The problematical nature of photon spin
Neil
problem concerning the spin of photons. The angular momentum of photons
(hbar each, regardless of energy) has been found to carry the
equivalent of literal angular momentum (Ref. the Richard Beth
experiment, reported 1936 in Physical Review.) Furthermore, the spin of
photons is presumed to be directed either parallel or anti-parallel to
the direction of propagation (S = +/-hbar*unitk). There can be some
ambiguity about this for a single photon, but as N increases, the
classical situation is approached of a circularly polarized propagating
wave with S depending upon E cross B.
Consider the case of pulses of C-Pol light emitted in opposite
directions from a source. If both are right-handed (relative to their
respective directions, of course), then the net angular momentum of the
system remains unchanged. However, suppose that we are observing this
emitter from a reference frame in which the emitter is moving rapidly
in a direction perpendicular to its rest-frame directions of
propagation of the photons. For us, the two beams are tilted in the
direction of emitter motion, forming a "V" shape. If the spin of
photons actually must be parallel to the direction of propagation -
which is relative to the state of motion - then their spins are tilted
relative to their original orientations and no longer cancel each other
out. Specifically, there is now a component of angular momentum in the
direction of motion. This is not consistent with the conservation of
angular momentum! We started out with zero net angular momentum, and
that should be agreed upon by all observers. (We could collect the spin
from the photons.)
What makes this problem even more challenging is that the
perpendicular component of angular momentum is relativistically
transformed to be gamma factor times more than the rest component.
(Verify this for yourself by directly using the velocity addition
formulae and insert into S = sum of gamma*m*v*r in the perpendicular
case where the orientation and calculations are simple.) Hence, not
only the direction but the magnitude of the photon spins will not be
what we need to fix things up.
Does anyone have any ideas about this? Is light more complicated
than we thought?
Neil Bates
* Sent from RemarQ http://www.remarq.com The Internet's Discussion Network *
The fastest and easiest way to search and participate in Usenet - Free!
Gerard Westendorp
quantum mechanical amplitudes, but I realised, as you also pointed
out, that the problem also occurs classically.
Lets say the source is at rest, and the observer is moving. The
velocity is in the x-direction, and the photons are going in the z
direction of the rest frame.
So I started writing down the E an B fields for right-circular
polarised waves in 2 opposite directions. You can just
Lorentz-transform these fields, (formula for example in Feynman
lectures on Physics), and then you know all the fields in the moving
reference frame. So you should be able to determine exactly if there
is a problem or not.
Apart from angular momentum, linear momentum is also tricky. The
Poynting vectors in the moving frame are bigger than in the rest
frame. Also they have a component in the x-direction in the moving
frame, but not in the rest frame. The (linear) momentum flux is equal
to the Poynting vector, divided by c^2. So In the moving frame, it
looks as if the source is shedding x-momentum, and should be
accelerated. But relativity also allows for another option: The rest
mass of the source is decreasing (it is shedding energy) Therefor, the
apparent momentum decreases, although the speed remains constant. I
haven't checked if this works out quantitatively, but it should,
otherwise we are in big trouble.
As for angular momentum, I tried writing down the transformed
E and B fields, projected onto the yz-plane of the moving
observer, but they seem to rotate in *oppostie* directions,
instead of the same direction, which is what you would expect if
they are both right-circular polarised. I think the moving observer
is seeing linearly polarised waves in the yz-plane. I suspect that a
Lorentz-transform does not tilt the angular momentum vector.
As for the energy flux, it is also tricky that the Poynting vector is
bigger in the moving frame than in rest frame. But here, there are
quite a few escape routes for relativity, eg the 4-momentum looks
different anyway to both obsevers anyway.
I suspect this all works out if you write it down properly.
--
I
C U
A N D
G O O D
_ L_ U_ C_K!
\_\ |________|
\_\|________|
\ _________/
\________/
\______/_______Gerard________(http://www.xs4all.nl/~westy31/)
Neil
Your interest in this curious paradox is much appreciated, but I do not
think that you have solved it. I consider this problem to be actually -
not just apparently - paradoxical, and a genuine challenge to our
understanding of light. It is not just a student-type problem that we
can resolve painlessly in terms of known concepts. That does not mean
that it cannot be resolved at all (in the sense that angular momentum
would really not be conserved), but not "painlessly," meaning that
something has to give - about light. (On the other hand, true adherence
to the principle that science is not finished means not assuming that
any presumed law must be a genuine absolute - remember what happened a
century ago!)
Gerald wrote:
<snip>
> As for angular momentum, I tried writing down the transformed
> E and B fields, projected onto the yz-plane of the moving
> observer, but they seem to rotate in *oppostie* directions,
> instead of the same direction, which is what you would expect if
> they are both right-circular polarised. I think the moving observer
> is seeing linearly polarised waves in the yz-plane. I suspect that a
> Lorentz-transform does not tilt the angular momentum vector.
<snip>
> I suspect this all works out if you write it down properly.
I am not sure what Gerald is getting at concerning the direction of
spin of the radiation as seen by the moving observer. To say "rotate in
*opposite* directions" can be ambiguous, since circular handness is
defined relative to direction of propagation. Hence, two right-handed
photons moving in opposite directions *ought* to spin in "opposite
directions" rather than the same direction - defined *absolutely*. It
is better to directly refer to the orientations of the spins as vectors
per se and jettison confusing talk about rotational sense. It is not
clear to me how to justify his point about linear polarization, or its
relevance to this particular problem.
The truly serious issue consists of two parts: (1.) The spin of photons
is supposed to be parallel to the direction of propagation. This is
part of accepted QM, but it should also be consistent with classical
EM. Gerald says "I suspect that a Lorentz-transform does not tilt the
angular momentum vector." If that is true, then QM will be
contradicted. However, I think that the spin vector should be tilted
since now the wave fronts are tilted, and the orientations of E and B
are within that plane. Thus the spin should be perpendicular to the new
orientation of the wave front, not the same as it was before. But then,
we have the original problem for conservation of angular momentum.
Even more serious is issue (2.): the transformation of the magnitude of
the perpendicular spin component, which is multiplied by the gamma
factor. (Intuitive boost: consider that the time interval required to
apply a given exertion of torque is considered to be longer by the
observer for whom the target is in relative motion.) Mr. Westendorp
does not address this problem. Note: if we want to show the problem of
magnitude, then instead of sending off two right-photon bursts, we send
off bursts of the same absolute spin (thus opposite circular sense).
Then we have exerted net torque on the emitter, and the problem is now
that the emitter AM for the moving observer is gamma times the rest
value. It is very difficult to imagine that somehow the photon pulses
will now have a spin of gamma*N*hbar, in violation of all known QM and
EM theory.
Perhaps a graphic representation would help, showing only one "leg" of
the V-shaped configuration of the beams when the emitter is moving past
the observer:
^
|
/
/
/
/
*** - - - -> velocity of emitter
The vector shown at the end of the diagonal light path is the
orientation that the photon's spins must have in order to conserve
angular momentum. This behavior is not what we expect for light in
terms of previous understanding (spin parallel to direction of
propagation). Furthermore there is the issue of the necessary
magnitude of the spin in this reference frame. No, we cannot assume
that this all works out if you write it down properly.
Vesselin G Gueorguiev
[...]
> The truly serious issue consists of two parts: (1.) The spin of photons
> is supposed to be parallel to the direction of propagation. This is
> part of accepted QM, but it should also be consistent with classical
> EM. Gerald says "I suspect that a Lorentz-transform does not tilt the
> angular momentum vector." If that is true, then QM will be
> contradicted. However, I think that the spin vector should be tilted
> since now the wave fronts are tilted, and the orientations of E and B
> are within that plane. Thus the spin should be perpendicular to the new
> orientation of the wave front, not the same as it was before. But then,
> we have the original problem for conservation of angular momentum.
You have not included the angular momentum of the source with respect to
the observer, aren't you?
[...]
Paul Colby
expect that the angular momentum in the field is a
Lorentz invariant? From field theory the angular
momentum appears as an antisymmetric tensor with
the usual three components of angular momentum
being the generators of rotation, Rx=M23, Ry=M31,
Rz=M12. The three time components,
Bx=M01,By=M02,Bz=M03, are the boost generators.
I'm not sure these are observables but they will
certainly mix with the space components upon
Lorentz Transformation. This would imply that a
zero angular momentum configuration in one frame
would transform into a non-zero one in some other
frame because Bx,By or Bz are non-zero. Likewise
for the momentum of your two photons. In the frame
where the momentum vanishes, the total energy
doesn't. So when I transform I get a net momentum
because the time component mixes into the space
component.
Regards
Paul Colby
Neil <neil_delv...@hotmail.com.invalid>
wrote in message
news:01906254...@usw-ex0101-001.remarq.com..
John Baez
Paul Colby <Paul....@worldnet.att.net> wrote:
>At the risk of sounding real stupid, does one
>expect that the angular momentum in the field is a
>Lorentz invariant?
Rotations don't commute with Lorentz boosts, so
angular momentum, being the generator of rotations,
is not invariant under Lorentz boosts. Instead,
it gets mixed up with the boost generators. Just
like you said:
>From field theory the angular
>momentum appears as an antisymmetric tensor with
>the usual three components of angular momentum
>being the generators of rotation, Rx=M23, Ry=M31,
>Rz=M12. The three time components,
>Bx=M01,By=M02,Bz=M03, are the boost generators.
>I'm not sure these are observables but they will
>certainly mix with the space components upon
>Lorentz Transformation.
Yeah, they are observables, and they get mixed with
the boost generators in a nice simple way when you
do a Lorentz transformation.
>This would imply that a
>zero angular momentum configuration in one frame
>would transform into a non-zero one in some other
>frame because Bx,By or Bz are non-zero.
Righto. A rock at rest has no angular momentum.
A rock moving along a straight line does - unless
it happens to be moving on a line through the origin.
If you think about this long enough you can guess
the physical meaning of the Lorentz boost generators.
Textbooks rarely talk about these observables -
probably `cause the people who wrote the textbooks
learnt their stuff from other books that left out
this information. But us folks here have talked about
these observables a lot! We just sit out here on
our rocking chairs on the front porch, and people
keep walking by and asking about Lorentz boost
generators, and we keep answering... it's like
Nietzsche's eternal recurrence of the same.
After a while it gets boring, so I'll just reach
into my files and pull out some stuff for you...
okay? If you dig deeper with DejaNews you may
reach even more ancient strata of discussion on
this issue, full of arcane wisdom.
..............................................................
Newsgroups: sci.physics,sci.math
Subject: Re: Noethers Theorem and the Poincare Group
From: ba...@galaxy.ucr.edu (John Baez)
References: <4o1019$r...@darkstar.UCSC.EDU>
Organization: University of California, Riverside
In article <4o1019$r...@darkstar.UCSC.EDU> drac...@cats.ucsc.edu
(Jacob Alexnder Mannix) writes:
>what do those
>funny lorentz "rotations" in the x-t, y-t, z-t planes (the boosts)
>give rise to as conserved quantities?
This question comes up repeatedly on sci.physics because for some stupid
reason most textbooks are lazy to discuss these conserved quantities.
Hopefully Matt McIrvin and others will repost the enlightening replies
they gave to this question the last time it rolled around. And maybe I
can get up the energy to work this into a FAQ for the next time it rolls
around.
Let me just say this:
It is amusing to note that analogous observables also show up in
nonrelativistic mechanics: not from the Lorentz boosts, of course, but
from the "Galilei boosts". The Galilei group is the limit of the
Poincare group as c -> infinity, and it's generated by spacetime
translations, spatial rotations, and "boosts" of the form
t -> t
x -> x + vt
y -> y
z -> z
and similarly for y and z.
If we consider a single nonrelativistic free particle --- in
1-dimensional space, to keep life simple --- and describe its state by
its position q and momentum p at t = 0, we see that the Galilei boost
t -> t
x -> x + vt
has the following effect on its state:
p -> p + mv
q -> q
In other words, a Galilei boost is just a translation in momentum space.
In nonrelativistic quantum mechanics this should be familiar, though
somewhat disguised. Here it is a commonplace that the momentum
observable p generates translations in position space; in the
Schroedinger representation it's just -i hbar d/dx. But by the same
token the position observable q generates translations in momentum
space. As we've seen, a translation in momentum space is just a Galilei
boost. So basically, the generator of Galilei boosts is the observable
q.
Ugh, but there is a constant "m" up there in the formula for a Galilei
boost. So I guess the observable that generates Galilei boosts is
really mq. If we generalize this to a many-particle system we'd get
the sum over all particles of their mass times their position, or in
other words: the total mass times the center of mass.
Now this seems weird at first because it's not a conserved quantity!
Wasn't Noether's theorem supposed to give us a conserved quantity?
Well, it does, but since our symmetry (the boost) was explicitly
time-dependent --- it involved "t" --- our conserved quantity will also
be explicitly time-dependent. What I was just doing now was working out
its value at t = 0.
If we work out its value for arbitrary t I guess we get: the total mass
times the center of mass minus t times the total momentum.
Using the fact that total mass is conserved we can turn this conserved
quantity into something perhaps a bit simpler: the center of mass minus
t times the velocity of the center of mass.
Something similar applies to the relativistic case.
> Also, I know that SO(3,1) (lorentz) can be (locally) given
>up in favor of SL(2,C) (which makes commutator calculations MUCH
>easier), is there something similar for Poincare?
As you note the Poincare group is the semidirect product of SO(3,1) and
R^4, where SO(3,1) acts on R^4 in the obvious way. That means that
the double cover of the Poincare group is the semidirect product of
SL(2,C) and R^4, where SL(2,C) acts on R^4 in the obvious way. What's
the "obvious way"? Well, you take your element of SL(2,C), turn it into
an element of SO(3,1), and let that act on R^4. But there is a slicker
description of this "obvious way", which is probably what you are
looking for (perhaps unbeknownst to yourself). Think of R^4, or more
precisely Minkowski spacetime, as the 2x2 hermitian matrices. Then an
element g of SL(2,C) acts on such a matrix T by:
x -> gTg^t
where g^t is the transpose.
The details of this are described on pp. 182-183 of my book with Javier
Muniain, "Gauge Fields, Knots and Gravity". Note a typo in Exercise 29;
it should be rho(g)T = gTg^t.
>Or do only semi-simple Lie groups have nice universal covers?
All Lie groups have universal covers; as for nice, well, it depends on
what you consider nice, but the Poincare group is damn nice.
