Does an electric charge curve spacetime ?

[nrqed]

The weight of a charged particle will depend on the spacetime curvature. If there is no spacetime curvature then the weight of the charge will not tell you if you are in a curved spacetime. But if there is spacetime curvature then the weight of the charge will be different given the same local acceleration. This is due to the fact that a charged particle does not follow a geodesic in spacetime. The field of the charge is not localized and thus the field can "feel out" the surrounding spacetime and is thus not a locall phenomena. Clifford Will wrote a paper on the weight of a charged particle in a Scharzchild spacetime.
It turns out that the weight is a function of charge (I think I recall that the charge weighed less but am not sure). Thus if one is in a box in a Schwarzschild spacetime then you can tell if you're not in an accelerating box in flat spacetime by using a charged particle and weighing it. This is not cheating the equivalence principle since the fields are not local and the equivalece principle, when applied to a curved spacetime, is a local phenomena and the field of a charge is not a local phenomena. Think of this as using the field of a charge to probe spacetime for curvature.


Pete

Very interesting point.

But then I got thinking and it seems to me that the wwight of *anything* will also depend on the curvature because, as far as I can tell, there is no way (even as a thought experiment) to measure the weight of anything in a purely local way (again, afaik). Any weight measurement involves a displacement (of a spring for example) so is not purely local. Therefore, in principle, any weight measurement will distinguish a curved spacetime from a noncurved one.

Unless I am missing something.

Pat

 

[pervect]

Let's say the charge is surrounded by detectors (let's say photomultipliers). Either the detectors will register a count can't be observer dependent. Either they click or they don't. But you are saying that some obersever could see the detector click while no radiation is emitted by the charge.

So I guess that from the point of view of that observer, the detector would be picking up vacuum fluctuations? Is that the usual explanation?

If this is correct, then it seems to me an amazing fact that such a simple consideration (thinking about a charge in the context of GR) leads to a quantum physics concept (quantum fluctuations)! It's as if trying to marry GR and E&M points to the need for quantum physics, and I have never seen things presented this way.

Pat

The phenomenon of radiating charges can be understood entirely classically. In fact, it's probably a bit easier to understand that way. One doesn't think of a wave as having a definite location, like a particle does. When you think of the wave as having particle properties, of arriving in a "packet" that is either there or not there, you are introducing the quantum element into an otherwise classical problem.

In quantum terms, I would describe Unruh radiation as particles coming from the vacuum, but I would not describe the radiation of accelerating charges in that way. There are some similarities in the mechanism, though. What's happening (in my interpretation, anyway) is that photons that are virtual from one viewpoint are no longer virtual from another, so that the virtual photons that produce the columb force in one coordinate system are seen as "real" (non-virtual) photons in another coordinate system. So the photons arise not from the charge and not the vacuum for the problem of radiating charges - but the mechanism is similar to Unruh radiation in that photons that are virtual in one coordiante system are real in another.

 

[Garth]

. What's happening (in my interpretation, anyway) is that photons that are virtual from one viewpoint are no longer virtual from another, so that the virtual photons that produce the columb force in one coordinate system are seen as "real" (non-virtual) photons in another coordinate system. So the photons arise not from the charge and not the vacuum for the problem of radiating charges - but the mechanism is similar to Unruh radiation in that photons that are virtual in one coordiante system are real in another.

Thank you for that pervect.

So the inertial observer freely falling past a supported electric charge receives real photons, whereas for the supported observer they are only virtual?
Is the inertial observer then drawing (real) energy from the false vacuum?

If so, would it be that for the supported observer, co-moving with the Centre of Mass (the 'Machian' frame) of the gravitating body, energy is conserved; whereas for the inertial freely falling observer it is not?

That would be my understanding of the situation, concordant with the principles of SCC.

