Angular momentum of electromagnetic and gravity radiation
Jaak Suurpere
intrinsic angular momentum of h/2pi.
This means that an assembly of photons can have parallel or
nonparallel spins and that the angular momentum of that assembly is at
most E/2*pi*nu.
It is possible to work out from Maxwell equations alone, without
making any assumptions about fundamental nature of light speed or
quantization of light, what possible states of polarization can
electromagnetic wave have. Circular or elliptical or linear.
Is it possible to derive correct value for the angular momentum of
electromagnetic radiation from Maxwell equations alone?
Is it possible to derive the value from Maxwell equations and special
relativity, without assuming quantization of light?
A graviton is said to have spin 2.
This should mean that depending on the polarization state, a wave of
gravity can have angular momentum up to E/pi*nu.
Is it possible to derive an expression for angular momentum of a wave
of gravity depending on state of polarization, leading to that bound
of angular momentum, by general relativity solution for continuous
gravity field/spacetime waves, without making assumptions about
quantization?
Michael Moroney
>A graviton is said to have spin 2.
(Distantly) related:
I remember reading something about how certain properties of a force
depended on the spin of the force carrier, but I forgot what they were.
So I'll ask: How would the universe be different, if the graviton had
spin 1? Spin 0? Spin 3 or more? I do remember that half-integer spin
force carriers were ruled out due to the Pauli Exclusion Principle.
-Mike
Y.Porat
acording to my understanding of any attraction maker
it must be free of any biased property in space related.
the attraction force is created by something that is fully symetrical
at the 3d space.
(or at least in a 2 D plan)
iow without symetry particle A will not be attracted
on the shortest line that is connecting it to the attracting
particle B
all the best
Y.Porat
---------------------
Jaak Suurpere
The problem is that my question requires ruling out quantization and
Pasuli exclusion principle.
I understand that the properties of gravitation are no just
assumptions about particles, but, in theory of general relativity,
fundamental to spacetime. If the Maxwell equations required photons to
have spin of 1/2 or worse, 2/3 - what could quantum mechanics do about
it?
Bilge
>intrinsic angular momentum of h/2pi.
>
>This means that an assembly of photons can have parallel or
>nonparallel spins and that the angular momentum of that assembly is at
>most E/2*pi*nu.
>
>It is possible to work out from Maxwell equations alone, without
>making any assumptions about fundamental nature of light speed or
>quantization of light, what possible states of polarization can
>electromagnetic wave have. Circular or elliptical or linear.
>
>Is it possible to derive correct value for the angular momentum of
>electromagnetic radiation from Maxwell equations alone?
No. You can obtain a relationship for the angular momentum of the
wave, however:
L = (1/4\pi c)\integral d3x r x (E x B)
A problem in chapter 7 of jackson is to prove that this gives the
result:
L = (1/4\pi c)\integral d3x [ (E x A) + \sum E_j (r x \nabla)A_j ]
and which the term containg the sum contains the angular momentum
and the term E x A is associated with the photon spin. You can prove
this result as follows:
r x (E x B) = r x (E x (\nabla x A))
Using vector identities here would be really messy, so write the cross
products in terms of levi-civita symbols:
r x (E x (\nabla x A)) = e_qms r_q e_nkm E_n e_ijk \nabla_i A_j
(e_qms e_nkm e_ijk) r_q E_n \nabla_i A_j
The levicita symbols may reduced:
e_qms e_nkm e_ijk = -e_qms e_nmk e_ijk
= -e_qms (\delta_in\delta_jm - \delta_im\delta_nj)
= e_qis\delta_nj - e_qms\delta_in
giving:
r x (E x B) = [ (e_qis\delta_nj - e_qms\delta_in ] r_q E_n \nabla_i A_j
= E_j e_qis r_q \nabla_i A_j - E_i e_qms r_q \nabla_i A_j
The first above is the term: E_j (r x \nabla)A_j in the integral. The
second term can be simplified by noting:
r_q \nabla_i A_j = \nabla_i (r_q A_j) - A_j \nabla_i r_q
= \nabla_i (r x A) - A_j \delta_iq
so that:
E_i e_qms r_q \nabla_i A_j = E . \nabla (r x A) - E x A
E . \nabla (r x A) = 2 (E . \nabla)B = 0
since E and B are orthogonal.
Now as to getting the equivalent quantum mechanical expression, you can't.
The quantization of the z component makes this impossible. Classically,
there is no restriction on how preciseky you can measure the angular
momentum of a photon. Quantum mechanics allows you to measure only one
component of the spin, such as the z component. In that case, the other
components are indeterminate and for a multipole field, you get different
ratios for the square of the angular momentum divided by the square of
the energy.