From: to...@ugcs.caltech.edu (Toby Bartels)
Newsgroups: sci.physics
Subject: Re: Noethers Theorem and the Poincare Group
Date: 24 May 1996 17:49:49 GMT
Organization: California Institute of Technology, Pasadena, CA USA
Lines: 20
Message-ID: <4o4srt$o...@gap.cco.caltech.edu>
References: <4o1019$r...@darkstar.ucsc.edu>
Jacob Alexnder Mannix <drac...@cats.ucsc.edu> wrote in part:
>I have the infinitesimal rotations in the x-y, y-z, and z-x planes,
>giving me conservation of each component of angular momentum, and by
>now, you probably see what I am leaving for last, but what do those
>funny lorentz "rotations" in the x-t, y-t, z-t planes (the boosts)
>give rise to as conserved quantities? Does it coorespond to some
>sort of relationship between the momentum and energy, like
>E^2 = (pc)^2 - (mc^2)^2 give or take a minus sign or power of 2, or
>something completely strange?
Symmetry under rotation in the xy plane corresponds to conservation of
L_z = r_x p_y - r_y p_x (where r and p are linear position and momentum).
In the same way, symmetry under hyperbolic rotation in the tx plane
corresponds to conservation of r_t p_x + r_x p_t = t p_x - x E.
In other words, the system's speed in the x direction is p_x / E.
-- Toby
to...@ugcs.caltech.edu
From: to...@cc.usu.edu (Charles Torre)
Newsgroups: sci.physics
Subject: Re: Noethers Theorem and the Poincare Group
Message-ID: <1996May23.1...@cc.usu.edu>
Date: 23 May 96 12:01:28 MDT
References: <4o1019$r...@darkstar.UCSC.EDU>
Organization: Utah State University
Lines: 50
In article <4o1019$r...@darkstar.UCSC.EDU>,
drac...@cats.ucsc.edu (Jacob Alexnder Mannix) writes:
> I have a question concerning a physical interpretation of some of
> the conserved quantities associated with the Poincare group (Lorentz
> boosts, rotations in 3-space, and translations).
...
> but what do those
> funny lorentz "rotations" in the x-t, y-t, z-t planes (the boosts)
> give rise to as conserved quantities?
This is a good question which comes up with some regularity and is rarely given
much treatment in the texts I am familiar with. I will make a brief attempt to
explain it.
From the point of view of
relativity there is no invariant way to separate the boosts from the rotations
because, in effect, different inertial observers will not agree on the
definitions of the various x-t, x-y, etc. planes. So, ultimately, one should
simply view the 6 conserved quantities associated with the boosts and rotations
as the relativistic form of angular momentum. This is completely analogous to
the fact that there is no invariant way to separate energy from momentum -
different inertial observers will see different mixtures-and so one speaks only
of the "4-momentum". Still, this does not quite answer the question. In a
given inertial reference frame I can identify 3 of the six conservation laws,
the ones coming from spatial rotational symmetry in the given frame, as
corresponding to familiar notions of angular momentum. What of the other 3
coming from boosts? Actually this question can be asked already in Newtonian
physics which employs "Galilean relativity". One can show that a Galilean
invariant Lagrangian will be symmetric (up to a total time derivative) with
respect to boosts (the non-relativistic boosts, not Einstein's). The
corresponding conserved quantity is an example of an "explicitly time dependent
constant of the motion" and has the physical interpretation of the initial
position of the center of mass of the system. So, one can view the extra 3
conservation laws associated with boosts in a Poincare invariant theory as
being a relativistic generalization of this. The Einstein-relativistic
generalization of Galilean-relativistic angular momentum and initial center
of mass is a bit complicated since the Galilean version of these quantities
involves, in effect, a preferred notion of simultaneity, which is of course
absent in Einstein relativity. You can usually find discussion of bits and
pieces of these issues in a mechanics book which treats special relativity,
such as Goldstein and/or which bothers to discuss Galilean
relativity such as Landau and Lifshitz.
Charles Torre
Department of Physics
Utah State University
Logan, UT 84322-4415 USA
From: to...@ugcs.caltech.edu (Toby Bartels)
Newsgroups: sci.physics
Subject: Re: Noethers Theorem and the Poincare Group
Date: 24 May 1996 18:11:15 GMT
Organization: California Institute of Technology, Pasadena, CA USA
Lines: 43
Message-ID: <4o4u43$p...@gap.cco.caltech.edu>
References: <4o1019$r...@darkstar.ucsc.edu>
Jacob Alexnder Mannix <drac...@cats.ucsc.edu> wrote in part:
>One of my professors put it as a conserved
>current, but I don't think I want to get into what a current means
>for system of classical particles in spacetime.
It doesn't (unless you want to express it in terms of delta functions).
If you have a system of particles, that a quantity is conserved
means that the sum of this quantity for all particles is conserved.
OTOH, if you have a system of fields, that a quantity is conserved
means that the integral of its density over all space is conserved.
This can be expressed locally and more elegantly in terms of a current.
This isn't difficult to see; I'll use an example that may be familiar to you.
Electric charge is a conserved quantity; d Q / d t = 0.
But if your charge is spread out into a field,
you'll measure it with rho (the charge density)
and the 3vector j (the current density).
(These can be put into a 4vector (rho, j),
so density is the time component of current density.)
You may have seen charge conservation expressed as @ rho / @ t + div j = 0
(where `@' indicates partial differentiation and `div' means divergence).
Integrating this equation over 3D volume,
you get d Q / d t + (integral over all space) (div j) = 0.
Gauss's theorem tells you (integral over all space) (div j)
is just the integral over space's boundary of j.
Since space's boundary is at infinity, and j = 0 at infinity,
integrating @ rho / @ t + div j = 0 gives you d Q / d t.
So @ rho / @ t + div j is a local equation for conservation of charge.
Now, if you remember that rho is the time component j_t of the current density,
you can say @ j_t / @ t + @ j_x / @ x + @ j_y / @ y + @ j_z / @ z = 0.
This equation is essentially div j = 0, only now j and div are 4 dimensional.
This makes the 4 dimensional j a conserved 4D current density.
Just as j's time component measures the charge density
and j's space components measure the flow of the charge,
you can keep track of energy density and energy flow in a field
with a 4 dimensional conserved energy current.
The same thing goes for momentum, angular momentum, and even t p_x - x E.
-- Toby
to...@ugcs.caltech.edu
From: to...@cc.usu.edu (Charles Torre)
Newsgroups: sci.physics
Subject: Re: Noethers Theorem and the Poincare Group
Message-ID: <1996May24.0...@cc.usu.edu>
Date: 24 May 96 08:51:45 MDT
References: <4o290j$r...@www.oracorp.com>
Distribution: world
Organization: Utah State University
In article <4o290j$r...@www.oracorp.com>, da...@racorp.com (Daryl McCullough)
writes:
> colu...@pleides.osf.org (Michael Weiss) writes:
>
>>Time translation symmetry corresponds to conservation of energy.
>
> Actually, I don't understood in what sense conservation of energy
> is an instance of Noether's theorem. The auto-didactic way that I
> understand Noether's theorem is this:
>
> Suppose that we have a parameterized family of paths x(l,t)
> such that L(x,v) is invariant under changes of the parameter l.
> In that case, we can write:
>
> dL/dl = @L/@x dx/dl + @L/@v dv/dl = 0
>
> (where I am using @ to mean partial derivative)
>
> Using the Lagrangian equation of motion @L/@x = d/dt (@L/@v),
> together with v = dx/dt and interchanging the order of differentiation
> gives:
>
> (d/dt (@L/@v)) dx/dl + @L/@v (d/dt (dx/dl)) = 0
>
> or
> d/dt ( @L/@v dx/dl) = 0
>
> Thus the quantity Q = @L/@v dx/dl is conserved (it's value is constant).
>
> Now, the problem with applying this to time symmetry is this: the
> Lagrangian is *not* invariant under translations in time! In the
> case where l = t, we don't get dQ/dt = 0, we get instead
>
> dQ/dt = dL/dt
>
> (where Q = @L/@v dx/dt)
> So, we can still get an invariant quantity by letting Q' = Q - L,
> but it seems to me that this is not an instance of Noether's theorem.
Your point is well-taken, but Noether's theorem IS meant to handle this case.
In one version of the theorem a "symmetry" of the Lagrangian is an
infinitesimal transformation that changes the Lagrangian by a total derivative
of a function. In this case, the conserved quantity includes that function in
its definition as you surmised above. I think these symmetries are sometimes
called divergence symmetries. I also seem to recall that if one formulates
the variational problem in terms differential forms on "jet space" then the
divergence symmetry is just a traditional symmetry of the Lagrangian n-form,
where n is the dimension of the space of independent variables. By
"traditional" symmetry I mean: view the infinitesimal symmetry as a sort of
vector field in jet space (the space of independent variables, dependent
variables, all their derivatives), the Lagrangian n-form has vanishing Lie
derivative along this vector field.
Charles Torre
Department of Physics
Utah State University
Logan, UT 84322-4415 USA
From: to...@ugcs.caltech.edu (Toby Bartels)
Newsgroups: sci.physics
Subject: Re: Noethers Theorem and the Poincare Group
Date: 24 May 1996 18:30:44 GMT
Organization: California Institute of Technology, Pasadena, CA USA
Lines: 10
Message-ID: <4o4v8k$q...@gap.cco.caltech.edu>
References: <4o1019$r...@darkstar.ucsc.edu> <4o4u43$p...@gap.cco.caltech.edu>
I <to...@ugcs.caltech.edu> wrote in small part:
>Integrating @ rho / @ t + div j = 0 gives you d Q / d t.
More than that, it gives you d Q / d t = 0.
^^^^
-- Toby
to...@ugcs.caltech.edu
Neil
Thank you for taking an interest in issues raised by discussions of the
photon spin question. The theoretical underpinnings of angular
momentum, etc. are definitely relevant to work on the problem, and you
have provided much useful material. Would you mind giving your thoughts
on the original specific problem of the conservation of spin angular
momentum of source plus photons as seen in different reference frames?
I must apologize for not always being perfectly clear in my original
posting of this problem, which may have contributed to
misunderstandings and my subsequently feeling the need to reiterate -
perhaps excessively - certain points in various ways. First, the source
starts with zero spin angular momentum in its rest frame. (That would
usually be assumed unless stated otherwise, but in a case like this it
is better to be explicit.) Furthermore, we define the problem in terms
of "spin," a rough definition of which is that component of angular
momentum which is independent of a reference point (but not necessarily
of observer velocity!) In mechanics, spin is caused by the application
of torque. The case of orbital-type angular momentum L is of course
quite different:
> ...
> Righto. A rock at rest has no angular momentum.
> A rock moving along a straight line does - unless
> it happens to be moving on a line through the origin.
> If you think about this long enough you can guess...
A point: most basic textbook discussions of AM do not talk enough about
this distinction.
So, we start with a source with zero spin S, presumably agreed upon by
all observers since the source is in effect a generic object - like a
rock - of mass m. (I unfortunately referred at one point to S in a
context that could have confused it with the Poynting vector, which is
also called S. BTW, here I mean the ordinary macroscopic three-vector
spin of paired r cross p or its convertible equivalent.) After emission
of pulses of right-hand photons in opposite directions, the S of the
source is still zero, and the combined S of the photons cancels out in
the source rest frame - but not in a frame in which the source is
moving perpendicular to the direction of the pulses. We started with
zero spin and end up with a component of S in the direction of motion -
if we don't adjust our expectation that photon spin S is aligned with
the direction of propagation.
I unfortunately rather carelessly transitioned to a discussion of the
velocity-transformation of spin. First of all, I neglected to explain
that this second problem only makes sense if photon emission is done in
such a way that there *is* spin transferred to the source - unlike in
the first posing of the problem. Another thing I did not make clear in
that section was the distinction between the issues of transforming the
spin of the source versus transforming the spin of the photons. I have
derived the transformation for literal material bodies revolving around
a center, and leave it to others to address the issue of how to
reconcile that with the presumed nature of the spin of photons.
The effort to solve this little problem has elicited much interest and
discussion of basic issues. To me, that makes it worthwhile whatever
the ultimate resolution is.
Neil Bates
In article <82k7p8$n...@charity.ucr.edu>, ba...@galaxy.ucr.edu (John
Baez) wrote:
> ...
Vesselin G Gueorguiev
Neil wrote:
[...]
> So, we start with a source with zero spin S,..... After emission
> of pulses of right-hand photons in opposite directions, the S of the
> source is still zero, and the combined S of the photons cancels out in
> the source rest frame - but not in a frame in which the source is
> moving perpendicular to the direction of the pulses. We started with
> zero spin and end up with a component of S in the direction of motion -
> if we don't adjust our expectation that photon spin S is aligned with
> the direction of propagation.
[...]
If the source is moving in x direction and the photons are emitted
in z direction we have:
S=[S_0,0,0,S_z] and P=[P_0,0,0,P_z] and a Lorentz boost in x direction is:
| 1 v 0 0 |
gamma(v)|-v 1 0 0 | where gamma(v)=1/(1-v^2) in c=1 units
| 0 0 1 0 |
| 0 0 0 1 |
So S and P does not change! Thus for any v along x axis the spin S
is aligned with P as it was at v=0!
Where is this "component of S in the direction of motion"?
Neil
First, I point out that my graphic (second message) did not come out
quite right (loss of the spaces before the lines showing the
propagation of the light beam.) Just imagine that the line coming from
the three asterisks is diagonal, and keep the spin vector at the top
the same. Then you can readily visualize the peculiarity of photon spin
being at an angle relative to the motion of the beam.
In response to Vesselin Gueorguiev:
Yes, I do take the angular momentum of the source with respect to the
observer into account. Indeed, it is the velocity-transformation of the
perpendicuar angular momentum of the source which makes the problem so
difficult: No orientation of the photon spins can square the account,
since the magnitude of the perpendicular AM left in the source is gamma
times the rest value. Remember that there are two ways to work this
Gedanken experiment:
(1.) Start with a source imagined to have "zero" spin angular momentum
(although it is the changes that matter.) Emit two "right-hand" photon
pulses in opposite directions. That should leave the AM of the source
at zero (see for example Feynman Lectures Vol. 3). Then change to the
frame in which the source is moving. The light is taking diagonal paths
as explained earlier. The AM of the source should still be zero in this
frame (I will reply briefly to Mr. Colby on this matter later.) If
total AM is to be conserved, then the photon spins need to keep their
original orientations, which are now no longer parallel to the
propagation vector k.
