Re: Re:Does the Electromagnetic field have a Gravitational field?

rwmw_at_telus.net
Date: 01/12/05 

Date: Wed, 12 Jan 2005 20:01:19 +0000 (UTC)

tessel@tum.bot wrote:
> To immediately answer the question in the subject line: of course it does!

Thank you for your long and thoughtful reply. However your example of
an exact electrovac wave is conformally flat and has no gravitational
field. I'm not sure I buy the dipole argument that says the
gravitational field should not be there. I rather think that all
conformally flat solutions are unphysical. (There was a rumour that the
renaissance of general relativity around 1960 occured because US
airforce generals were interested in conformally flat materials that
could be launched effortlessly into space!!!)It seems to me that all
real objects ought to have a gravitational field,i. e. be surrounded by
a non vanishing Weyl curvature.

>
> Rob Woodside <rwmw@telus.net> wrote:
>
> > Since the electromagnetic field carries energy and momentum with
> > Maxwell's stress-energy tensor, one thinks of it as a real thing
capable
> > of gravitating. Currently Einstein Maxwell theory,
> ^^^^^^^^^
> Is that a pun?

I wish I were that bright. Actually I have a predjudice for the source
free Maxwell equations with the sources excised, leaving horrible
space-time boundaries. I think it easier to deal with boundaries than
currents.
>
> (The term "Einstein-Maxwell solutions" is sometimes used to refer to
> Lorentzian spacetimes equipped with suitable tensor fields describing
a
> charge/current distribution and an EM field, such that a curved
spacetime
> version of the Maxwell equations are satisfied, and such that the EFE
is
> also satisfied, where the only term contributing to the
> stress-momentum-energy tensor comes from the energy-momentum of the
EM
> field itself.

That gives the electrovac solutions. Only if you excise the currents do
you get these solutions. Otherwise you must insert into Einstein's
equations a stress-energy tensor for the currents as well as the
Maxwell stress energy tensor for the electromagnetic field. It is the
ignorance of this current stress-energy tensor that prompts me to cut
it out and be left with electrovac solutions on a manifold with strange
boundaries.

> But as FrediFizzx <fredifizzx@hotmail.com> guessed, the
> gravitational effects of man-made fields are unfortunately far too
small
> to measure in the lab, and even astrophysical EM fields generally
> gravitate too weakly to noticeably affect the gravitational field,
> although there are interesting proposed exceptions to this rule, like
> cosmological magnetic fields.)

I completely agree that this would be a sod to find experimentally and
probably wouldn't occur in our lifetimes. However, there might be
cosnological or galactic or magnetar effects that might be noticeable
>
> But seriously, if you meant to imply that what gtr says about how
Weyl
> curvature couples to Ricci curvature is somehow subject to
interpretation
> (or even subject to change), that certainly isn't true. Gtr might
well be
> wrong about what it says here, but there is no doubt what it does
say.
>
Bingo! Decomposing the full curvature by trace yields: R = C + E + G
where C is the totally traceless Weyl tensor, E has a single trace that
is the traceless part of the Ricci tensor, G has two traces, the second
being the curvature scalar. One can then beat the full 2nd Bianchi
identities into: Div E = Div C + Div G, where the divergence is on the
first index in MTW convention. On excising the currents, the electrovac
situation gives:
E = (k/2)(FF +*F*F) and G = 0 as the Maxwell stress-energy tensor has
no trace and k is given elsewhere in this thread and F is the
electromagnetic field two form and *F its Hodge dual. In general Div E
will not vanish and so it must be balanced by a non zero Div C. Since E
is expressed in terms of F, shouldn't the piece of C with a non zero
divergence also be expressed in terms of F to ensure the identity??? So
I am really asking if there is a UNIQUE gravitational field for the
electromagnetic field that is expressible in terms of the electric and
magnetic field at each point.

