compute the flux
of angular momentum entering some closed volume around an
object, to find the torque on this object. For instance, to
find the z-component of torque (meaning $ij=12$), integrate
over the in-flux given by the vector $f^k = M^{k12}$.
-- Jos
Dr. Jozef R. Bergervoet Integrated Transceivers
Philips Research Laboratories, Eindhoven, The Netherlands
Building WAY5 83 Phone: +31-40-2742643
E-mail: Jos.Ber...@philips.com FAX: +31-40-2744657
Gerard Westendorp
realize that I don't even understand how linear momentum works.
If a plane wave with linear momentum density (EXB)/c hits
a lossy conductor, the momentum of the EM wave should get
transferred to the lossy conductor. But if I look at actual forces,
I can see no force working in the right direction.
It seems that electrons will only be accelerated in the
transverse direction, parallel to the E fields.
Gerard
Steve Carlip
> If a plane wave with linear momentum density (EXB)/c hits
> a lossy conductor, the momentum of the EM wave should get
> transferred to the lossy conductor. But if I look at actual forces,
> I can see no force working in the right direction.
> It seems that electrons will only be accelerated in the
> transverse direction, parallel to the E fields.
It's a second order effect. Electrons are accelerated parallel
to E, which gives them a velocity perpendicular to B. The B
fields then give an acceleration in the direction vxB, which
is parallel to ExB. For a periodic plane wave, the displacement
in the E direction averages to zero, but the displacement in the
ExB direction adds over the whole period, so there's a net
momentum transfer in the direction of the wave.
Steve Carlip
Gordon D. Pusch
Think about the electric current those moving electrons represent,
and then think about how they interact with the wave's magnetic field,
which is at right angles to the electric field, and also to those
electron's velocities...
-- Gordon D. Pusch
perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;'
Timo Nieminen
And then you have electrons with velocity perpendicular to the direction of
propagation, and perpendicular to the magnetic field.
No lossy conductor needed, either. Lossless dielectric will give you a
radiation pressure force as well. Consider the forces acting on a small
oscillating dipole in an oscillating field.
Roland Franzius
Gerard Westendorp wrote:
I assume you know that linear em field momentum density is defined ~ E x
B in order to save the law of total momentum conservation if em
charges, currents and fields have to be incorporated. This is somehow
self explaining by regauging the mechanical momentum p -> p- e/c A and
considering the momentum current conservation.
Perhaps the mean field picture may clarify the QM transfer process:
Consider a metal plate carrying an alternating surface current parallel
to the surface in order to absorb and reflect a linear polarized em
wave. The surface current density defines the outer tangent H field. The
collection of one particle states making up the current state feels the
B- field as a mean field with some penetration depth and is answering by
a linear response Hall voltage perpendicular to the surface. The E-field
corresponding to the Hall voltage gradient is the kick transferring the
field momentum onto the lattice of positively charged ions and thereby
to the mass of the solid. This is the QM picture of the classical light
pressure.
--
Roland Franzius
Jos Bergervoet
> Thinking about torques exerted by EM waves, I
> realize that I don't even understand how linear momentum works.
As with angular momentum: the force exerted is just the in-flux of
momentum into a surface around an object. So you use the space-space
components of the energy-momentum tensor.
> It seems that electrons will only be accelerated in the
> transverse direction, parallel to the E fields.
Take the simple case of a plane wave, normally incident on a resistive
slab (infinite in extent, but of zero thickness).
As a first order effect you will indeed see a current in the
E-direction. But then the Lorentz force on this current, due to the
B-field, is in the direction normal to the slab. This is second order
in fieldstrength, as it should be (number of impinging photons is
proportional to wave enegrgy, hence to the square of the amplitude).
-- Jos
t...@rosencrantz.stcloudstate.edu
Gerard Westendorp <wes...@xs4all.nl> wrote:
>If a plane wave with linear momentum density (EXB)/c hits
>a lossy conductor, the momentum of the EM wave should get
>transferred to the lossy conductor. But if I look at actual forces,
>I can see no force working in the right direction.
>
>It seems that electrons will only be accelerated in the
>transverse direction, parallel to the E fields.