(2.) Variation: Send a right-hand beam in the "up" direction and a
left-hand beam in the "down" direction. Now the total emitted AM is
2N*hbar in the up direction, where N is the number of photons in each
beam. That must exert a torque on the source, giving it rest-frame AM
of the same magnitude and opposite direction. (But the net AM of source
plus photons is still zero.) If one transforms to the relatively moving
frame, then the perpendicular AM of the source should be transformed
(by *direct* definition of AM as r cross p combined with relativistic
transformation of momentum!) to be gamma times the rest value. The
combined AM of the photons (as we know them), whatever their
orientation, and this new value cannot cancel out to get the orginal
net zero AM before emission.
I do not know how to square the above analysis with the approach of
Paul Colby in his recent post on this problem. Many different ways
exist to "formulate" energy, momentum, etc. However, let us remember
that in the final analysis, they must give answers consistent with the
humble basic definitions of the quantities. (Yes, I am aware of the von
Laue energy current, etc., but I don't think such adjustments have a
role here.) As a final thought: isn't it interesting that this simple
little problem (?!) has generated such perplexity and variation of
attempted solutions from respondents? I have not seen it discussed
previously (but if you find references let us know), and it certainly
is not the sort of thing that someone just "puts to bed" in a clear,
confident way mutually agreed on by most in the end. Perhaps it will
become a famous paradox.
Neil Bates
Neil
<vess...@baton.phys.lsu.edu> wrote:
...
> > zero spin and end up with a component of S in the direction of
> >motion -
...
> So S and P does not change! Thus for any v along x axis the spin S
> is aligned with P as it was at v=0!
> Where is this "component of S in the direction of motion"?
Perhaps the best way to characterize the nature of this problem about
photon spin and to explain why it matters is a quote from Richard
Feynman, in The Feynman Lectures in Physics, vol. II (p. 17-5):
"A paradox is a situation which gives one answer when analyzed one way,
and a different answer when analyzed another way, so that we are left
in somewhat of a quandary as to actually what should happen."
In this case the quandary is based on the difference in our
understanding and analysis of the angular momentum of material bodies,
versus our understanding of the spin of photons as provided by optics
and quantum mechanics. The analysis provided by Mr. Gueorguiev is
ultimately rooted in the momenta and positions of material particles as
found in different reference frames. Consider a spinning top, and then
imagine it to be moving rapidly for us in a direction perpendicular to
the spin axis. Indeed, we expect the spin as we define it to remain in
its original orientation, even at relativistic velocities. There is no
component of S in the direction of motion, and no problem for
conservation of angular momentum either.
However: although it would be reasonable to *expect* that the same
transformations would apply to the spin of beams of light, that is not
what optics and quantum mechanics tell us. An illustration circa page
275 of Optics by Eugene Hecht is graphically clear about the nature of
photon spin as we think we know it: it shows the spin vector parallel
(or antiparallel) to the direction (unit k) of the propagation of
light. This is of course what creates right-hand or left-hand
circularly polarized light. Feynman calls this the "screwy" behavior of
light, made so by its having no rest mass. This situation is completely
in accord with all that we know (or think we know) about the optics of
circularly polarized light, and accords with both quantum mechanics and
the effect of the E and B fields on electrons in media, etc. I know of
no exceptions deriving from motion of the source.
Now: consider that the light emitted by the moving source in the
original problem is tilted at an angle relative to its rest-frame
propagation along the z axis, and is now leaning towards positive x.
This angle can come close to 90 degrees from the z axis if the source
is moving rapidly. The light from the other beam is also tilted, as a
mirrored angle with respect to the x axis. If the spin of light is as
Hecht and Feynman and every physics reference I know of describes it,
that spin should be aligned with these directions of propagation for
the observer for whom the source is in motion. There should then indeed
be a component of spin along x, the direction of motion, as seen by
simple inspection of the geometry of the situation. In that case, the
angular momentum of the total system changed without external
interactions, since the spin of the source is still zero.
On the other hand, if the spin of a light beam works like the spin of a
top or bullet, then the spin will still be pointing along the z axis in
the moving frame. (Refer to the illustration in my reply to Gerald
Westendorp if your viewer shows it correctly.) The spin of the light
will be tilted at an angle that can approach 90 degrees relative to its
line of propagation. Then angular momentum is conserved, but this
situation does not accord with our understanding of light. Imagine how
peculiar it would sound for an optical physicist to say something like:
"We are now receiving circular light, with the spin tilted 73 degrees
from the line of propagation, and oriented at an projected angle of 23
degrees in the perpendicular plane. Well, it must have come from a
rapidly moving source." I don't think any of us has ever heard of
anything like a "tilt angle" of the spin of polarized light or photons
- it is always either plus or minus spin, given that it is completely
circularly polarized.
So there is the main problem: the spin of light does not act like the
spin of a bullet, as far as we know. It is a paradox according to
Feynman's excellent definition: if we try to conserve angular momentum
in different frames, then we get a behavior of light which is peculiar
according to optics and quantum mechanics. (There may be a problem
concerning the magnitude of the spin and not just its direction, but
perhaps we need to resolve the issue of direction first.) But what if
light is more complicated than we thought, as I asked in my original
post? That would be very interesting.
Gerard Westendorp
I will choose units so that |E| = |B| = 1 fot the observer emitting the
.photons:
(gamma=1/sqrt(1-v^2/c^2)
emitter:
photons up photons down
Ex +cos(wt) +cos(wt)
Ey +sin(wt) -sin(wt)
Ez 0 0
Bx -sin(wt) -sin(wt)
By +cos(wt) -cos(wt)
Bz 0 0
receiver:
photons up photons down
Ex +cos(wt) +cos(wt)
Ey +sin(wt)*gamma -sin(wt)*gamma
Ez +cos(wt)*gamma*v -cos(wt)*gamma*v
Bx -sin(wt) -sin(wt)
By +cos(wt)*gamma -cos(wt)*gamma
Bz -sin(wt)*gamma*v +sin(wt)*gamma*v
So looking at the components of E and B in the receiver yz-frame, we
can see that both photon beams are rotating in the *same* direction.
This is contrary to my earlier mail, which was once again wrong.
But that means I have to be on the side of Niel: Yes, the receiver is
getting angular momentum in the x-direction. If you let the fields of
the 2 photon beams act on two seperate (non perfect) conductors, each
conductor will feel a net torque, and start rotating. Both will rotate
in the same direction. But the emitter is not rotating. So were does
the AM come from?
Gerard.
Vesselin G Gueorguiev
Neil wrote:
>
> In article <384F0A6C...@phys.lsu.edu>, Vesselin G Gueorguiev
> <vess...@baton.phys.lsu.edu> wrote:
>
> ...
> > > zero spin and end up with a component of S in the direction of
> > >motion -
> ...
> > So S and P does not change! Thus for any v along x axis the spin S
> > is aligned with P as it was at v=0!
> > Where is this "component of S in the direction of motion"?
>
> Perhaps the best way to characterize the nature of this problem about
> photon spin and to explain why it matters is a quote from Richard
> Feynman, in The Feynman Lectures in Physics, vol. II (p. 17-5):
> "A paradox is a situation which gives one answer when analyzed one way,
> and a different answer when analyzed another way, so that we are left
> in somewhat of a quandary as to actually what should happen."
Thanks for this reference and the definition of a paradox, which I would
call inconstant analysis or conclusions.
> In this case the quandary is based on the difference in our
> understanding and analysis of the angular momentum of material bodies,
> versus our understanding of the spin of photons as provided by optics
> and quantum mechanics.
I strongly disagree with that statement. My analysis are relativistic
and so is QFT, QM and so... (till gravity proves important)
> The analysis provided by Mr. Gueorguiev is
> ultimately rooted in the momenta and positions of material particles as
> found in different reference frames.
It is more then that, it is SR and Lorentz transformations which tell us
how vector & tensor fields are transformed.
> Consider a spinning top, and then
> imagine it to be moving rapidly for us in a direction perpendicular to
> the spin axis. Indeed, we expect the spin as we define it to remain in
> its original orientation, even at relativistic velocities. There is no
> component of S in the direction of motion, and no problem for
> conservation of angular momentum either.
Did I say anything about S or P, I did not assigned them to any specific object.
Look again:
S=[S_0,0,0,S_z] and P=[P_0,0,0,P_z] and a Lorentz boost in x direction is:
| 1 v 0 0 |
gamma(v)|-v 1 0 0 | where gamma(v)=1/(1-v^2) in c=1 units
| 0 0 1 0 |
| 0 0 0 1 |
So S and P does not change! Thus for any v along x axis the spin S
is aligned with P as it was at v=0!
Where is this "component of S in the direction of motion"?
Maybe I should also have asked:
Where is this "component of P in the direction of x-motion"?
Probably, this would have save me from having to write this post :)
> However: although it would be reasonable to *expect* that the same
> transformations would apply to the spin of beams of light, that is not
> what optics and quantum mechanics tell us. An illustration circa page
The Lorentz transformations apply to any vector & tensor quantities!
> 275 of Optics by Eugene Hecht is graphically clear about the nature of
> photon spin as we think we know it: it shows the spin vector parallel
> (or antiparallel) to the direction (unit k) of the propagation of light.
Right, and the direction of the propagation of light is given by its space
part of its 4-momentum vector! In the case above it is {0,0,P_z}
> This is of course what creates right-hand or left-hand
> circularly polarized light. Feynman calls this the "screwy" behavior of
> light, made so by its having no rest mass. This situation is completely
> in accord with all that we know (or think we know) about the optics of
> circularly polarized light, and accords with both quantum mechanics and
> the effect of the E and B fields on electrons in media, etc. I know of
> no exceptions deriving from motion of the source.
Good!
> Now: consider that the light emitted by the moving source in the
> original problem is tilted at an angle relative to its rest-frame
> propagation along the z axis, and is now leaning towards positive x.
How would that be? I have show you that is you have a photon moving
in z direction, it will be moving in the same direction in any other
frame that is moving in x direction relative to the original frame!
Just look again:
S=[S_0,0,0,S_z] and P=[P_0,0,0,P_z] and a Lorentz boost in x direction is:
| 1 v 0 0 |
gamma(v)|-v 1 0 0 | where gamma(v)=1/(1-v^2) in c=1 units
| 0 0 1 0 |
| 0 0 0 1 |
So S and P does not change! Thus for any v along x axis the spin S
is aligned with P as it was at v=0!
Where is the non-zero component of P in the x-direction which would be
required if the photon was propagating in that direction?
Vesselin G Gueorguiev
Gerard Westendorp wrote:
>
> I will finish my job of spelling out E- and B fields for both obswervers:
>
> I will choose units so that |E| = |B| = 1 fot the observer emitting the
> .photons:
> (gamma=1/sqrt(1-v^2/c^2)
>
> emitter:
> photons up photons down
> Ex +cos(wt) +cos(wt)
> Ey +sin(wt) -sin(wt)
> Ez 0 0
>
> Bx -sin(wt) -sin(wt)
> By +cos(wt) -cos(wt)
> Bz 0 0
>
> receiver:
> photons up photons down
> Ex +cos(wt) +cos(wt)
> Ey +sin(wt)*gamma -sin(wt)*gamma
> Ez +cos(wt)*gamma*v -cos(wt)*gamma*v
>
> Bx -sin(wt) -sin(wt)
> By +cos(wt)*gamma -cos(wt)*gamma
> Bz -sin(wt)*gamma*v +sin(wt)*gamma*v
>
These don't look right to me! You have to use the Faraday tensor:
| 0 -Ex -Ey -Ez |
F_{ij}= | Ex 0 Bz -By |
| Ey -Bz 0 Bx |
| Ez By -Bx 0 |
and the Lorentz transformation:
| 1 v 0 0 |
L^i_j=gamma(v)| v 1 0 0 | where gamma(v)=1/(1-v^2) in c=1 units
| 0 0 1 0 |
| 0 0 0 1 |
transform F as a tensor i.e. F'_{ij}=L_i^k L_j^l F_{kl}
and deduce E' & B' in the new frame.
You should get that Ex'=Ex & Bx'=Bx but
y-z projected part is E'=gamma(v)(E+[v,B]) and B'=gamma(v)(B-[v,E]), where
[v,B] is the vector product of v which is in x-direction & y-z part of B.
For example:
Ez'=gamma(v)(Ez+v*Bx) you don't have such expressions for E' & B'
Gerard Westendorp
> Just look again:
>
> S=[S_0,0,0,S_z] and P=[P_0,0,0,P_z] and a Lorentz boost in x direction is:
>
> | 1 v 0 0 |
> gamma(v)|-v 1 0 0 | where gamma(v)=1/(1-v^2) in c=1 units
> | 0 0 1 0 |
> | 0 0 0 1 |
>
> So S and P does not change! Thus for any v along x axis the spin S
> is aligned with P as it was at v=0!
>
> Where is the non-zero component of P in the x-direction which would be
> required if the photon was propagating in that direction?
If I operate your matrix on your vector, I get:
P'= gamma(v)[P_0,-vP_0,0,P_z]
So the component of P' in the x-direction is -vP_0
Gerard
Gerard Westendorp
Vesselin G Gueorguiev wrote:
> Gerard Westendorp wrote:
> >
[..]
> For example:
> Ez'=gamma(v)(Ez+v*Bx) you don't have such expressions for E' & B'
Ez =0 in the first frame.
so
Ez'=gamma(v)(Ez+v*Bx)
becomes
Ez'=gamma(v)*v*Bx
as in my note.
Gerard
Marc Bellon
}-
}-and the Lorentz transformation:
}-
}- | 1 v 0 0 |
}-L^i_j=gamma(v)| v 1 0 0 | where gamma(v)=1/(1-v^2) in c=1 units
}- | 0 0 1 0 |
}- | 0 0 0 1 |
Here is your error: this is not a Lorentz transformation,
which is
| gamma gamma*v 0 0 |
|gamma*v gamma 0 0 |
| 0 0 1 0 |
| 0 0 0 1 |
The last two diagonal elements are 1, they must not be multiplied by
gamma.
Marc Bellon.
Marc Bellon
[ Other parts skipped]
}-Now: consider that the light emitted by the moving source in the
}-original problem is tilted at an angle relative to its rest-frame
}-propagation along the z axis, and is now leaning towards positive x.