> > like the rest of general relativity, puts no constraints on the
Weyl
> > conformal piece of curvature that is the gravitational field.
>
> Careful! (I discussed this very point in a recent post in which I
offered
> an overview of the EFE, and we have discussed the same point on
numerous
> past occasions in this group...)
>
> I think you are thinking of the fact that the Riemann tensor is the
sum of
> its "traceless part", the Weyl tensor, and its "traced part", a
fourth
> rank tensor built from the Ricci tensor (or equivalently, from the
> Einstein curvature tensor). And of course the EFE directly
characterizes
> the latter, without explicitly mentioning the former. This is all
true---
> but it doesn't mean that the EFE "puts no constraints on the Weyl
tensor"!
> If that were true, long range gravitational interactions would be
> impossible, because the immediate presence of matter here and now
(which
> according to the EFE creates Ricci curvature here and now) could not
curve
> up a surrounding vacuum region, because the Weyl curvature would not
> "couple" to Ricci curvature. Similarly, suitably wiggling matter here
> could not create gravitational radiation capable of propagating
through a
> vacuum.
>
> In fact, the EFE plus the differential Bianchi identity implies a
> differential equation which -does- couple Weyl curvature to Ricci
> curvature. This "propagation equation" is discussed in various
places,
> including the monograph by Hawking and Ellis and section 4.4 of the
> excellent new textbook:
>
> author = {Sean Carroll},
> title = {Spacetime and geometry: an introduction to general
relativity},
> publisher = {Addison-Wesley},
> year = 2004}
>
> The equation I am referring to is (4.90) in that book, which I highly
> recommend to anyone who is unfamiliar with the relationship between
Weyl
> and Ricci curvature in gtr.
>
What you say is quite correct, but there are no explict demands put on
C. C is what ever it needs must be to permit a solution to Einstein's
equations and consequently the full 2nd Bianchi identities. If
conformally flat solutions are not ruled out as unphysical, there will
be matter with no gravitational field. I am asking the Einstein Maxwell
equations to produce solutions that have a piece of the Weyl tensor
written explicitly in terms of the electromagnetic field. I have no
qualms about removing conformally flat solutions, but my concern is
that there may be physical solutions to Einstein-Maxwell equations that
do not have the prescribed Weyl componet
CF = k([3/2][FF-*F*F]-[1/4]del*f1 +[1/4]eta*f2), where del is a doubly
anti symmetric kronecker delta, f1 is the invariant [-2(E^2 -{cB}^2),
eta is the permutation tensor and f2 is the invariant 4cE*B where E and
B are the Electric and Magnetic field intensities. This CF is found
simply by asking for a tensor quadratic in F that has the same
symmetries as a Weyl tensor and fixing the coupling constant with the
Riessner Nordstrom solution.