Classically, all of the angular momentum may be taken along a single
axis, so for (L, L_z) having values (l,m), you get:
(mhbar)^2/E^2 = (m/w)^2
while quantum mechanically, the angular momentum squared is l(l+1),
giving:
l(l+1)hbar^2/E^2 = l(l+1)/w^2
>Is it possible to derive the value from Maxwell equations and special
>relativity, without assuming quantization of light?
>
>A graviton is said to have spin 2.
>
>This should mean that depending on the polarization state, a wave of
>gravity can have angular momentum up to E/pi*nu.
>
>Is it possible to derive an expression for angular momentum of a wave
>of gravity depending on state of polarization, leading to that bound
>of angular momentum, by general relativity solution for continuous
>gravity field/spacetime waves, without making assumptions about
>quantization?
Sure. Also, like the photon, there are only two polarization states,
since gravity propagates at c.
Y.Porat
>
>
>
> L = (1/4\pi c)\integral d3x r x (E x B)
>
> A problem in chapter 7 of jackson is to prove that this gives the
> result:
>
>
>
>
> axis, so for (L, L_z) having values (l,m), you get:
>
> (mhbar)^2/E^2 = (m/w)^2
>
> while quantum mechanically, the angular momentum squared is l(l+1),
> giving:
>
> l(l+1)hbar^2/E^2 = l(l+1)/w^2
>
>
> >
> >
> >of gravity depending on state of polarization, leading to that bound
> >of angular momentum, by general relativity solution for continuous
> >gravity field/spacetime waves, without making assumptions about
> >quantization?
you cannot derive any expression of gravity
on the basis of curved space time!
do you know why ?
because space is not curved.
>-----------------
> since gravity propagates at c.
is there experimental prove or even indications
that gravity propagates at c ?
TIA
Y.Porat
-----------------
Bilge
>is there experimental prove or even indications
>that gravity propagates at c ?
Yeah. I did 10 of those experiments last week and 15 the week
before. Send a cashier's check for $100.00 each and I'll email
the results.
Michael Moroney
> Y.Porat:
Seriously, there was an announcement a couple months ago that did measure
the speed of gravity as being c. It had something to do with the
occulting of a star by Jupiter with Jupiter also acting as a gravitational
lens, or something like that.
-Mike
Y.Porat
Thank you Bilge for the information.
anyway do you delude youself that anyone will pay you
100 $ for more information?
btw i dont know you good enough to know
if you saied that seriously or was joking
did i for instance ever demanded here any payment for information.
only a sacker will pay you.
because your information is worth nothing unless it is widely known
so you are in a trap (:-)
2 i would like to hear from other members if Bilg's above
information is confirmed.
TIA
Y.Porat
---------------
Gordon D. Pusch
From Ned Wright's cosmology page,
<http://www.astro.ucla.edu/~wright/cosmolog.htm>:
"7 Jan 2003 - Kopeikin and Fomalont claim
<http://www.nrao.edu/pr/2003/gravity/>
to have measured the speed of gravity by observing the deflection of
radio waves from a quasar by the gravity of Jupiter. They found that
the speed of gravity was equal to the speed of light, as predicted
by general relativity. Cliff Will disputes this claim,
<http://arxiv.org/abs/astro-ph/0301145>
stating that the deflection does not depend on the speed of gravity.
[...More precisely, Will claims that, contrary to Kopeikin's claim, the
deflection does not depend on the ratio of the speed of Jupiter
to the speed of gravity to first order, but only in second order,
so that nothing can be learned about the speed of gravity from
Kopeikin's experiment -- /gdp]
But nobody disputes the fact that Kopeikin and Fomalont's data agree
with general relativity. The only dispute is over what a theory with
the speed of gravity different than the speed of light would predict."
-- Gordon D. Pusch
perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;'
Y.Porat
> the speed of gravity different than the speed of light would predict."
>
>
> -- Gordon D. Pusch
Thank you Moroney and Pusch
i understand it is not easy to prove.
anyway another important question of mine
(that i posted to a similar thread in sci.physics)
is :
is the attractive force on the closer side of the atracted objest
is stronger than the rare side?
and if is there any experimental evidence.
TIA
Y.porat
luke
> Jaak Suurpere:
> >A photon has spin 1. This means that a photon with energy h*nu has
> >intrinsic angular momentum of h/2pi.
> >
> >This means that an assembly of photons can have parallel or
> >nonparallel spins and that the angular momentum of that assembly is at
> >most E/2*pi*nu.