}-This angle can come close to 90 degrees from the z axis if the source
}-is moving rapidly. The light from the other beam is also tilted, as a
}-mirrored angle with respect to the x axis. If the spin of light is as
}-Hecht and Feynman and every physics reference I know of describes it,
}-that spin should be aligned with these directions of propagation for
}-the observer for whom the source is in motion. There should then indeed
}-be a component of spin along x, the direction of motion, as seen by
}-simple inspection of the geometry of the situation. In that case, the
}-angular momentum of the total system changed without external
}-interactions, since the spin of the source is still zero.
I think that the whole problem is that you overlooked the fact that
the full angular momentum reduces to the spin of the particle only in
the direction of propagation. To obtain an angular momentum of h/2pi
(the magnitude of the spin angular momentum) in a direction orthogonal
to the propagation requires only a shift of lambda/2pi (where lambda is
the wavelength). As you may know, locating a photon to less than its
wavelength is perilous. Conservation of angular momentum is than possible
due to a small shift in the relative positions of the photons, which
is absolutely impossible to control.
There has been a rather interesting proposition for the definition of
the position operator from the Lorentz boost generators:
H. Bacry, The position operator revisited, Annales de l'I.H.P.,
Theoretical Physics, vol. 49, p 245.
The essential idea is that the boost operator at time 0 contains
the position time the energy, so that you can define the position
to be the boost devided by the energy. The position thus obtained
has the interesting property that the commutation of two coordinates
is proportional to the spin and this nicely reflects the limits to the
focusing of light of a given wavelength. I think that the spin dependent
modification of the position is just what is needed for the total angular
momentum along the direction of the boost to remain zero.
(Other interest of this definition is that this position operator
has not the zitterbewegung of the naive position operator for Dirac
particles and can be used for a quick derivation of the correct spin
orbit coupling in an atom).
In the form of exotic position operators, you can also read
with interest the following paper (which has the advantage of
being readily accessible through the archives):
Paper: quant-ph/9610004
From: Marc Jaekel <Marc....@lpt.ens.fr>
Date: Thu, 3 Oct 1996 12:46:45 +0200 (MET DST) (8kb)
Date (revised): Mon, 16 Jun 1997 12:03:12 +0200 (MET DST)
Title: Mass as a Relativistic Quantum Observable
Authors: Marc-Thierry Jaekel (Laboratoire de Physique Th\'eorique de l'Ecole
Normale Sup\'erieure) and Serge Reynaud (Laboratoire Kastler-Brossel)
Comments: 5 pages, revised version
Report-no: LPTENS 96/53
Journal-ref: Europhys. Lett. 38 (1997) 1
This again gives positions which depend in some way on the spin
of the particle.
Marc Bellon.
t...@rosencrantz.stcloudstate.edu
> (1.) Start with a source imagined to have "zero" spin angular momentum
> (although it is the changes that matter.) Emit two "right-hand" photon
> pulses in opposite directions. That should leave the AM of the source
> at zero (see for example Feynman Lectures Vol. 3). Then change to the
> frame in which the source is moving. The light is taking diagonal paths
> as explained earlier. The AM of the source should still be zero in this
> frame (I will reply briefly to Mr. Colby on this matter later.) If
> total AM is to be conserved, then the photon spins need to keep their
> original orientations, which are now no longer parallel to the
> propagation vector k.
The key here is that photons don't have to be in states of definite
angular momentum. A photon can be in a superposition of positive-helicity
and negative-helicity states. In the boosted reference frame, both
of the photons are in such superpositions. Anybody in such a frame
who tried to measure the helicity of one of the photons would get
either +1 or -1 (that is, she would find the photon had spin
either parallel or antiparallel to its propagation direction), but
you can't predict which in advance.
- -Ted
Pierre Asselin
>The key here is that photons don't have to be in states of definite
>angular momentum. A photon can be in a superposition of positive-helicity
>and negative-helicity states. In the boosted reference frame, both
>of the photons are in such superpositions.
That's what I tought too, but I transformed the E and B fields and got
pure circularly polarized states for both photons.
--
--Pierre Asselin, Westminster, Colorado
l...@netcom.com
Gerard Westendorp
Last time I said I had calculated the E and B fields:
gamma=1/sqrt(1-v^2/c^2)
emitter:
photons up photons down
Ex +cos(wt) +cos(wt)
Ey +sin(wt) -sin(wt)
Ez 0 0
Bx -sin(wt) -sin(wt)
By +cos(wt) -cos(wt)
Bz 0 0
receiver:
photons up photons down
Ex +cos(wt) +cos(wt)
Ey +sin(wt)*gamma -sin(wt)*gamma
Ez +cos(wt)*gamma*v -cos(wt)*gamma*v
Bx -sin(wt) -sin(wt)
By +cos(wt)*gamma -cos(wt)*gamma
Bz -sin(wt)*gamma*v +sin(wt)*gamma*v
But I forgot that the argument in the sin and cos functions
should not be just (wt).
In the emitter frame, the argument for the photons-up should
be (wt-kz). The argument or the photons-down should be
(-wt+kz). But I worked away the minus-signs to the outside of
the sin and cos functions, so we get:
emitter:
photons up photons down
Ex +cos(wt-kz) +cos(wt-kz)
Ey +sin(wt-kz) -sin(wt-kz)
Ez 0 0
Bx -sin(wt-kz) -sin(wt-kz)
By +cos(wt-kz) -cos(wt-kz)
Bz 0 0
receiver:
The argument (wt-kx) should be transformed as the scalar product
of 2 four-vectors. It is now:
A = gamma^2(wt(1+v^2)+wvx + kz)
photons up photons down
Ex +cos(A) +cos(A)
Ey +sin(A)*gamma -sin(A)*gamma
Ez +cos(A)*gamma*v -cos(A)*gamma*v
Bx -sin(A) -sin(A)
By +cos(A)*gamma -cos(A)*gamma
Bz -sin(A)*gamma*v +sin(A)*gamma*v
The argument A is not just a function of t, but also
of z. So there is not just one set of E and B crossing
the yz-plane. This makes things a lot less nice.
Gerard
Neil
<bel...@lpthe.jussieu.fr> wrote:
> I think that the whole problem is that you overlooked the fact that
> the full angular momentum reduces to the spin of the particle only
> in the direction of propagation. To obtain an angular momentum of
.....
Marc:
My thought experiment was intended to involve pulses of many photons,
which in the net consideration of momentum and angular momentum, etc.,
reduce to the classical circularly polarized wave. (Pardon me if I did
not make that clear before.) I admit to not having checked out your
references yet, but just on the basis of general considerations, your
attempted solution (like that of Ted) apparently relies on not
considering the summed up effect of measuring many photons (or: not
considering the pulse as a whole, as when we allow it to be absorbed,
and then measure the rotational AM transferred to the absorber.) The
description of classical beams of circularly polarized light is
understood - or at least we think it is. When such a beam is absorbed
in toto by a target some distance from the source in my thought
experiment, it should transfer the expected AM of N*hbar to the target
- that means parallel to the beam. Now we have a component N*hbar*v/c
of macroscopic rotational AM of the absorber in the direction of source
motion, which does not add up to what we started with (zero AM,
considered either as rotational AM or along the line of source motion.)
Furthermore, there are issues concerning the velocity-transformed
magnitude of the effective source rotational AM in version two of the
problem (discussed in various threads here), where photon pulses were
emitted so AS to put the source into rotation. Then we have to worry
about how to get enough spin magnitude out of the photons, and not just
the direction.
The quantum issues are interesting, but what all of the photons en
masse should effectively add up to is what makes this whole thing
problematical - and I think it really is, and not just a matter of
someone not recognizing a known correction or subtlety of quantum
mechanics, etc. I have little clue what the resolution may be, but I am
reasonably confident that I have correctly framed the issues.
Neil Bates
Neil
<vess...@baton.phys.lsu.edu> wrote:
> Neil wrote:
...
> > The analysis provided by Mr. Gueorguiev is ultimately rooted in the
> > momenta and positions of material particles as found in different
> > reference frames.
> It is more then that, it is SR and Lorentz transformations which
> tell us how vector & tensor fields are transformed.
....
We must first be very careful to note whether we are dealing with
4-vectors, or 3-vectors of the sort which transform according to
various hard-to-predict formulas (for example, acceleration according
to a_x = gamma^-3 a_0x, but f_x = f_0x, where x is the direction of
motion.) Conventional rotational angular momentum is an axial
three-vector (actually, it is better represented as a tensor anyway:
J_ik = r_i p_k - r_k p_i. How covariant that is, I do not know.) Its
transformation is highly problematical in many ways (consider for
example the Thomas precession), but the mechanical rotational AM of a
non-accelerated body can be shown to transform such that the parallel
components are equal in all frames, but the perpendicular components
are increased to be gamma times the rest value (this can be checked by
direct calculation of paired r cross p in the different frames.) One
may use whatever mathematical entities one wants to during analysis,
but eventually they must be converted back into the direct angular
momentum, etc., that could be given to a resting flywheel - one must
cash in the chips.
Of course, if the spin of the photons transformed in the same way as
the rotational AM of the source, we would have no problem. The two
types of spin ought to transform the same way, but the fact known from
optics that they don't is the very thing that makes the whole issue a
true contradiction - in the Feynman sense - concerning our
understanding of light! (Note: the pulses contain many photons, and QM
ambiguities involving single photons are not helpful.)
.....
> So S and P does not change! Thus for any v along x axis the spin S
> is aligned with P as it was at v=0!
> Where is the non-zero component of P in the x-direction which
> would be required if the photon was propagating in that direction?
I think your mistake is that you did not include the energy component
for defining p_x:
p_x = gamma*(p'_x + vE'/c^2)
The p'_x of the photons in the source frame are zero, but they get a
component in the x direction from their energy. Also consider
aberration of starlight, and Einstein's original illustration of time
dilation: light is emitted along the z axis of a moving source, is
reflected off a mirror carried at right angles to the motion, and
returns to the source. The light follows a saw-tooth path, the path
length of which is equal to the gamma factor times the distance
traveled by the source. One can even use the relativity of simultaneity
to show that the wave front which is parallet to x in the rest frame is
tilted towards z because the time setting in the rear of the moving
frame is ahead of that in the front: just enought to get the sort of
propagation given in the Einstein example. The spin paradox is indeed
real.
Neil Bates
t...@rosencrantz.stcloudstate.edu
Neil <neil_...@hotmail.com> wrote:
>My thought experiment was intended to involve pulses of many photons,
>which in the net consideration of momentum and angular momentum, etc.,
>reduce to the classical circularly polarized wave. (Pardon me if I did
>not make that clear before.)
Do you mean to say that your paradox can be expressed in terms purely
of electromagnetic fields, without reference to quantum physics? If
so, it'd be very helpful if you did so. From what I've seen, your
argument depends on the fact that a photon's spin is parallel or
antiparallel to its propagation direction, which is a fact from
quantum physics, not classical electromagnetism.
>The
>description of classical beams of circularly polarized light is
>understood - or at least we think it is. When such a beam is absorbed
>in toto by a target some distance from the source in my thought
>experiment, it should transfer the expected AM of N*hbar to the target
>- that means parallel to the beam. Now we have a component N*hbar*v/c
>of macroscopic rotational AM of the absorber in the direction of source
>motion, which does not add up to what we started with (zero AM,
>considered either as rotational AM or along the line of source motion.)
Can you explain again where that N hbar v/c comes from? I'm not sure
I get it.
As I understand your argument, it goes like this. Please correct me
if I've got it wrong.
A pair of light beams propagate in the + and - x directions. Both
have positive helicity and equal intensity, so the total angular
momentum is zero. Look at the light beams from an inertial reference
frame that is moving with respect to the initial frame in, say, the y
direction. In this frame, the beams will be propagating diagonally
(their momenta will have components in both the x and y directions).
They'll therefore have angular momenta with y components, so the total
angular momentum won't be zero.
I agree with this up until the last sentence. I claim that the
angular momenta of the beams will still point in the + and - x
directions, even in the boosted reference frame. I don't see how this
leads to any sort of paradox. Classically, there is no rule (that I
know of) that says that angular momentum has to be parallel to
propagation direction. Quantum mechanically, each beam consists of a
bunch of photons that are in superpositions of + and - helicity
states.
>The quantum issues are interesting, but what all of the photons en
>masse should effectively add up to is what makes this whole thing
>problematical - and I think it really is, and not just a matter of
>someone not recognizing a known correction or subtlety of quantum
>mechanics, etc.
I don't quite understand what you mean here. It sounds like you mean
that you've discovered a logical inconsistency in classical
electromagnetism. If so, then I'm quite confident that you're wrong.
Classical field theory is very well-understood mathematically, and
conservation of angular momentum can be rigorously proved in it; if
you think you've found a situation where it doesn't hold, then you've
made an error.
-Ted
Toby Bartels
>Vesselin G Gueorguiev <vess...@baton.phys.lsu.edu> wrote:
>> | 1 v 0 0 |
>>L^i_j=gamma(v)| v 1 0 0 | where gamma(v)=1/(1-v^2) in c=1 units
>> | 0 0 1 0 |
>> | 0 0 0 1 |
> | gamma gamma*v 0 0 |
> |gamma*v gamma 0 0 |
> | 0 0 1 0 |
> | 0 0 0 1 |
Also, Vesselin's definition of gamma misses a square root.
gamma = 1/sqrt(1 - v^2) != 1/(1 - v^2).
Once you have the correct gamma, you can choose the correct matrix
by remembering that the matrix must have determinant 1.
-- Toby
to...@ugcs.caltech.edu
Vesselin G Gueorguiev
> The argument (wt-kx) should be transformed as the scalar product
> of 2 four-vectors. It is now:
> A = gamma^2(wt(1+v^2)+wvx + kz)
A scalar product doesn't change under Lorentz transformations:
p^i g_ij x^j= p'^i g_ij x'^j where p=hbar [w,k]
These are little things that make me doubt your calculations, but
still I have no time and motivation to do them myself.
[...]