Garth

 

[pervect]

Google turns up

http://prola.aps.org/abstract/PRD/v29/i6/p1047_1

The nature of the interaction between a quantum field and an accelerating particle detector is analyzed from the point of view of an inertial observer. It is shown in detail for the simple case of a two-level detector how absorption of a Rindler particle corresponds to emission of a Minkowski particle. Several apparently paradoxical aspects of this process related to causality and energy conservation are discussed and resolved.

unfortunately I don't have access to the full article :-(. But from what I can recall reading in Wald about the general topic of Unruh radiation (aka acceleration radiation), there is some mapping between positive and negative frequencies when one goes from the Rindler coordinate system of an accelerating observer to the Minkowski coordinate system of an inertial observer. So the terse result quoted in the abstract above is more or less what I'd expect, one observer sees a negative frequency particle as being absorbed, another observer sees a positive frequency particle as being emitted.

 

[Chronos]

Reflecting upon the original question, a simple experiment. Project a tightly focused laser on a sufficiently distant, sensitive target. Place a strong electromagnet near the source and in the path of the beam. Power up the electromagnet and rapidly alternate polarity and see if focal spot moves.

 

[Antonio Lao]

If theorists (starting with A. E.) can make a theory about spacetime curvature caused by mass (GR), couldn't there be a similar theory where some spacetime curvature is caused by electric charges?

If we view mass as the sum of equal amount of positive and negative electric charges then we can say that the curvature is caused by electric charges (equal amount of pluses and minuses).

It is a fact that black holes cause spacetime curvature. But it is also a fact that there are black holes with electric charges therefore we can also say that electric charges indirectly affect curvature. The question really is how much charges must be needed to be localized in a small region of spacetime in order to be comparable in strength with relativistic mass located at the same region.

 

[Gonzolo]

Reflecting upon the original question, a simple experiment. Project a tightly focused laser on a sufficiently distant, sensitive target. Place a strong electromagnet near the source and in the path of the beam. Power up the electromagnet and rapidly alternate polarity and see if focal spot moves.

Ok, but I would have to know what to expect as a result, before actually investing time in doing this. What strength of electromagnet to use? And how much of a distance shift should I expect if there is curvature (assuming there is a theory out there that predicts it). I would expect someone to have tried similar things already, so there might relevant litterature on the subject.

I have yet to find the time to formally introduce myself to general relativity, diff. goem. and these curvature theories, so if you can tell me and calculate an expected shift as a function of field strength, I'll see if its possible for me to do this.

Just to make it clear to all (I don't follow everything that has been said in the thread), I'm looking for an electric curvature that is independant of mass (and I would rather not bother with magnetic effects if possible). If an electric field curves a massless particle, that's pretty much up the alley I'm loitering.

 

[Antonio Lao]

Assuming magnetic field is zero, then the localized charge needed for unit electric field along one component of the energy-momentum tensor is given by

q= \frac{8\pi T_{11}}{r}

as r approaches zero-point region, the charges needed approaches infinity. This type of disparity in the charge polarization cannot be found anywhere in the universe and also is technologically impossible to construct. Not even the infinite energy of spontaneous vacuum polarization can be harnessed in an effective time period. Because of the infinitesimal time interval and the Heisenberg's Uncertainty Principle, the infinite energy from the vacuum polarization of electric charges cannot be used to effect the curvature of spacetime. Therefore, spacetime is flat if devoid of mass but not devoid of polarized electric charges.

This seems to indicate that at a deeper level, charge and mass are really the composites of a positive and a negative spacetime curvatures where the curvature is given as the inverse of Planck length. These two structures of curved spacetime cannot be separated in order to create an electric potential comparable to the gravitational potential which causes spacetime curvature in the large domain of GR. GR is really a macroscopic theory while the infinite curvature of spacetime is a theory of quantum gravity.

 

[Gonzolo]

Thanks Antonio. "Therefore, spacetime is flat if devoid of mass but not devoid of polarized electric charges." If this is according to regular GR, and everyone agrees, I'm pretty happy.