Once the electrons start moving, the B field will exert forces
on them, right?
-Ted
Eric A. Forgy
The incident plane wave induces a current density J on the surface of
the conductor. I believe it is the magnetic force F = q v x B which is
repronsible for the linear momentum transfer along the direction of
propagation.
Eric
Steve McGrew
In a medium where the electron motion is not in phase with the
E field in the wave at a particular frequency, in what ways would the
momentum transfer be different?
Steve
Gerard Westendorp
t...@rosencrantz.stcloudstate.edu wrote:
> In article <3B8D69C5...@xs4all.nl>,
> Gerard Westendorp <wes...@xs4all.nl> wrote:
> >I think when a big circularly polarized wave hits a small conductor,
> >the conductor is torqued, and a "shadow" of opposite A.M. is cast
> >in the wake of the conductor. This takes care of the locality
> >problem: The wave's original AM remains far away on the fringes of the
> >wave, while the new opposite AM is created on the shadow fringes.
> >Apparently, we must distinguish
> >between "ability to exert torque" , which is a carried locally
> >by the plane wave, and "angular momentum density" which is carried
> >far away on the fringes of the wave.
> Well, of course we must! Those are utterly different things.
> Anything can exert torque, regardless of whether or not it has angular
> momentum. I don't have any angular momentum right now, but I can
> exert torque on things!
Yes, but the linear momentum of a photon and the energy of a photon
are carried locally by the photon, and transferred to objects with
which absorb them. The question is if angular momentum is
different. If you use the expression rxExB as am density, it would
seem to be the case. But rxExB is not the only expression that
fulfills the requirements for am. density. Judging by the use of
exclamation marks in your replies, you don't understand that
some of us like to know if there is a way to write down am. density
such that it relates more directly to for example the idea of
photons with spin being absorbed, and transferring all of their
conserved quantities to the absorbing object.
Maybe this is not possible, but if not, it is far from obvious,
and I for one would like to understand why.
Gerard
Gerard
Gerard Westendorp
>
> In sci.physics.electromag Gerard Westendorp <wes...@xs4all.nl> wrote:
>
> > Thinking about torques exerted by EM waves, I
> > realize that I don't even understand how linear momentum works.
>
> As with angular momentum: the force exerted is just the in-flux of
> momentum into a surface around an object. So you use the space-space
> components of the energy-momentum tensor.
Interesting that you say "as with angular momentum". I agree with
this, but if you use rxExB as angular momentum density, it would
not work out. This is why I am trying to work out an alternative
expression for am. density.
[..]
> As a first order effect you will indeed see a current in the
> E-direction. But then the Lorentz force on this current, due to the
> B-field, is in the direction normal to the slab. This is second order
> in fieldstrength, as it should be (number of impinging photons is
> proportional to wave enegrgy, hence to the square of the amplitude).
That makes sense. Indeed, as you a force just proportional to E
could never be correct for the above reason.
Gerard
t...@rosencrantz.stcloudstate.edu
Gerard Westendorp <wes...@xs4all.nl> wrote:
>Yes, but the linear momentum of a photon and the energy of a photon
>are carried locally by the photon, and transferred to objects with
>which absorb them.
Well, if you're really talking about photons, then we're talking about
different things. I was talking about classical electromagnetic
waves. I'm honestly not sure I know how much of a difference that
makes, so if you really want to talk about the quantum case I'll bow
out and let others take over. I understand classical electrodynamics
much better than quantum electrodynamics!
>The question is if angular momentum is
>different. If you use the expression rxExB as am density, it would
>seem to be the case. But rxExB is not the only expression that
>fulfills the requirements for am. density.
I agree. You can add any locally conserved vector field.
The same is true for momentum, of course.
>Judging by the use of
>exclamation marks in your replies, you don't understand that
>some of us like to know if there is a way to write down am. density
>such that it relates more directly to for example the idea of
>photons with spin being absorbed, and transferring all of their
>conserved quantities to the absorbing object.
Or maybe I'm just an enthusiastic guy!