Vesselin G Gueorguiev
my previous posts on this topic. I made a few very wrong statements
and technical errors which I shouldn't have done. So let me
try to make up for all this starting with correcting the
Lorentz transformation as noted by Marc Bellon.
| k(v) k(v)v 0 0 |
L(v)= | k(v)v k(v) 0 0 | where k(v)=1/(1-v^2) in c=1 units
| 0 0 1 0 |
| 0 0 0 1 |
Consider the rest frame of the emitter, before emission:
P_emitter=Pem=M[1,0,0,0] 4-momentum
S_emitter=Sem=Se[1,0,0,0] 4-spin
after photon emission:
Sem=Se'[1,0,0,0] & Pem=(M-2E)[1,0,0,0]
S1=[S1_0,0,0,S1_z] & P1=E[1,0,0,+1]
S2=[S2_0,0,0,S2_z] & P2=E[1,0,0,-1]
we could start off right now with 4-spin conservation which will
lead us to Se=Se'+S1_0+S2_0 and since Se=0 is a natural choice together
whit S1_0=S2_0=S0 and hbar=1 which would mean that S1_z=1, S2_z=-1 we may
have:
Sem=[-2S0,0,0,0] & Pem=(M-2E)[1,0,0,0]
S1=[S0,0,0,+1] & P1=E[1,0,0,+1]
S2=[S0,0,0,-1] & P2=E[1,0,0,-1]
But before doing that let's see what happen after a Lorentz
transformation with boost along x:
Sem'=Se'k(v)[1,v,0,0] & Pem'=(M-2E)k(v)[1,v,0,0]
S1'=[k(v)S0,k(v)vS0,0,+1] & P1'=E[k(v),k(v)v,0,+1]
S2'=[k(v)S0,k(v)vS0,0,-1] & P2'=E[k(v),k(v)v,0,-1]
So if we want the 3D part of the 4spin & 4-momentum to be aligned
we should have:
" P1'/S1' " =constant=E/(+1)=E/S0 ==> S0=+1
" P2'/S2' " =constant=(-E)/(-1)=E/S0 ==> S0=+1
Conserving the 4spin now gives Se=Se'+2 ==> Se=-2 if Se=0
Therefore in the rest frame after emission:
Sem=[-2,0,0,0] & Pem=(M-2E)[1,0,0,0]
S1=[1,0,0,+1] & P1=E[1,0,0,+1]
S2=[1,0,0,-1] & P2=E[1,0,0,-1]
Thus 4spin & the 4-momentum of the photons will always be aligned, and
since Sem+S1+S2=0 it will stay ZERO in any coordinate system.
PS. If -2 in the 4spin of the source is disturbing, then you should
try 4x4 anti-symmetric tensor as generalization of the spin since the
spin is actually a pseudo vector, as the magnetic field is.
Douglas A. Singleton
>
>> (1.) Start with a source imagined to have "zero" spin angular momentum
>> (although it is the changes that matter.) Emit two "right-hand" photon
>> pulses in opposite directions. That should leave the AM of the source
>> at zero (see for example Feynman Lectures Vol. 3). Then change to the
>> frame in which the source is moving. The light is taking diagonal paths
>> as explained earlier. The AM of the source should still be zero in this
>> frame (I will reply briefly to Mr. Colby on this matter later.) If
>> total AM is to be conserved, then the photon spins need to keep their
>> original orientations, which are now no longer parallel to the
>> propagation vector k.
>
The "paradox" needs some simplifying and cleaning up.
First you're taking a single system and then looking at
some particular quantity (angular momentum) in two
different frames and comparing these quantities. This
is not such a hot idea. For example I could show that
momentum is not conserved in this way. In the rest frame
the system will have zero momentum. Now boost to some other
frame where the particle is not at rest. Now it has a
non-zero momentum. If you comapre the momenta of these
two differnet frames there not equal, but this does
not imply that momentum conservation doesn't hold. You're
kind of doing the same thing.
Second, you should try to get rid of the initial source of
the photons, since people will always tend to suspect that
in some way the extra angular momentum is hiding there.
You can easily do this and at the same time make the "paradox"
much more realistic by considering a pi^0 which does in fact
decay into 2 photons ~99% of the time. If the original
particle is gone then there's no way it can carry any
angular momentum.
OK how about this restatement : A pi^0 zips by you
in the z-axis and decays into 2 photons, which
move off at some angle +@ and -@ w.r.t the z-axis.
The larger the initial momentum of the pi^0 the smaller
the angle @. Now in any reference frame the pi^0
has 0 angular momentum. Yet the photons must have their
angular momentum along the direction of motion and
so it appears as if some angular momentum has developed
along the z-axis.
This has the advantage of not requring that you compare
some quantity coming from two different frames, and
tehre's no way the original particle can hide anything
since it stops existing.
Also in writing this up I think I see a problem. In your
statement you seem to indicate that the photons *total*
angular momentum points along its direction of motion.
One can say that there is a projection of +hbar along
its direction of motion, but this isn't the total
spin of the photon. The photon being spin 1 has
a total angular momentum of sqrt(2) hbar. The total
angular momentum may be zero even though the projections
along two different, non-orthogonal directions may not cancel.
In any case fixing the photons angular momentum to be strictly
along some particular axis seems to violate the commutation
algebra for angular momentum --> [S_x , S_y] = i hbar S_z
(if S_x = S_y = 0 there's no way this will work. I'm being a
little -- or a lot -- careless since setting operators = 0
is fishy sometimes).
Hope this helps,
Doug
Vesselin G Gueorguiev
Vesselin G Gueorguiev wrote:
>
> I own an apology to Neil, Gerald, and all other readers for
> my previous posts on this topic. I made a few very wrong statements
> and technical errors which I shouldn't have done. So let me
> try to make up for all this starting with correcting the
> Lorentz transformation as noted by Marc Bellon.
>
> | k(v) k(v)v 0 0 |
> L(v)= | k(v)v k(v) 0 0 | where k(v)=1/(1-v^2) in c=1 units
> | 0 0 1 0 |
> | 0 0 0 1 |
Woops! I didn't write it correctly again, as Topy notices there is sqrt:
k(v)=1/sqrt(1-v^2)
Probably, I should shut up till I get in a better shape & concentration.
Vesselin G Gueorguiev
"Douglas A. Singleton" wrote:
>
> ... try to get rid of the initial source of
> the photons, since people will always tend to suspect that
> in some way the extra angular momentum is hiding there.
> You can easily do this and at the same time make the "paradox"
> much more realistic by considering a pi^0 which does in fact
> decay into 2 photons ~99% of the time. If the original
> particle is gone then there's no way it can carry any
> angular momentum.
So much for may spin as 4-vector, it goes in flames!
( or in a "graviton" :-) :-)) Just kidding!
Any one working on gravity B experiment who cares to explain
how spin is presented as 4x4 anti-symmetric matrix?
[...]
Neil
Only rarely would I ask for correction to a previous message.
The word "absolute" was mispelled below in the last paragraph:
Finally, the statement "if you think you've found a situation where
[conservation of angular momentum] it doesn't hold, then you've made an
error." is ill-advised for a number of reasons. First, it is not in
keeping with the spirit of inquiry to have unchallengeable absolute
certainties, especially about "negative" claims. (Remember the
conservation of parity - it was so reasonable to assume.) Second, even
if the conservation law does hold, then something else will have to
change, such as the description of light (such as your proposed and
unheard-of "tilted" spins) - that would be news itself. Last, critics
who claim errors should actually produce the error and expose it, not
just reason in the "must have" mode. I do not think that you have
succeeded in doing so, and the paradox still stands - strange as it is.
Neil
* Sent from RemarQ http://www.remarq.com The Internet's Discussion Network *
The fastest and easiest way to search and participate in Usenet - Free!
[Moderator's note: It's perfectly OK to post corrections to one's
previous posts. My personal opinion is that correcting spelling
errors isn't usually worth the bother unless the error significantly
obscured your meaning, but it's up to you. -TB]
Marc Bellon
}-
}-In article <ygcyaax...@clet.lpthe.jussieu.fr>, Marc Bellon
}-<bel...@lpthe.jussieu.fr> wrote:
}-
}-> I think that the whole problem is that you overlooked the fact that
}-> the full angular momentum reduces to the spin of the particle only
}-> in the direction of propagation. To obtain an angular momentum of
}-.....
}-
}-Marc:
}-
}-My thought experiment was intended to involve pulses of many photons,
}-which in the net consideration of momentum and angular momentum, etc.,
}-reduce to the classical circularly polarized wave. (Pardon me if I did
}-not make that clear before.) I admit to not having checked out your
}-references yet, but just on the basis of general considerations, your
}-attempted solution (like that of Ted) apparently relies on not
}-considering the summed up effect of measuring many photons (or: not
}-considering the pulse as a whole, as when we allow it to be absorbed,
}-and then measure the rotational AM transferred to the absorber.) The
}-description of classical beams of circularly polarized light is
}-understood - or at least we think it is. When such a beam is absorbed
}-in toto by a target some distance from the source in my thought
}-experiment, it should transfer the expected AM of N*hbar to the target
}-- that means parallel to the beam. Now we have a component N*hbar*v/c
}-of macroscopic rotational AM of the absorber in the direction of source
}-motion, which does not add up to what we started with (zero AM,
}-considered either as rotational AM or along the line of source motion.)
No, my solution does not have anything to do with large N (classical) or
quantum behavior.
The point is simply that for a zero mass particle, only one component
of spin has a meaning, the one along the direction of propagation.
This is what make possible the existence of only 2 spin states for the
photon, and not the three spin states of a massive spin-1 particle.
The distinction between an intrinsec (spin) component and an orbital
component of the angular momentum is not invariant. You cannot ask
for their independent conservation. The orbital angular momentum is of
the order of the spin angular momentum for impact parameters of the
order of the wavelength, that is distances much less than the natural
extension of a particle in free space.
[snip]
}-The quantum issues are interesting, but what all of the photons en
}-masse should effectively add up to is what makes this whole thing
}-problematical - and I think it really is, and not just a matter of
}-someone not recognizing a known correction or subtlety of quantum
}-mechanics, etc. I have little clue what the resolution may be, but I am
}-reasonably confident that I have correctly framed the issues.
I really think that you are somewhat considering your photons as
you would consider massive particles, and this is not possible.
There is a fundamental difference between zero mass and non-zero mass.
}-Neil Bates
}-
Marc Bellon.
PS: I shall be away from my computer during this Christmas period,
and I fear that article posted now would have disappeared from the
news server by the time I am back. So please, if you have further
questions, think of sending me a copy by mail. I shall put sender
in the follow-ups, but it may not be acknowledged by every news reader.
Harry Johnston
> My thought experiment was intended to involve pulses of many photons,
> which in the net consideration of momentum and angular momentum, etc.,
> reduce to the classical circularly polarized wave. (Pardon me if I did
> not make that clear before.) I admit to not having checked out your
> references yet, but just on the basis of general considerations, your
> attempted solution (like that of Ted) apparently relies on not
> considering the summed up effect of measuring many photons (or: not
> considering the pulse as a whole, as when we allow it to be absorbed,
> and then measure the rotational AM transferred to the absorber.) The
> description of classical beams of circularly polarized light is
> understood - or at least we think it is. When such a beam is absorbed
> in toto by a target some distance from the source in my thought
> experiment,
You do realise that the light will not necessarily be circularly
polarised in the frame of reference of the observer? I wouldn't
expect the polarisation state to be conserved by a Lorentz boost.
Harry.
---
Harry Johnston, om...@ihug.co.nz
One Earth, One World, One People
Vesselin G Gueorguiev
>
> In article <38545C10...@phys.lsu.edu>, Vesselin G Gueorguiev
> <vess...@baton.phys.lsu.edu> wrote:
>
> > Neil wrote:
> ...
> > > The analysis provided by Mr. Gueorguiev is ultimately rooted in the
> > > momenta and positions of material particles as found in different
> > > reference frames.
> > tell us how vector & tensor fields are transformed.
> ....
>
> We must first be very careful to note whether we are dealing with
> 4-vectors, or 3-vectors of the sort which transform according to
> various hard-to-predict formulas (for example, acceleration according
> to a_x = gamma^-3 a_0x, but f_x = f_0x, where x is the direction of
> motion.)
I didn't get you example. 4-acceleration has coordinates (0,a_x,a_y,a_z)
in the co-moving frame of the accelerated object.
> Conventional rotational angular momentum is an axial
> three-vector (actually, it is better represented as a tensor anyway:
> J_ik = r_i p_k - r_k p_i. How covariant that is, I do not know.)
That is right! I don't know either how the spin should be written as 4x4
antisymmetric tensor, but sure that's must be the way to go.
However, there is the position operator for a photon which makes difficult
to work with J_ik = r_i p_k - r_k p_i in that case.
For point source object at the origin we get J_ik=0 in its rest frame and
thus it would stay this way in any Lorentz frame.
> Its transformation is highly problematical in many ways
Not problematical, if one gets the right generalization! Tensor or vector?
> (consider for
> example the Thomas precession), but the mechanical rotational AM of a
> non-accelerated body can be shown to transform such that the parallel
> components are equal in all frames, but the perpendicular components
> are increased to be gamma times the rest value (this can be checked by
> direct calculation of paired r cross p in the different frames.)
I don't know, but probably would find time to write all the stuff down, but
if you do it I will try to follow your calculations.
> One may use whatever mathematical entities one wants to during analysis,
> but eventually they must be converted back into the direct angular
> momentum, etc., that could be given to a resting flywheel - one must
> cash in the chips.
Right!
> Of course, if the spin of the photons transformed in the same way as
> the rotational AM of the source, we would have no problem. The two
> types of spin ought to transform the same way,
Yes.
> but the fact known from
> optics that they don't is the very thing that makes the whole issue a
Which fact? I don't understand what do you mean?
> true contradiction - in the Feynman sense - concerning our
> understanding of light! (Note: the pulses contain many photons, and QM
> ambiguities involving single photons are not helpful.)
>
> .....
> > So S and P does not change! Thus for any v along x axis the spin S
> > is aligned with P as it was at v=0!
> > Where is the non-zero component of P in the x-direction which
> > would be required if the photon was propagating in that direction?
>
> I think your mistake is that you did not include the energy component
> for defining p_x:
No, my mistake was that I didn't do the calculations, instate I was putting
my energy to lay down the mathematical formalism but I did used,
[now I did, and there's a post on this which considers the spin as 4-vector]
[...]
Gerard Westendorp
[..]
> Classical field theory is very well-understood mathematically, and
> conservation of angular momentum can be rigorously proved in it; if
> you think you've found a situation where it doesn't hold, then you've
> made an error.
It is perhaps even easier to prove that if a covariant quantity is
conserved in one frame, it must be conserved in another.