 

[pervect]

Thanks Antonio. "Therefore, spacetime is flat if devoid of mass but not devoid of polarized electric charges." If this is according to regular GR, and everyone agrees, I'm pretty happy.

I'm not quite sure what this is supposed to mean. Certainly charge affects the space-time geometry, as per the Reissner-Nordstrom metric

http://encyclopedia.thefreedictionary.com/Reissner-Nordstr%F8m

Let me try and explain this in English.

The charge on a black hole, Q, affects the metric, or if you prefer, the curvature, of space-time. This means that _uncharged_ particles, following a geodesic, will not follow the same path near a charged black hole as they will near an uncharged one with the same mass.

I should also add that it is usually believed that Q < M (in geometric units) for the above metric.

Note that the stress-energy tensor (and hence the Einstein and Ricci tensors) are _not_ zero as they are with the Schwarzschild solution, because the electric field E acts as a source term in Einstein's field equations. I.e. electric fields generate curvature

G^a{}_{ b} = 8 \pi T^a{}_{ b} = {\begin{array}{cccc}<br /> - {\displaystyle \frac {Q^{2}}{r^{4}}} &amp; 0 &amp; 0 &amp; 0 \\ [2ex]<br /> 0 &amp; {\displaystyle \frac {Q^{2}}{r^{4}}} &amp; 0 &amp; 0 \\ [2ex]<br /> 0 &amp; 0 &amp; {\displaystyle \frac {Q^{2}}{r^{4}}} &amp; 0 \\ [2ex]<br /> 0 &amp; 0 &amp; 0 &amp; - {\displaystyle \frac {Q^{2}}{r^{4}}} <br /> \end{array}}<br />

Thus the various components of T^a _b are proportional to the square of the electric field E, as E=Q/r^2, which is what one would expect.

I'm afraid I can't make any sense of Antonio's second post, I'm not sure what he's trying to say.

 

[Antonio Lao]

I'm afraid I can't make any sense of Antonio's second post, I'm not sure what he's trying to say.

what i mean, without going into a lot of mathematical jargons, is that infinite mass of GR leads to a singularity but infinite spacetime curvatures of positive and negative structures from polarization of localized gauge fields (hence structures for electric charge, weak charge, and color charge) lead to a viable theory of quantum gravity.

Note:

GR deals with only one singularity but a workable quantum theory of gravity will have to deal with two singularities for both positive and negative curvature.

 

[selfAdjoint]

AFAIK ther's no infinite mass in GR. The singularity inside a black hole is a finite mass confined in zero volume; infinite density.

 

[Antonio Lao]

selfAdjoint,

Thanks for your clarification.

 

[Gonzolo]

"This means that _uncharged_ particles, following a geodesic, will not follow the same path near a charged black hole as they will near an uncharged one with the same mass."

This is interesting. Are these charges necessarily paired up as dipoles? Or does it include the case of a net charge (+ or -)? This would apply no only to black holes, but also to a body with finite and measurable volume (i.e. a balloon or static-charged balloon), wouldn't it?

 

[pervect]

"This means that _uncharged_ particles, following a geodesic, will not follow the same path near a charged black hole as they will near an uncharged one with the same mass."

This is interesting. Are these charges necessarily paired up as dipoles? Or does it include the case of a net charge (+ or -)? This would apply no only to black holes, but also to a body with finite and measurable volume (i.e. a balloon or static-charged balloon), wouldn't it?

The metric I gave the URL for (Reisner-Nordstrom, abbreviated as R-N from here on) describes a charged black hole, a black hole with a net charge. Physically, if the BH has a positive charge, the negative charges would have to be located somewhere. Such charges are _not_ included in the metric I gave, however.

You would use the R-N metric to describe the external gravitational field of a highly charged spherically symmetric body, just as you would use the Schwarzschild metric to describe the gravitational field of an uncharged spherically symmetric body.