My overuse of exclamation points may well be a stylistic flaw in my
writing, but I don't think that it has anything to do with this
particular point. The point I was making in an exclamatory way is
that having angular momentum is not a requirement for exerting torque.
To get back to the main issue, as far as I know, rx(ExB) is the only
form for angular momentum density that is defined locally, is
gauge-invariant, and transforms the way an angular momentum density
should under Lorentz transformations. Those seem to me to be
extremely nice criteria for an angular momentum density in classical
electromagnetism. If you're willing to give these up, you can come up
with other self-consistent ways of describing the physics. I have no
argument with that at all.
Personally, I'm not much interested in these alternate descriptions.
In particular, giving up gauge invariance seems to me to be a cure
much worse than the disease. (This is not surprising, since I'm not
sure what the "disease" is -- the standard picture works fine as far
as I'm concerned!) (There go those exclamation points again.)
If your complaint is that the standard recipe for angular momentum
density is in some sense "not physically real," then exchanging
it for something gauge-dependent can only make matters worse,
as far as I'm concerned. But in any case, I cheerfully concede
that such alternate descriptions exist, and if it makes you happy
to think about them, be my guest!^H.
My only purpose in my posts in this thread is to explain what's going
on within the context of standard classical electrodynamics, so that
people can see that things do work out properly in the standard
picture. That's not to say that other ways of describing
the physics don't exist as well.
-Ted
Harry Johnston
> Yes, but the linear momentum of a photon and the energy of a photon
> are carried locally by the photon, and transferred to objects with
> which absorb them. The question is if angular momentum is
> different.
I don't see that it is. The angular momentum of each individual
photon is carried locally by the photon and transferred to any object
which absorbs it. It is only when you calculate the expectation value
and sum over all photons that you get a zero angular momentum density
inside the light beam.
> If you use the expression rxExB as am density, it would
> seem to be the case. But rxExB is not the only expression that
> fulfills the requirements for am. density.
I take it actually *being* the AM density isn't one of your
requirements? ;-)
Seriously, I don't think there is any room for ambiguity here. The
(total) linear momentum density is ExB so the (total) angular momentum
density is, by definition, rxExB.
> Judging by the use of
> exclamation marks in your replies, you don't understand that
> some of us like to know if there is a way to write down am. density
> such that it relates more directly to for example the idea of
> photons with spin being absorbed, and transferring all of their
> conserved quantities to the absorbing object.
This should be possible; essentially, all you need is to write the
angular momentum for each photon separately, instead of just looking
at the total.
To do this properly would require a full QFT treatment, but if you
just want a rough picture of what is happening, you can imagine the
photons as little spinning disks [1] of an arbitrarily chosen
infinitesimal radius, oriented so the axis of spin is in the direction
of travel. You should write the usual formula for the angular
momentum of a spinning disk. The "mass density" and "angular
velocity" should be chosen as a function of radius to make the total
mass and angular momentum come out right. We then take the radius to
zero at the end of the calculation.
When you want to find the total angular momentum at a given point in
space, you have to integrate over the photon probability density and
multiply by the number of photons. If the wavefunction is that of a
plane wave within the chosen disk radius, then the probability density
is constant in the plane parallel to the disk, so you just need to
integrate over position in that plane.
It is reasonably easy to see that this must be zero, because for every
point on every disk there is exactly one other disk whose motion at
that point is equal and opposite. If you like you can do the integral
explicitly; it shouldn't be too hard.
Wherever the probability varies the disks won't cancel one another out
any more, so there will be net angular momentum. As we take the
radius of the disks to zero, the net angular momentum will be
localised to the regions where the probability varies.
When a photon hits something the wavefunction collapses and all of
that photon's angular momentum is transferred to the object. Behind
the object the wavefunction (of the remaining photons) will have a
shadow in it, so there will be an equal and opposite angular momentum
at the edge of the shadow, as required.
Harry.
[1] Disks rather than spheres; this is really just to make the sums
easier, but you can justify it by noting that spheres travelling at
the speed of light would be squashed flat anyway by an infinite
Lorentz contraction. ;-)
---
Harry Johnston, om...@ihug.co.nz
One Earth, One World, One People
Jos R Bergervoet
>> As with angular momentum: the force exerted is just the in-flux of
>> momentum into a surface around an object. So you use the space-space
>> components of the energy-momentum tensor.