Say T1, T2, T3, T4 are covariant quantities such that:
T4 = T1 + T2 + T3
Because the Lorentz transformation is linear:
T4' = T1' + T2' +T3'
A special case of which is:
AM_emitter_before = AM_emitter_after
+ AM_light_up+ AM_light_down
So if angular momentum is a covariant quantity, and it
is conserved in the emitter frame, it must be conserved
in the receiver frame.
The problem is what the covariant way for writing down
the angular momentum is.
I you say it is a vector parallel to the motion of the photon,
you get a funny answer, so that is probably wrong. But what
is the correct way of writing down angular momentum,
for example, what is the angular momentum of a given
set of E and B fields?
Gerard
Gary
t...@rosencrantz.stcloudstate.edu wrote:
> Neil <neil_delv...@hotmail.com.invalid> wrote
>
> > (1.) Start with a source imagined to have "zero" spin angular
momentum
> > (although it is the changes that matter.) Emit two "right-hand"
photon
> > pulses in opposite directions. That should leave the AM of the
source
>
> The key here is that photons don't have to be in states of definite
> angular momentum. A photon can be in a superposition of positive-
helicity
> and negative-helicity states. In the boosted reference frame, both
> who tried to measure the helicity of one of the photons would get
> either +1 or -1 (that is, she would find the photon had spin
> either parallel or antiparallel to its propagation direction), but
> you can't predict which in advance.
What's QM got to do with it? The paradox still stands considering
classical EM fields alone.
Interesting thread. I wish I had more time to think about this one.
No one seems to be thinking clearly. A quick look at the mathematics in
another branch of this thread has got me thinking that *maybe* just
*maybe* Vesselin G Gueorguiev has got it figured out in one of his
later posts.
------------------------
Gary Pajer
Sent via Deja.com http://www.deja.com/
Before you buy.
Vesselin G Gueorguiev
[...]
> Interesting thread. I wish I had more time to think about this one.
> No one seems to be thinking clearly. A quick look at the mathematics in
> another branch of this thread has got me thinking that *maybe* just
> *maybe* Vesselin G Gueorguiev has got it figured out in one of his
> later posts.
No, actually there was a post proposing pi^0->2 gammas which clearly
shows that my 4-spin analysis are no good and one needs to go to
4x4 antisymmetric matrices. For example, we could assign a 4-spin to the
pion to be [2,0,0,0] which will be cleared after decay in 2 gammas,
but then there is a problem with any Lorentz boost that will produce
3D angular momentum to the pion. It is true that in most coordinate
systems we will observe angular momentum =b_{inf} p where b is the impact
parameter but that is not what one gets after boosting [2,0,0,0]
So let me try to construct the anti-symmetric 4x4 matrix by analogy
with E&M we need some vector to play the role of E.
In E&M for boost along x-axis we have:
Ex'=Ex & Bx'=Bx,
y-z projected part of E and B are transformed as:
E'=k(v)(E+[v,B]) and B'=k(v)(B-[v,E]),
where [v,B] is the vector product of v which is in x-direction
and k(v)=1/sqrt(1-v^2) in c=1 units.
Now one realizes that if we have B as the angular momentum J then
transforming it from the rest frame where J=0 to a new frame
gives J'=gamma(v)(0-[v,E]) compare with the L=-[P,R] we find our
vector E up to a momentum vector ([p,p]=0). Thus we have:
E->m R (mass*position),
B->S spin then we have:
| 0 -m*x -m*y -m*z|
J_{ij}= | m*x 0 Jz -Jy |
| m*y -Jz 0 Jx |
| m*z Jy -Jx 0 |
this make sense for a massive particle and seems to work.
However, there is a problem with the position operator of the photon
which cannot be cured by its zero mass. We need an equivalent of
L=-m[v,R] in the case of a photon.
Note: if we place a particle with non zero spin at R=0 and then view
it from a x-boosted coordinate system we get:
x'=x=0 & Sx'=Sx,
x-y plane: m*r'=k(v)[v,S] and S'=k(v)S,
say we have the particle spin along the y axis then it seems that our
particle should have new position 3-vector [0,0,k(v)vS/m].
But we know that when boosted we should get it moving along x axis,
thus its momentum along the x axis should change, therefore maybe we
really need to make:
E->m R+P,
B->S spin
but then we still have P'x=Px?!
I don't know how much sense does this all make since spin is a Quantum
property and has no classical analog and all we did here was for point
particle.
Douglas A. Singleton
Vesselin G Gueorguiev <vess...@baton.phys.lsu.edu> wrote:
[stuff deleted]
>
>So let me try to construct the anti-symmetric 4x4 matrix by analogy
>with E&M we need some vector to play the role of E.
>In E&M for boost along x-axis we have:
It might also be worthwhile to think about the
Pauli-Lubanski pseudo-vector
W_u = -(1/2) e_{uvps} J^{up} P^s
where P^s is the 4-momentum, J^{up} is the antisymmetric
tensor whose elements include angular momenta and boosts.
e_{uvps} is the 4D Levi-Civita symbol
W_u satisfies the following properties :
(a) W_u P^u = 0
(b) W_u W^u = -m^2 s(s+1)
(Note that this seems to imply a value of sqrt{2} \hbar
for the *total* spin of the photon as opposed to the
projection which takes values of +-hbar. I think, but am
not sure, that this may have something to do with giving
a *physical* answer to the problem, since I think the
original poster had the total angular momentum being
entirely along the direction of propagation and having
only a value hbar not sqrt{2} hbar).
>From a more mathematical persceptive :
In (b) s=spin. P_u P^u = m^2 and W_u W^u are Casimir
invariants of the system. In any case W_u W^u is
a scalar under normal Lorentz transformations and
so it shouldn't matter how one boosts around, it should
still stay the same. Thus if s=0 (i.e. W_u W^u = 0) in
one frame it will equal zero in any frame. The original
poster might like to verify this for himself by making
up a few specific cases and explicitely checking them.
Also to anticipate a little bit
one might make the mistake of saying "But W_u W^u can be
zero in the end but still have s =/= 0 since photons
are mass zero i.e. m=0." Just remember that m^2 = P_u P^u
which is an invariant as well -- "m" in the above is not the
photon mass, or sum of the photon masses or anything like
that.
In the re-statement of the paradox where one considers
the physical process pi^0 --> 2 gamma "m" would be the rest
mass of the pi^0.
All of this can be found in more detail "Quantum Field
Theory' by Lewis Ryder section 2.7. Also P. Ramond's
"Field theory : A modern primer" discusses the Pauli-Lubanski
pseudovector in chapter 1
Unfortunately I won't be able to follow this thread
for the next 8-9 days.
Doug
[construction of 4x4 spin matrix deleted]
par...@nmt.edu
In article <164c000c...@usw-ex0101-005.remarq.com>,
Neil <neil_...@hotmail.com> wrote:
>Finally, the statement "if you think you've found a situation where
>[conservation of angular momentum] it doesn't hold, then you've made an
>error." is ill-advised for a number of reasons.
No, I think there are many folks lurking out there who would agree
that Ted Bunn gave excellent advice here! Lack of a response you
like doesn't mean there isn't one out there. It's like laws of
thermodynamics: if somebody claims to have found a way around
them, you can be *almost* certain that they haven't. The details
may be messy -- which will keep a lot of people from wading through
the argument to find the precise error.
For instance, I don't think I could write a quick response explaining
why Maxwell's demon can't be made to make heat run from a cold room
to a hot room. (Maxwell's demon is a famous thermodynamics paradox
whereby a little gremlin can open a door when there is a fast
particle coming, then close it again when there is a slow particle
coming.)
> Second, even
>if the conservation law does hold, then something else will have to
>change, such as the description of light
>... and the paradox still stands - strange as it is.
No. It's not really a crisis for physics, just a nifty paradox.
Part of the reason it's hard to answer in the way that you would
like is that light is tricky to handle under boosts. The
polarization 4-vector in one frame may be purely spatial
(0,ex,ey,ez), where |ex|^2 + |ey|^2 + |ez|^2 = 1
-- but it acquires a *timelike* component when boosted! What
does a timelike polarization look like? The resolution is a
bit complicated, but it's well-known. In quantum field theory,
one has to worry about theories that are relativistic from
the get-go, so the photon polarizations are a bit of a
headache. It's a funny situation where you end up throwing
out some polarizations somewhere along the line as unphysical,
or perform other sneaky but perfectly well-understood and
valid tricks -- like defining equivalence classes on the
space of polarizations.
But the whole issue of *photons* is a red herring here! I'll
write some easier-to-answer but related paradoxes, and let *you*
answer your own paradox by figuring them out.
You've made a few misinterpretations of quantum mechanics, for
starters! The photons are *not* to be thought of as definitely
circularly polarized along (or anti-along) their directions of
travel. As Ted Bunn pointed out, they can be in superpositions
of the circular polarizations, being linearly or elliptically
polarized or whatever.
But it's actually worse than that. The photons are in *mixtures*
of superpositions of such states! And that brings us to the
first new puzzle (Paradox 1 is your photon puzzle, for
numerological purposes):
***************************************************************
PARADOX 2: A spin-0 particle at rest decays into two photons.
Before decay, there is one piece of information in the angular
momentum: it's zero! After decay, there is the angular
momentum of one photon, and the other too! There's a bunch
of information in the photons' momenta, also, that wasn't there
before. Where did all this extra information come from?
***************************************************************
In fact, we put so much faith in conservation of angular
momentum that it's typically *how* we solve the problem:
given one photon's spin (in the above scenario), what will
the other one be?
How would we solve Paradox 1? Usually, by transforming
to a frame moving along with the original particle; then
it becomes like Paradox 2! If you measure the photons'
circular polarizations using detectors that are also at
rest in this frame, you know the answer: they cancel
precisely. But in your paradox, the detectors will be
moving with respect to the rest frame - and they are
answering *DIFFERENT* "what is the photons' polarization?"
questions than detectors at rest do.
Which brings us to:
***************************************************************
PARADOX 3: A spin-0 particle at rest decays into an
electron-positron pair. A Stern-Gerlach apparatus is set
up to measure each particle's spin, but one detector is
horizontally oriented, and the other is vertically oriented.
No matter which of the 4 possible outcomes (up-left,up-right,
down-left, or down-right), THE SPINS WILL NOT ADD TO ZERO.
What's going on here?
***************************************************************
(I like this version; it would make a nice quantum mechanics
test question.)
If you have trouble figuring out Paradox 3, forget quantum
mechanics for a minute and try:
***************************************************************
PARADOX 4: A ballistic pendulum is a lump of clay hanging from
a string. A bullet is fired at it, and the bullet sticks into
the clay. (By measuring the maximum height to which the
combined clay-bullet system rises, one can figure the original
momentum of the bullet if the masses are known.) But there's
momentum in the bullet before, and NO momentum left when the
pendulum is at its maximum height. Where did the momentum go?
***************************************************************
If you have trouble with Paradox 4, try the "donkey and cart"
paradox being discussed in sci.physics now. If you work out
these easier paradoxes first and go backwards to the harder
ones, I'll bet you can figure them out -- and in the process
see why nobody else seems concerned that there's a major
problem with light here. It's a fun paradox, but nothing
more!
John Baez
sci.physics.research! I think solving lots of paradoxes like this
is a great way for people to learn that many things which initially
seem paradoxical can eventually be understood. If you don't understand
that, whenever you trip over something puzzling you start worrying
that the whole framework of physics is collapsing.
Taylor and Wheeler's "Spacetime Physics" is a good source of special
relativity "paradoxes" and their solutions, which makes it a great way
to learn relativity. There should be a similar book for quantum theory -
I don't know if there is! But there are these:
Surprises in theoretical physics, Rudolf Peierls. Princeton, N.J.:
Princeton University Press, 1979.
More surprises in theoretical physics, Rudolf Peierls. Princeton, N.J.:
Princeton University Press, c1991.
In article <1999121808...@white.nmt.edu>, <par...@nmt.edu> wrote:
>***************************************************************
>PARADOX 3: A spin-0 particle at rest decays into an
>electron-positron pair. A Stern-Gerlach apparatus is set
>up to measure each particle's spin, but one detector is
>horizontally oriented, and the other is vertically oriented.
>No matter which of the 4 possible outcomes (up-left,up-right,
>down-left, or down-right), THE SPINS WILL NOT ADD TO ZERO.
>What's going on here?
>***************************************************************
>
>(I like this version; it would make a nice quantum mechanics
>test question.)
I like this one too. I remember it used to bug me immensely -
where did the heck did conservation of angular momentum go???
I won't give away the answer, because it's much better to figure
out this kind of thing oneself.
But for the experts, I should note that the answer is related to
the Wigner-Araki-Yanase theorem - see
http://math.ucr.edu/home/baez/week33.html
t...@rosencrantz.stcloudstate.edu
You're right that my analysis is wrong. Sorry about that. I retract
most of the content of my posts on this subject.
I can't see how to do a proper analysis without some detailed and
messy calculations, which I don't have time to do right now, but I'll
try to think about it when I'm on a plane in a few days. In the mean
time, I'll make some general comments about the (small) portion of my
earlier post that I still agree with, and your response to that
portion.
>However, I must disagree with: "Classical field theory is
>very well-understood mathematically, and conservation of angular
>momentum can be rigorously proved in it."
I promise that this is true. The fact that angular momentum is
conserved in Maxwellian electrodynamics is an honest-to-goodness
theorem. You can look it up in textbooks.
Actually calculating the angular momentum in a particular situation
can be messy, and the way the angular momentum transforms when you
shift reference frames can be very complicated (which is why I screwed
up so badly in my analysis). But even if I can't (or don't feel like)
calculating the angular momentum in a particular situation or in a
particular frame, I can be 100% confident that it'll be conserved --
that's the great thing about theorems!
>Finally, the statement "if you think you've found a situation where
>[conservation of angular momentum] it doesn't hold, then you've made an
>error." is ill-advised for a number of reasons. First, it is not in
>keeping with the spirit of inquiry to have unchallengeable pabsolute
>certainties, especially about "negative" claims.
What I meant by that sentence was something like this: If you think
you've found a situation *in classical electromagnetism as described
by the Maxwell equations* where angular momentum isn't conserved, then
you've made a mistake. That is, it was meant to be a statement
about a particular mathematical formalism (classical electromagnetic
theory), not a statement about nature.