You'd model the electric field of such a black hole as Q/r^2, (just like the columb field) in the R-N coordinates.

This is a purely classical solution - it doesn't include quantum effects, such as particle pair production from intense electric fields (the Schwinger critical field limit).

 

[Gonzolo]

Thanks, this is just about my level of talk when it comes to GR.

(although, I would tend to believe it is in great part because I haven't chosen to completely dive into it yet...the 1916 paper is the only formal litterature I have, and I can confirm that it is not the best for an introduction to the subject...)

 

[pervect]

what i mean, without going into a lot of mathematical jargons, is that infinite mass of GR leads to a singularity but infinite spacetime curvatures of positive and negative structures from polarization of localized gauge fields (hence structures for electric charge, weak charge, and color charge) lead to a viable theory of quantum gravity.

This looks like a case where the mathematical jargon might be more comprehensible than the English :-).

From what I can make out you are not doing an analysis based on General Relativity. Exactly what it's based on isn't too clear at this point - here's an opportunity for a simple description, such as - string theory, M theory, loop quantum gravity, or none of the above.

 

[Antonio Lao]

pervect,

Can You help me visualize the various intersections of a surface of positive curvature and a surface of negative curvature?

 

[pervect]

I'll assume that the answer is "other theory" until I hear different.

The usual visual aid/3d representation for/of a negative curvature is a saddle surface, for a positive curvature a sphere. So I guess you'd have to visualze a sphere intersecting a saddle surface, though I"m not sure why or what you are visualizing.

 

[Antonio Lao]

pervect,

Since both the saddle and sphere are both described by some sort of parametric equations of differential geometry (a subject of which I have little knowledge of), their intersections can become simultaneous solutions of their interactions when the parameters are quantitified as time, mass, force, or energy functions. Can I take this for granted? I am looking for some minimum geodesics. Perhaps, a geodesic that is equivalent to the Planck length?

 

[raodhananjay]

I am a statistician, I wanted to know whether light is a wave or a particle ? And what is light in string theory ?

 

[selfAdjoint]

I am a statistician, I wanted to know whether light is a wave or a particle ? And what is light in string theory ?

Short answers. Yes. and The Same.

Longer answer, since early in the twentieth century light has been viewed as having both particle and wave properties. Which property you see depends on what experiment you do; if your experiment concerns, say, frequency or interference you will detect waves, and if you do an experiment based on momentum, you will detect particles. This idea actually came out of experiments, not theory, and it was only later includied in the growing theory of quantum mechanics.

String theory builds particles at the wave level out of string vibrations, so it basically supports this wave/particle duality.

 

[juju]

This means that _uncharged_ particles, following a geodesic, will not follow the same path near a charged black hole as they will near an uncharged one with the same mass.

This should, in theory, be testable using neutral particles in a large electric field. If it is true for black holes it should be true everywhere.

Here's another thought.

The curvatures (positive and negative) do not need to be thought of as being in the same direction as the gravitational curvature. They can be looked at as being orthogonal to it.

They can also be looked as a twisting of space/time, rather than a curvature.

juju

 

[gptejms]

I have a question:-from the little bit that I know,any physical field except the gravitational is a part of T_{mu,nu} and causes curvature.Has any experiment
ever been done which shows that the electromagnetic field causes space-time
curvature?Are static fields from charges also included in T_{mu,nu}?

 

[raodhananjay]

Thanks for the reply. Sorry I would love fire in some more so please ...
In the duality priniciple , how does it explain an energy dissipation ?

Another question is as how does Light travel continously for an infinitely long distance (say sun to Earth )?

Atmosphere starts at some good distance above Earth's surface then in absence of vacuum , shouldn't the light get decayed ?
If you answer that its too powerful so it reaches us then shouldn't be too bright and hot above Earth atmosphere.

 

[Nereid]

Welcome to Physics Forums raodhananjay!