> > Why doesn't the electromagnetic field carry a piece of the
conformal
> > tensor explicit in the electromagnetic field just as it carries
> > Maxwell's stress-energy tensor which is explicit in the
electromagnetic
> > field?
>
> Not sure I understood the question, but roughly speaking, if you
suddenly
> vary a distribution of charges, which is concentrated in some compact
> region, in such a way that the distribution of EM field energy varies
> aspherically and rapidly, then you can expect to create outgoing
> gravitational radiation, which will accompany the outgoing EM
radiation.
> (The fact that the gravitational wave "signal" and EM wave "signal"
move
> in lockstep, at least in a electrovacuum, is the best way to
understand
> what we mean by saying that in gtr, "EM and gravitational radiation
> propagate at the same speed".)
>
> To avoid any possible misunderstanding, I'll add a caveat: in gtr,
under
> suitable circumstances it -is- possible to create an EM wave which is
not
> accompanied by any gravitational radiation: the gravitational effect
of
> such an EM wave is "pure Ricci". In such a wave, the immediate
presence
> of EM field energy-momentum arriving at each event results in Ricci
> curvature there and then, but without any accompanying Weyl
curvature.
> This follows, essentially, from the fact that the strongest EM
radiation
> is -dipole- radiation, but the strongest gravitational radiation is
> -quadrupole- radiation.
>
> Here is a simple explicit exact solution exhibiting the
propagation--- not
> the generation--- of such a wave. It is called the "SG17 EM wave"
(type
> 17 in the classification of Sippel/Goenner, with a suitable EM vector
> potential thrown into the mix). In a "harmonic chart", it can be
written:
>
> ds^2 = -phi du^2 - 2 du dv + dx^2 + dy^2,
>
> phi = m (x^2 + y^2)/u^2,
>
> A = sqrt(m) log(u) @/@x
>
> 0 < u < infty, -infty < v,x,y < infty
>
> Here, phi defines the gravitational field, and A (the EM vector
potential)
> defines the EM field.
>
> To study this solution, it is convenient to define a frame field and
to
> compute various scalars and tensorial quantities wrt this frame.
This
> corresponds to choosing a family of ideal observers (one at each
event)
> and studying their physical experience in detail. Equivalently, one
could
> choose an NP tetrad and compute with the Newman-Penrose formalism,
but the
> idea of a frame field is probably easier to grasp. For example, one
> possible frame field is
>
> e_1 = 1/sqrt(2) (@/@v + @/@u - phi/2 @/@v) (timelike)
>
> e_2 = 1/sqrt(2) (@/@v - @/@u + phi/2 @/@v) }
> }
> e_3 = @/@x } (spacelike)
> }
> e_4 = @/@y }
>
> where these vector fields have unit squared length and are mutually
> orthogonal (hence an alternative term, ONB for "orthonormal basis"),
and
> where not all structure constants vanish (hence another alternative
term,
> "anholonomic basis").
>
> (In the past, we've studied the interpretation of this particular
frame,
> in terms of the appearance of the light cones in our chart. We've
also
> studied this solution using another chart and an inertial frame---
the
> frame above is not inertial, so it's rather artificial, but it serves
to
> make the point.)
>
> This gives the following EM field (as measured by our ideal
observers):
>
> E = sqrt(m/2)/u e_3 B = -sqrt(m/2)/u e_4
>
> This "null" EM field gives rise to a stress-momentum-energy tensor
(as
> measured by our observers):
>
> m [ 1 1 0 0 ]
> T^(ab) = -------- [ 1 1 0 0 ]
> 8 pi u^2 [ 0 0 0 0 ]
> [ 0 0 0 0 ]
>
> Of course, this "immediate presence" of EM field energy-momentum
should
> give rise to Ricci curvature at each event. And indeed, the EFE is
> satisfied, so we have an "exact Einstein-Maxwell solution", or more
> precisely, a "null electrovacuum". However, the Weyl tensor
-vanishes-, so
> this is a -conformally flat- spacetime--- there is no accompanying
> gravitational radiation.
>
> In the past, we've extensively discussed pp waves in this group.
We've
> contrasted the tidal acceleration of initially static test particles
when
> "hit" with a suddenly arriving SG17 EM wave versus say an EK4
> gravitational wave (this is the one which in a suitable limit gives
the
> Aichelburg/Sexl "ultraboost" of a Schwarzschild object, which we have
also
> frequently discussed here in great detail--- see for example the
current
> thread titled "Mass increase and relativistically moving black
holes").
> We have described interesting optical effects summarized by the
slogan,
> "you can't see the wave train enter the station, but you can see it
> depart"--- namely, by looking at the distortion of the -optical
> appearance- of objects between yourself and the departing wavefronts!
> (Exercise: what kind of distortion would you expect for the SG17
example
> above? Bear in mind that the Weyl tensor of this spacetime vanishes,
but
> be careful!) We've discussed the importance of the observation that
-all-
> the polynomial scalar curvature invariants vanish identically for pp
> waves, which is analogous to the fact that in Maxwell's theory of EM,
both
> polynomial invariants of F_(ab) can vanish without having the field
> vanish--- this is just what we mean by saying that we have a "nonzero
but
> null" EM field above. And so forth.
>
Thank you very much for these. The null field where the invariants f1
and f2 vanish is particularly simple and nasty. Here Div E = Div CF = 0
and so the existence argument from the full 2nd Bianchi identity breaks
down. The CF mirrors the polarization or phase of the electromagnetic
radiation. Radiation is such a problem in classical physics that I'm
not sure how far one can go with this. There is a tremendous gulf
between null and non null electromagnetic fields. Never the less one
would expect null and non null field to couple to curvature in the same
way, provided that coupling is unique.

> You can look for those past threads (search under keywords); for the
SG
> classification of nonvacuum pp waves by their Killing vector fields,
see
>
> author = {R. Sippel and H. Goenner},
> title = {Symmetry Classes of pp-waves},
> journal = {Gen. Rel. Grav.},
> volume = 18,
> year = 1986,
> number = 12,
> pages = {1229--1243}}
>
> where you will also find a citation to an earlier paper by Ehlers and
> Kundt classifying vacuum waves by their symmetries. (In the threads
in
> this group, some minor oversights in these papers was corrected.)

Somehow my 16 year old son has developed elitist tendencies as
"highborn" is his google name and I wasn't able to change it when I
posted:
> highborn@gmail.com commented:
>
> > It will be a long while before
>
> [curvature due to the energy of an EM field]
>
> > is directly detected. However, some important matters of principle
would
> > be sorted out if the electromagnetic field did carry a unique
> > gravitational field.
>
> Why do you believe the EM field does not induce a "unique"
(well-defined?)
> gravitational field? We -are- all discussing gtr here, correct? Are
you
> perchance thinking of factor ordering problems? If so, can you be
more
> explicit?
>
> "T. Essel"
I do believe that the E/M field does have the unique gravitational
field CF given above. Unfortunately three referees do not agree or
think that it is unpublishable!!! The examples of spherical and
rotating black holes prove nothing to them. It is difficult for me to
see through the sarcasm and ignorance of the referee reports and I
posted this here in hopes of finding out the real objections. Thank you
very much for taking the time and effort to respond.
> (spelunking somewhere in cyberspace)
Enjoy! Rob Woodside

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