> >
> >It is possible to work out from Maxwell equations alone, without
> >making any assumptions about fundamental nature of light speed or
> >quantization of light, what possible states of polarization can
> >electromagnetic wave have. Circular or elliptical or linear.
>
> Not really.
>
> >
> >Is it possible to derive correct value for the angular momentum of
> >electromagnetic radiation from Maxwell equations alone?
>
> No. You can obtain a relationship for the angular momentum of the
> wave, however:
>
>
> L = (1/4\pi c)\integral d3x r x (E x B)
>
> A problem in chapter 7 of jackson is to prove that this gives the
> result:
>
Great questions Jaak!
Thanks for Jackson ref. and solution Bilge.. Sorry I don't have my
Jackson on me now but I will look... just to show you somebody is
trying to pay attention, shouldn't the very last term below be
e_qjs\delta_in ?
> The levicita symbols may reduced:
>
> e_qms e_nkm e_ijk = -e_qms e_nmk e_ijk
>
> = -e_qms (\delta_in\delta_jm - \delta_im\delta_nj)
>
> = e_qis\delta_nj - e_qms\delta_in
>
Moving on..
The r in that formula for L is confusing me a bit.. that's because we
are calculating L about some point. No "r" in h/2pi.. that is in
photon's frame?
Circular polarization doesn't imply angular momentum I don't think..
it can be expressed as a sum of phase shifted linear polarization.
Perhaps the quantization is due to the nature of the mechanism
generating the light, namely electron energy level changes in atoms..
Could somebody please prove me wrong by showing how a radio wave from
a linear antenna measurably carries angular momentum h/2pi or is
quantized at all?
Bilge
\
>> = e_qis\delta_nj - e_qms\delta_in
>>
Yeah, it looks like it should. Since I moved things around, after
I entered them to clean up the minus signs, I could have retyped the
indicies incorrectly.
>Moving on..
>
>The r in that formula for L is confusing me a bit.. that's because we
>are calculating L about some point. No "r" in h/2pi.. that is in
>photon's frame?
The angular momentum doesn't reference a point, since the integral
is over the volume. It might have been confusing to use `r' and
d3x (jackson uses x rather than r), but, lacking the use of emboldened
characters, I thought it would be more confusing to write something
like:
\integral d3x x x (E x B)
>Circular polarization doesn't imply angular momentum I don't think..
>it can be expressed as a sum of phase shifted linear polarization.
It does imply an angular momentum. Linear polarization is just a
coherent superposition of two circularly polarized photons:
|x> = (1/sqrt(2))[|+> + |->]
|y> = (i/sqrt(2))[|+> - |->]
>Perhaps the quantization is due to the nature of the mechanism
>generating the light, namely electron energy level changes in atoms..
Actually, not quite. The quantized _energy_ is due to the quantization
of the energy levels, however, the quantized angular momentum of the
photon is responsible for the selection rule \Delta L = 1. For example,
the transition 2p->1s is \Delta L = 1 and is an allowed transition.
The 2s->1s is a strictly forbidden transition and must decay by
a two photon process through a virtual p state (or actually, the virtual
p state is just the largest factor - a sum over all of the possible
intermediate states is more accurate).
The energy cannot really be quantized, since I can perform an
arbitrary lorentz boost to a frame in which that energy is
whatever I wish. On the other hand, the helicity of a photon
is lorentz invariant. No lorentz transform can change the
helicity of a photon, so the spin is truly quantized.
>Could somebody please prove me wrong by showing how a radio wave from
>a linear antenna measurably carries angular momentum h/2pi or is
>quantized at all?
See part (b) to the problem for which I worked part (a). Certainly,
you cannot justify the quantization. That is due to the quantum mechan-
ical commutation relations.
Michael Moroney
>is the attractive force on the closer side of the atracted objest
>is stronger than the rare side?
Yes, to the extent that the near side is nearer so an inverse square law
(gravitation, electric charge) will be stronger. How much depends on the
distance and the delta distance.
>and if is there any experimental evidence.
Get a magnet and some iron filings. Note that the magnet will attract
the iron filings.
Comb your hair briskly on a dry day. Bring the comb near some tiny bits of
paper. Notice the (electrically charged) comb will attract the
electrically neutral paper bits.
--
-Mike
Y.Porat
thanx Moroney
anyway what i mean is much more delicate
ie beyond the inverse law
ie an experiment that will neutralise the distance effect
lets take for instance the earth and sun:
is there any experimental evidence that
g at day time is somewhat bigger than at night time .......
at exactly the same geographys spot (at the same 'square meter'
i supose that at this distance between sun and earth
the difference in distance should be negligable.
and even not it can be neutralised by a simple calculation.
TIA
Y.Porat
-----------------------