I thought that was reasonably clear from the context, but if it
wasn't, then I apologize. Note that this meaning is the one that's
relevant to our current discussion. You're not claiming to have done
an experiment to show that angular momentum isn't conserved; you're
claiming to have analyzed a particular thought experiment in classical
electromagnetism and shown that angular momentum is not conserved.
We're talking about theory, not experiment here.
This distinction makes a huge difference. If I claimed to have
absolute certainty that angular momentum was conserved in nature, your
criticism would be absolutely right. We're never absolutely sure
about anything in nature; we always have to remain at least a bit
skeptical, since some future experiment could always surprise us.
But we can be absolutely certain about statements about mathematical
formalisms. Suppose someone came up to you and said that they'd found
the largest prime number. You could state with complete confidence
that they'd made a mistake. My statement above is in the same
category. Both are statements about mathematical formalisms for which
rigorous theorems can and have been proved. It's a theorem that there
are infinitely many primes, and it's a theorem that the Maxwell
equations conserve angular momentum.
So I stand by my statement, with the clarification above: if you think
you've discovered a situation in which the Maxwell equations violate
angular momentum conservation, then you've made a mistake.
-Ted
Gerard Westendorp
> It might also be worthwhile to think about the
> Pauli-Lubanski pseudo-vector
>
> W_u = -(1/2) e_{uvps} J^{up} P^s
This vector looks interesting, but I think there is
a problem with any vector describing angular
momentum.
With a vector, you Lorentz transform the spatial
components to zero, by choosing a suitable frame.
But that would not fit in with the concept of spin, which
has a definite spatial direction.
Gerard
Vesselin G Gueorguiev
> With a vector, you Lorentz transform the spatial
> components to zero, by choosing a suitable frame.
That is true only if the vector is *time-like*,
if the vector is space-like then we can find a
Lorentz frame where all components are zero, except
one special component. (That's a theorem in SR)
I just realized that for light-like vectors the theorem
should say that there is a Lorentz frame where the
vector looks like this: something*[1,0,0,1]. Since I don't
remember this as part of the theorem I could be wrong, but
it seems that with simple linear algebra one should be able
to prove it.
[Moderator's note: it's true. - jb]
Gerard Westendorp
An important 4-vector which cannot be Lorentz-transformed
so that its spatial components are zero, is one with a t-component
of zero.
Gerard Westendorp wrote:
> Douglas A. Singleton wrote:
> > It might also be worthwhile to think about the
> > Pauli-Lubanski pseudo-vector
> >
> > W_u = -(1/2) e_{uvps} J^{up} P^s
> This vector looks interesting, but I think there is
> a problem with any vector describing angular
> momentum.
> With a vector, you Lorentz transform the spatial
> components to zero, by choosing a suitable frame.
Gerard Westendorp
Two evil Vodons are sentenced to death.
They are tied together by their feet, and set to collide at
high speed with planet Earth. However, they manage to untie
themselves, and they decide to escape. In fact they do not
only escape, but take a terrible revenge. They decide to generate
a paradox, thereby creating an anomaly in the fabric of space time,
which would destroy the earth.
Instead of just jumping away from each other in opposite directions,
each avoiding the earth, they also push themselves in an equal and
opposite pirouette. The Earth people look at this through
their telescopes in amazement, they see angular momentum in the
direction of the velocity of the Vodons being created from nothing!
But this paradox can be completely checked by explicit calculation. I
started doing this, and it is quite messy. But when I started working it
out, I found out why there is no paradox.
The paradox comes from the fact that you think the axes of rotation
coincides with the direction of motion. This is not true.
Suppose the Vodons are doing a pirouette around their length axes.
This axes is defined by the difference between the vector defining
the head, and the vector defining the feet. Say:
FEET := (t,0,0,ut)
HEAD := (t,0,0,wt+L)
AXES: = HEAD-FEET = (0,0,0,L)
Lorentz transform:
FEET := (gamma(t-vx),gamma*vt,0,ut)
HEAD := (gamma(t-vx),gamma*vt,0,wt+L)
AXES: = HEAD-FEET = (0,0,0,L)
So the orientation of the axis of rotation remains unchanged after a
Lorentz transform. In the rest frame, the Vodons are doing a pirouette
in the direction of their motion. But in the Earth frame, the Vodons
are doing a "Crab walk". They are doing a pirouette, and moving also
sideways.
So, as Neil in his 2nd post suggested:
^
|
/
/
/
/
***- - - -> velocity of emitter
This is the correct picture!
Gerard
Ralph E. Frost
> I don't know how much sense does this all make since spin is a Quantum
> property and has no classical analog and all we did here was for point
> particle.
Are you saying you do not see any objects in the classical environment
that exhibit spin or oscillation?
Just a thought.
--
Best regards,
Ralph E. Frost
http://www.dcwi.com/~refrost/index.htm ..Feeling is believing..
Vesselin G Gueorguiev
>
> Vesselin G Gueorguiev wrote:
>
> > I don't know how much sense does this all make since spin is a Quantum
> > property and has no classical analog and all we did here was for point
> > particle.
>
> Are you saying you do not see any objects in the classical environment
> that exhibit spin or oscillation?
No. We have the angular momentum L~ p x r which is a good classical object.
What I was saying is that the internal spin does not have classical equivalent.
Take for example S=1/2 particle, there is no way to have it in a usual
classical mechanics since all representations of so(3) in the Fun(R^3) are
integer. Therefore, we have to attach internal degree of freedom to get S=1/2.
JMcmi24918
statistics. yet the interesting part is the lack of 3 azimuthal spin angular
momenta. photons only have 2! why? relativistic invariance. photons move
exactly at speed c. a third "z " component would imply photon mass (which
experimentally < 10^-45 kg) actually a good discussion of the issue of photon
mass (and the proca lagrangian and alteration of maxwell's equations for
massive photons) is found , where else(?), in jackson's book Classical
Electrodynamics
.
dr j.f. mcmillan
jmcm...@chaminade.edu
Gerard Westendorp
density of the E an d B fields.
Could someone hive me a hint on how to calculate angular momentum
density of E and B fields using Noether's theorem and the Lagrangian
F_mu_nu F^mu^nu?
I know how to do the trick with Noether if the fields transform under
a transformation of a set of phi's like
phi_a -> phi_b
phi_b -> -phi_a
But what about a transformation that involves fields (Ax,Ay,Az)
*and* coordinates (x,y,z) both changing under rotation?
Gerard
Paul Colby
news:1415c574...@usw-ex0105-037.remarq.com...
> In article <82ep2n$jq2$1...@bgtnsc02.worldnet.att.net>, "Paul Colby"
> <Paul....@worldnet.att.net> wrote:
>
> > At the risk of sounding real stupid, does one
> > expect that the angular momentum in the field is a
> > Lorentz invariant? From field theory the angular
> > momentum appears as an antisymmetric tensor with [...]
>
> I'm sorry that this is so belated, but the point needs to be made (I
> did respond earlier, but somehow that didn't post.)
>
> There is indeed no reason to expect that the angular momentum in the
> field is a Lorentz invariant. I believe, though, that you have
> misunderstood the point of the problem. I was not expressing amazement
> at the very fact of the spin angular momentum of the light pulses being
> oriented differently in the new frame. The whole point of the paradox
> is that if the spins are tilted diagonally in the frame where the
> source is moving, then they don't add up to zero anymore relative to a
> reference point along the line of motion of the source. (This last
> neccessity also seems to confuse some correspondents.)
>
This is a result of the space components of the angular momentum tensor,
M(u,v), mixing with the time components as a result of the Lorentz
transformation. In one frame the space components are zero. In another,
they ain't. I don't see a paradox since
M'(1,3)=L(1,0)*L(3,1)*M(0,1) + ... != 0
Also remember that the source is involved as well. The total M(u,v) of the
source and radiation is conserved before and after the reaction.
> If you moved a gyroscope along at high velocity, with its spin axis
> perpendicular to the line of motion, you wouldn't think of the new spin
> as tilted, would you?
>
Yes, I would. I would transform the M(u,v) at rest to the moving frame.
What I would find is that the space components are no longer along the axis
of the gyroscope in the moving frame.
> Yet most people agree that the spin of the photon pulses is along the new
lines of
> motion. Isn't that problematical?
>
No, I think careful use of the transforms remove the problems. (If you
think the classical case is confusing try the quantum.
The M(u,v) are now operators that don't commute. In the frame in which the
spins are collinear the spin components commute. In the frame moving
relative to this one the spin components are along different directions and
therefore projections of M(u,v) along these directions don't commute.)
--
Regards
Paul Colby
Paul....@worldnet.att.net
Gerard Westendorp
Douglas A. Singleton wrote:
[..]
> Both of these lead to a general field angular momentum
> tensor given by
>
> M^{uv} = Int(T^{0u} x^v - T^{0v} x^u ) d^3 x
>
[..or..]
> L = Int (r x (E x B)) d^3 x
This looks pretty sensible, but I am having a problem.
ExB = Poynting vector = (0,0,p)
so
L = Int (yp , -xp, 0) d^3 x
But this is zero over a symmetric interval of space! Also,
the direction is wrong. It seems
something is wrong. (could be me though)
Actually, the fact that the Poynting vector is constant over space
is not completely obvious:
*Linear pol.* waves:
Ex = cos(t-z)
By = cos(t-z)
->. P = (0,0,cos^2(t-z) )
T00 = cos^2(t-z).
So the Poynting vector is kind of 'pulsed', but positive definite
in the z-direction.
The energy is not distributed smoothly over space-time.
*Circular pol.* waves:
Ex = cos(t-z) Ey = sin(t-z)
By = cos(t-z) Bx = -sin(t-z)
->. P = (0,0,1 )
T00 = 1
So here, the energy is distributed smoothly over space-time.
Still working at a way of expressing a.m. of the em field,
Gerard
Douglas A. Singleton
[E&M field angular momentum formulas deleted]
>
>This looks pretty sensible, but I am having a problem.
>
>
> ExB = Poynting vector = (0,0,p)
>
>so
>
> L = Int (yp , -xp, 0) d^3 x
>
>But this is zero over a symmetric interval of space! Also,
>the direction is wrong. It seems
>something is wrong. (could be me though)
In the above I bet you're using linear polarized E and B fields ?
If you are then the fact that you get 0 is what you expect. In the
end a linear polarized light beam is an equal combination of
left and right circularly polarized light. This means that it's
an equal combination of left and right handed photons, which have
their spins pointing in opposite directions thus yielding L=0.
>Actually, the fact that the Poynting vector is constant over space
>is not completely obvious:
>
>*Linear pol.* waves:
>
> Ex = cos(t-z)
> By = cos(t-z)
>
> ->. P = (0,0,cos^2(t-z) )
> T00 = cos^2(t-z).
>
>So the Poynting vector is kind of 'pulsed', but positive definite
>in the z-direction.
>The energy is not distributed smoothly over space-time.
>
>*Circular pol.* waves:
>
> Ex = cos(t-z) Ey = sin(t-z)
> By = cos(t-z) Bx = -sin(t-z)
>
> ->. P = (0,0,1 )
> T00 = 1
>
>So here, the energy is distributed smoothly over space-time.
>
>
>Still working at a way of expressing a.m. of the em field,
>
OK maybe try this. Split the Poynting vector in two as
ExB = E^i grad(A^i) - (E*grad)A (since B=curl A)
In the above anything with subscripts just means a
component of the vector (E or A). (E*grad) means take
the dot product of E with the grad opertor which then
operates on the vector A. Now using this in
Int (r x (E xB)) d^3 x allows you to split L into
two terms one which has an interpretation as the orbital
angular momentum and the other having an interpretation
as the spin of the E&M wave. You'll need to do an integration
by parts and use the fact that div E = 0 (no charges). When you
do this you'll get L in terms of E and A. Then just use
an A for a circular wave e.g.
A= K(x-hat +- i y-hat) exp(iwt -iwz/c)
E=-@A/@t (@=partial)
Some parts of this (in particular the split of L
so that it can be written in terms of E and A) can be
found in Jackson (at least the 2nd Edition).
Hope this helps,
Doug
t...@rosencrantz.stcloudstate.edu
>Douglas A. Singleton wrote:
>> L = Int (r x (E x B)) d^3 x
>This looks pretty sensible, but I am having a problem.
>
>
> ExB = Poynting vector = (0,0,p)
So you're assuming that the Poynting vector always points in the z
direction. I guess that means you're talking about a plane wave
propagating in the z direction. Note that such a plane wave extends
forever in the x and y directions and carries infinite amounts of
momentum and energy. If you want the integrals to converge, you've
got to consider a finite-sized wave packet, and then the Poynting
vector will have x and y components somewhere.
>so
>
> L = Int (yp , -xp, 0) d^3 x
>
>But this is zero over a symmetric interval of space! Also,
>the direction is wrong. It seems
>something is wrong. (could be me though)
Try doing the calculation for a finite-sized wave packet. I
think that the computation isn't too messy if you assume
that the E and B fields look like this:
E = exp(ik (z-t)) f(x,y,z,t)
B = exp(ik (z-t)) g(x,y,z,t)
where f and g are assumed to be slowly varying in space
(|grad f / f| << k or something). What you'll find is that the x and
y components of L do indeed vanish if you set everything up to be
symmetric about the z axis, but that there is a nonzero z component to
L because of the small z components of E and B. (E and B must have
small z components now in order to satisfy the Maxwell equations,
since E and B now slowly vary in the x and y directions).
Furthermore, if you set things up so that the wave is circularly
polarized, you'll find that L is equal to the total momentum divided
by k, just as it should be.
To see how to set all of this up properly, take a look at one of the
problems in Jackson. I can look up which problem if you can't find
it, but I don't have my copy of Jackson with me at the moment.
-Ted
Gerard Westendorp
expression for the angular momentum of the electromagnetic field.
Specifically, I get zero for the case of a circularly polarised plain
wave. Douglas Singleton and Ted Rosencrantz gave some suggestions.
******** The following may be a bit long and boring, but I am coming to
he conclusion that the formula for angular momentum density
L = r X eps c (E X B)
Is not correct. Although it seems to be in various text books. I am
suggesting:
L = eps c (E X A)
In a next post, I will get back to the original paradox. *********
Douglas A. Singleton wrote:
[..]
> Then just use an A for a circular wave e.g.