>
> Interesting that you say "as with angular momentum". I agree with
> this, but if you use rxExB as angular momentum density, it would
> not work out.
Why not? The differential conservation law for the angular momentum
tensor ensures this:
\partial_\alpha M^{\alpha\beta\gamma} = 0,
And rxExB is exactly what we have for the angular momentum density
in this tensor (see Jackson, 2nd ed., 12.114-12.117).
So we now have: the _torque_ exerted is just the in-flux of angular
momentum into a surface around an object. And the flux of angular
momentum is of course given by the space-space-space components of
the tensor.
Example: let's compute the z-component of the torque. The k-component
of angular momentum density is in \epsilon_{kij} M^{0ij}, so the
z-component is the tensor component M^{012}, and the conserved current
associated with it is the 4-vector M^{\mu 12}. So the in-flux of the
z-component of angular momentum is given by the 3-vector M^{i12}.
Integrate this 3-vector over a closed surface, using Gauss' law,
and you'll find the z-component of the torque.
Toby Bartels
>The point I was making in an exclamatory way is
>that having angular momentum is not a requirement for exerting torque.
Any more than having momentum is a requirement for exerting force.
-- Toby
to...@math.ucr.edu
Gerard Westendorp
> Gerard Westendorp wrote:
> >> As with angular momentum: the force exerted is just the in-flux of
> >> momentum into a surface around an object. So you use the space-space
> >> components of the energy-momentum tensor.
> > Interesting that you say "as with angular momentum". I agree with
> > this, but if you use rxExB as angular momentum density, it would
> > not work out.
[...]
> So we now have: the _torque_ exerted is just the in-flux of angular
> momentum into a surface around an object. And the flux of angular
> momentum is of course given by the space-space-space components of
> the tensor.
I am not sure that you get a non-zero answer for a plane wave.
If I substitute a plane wave:
E = ( cos(z-t) ) B = ( -sin(z-t) )
( sin(z-t) ) ( cos(z-t) )
( 0 ) ( 0) )
I get for the energy-momentum tensor:
( 1 0 0 1 )
( 0 0 0 0 )
( 0 0 0 0 )
( 1 0 0 1 )
So I get a constant flux of z-momentum in the z-direction.
If I take the moment of this integrated over a volume with the
origin at its center, I still get zero.
Gerard
Gerard Westendorp
>
> Ted wrote in part:
>
> >The point I was making in an exclamatory way is
> >that having angular momentum is not a requirement for exerting torque.
>
> Any more than having momentum is a requirement for exerting force.
But there is reason to believe that for absorption of electromagnetic
fields, the resultant torque is due to absorption of angular momentum.
This is not obvious from classical physics, but from quantum mechanics
we know that circularly polarized photons carry angular momentum
h_bar per photon, and linear momentum h_bar*k per photon. So absorbing
a piece of a quasiplane wave, you would expect the torques, energy
transfer and momentum transfer (force) to be proportional to the flux
times the area of the "shadow" cast by the absorbing object.
I don't want to shift the discussion to quantum mechanics, but I think
given the right mathematical treatment, classical physics should give
the same result here. I know that it probably does, but I feel that
there must be a nice way to see this.
I am looking for a way to rewrite the term rxExB so
that it is more obvious that the torque is Area*absorbed flux.
Gerard
Gerard Westendorp
> Gerard Westendorp <wes...@xs4all.nl> wrote:
>
> > But rxExB is not the only expression that
> > fulfills the requirements for am. density.
>
> I take it actually *being* the AM density isn't one of your
> requirements? ;-)
>
> Seriously, I don't think there is any room for ambiguity here. The
> (total) linear momentum density is ExB so the (total) angular
> momentum density is, by definition, rxExB.
I dug up an old post, surprisingly it was from yourself:
> /->-\ /->-\ /->-\ /->-\ /->-\ /->-\ /->-\ /->-\
> | | | | | | | | | | | | | | | |
> | | | | | | | | | | | | | | | |
> \-<-/ \-<-/ \-<-/ \-<-/ \-<-/ \-<-/ \-<-/ \-<-/
>
> /->-\ /->-\ /->-\ /->-\ /->-\ /->-\ /->-\ /->-\
> | | | | | | | | | | | | | | | |
> | | | | | | | | | | | | | | | |
> \-<-/ \-<-/ \-<-/ \-<-/ \-<-/ \-<-/ \-<-/ \-<-/
>
> /->-\ /->-\ /->-\ /->-\ /->-\ /->-\ /->-\ /->-\
> | | | | | | | | | | | | | | | |
> | | | | | | | | | | | | | | | |
> \-<-/ \-<-/ \-<-/ \-<-/ \-<-/ \-<-/ \-<-/ \-<-/
This pictures how the integral of a field can be equal
to the integral of another field that is locally very
different from the first one.
Suppiose W is the vorticity, the strength of the circulation
you drew in your picture and G is the net flow, which is
concentrated on the fringes.
In particular, suppose the momentum density (G) can
be written as
G = curl(W)
then the angular momentum is
L = integral (rxG) = integral (r x curl x W)
= integral (W)
This is using partial integration, and W=0 at the
boundaries.
So here we have an example of ambiguity for am. density.
W is uniforly distributed, but rxG is concentrated on the
fringes
I know that not always of ExB can be written as a curl
of another vector field, but I suspect that if you split
the angular momentum into orbital and spin am, the spin
part always relates to a circulating ExB field, which
can (I conjecture) always be written as the curl of W.
We could call W the electromagnetic vortcity. I want
to find an expression for it in terms of fields E,B.
Gerard
Jos Bergervoet
> Jos R Bergervoet wrote:
>> So we now have: the _torque_ exerted is just the in-flux of angular
>> momentum into a surface around an object. ...
> I am not sure that you get a non-zero answer for a plane wave.
> If I substitute a plane wave:
> ...
> So I get a constant flux of z-momentum in the z-direction.
> If I take the moment of this integrated over a volume with the
> origin at its center, I still get zero.
Of course. If you have a pure plane wave, there obviously is no
object present, or it is completely transparent. No momentum is
absorbed from the wave in that case.
You could try to put in a sphere centered at the origin. Using
the expressions for Mie scattering, you might be able to compute
the tensor for the total field (plane wave plus scattered) and
you'll then find the force on the sphere. I think you will still
not find any torque if you use a perfectly conducting sphere (it
is too "slippery" without resistance) but I admit that I didn't
do the calculation.
You can also try an infinitisimal dipole scatterer at the origin.
It is fairly trivial to compute the scattered field in that case.
Or you can use the infinite resistive sheet of vanishing thickness
we discussed earlier (again, it should be resistive to experience a
torque from a circularly polarized wave, whereas for a simple force
you can just use a perfect conductor). For a normally inciding wave,
this case is really trivial to compute, but the "closed surface"
around the object now of course has to be generalized to a set of
two parallel planes on both sides of the sheet (for the perfectly
conducting sheet you need only one, in front of it).
-- Jos
Gerard Westendorp
>
> Ted wrote in part:
>
> >The point I was making in an exclamatory way is
> >that having angular momentum is not a requirement for exerting torque.
>
> Any more than having momentum is a requirement for exerting force.
Actually, it is interesting that electromagnetic force *can* be seen as
resulting from the absorption of electromagnetic momentum flux.
The momentum flux is the space part of the energy momentum tensor T_mn:
m_ij = -E_i E_j - B-i B-j + (E^2 + B^2)/2
So for example, a uniform field (E) in the x-direction has a momentum
flux:
m_xx = -1/2 E^2
A plate at (x=0) with charge Q has a field
E_plate = - Q (x < 0)
+ Q (x > 0)
So the absorbed momentum flux is:
-(1/2) ( (E-Q)^2 - (E+Q)^2 ) = E*Q
This is the right answer. I think it is true in general that all
electromagnetic forces can be calculated in this way. In the
above example, it is not very useful to do so, but that may
not be true in other cases.