> A= K(x-hat +- i y-hat) exp(iwt -iwz/c)
> E=-@A/@t (@=partial)
Ok, so
( 1 ) A = ( i ) exp(i(t-z)) ( 0 )
(setting c = w = k = 1)
so that
( i ) E = ( -1 ) exp(i(t-z)) ( 0 )
( -1 ) B = ( i ) exp(i(t-z)) ( 0 )
So:
( 0 ) T = ( 0 ) exp(2i(t-z)) ( -2 )
Actually, this is incorrect, because T does not change sign along the z
axis. The mistake arose because T is a non-linear expression. This means
that you can't use complex numbers as a trick to describe real-valued
fields.
Rather, you have to say:
( 1 ) ( 1 ) A = ( i ) exp(i(t-z)) + ( -i )
exp(-i(t-z)) ( 0 ) ( 0 )
( i ) ( -i ) E = ( -1 ) exp(i(t-z)) + ( -1 )
exp(-i(t-z)) ( 0 ) ( 0 )
( -1 ) ( -1 ) B = ( i ) exp(i(t-z)) + ( -i )
exp(-i(t-z)) ( 0 ) ( 0 )
( 0 ) ( 0 ) T = ( 0 ) exp(2i(t-z)) + ( 0 )
exp(-2i(t-z)) ( -2 ) ( -2 )
or:
( 0 ) T = ( 0 ) cos(2(t-z)) ( -4 )
This Poyting vector is the same as for linear polarised fields! And the
vector L= rXT is zero for the z-dircetion.
(Note that my formula for L gives
( 0 ) L = ( 0 ) ( -4 ) )
Ted Rosencrantz wrote:
[..]
> Try doing the calculation for a finite-sized wave packet. I
> think that the computation isn't too messy if you assume
> that the E and B fields look like this:
>
> E = exp(ik (z-t)) f(x,y,z,t)
> B = exp(ik(z-t)) g(x,y,z,t)
>
> where f and g are assumed to be slowly varying in space
> (|grad f / f| << k or something). What you'll find is that the
> x and y components of L do indeed vanish if you set everything
> up to be symmetric about the z axis, but that there is a nonzero
> z component to L because of the small z components of E and B.
> (E and B must have small z components now in order to satisfy
> the Maxwell equations, since E and B now slowly vary in the x
> and y directions).
I might as well use a dipole antenna, giving off circular polarised
radiation. This is certainly a solution to the Maxwell equations. The
electrons in the antenna are going round in circles. This motion is a
superposition of 2 linear dipole motions, (x and y), that ae 90 degrees
outof phase in time.
There is a trick to work out the fields at a large distance (far field).
This is from the Feynman Lectures on Physics. part I. The trick is:
1) construct a line called a 'line of sight' from the position (r) that
you want to know the fields. 2) project the *acceleration* vector of the
moving charge (the source) on the plain perpendicular to the line of
sight. 3) The far field E field is then given by:
E' = -q/(4 pi eps c^2 r) a'(t- r/c)
E'is the E projected on the plane orthogonal to the line of sight.
a'(t- r/c) is the projected and 'retarded' acceleration.
The acceleration vectors are going round in circles in the XY plane.
So on a sphere surrounding the source, looking in the z-direction, the
projected accelerations go round in ellipses, which look something like
this
- / \ | O | \ / -
All ellipses have the same retardation, so the E-fields all have the the
same direction projected onto the z axis. The E fields are "combed"
across the sphere, and the lengths of the vectors vary from place to
place. Note that in the plane of the circle the electrons move in, the
fields are *linearly* polarised.
If you are far away from the center, the gradients in the r-direction
will always be greater that in the other directions, This means that the
B-field will also be in the plane of the sphere, and orthogonal to the
E-Fields. But that means all the Poynting vectors are pointing outward
from the center.
Just as a corkscrew can appear to move foreword when it is rotated, a
corkscrew can also appear to rotate when moved. This is a bit like the
Poynting vectors in the circular polarised dipole source. The energy
flows only outward, but it has a screwy pattern, corresponding to the
retarded image of the source.
So we have a problem. We still get zero angular momentum density.
But even a circular polarised plane wave should carry angular momentum.
If you work out what an electron with mass (m) will do in such a field
r_dot_dot = ( -e / m) E r_dot = ( -e / m) int(E) r = (
-e / m) int(int( E)) L = (1/w ) e^2/m ( E X A)
L = (1/w ) e^2/(eps m c) eps c( E X A)
Basically, I am just guessing, using dimensional analyses, a hunch, and a
check for consistency with the facts, that
L_photonflux = eps c( E X A)
Gerard
[Moderator's note: Rosencrantz is my computer's name, not my name.
The two of us have not yet morphed into a single cyborg-like enitity.
Even if we do, I don't know if I'll take its name -- in this day and
age, lots of people are keeping their own names after such unions, as
my fiancee has reminded me. -TB]
Gerard Westendorp
and quotes unreadable. Here is the corrected version. Sorry for
possible confusion.
Last time I wrote in this thread I was having problems constructing an
expression for the angular momentum of the electromagnetic field.
Specifically, I get zero for the case of a circularly polarised plane
wave. Douglas Singleton and Ted Rosencrantz gave some suggestions.
[Moderator's note: "Ted Rosencrantz" is really Ted Bunn; I hope
I'm not giving any big secret away by revealing this. - jb]
********
The following may be a bit long and boring, but I am coming to
the conclusion that the formula for angular momentum density
L = r X eps c (E X B)
is not correct. Although it seems to be in various text books. I am
sugggesting:
L = eps c (E X A)
In a later post, I will get back to the original paradox.
*********
Douglas A. Singleton wrote:
[...]
> E=-@A/@t (@=partial)
Ok, so
so that
So:
or:
This Poynting vector is the same as for linear polarised fields! And the
vector L = rXT is zero for the z-direction.
Note that my formula for L gives
( 0 )
L = ( 0 )
( -4 )
Ted Rosencrantz wrote:
[...]
> Try doing the calculation for a finite-sized wave packet. I
> think that the computation isn't too messy if you assume
> that the E and B fields look like this:
>
> E = exp(ik (z-t)) f(x,y,z,t)
> B = exp(ik(z-t)) g(x,y,z,t)
>
> where f and g are assumed to be slowly varying in
> space > (|grad f / f| << k or something). What you'll find is that
> the x and y components of L do indeed vanish if you set everything up
> to be symmetric about the z axis, but that there is a nonzero z
> component to L because of the small z components of E and B.
> (E and B must have small z components now in order to satisfy
> the Maxwell equations, since E and B now slowly vary in the x
> and y directions).
I might as well use a dipole antenna, giving off circular polarised
radiation. This is certainly a solution to the Maxwell equations. The
electrons in the antenna are going round in circles. This motion is a
superposition of 2 linear dipole motions, (x and y), that are 90 degrees
out of phase in time.
There is a trick to work out the fields at a large distance (far field).
This is from the Feynman Lectures on Physics, part I. The trick is:
1) construct a line called a 'line of sight' from the position (r) that
you want to know the fields. 2) project the *acceleration* vector of the
moving charge (the source) on the plain perpendicular to the line of
sight. 3) The far field E field is then given by:
E' = -q/(4 pi eps c^2 r) a'(t - r/c)
E' is the E projected on the plane orthogonal to the line of sight.
a'(t - r/c) is the projected and 'retarded' acceleration.
The acceleration vectors are going round in circles in the XY plane.
So on a sphere surrounding the source, looking in the z-direction, the
projected accelerations go round in elipses, which look something like
this
-
/ \
| O |
\ /
-
All ellipses have the same retardation, so the E-fields all have the the
same direction projected onto the z axis. The E fields are "combed"
across the sphere, and the lenghts of the vectors vary from place to
place. Note that in the plane of the circle the electrons move in, the
fields are *linearly* polarised.
If yoy are far away from the centre, the gradients in the r-direction
will always be greater that in the other directions, This means that the
B-field will also be in the plane of the sphere, and orthogonal to the
E-Fields. But that means all the Poyting vectors are pointing outward
from the centre.
Just as a corkscrew can appear to move foreward when it is rotated, a
corkscrew can also appear to rotate when moved. This is a bit like the
Poynting vectors in the circular polarised dipole source. The energy
flows only outward, but it has a screwy pattern, corresponding to the
retarded image of the source.
So we have a problem. We still get zero angular momentum density.
But even a circular polarised plane wave should carry angular momentum.
If you work out what an electron with mass (m) will do in such a field
r_dot_dot = ( -e / m) E
r_dot = ( -e / m) int(E)
r = (-e / m) int(int( E))
L = (1/w ) e^2/m ( E X A)
L = (1/w ) e^2/(eps m c) eps c( E X A)
Basically, I am just guessing, using dimensional analyses, a hunch, and a
check for consistncy with the facts, that
L_photonflux = eps c( E X A)
and that
(1/w ) e^2/(eps m c)
has to do with the electron.
Gerard
t...@rosencrantz.stcloudstate.edu
> Last time I wrote in this thread I was having problems constructing an
> expression for the angular momentum of the electromagnetic field.
>
> Specifically, I get zero for the case of a circularly polarised plain
> wave. Douglas Singleton and Ted Rosencrantz gave some suggestions.
Egotist that I am, I'll skip ahead to where he mentions me again!
> Ted Rosencrantz wrote:
>
> [..]
>
> > Try doing the calculation for a finite-sized wave packet.
[...]
> I might as well use a dipole antenna, giving off circular polarised
> radiation.
[...]
> So on a sphere surrounding the source, looking in the z-direction, the
> projected accelerations go round in ellipses, which look something like
> this
>
> - - / \ | O | \ / -
I find this description a bit mystifying, but OK.
> If you are far away from the center, the gradients in the r-direction
> will always be greater that in the other directions, This means that the
> B-field will also be in the plane of the sphere, and orthogonal to the
> E-Fields. But that means all the Poynting vectors are pointing outward
> from the center.
Don't you want some "approximately"s sprinkled liberally through the
above? The B field points *approximately* tangent to the sphere, but
not quite. I claim that this matters, and that if you calculate the B
field at the next higher order of approximation, you'll get a
nonradial component to the Poynting vector and a nonzero total angular
momentum.
Let me urge you again to check out that problem in Jackson. In that
problem, which deals with an approximately plane-wave solution, you
have to look past the plane-wave approximation at the small variations
transverse to the propagation direction to get the angular momentum
right. I'd lay pretty good odds that the same is true for your
dipole-antenna problem.
> [Moderator's note: Rosencrantz is my computer's name, not my
> name.
Just in case anyone's confused by the somewhat silly moderator's note,
I'll explain that I am the same person as one of the spr moderators.
I use different login names for moderation duties and for posting my
own articles, but there's just one of me. My last name is Bunn.
-Ted
Gerard Westendorp
I don't live near a library, but I'll get hold of it somehow.
(btw, I hope you checked out the repost of my last post, which
clears some misunderstandings due to messed up formats)
Just some remarks in advance:
L = eps c (E X A) is gauge-dependent, but L itself is
dependent on the choice of r_0, the point from which
you do r-vectors. I was hoping to find an interesting
relation between these two convention dependencies.
Ted pointed out:
> The B field points *approximately* tangent to the sphere, but
> not quite.
As I said, I will read Jackson, but I think the approximation
becomes more and more accurate as you go to infinity, and in
the "far field" approximation, there are no "cross fields".
--
\______/_______Gerard________
http://www.xs4all.nl/~westy31/
Douglas A. Singleton
Gerard Westendorp <wes...@xs4all.nl> wrote:
>
>(btw, I hope you checked out the repost of my last post, which
>clears some misunderstandings due to messed up formats)
>
Yes. The reformated message made much more sense, and
I think it was mathematically correct, but the conclusion
was a little strange. This was actually my fault
since I should have mentioned the fishy business
with using infinite plane waves. Ohanian uses
infinite plane waves in his article to get at the
result, but he gives the proper warnings about
the gauge dependence etc. so its OK (i.e. if you "lie"
to someone and tell them you're "lying" this is OK).
I didn't so this.
>
> Ted pointed out:
>
>> The B field points *approximately* tangent to the sphere, but
>> not quite.
>
Now you also had the good idea to look at electric dipole
radiation in terms of angular momentum flow, but the bad
idea to only look at the radiation fields. I think you
need to keep a bit more than just the far field. This is
why I think this (the book that I have handy and will
reference in the following is "Electromagnetic Fields"
by R. Wangsness 2nd edition).
For electric dipole radiation the E and B fields are
E ~ [(1/r^2) - (1/r^3)] cos (@) r-hat +
[(1/r) + (1/r^2) + (1/r^3)] sin @ theta-hat
B ~ [(1/r) +(1/r^2)] sin@ phi-hat
(from pg. 477 eqs. 28-54 and 28-54 using spherical polar
coordinates and unit vectors (r-hat etc.) and leaving
off almost every constant , factor of i, and the
e^(ikr-iwt) oscillation part.
Now from Eq. 28-67 the momentum density is
E x B^* ~ [(1/r^2) + (1/r^5)] sin ^2 @ r-hat
-[(1/r^3) + (1/r^5)] sin (2@) theta-hat
To calculate momentum flux at infinity one would just
keep the first r-hat term. However for the
angular momentum density (r x (E X B^*)) the first
term gives zero since r x r-hat =0. However,
now the first term in the theta-hat part becomes
important since you have an extra power of r
in the angular momentum density compared to the
momentum density. This makes me think that if you
only use the far field E and B you'll get
zero, but for angular momnetum this isn't a
valid approx.
Hope this helps.
Doug
t...@rosencrantz.stcloudstate.edu
> Ted pointed out:
>
>> The B field points *approximately* tangent to the sphere, but
>> not quite.
>
>As I said, I will read Jackson, but I think the approximation
>becomes more and more accurate as you go to infinity, and in
>the "far field" approximation, there are no "cross fields".
It's true that the error you make in the Poynting vector goes to zero
as you go to infinity in this approximation. It even goes to zero
fast enough to make the momentum come out right when you integrate the
Poynting vector. But it doesn't go to zero fast enough to make the
angular momentum come out right. The reason is that the integral that
you have to do to calculate the angular momentum has an extra power of
r in it. That ends up meaning you have to go to the next order of
approximation to get it right.
If you don't have easy access to a library that has Jackson, I can
look over that problem and post the key features of it for you if you
like. If I recall correctly, it's not a terribly difficult problem
(by Jackson's standards), and once you set it up right you probably
won't have any trouble with it. I can't post it now, because my copy
of Jackson is at home and I'm not, but I can do it on Monday if I
remember.
-Ted