On the other hand, the Poyting vector, which is the momentum
*density*, is zero. This makes the momentum flux interpretation a
bit confusing.
Gerard
Toby Bartels
>Toby Bartels wrote:
>>[H]aving momentum is [not] a requirement for exerting force.
>Actually, it is interesting that electromagnetic force *can* be seen as
>resulting from the absorption of electromagnetic momentum flux.
Naturally.
But just because momentum is transferred from the field to the object
doesn't mean that the field had any momentum to begin with.
Similarly, if 2 people stand on roller skates and A pushes B,
then A will transfer momentum to B, and they'll move apart.
At the moment that the push begins, however, A has no momentum.
As the push continues, A will have *negative* momentum,
yet will continue to transfer *positive* momentum to B.
So A's positive force on B does indeed (if you want to see it that way)
result from a transfer of postive momentum from A to B,
but A still had only negative momentum, and no momentum at all to start with.
AB
AB
A B
A B
<--- --->
[Calculation cut -- I agree.]
>On the other hand, the Poynting vector, which is the momentum
>*density*, is zero. This makes the momentum flux interpretation a
>bit confusing.
I don't find it confusing, and the mechanical analogy helps me.
m_xx, the flux (I would say "flux density") of x momentum in the x direction
can easily be positive even if the x momentum density is 0 or negative,
just as in the mechanical example above.
(There, these densities are delta functions because of the discrete nature.)
-- Toby
to...@math.ucr.edu
Gerard Westendorp
[..]
> But just because momentum is transferred from the field to the object
> doesn't mean that the field had any momentum to begin with.
I think you are making the distinction between energy, which is
always positive, and momentum, which can also be negative? In the case
of energy, if energy is transferred from A to B, there must always
have been energy to start with in A.
But if you look just at global conservation, the same is not true
for momentum, you cannot judge the "potential to create a force"
by looking at the total momentum present. I think this is what
you are saying.
But maybe it is not so simple with momentum *flux*. In 1-D at least,
the momentum flux of electrostatic fields is -(1/2)E^2, which is
*always negative*! In order to have an electrostatic force, there
must always be some momentum flux to start with. (in 1-D)
I 3D, it is a bit more complex. The momentum flux of a coulomb
field is a 9-component tensor. For each of the 3 components (i) of
momentum, the flow m_ij is a vector field with components (j). This
field looks like a dipole, with the dipole aligned with the
i-direction.
This means momentum can be thought of being ejected from the
charge, and reabsorbed at the other end, with flow lines just like
those of a bar-magnet. Interestingly, the direction of the dipole
does not depend on the sign of the charge.
The momentum flux field is not additive: another nearby charge also
would have the same flow field if it were distant, but because
of the interfering E-fields, momentum flux is being created or
destroyed as the 2nd charge moves relative to the 1rst charge.
However, this creation/destruction is compensated by a change
in mechanical momentum of the moving charge. I am conjecturing
that if you could construct the flux lines of [mechanical+electric]
momentum, the flux lines would stay divergency-free.
> Similarly, if 2 people stand on roller skates and A pushes B,
> then A will transfer momentum to B, and they'll move apart.
> At the moment that the push begins, however, A has no momentum.
> As the push continues, A will have *negative* momentum,
> yet will continue to transfer *positive* momentum to B.
> So A's positive force on B does indeed (if you want to see it that way)
> result from a transfer of postive momentum from A to B,
> but A still had only negative momentum, and no momentum at all to start with.
>
> AB
> AB
> A B
> A B
> <--- --->
>
This example leaves out the details of the actual mechanism by which
A and B are pushed apart. A and B have gained energy, and this energy
must come from something. If the source of energy were electrostatic,
we would once again encounter momentum flux lines during the separation
process. The separation process may be a rapid phenomenon, so that it
looks like an instantaneous creation of momentum flux lines, but if you
zoom in to the actual separation process, I think you will see momentum flux
at work.
Conjecture:
The universe is filled with momentum flux lines, which may be deformed,
and which may be converted from electric to mechanical or other forms,
but which cannot be broken.
Gerard
Toby Bartels
>Toby Bartels wrote:
>>But just because momentum is transferred from the field to the object
>>doesn't mean that the field had any momentum to begin with.
>I think you are making the distinction between energy, which is
>always positive, and momentum, which can also be negative? In the case
>of energy, if energy is transferred from A to B, there must always
>have been energy to start with in A.
>But if you look just at global conservation, the same is not true
>for momentum, you cannot judge the "potential to create a force"
>by looking at the total momentum present. I think this is what
>you are saying.
I hadn't thought of this distinction between momentum and energy
quite so cogently, but yes, you're quite correct.
>But maybe it is not so simple with momentum *flux*. In 1-D at least,
>the momentum flux of electrostatic fields is -(1/2)E^2, which is
>*always negative*! In order to have an electrostatic force, there
>must always be some momentum flux to start with. (in 1-D)
I don't think that "to start with" is how I would phrase it.
Certainly you must have momentum flux to have a force.
This is because momentum flux *is* the force!
Specifically, if you integrate the momentum flux density
across a surface to get the momentum flux through that surface,
then this is *equal* to the force applied through the surface.
Momenum flux = transfer of momentum = application of force.
>In 3D, it is a bit more complex. The momentum flux of a coulomb
>field is a 9-component tensor. For each of the 3 components (i) of
>momentum, the flow m_ij is a vector field with components (j). This
>field looks like a dipole, with the dipole aligned with the
>i-direction.
Yes, this is the Maxwell stress tensor.
>The momentum flux field is not additive: another nearby charge also
>would have the same flow field if it were distant, but because
>of the interfering E-fields, momentum flux is being created or
>destroyed as the 2nd charge moves relative to the 1rst charge.
>However, this creation/destruction is compensated by a change
>in mechanical momentum of the moving charge. I am conjecturing
>that if you could construct the flux lines of [mechanical+electric]
>momentum, the flux lines would stay divergency-free.
The usual theorem is that, fixing a particular direction of momentum,
the divergence of the momentum flux density for that direction,
*plus* the time derivaitve of the momentum density for that direction,
is zero. This rule is preserved even in general relativistic contexts.
@m_xx/@x + @m_xy/@y + @m_xz/@z + @m_xt/@t = 0,
where m_xt is the density of momentum in the x direction.
See Section 12-3 of Goldstein (2nd edition).
>>Similarly, if 2 people stand on roller skates and A pushes B,
>>then A will transfer momentum to B, and they'll move apart.
>>At the moment that the push begins, however, A has no momentum.
>>As the push continues, A will have *negative* momentum,
>>yet will continue to transfer *positive* momentum to B.
>>So A's positive force on B does indeed (if you want to see it that way)
>>result from a transfer of postive momentum from A to B,
>>but A still had only negative momentum, and no momentum at all to start with.
>This example leaves out the details of the actual mechanism by which
>A and B are pushed apart. A and B have gained energy, and this energy
>must come from something. If the source of energy were electrostatic,
No, it's not electrostatic; at least, there's no external field.
They are people on rollerskates, one pushing the other.
The source of the energy is the chemicals stored in A's muscles.
The relevant electric fields never leave A's and B's bodies.
-- Toby
to...@math.ucr.edu
Danny Ross Lunsford
similar. Electrodynamics has a curious ambiguity in the definition of
energy flux - Feynman talked about it in his Lectures. The Poynting
vector and the energy density only make up a proper 4-vector (density)
in vacuo. The analogous situation in gravity is the lack of an EM tensor
for the gravitational field itself. The reason in both cases is a hidden
introduction of a surface integral when getting the conservation laws
from the respective variational principle. One usually just assumes that
this surface is far off and that in that far off place the contribution
it makes is negligible.
-ross
Gerard Westendorp wrote:
> Toby Bartels wrote:
> > Ted wrote in part:
> > >The point I was making in an exclamatory way is
> > >that having angular momentum is not a requirement for exerting torque.
> > Any more than having momentum is a requirement for exerting force.
> Actually, it is interesting that electromagnetic force *can* be seen as
> resulting from the absorption of electromagnetic momentum flux.
[unnecessary quoted text axed by angry moderator]