A suggestion if I may: your questions are good ones, but are not related to relativity, or to whether (and to what extent) electric charges curve spacetime - perhaps you could start a thread with these questions, in Quantum Mechanics?

In the duality priniciple , how does it explain an energy dissipation ?

It doesn't, directly; dissipation involves absorption (and, maybe, re-emission).

Another question is as how does Light travel continously for an infinitely long distance (say sun to Earth )?

As far as we can tell, there is no limit to how far light can travel. The furthest we've seen - and indeed, the further we can ever see - is the surface of last scattering, approx 13 billion light-years.

Atmosphere starts at some good distance above Earth's surface then in absence of vacuum , shouldn't the light get decayed ?

Some of the photons incident on the top of the Earth's atmosphere are indeed absorbed or scattered; astronomers measuring the apparent magnitude of a star include a correction - called 'air mass' - in their data reductions.

 

[pervect]

I have a question:-from the little bit that I know,any physical field except the gravitational is a part of T_{mu,nu} and causes curvature.Has any experiment
ever been done which shows that the electromagnetic field causes space-time
curvature?Are static fields from charges also included in T_{mu,nu}?

T_{mu,nu} = T_{\mu \nu} is the stress energy tensor, and as was previously mentioned, yes, the static electric field generates terms in the stress-energy tensor. As was also mentioned, the contribution is a trace-free one. This means that Baez's ball of perfectly electrical neutral coffee grounds

http://math.ucr.edu/home/baez/gr/outline2.html

don't change in volume (gravitate together) because of the electric field, as the trace of the stress-energy tensor determines R₀₀. However, the presence of the electric field does curve space-time.

 

[gptejms]

pervect,thanks for your answer.I repeat the second part of my question:-has any experiment ever been done which shows that the electromagnetic field causes space-time curvature?

 

[Nereid]

pervect,thanks for your answer.I repeat the second part of my question:-has any experiment ever been done which shows that the electromagnetic field causes space-time curvature?

If you don't mind, I'd like to ask a slightly different question ... with our current experimental capabilities, *could* we do an experiment which would show that an 'electromagnetic field causes space-time curvature'? In principle, what would an experiment to test the idea look like?

 

[gptejms]

Your question 'suggests' to me that with our present capabilities,we can't produce such high intensity electromagnetic fields---enough to cause any significant spacetime curvature.Can you give an order of magnitude calculation to give us an idea?An experiment to test the idea could look like this---if light bends in region of such a high intensity electromag. field(perhaps we could concentrate on just the magnetic field and try producing extraordinarily strong magnets)then we would know that the curvature has been produced.But I am sure there must be better ways around.

 

[pervect]

There are some relevant experimental tests, but they may not be very satisfying. For instance, Eotvos type tests comparing aluminum and gold.

Columb energy, (i.e. the energy in the electrostatic field), which is proportional to [nuclear charge]^2, amounts in a gold nucleus to .4 percent of the mass, but only .1 percent of the mass in an aluminum nucleus. Very sensitive experiments have been done to detect any differences in gravitational forces on aluminium and gold, (testing the principle of equivalence), but none have been found.

 

[Nereid]

Your question 'suggests' to me that with our present capabilities,we can't produce such high intensity electromagnetic fields---enough to cause any significant spacetime curvature.Can you give an order of magnitude calculation to give us an idea?An experiment to test the idea could look like this---if light bends in region of such a high intensity electromag. field(perhaps we could concentrate on just the magnetic field and try producing extraordinarily strong magnets)then we would know that the curvature has been produced.But I am sure there must be better ways around.

Let's think about this from the PoV of what the most intense EM fields and greatest spacetime curvature is, in the present universe, and later we'll examine whether any of these are amenable to tests - even in principle - of well-formed hypotheses. OK?

First, the most intense magnetic fields, in the present universe, are likely to be http://solomon.as.utexas.edu/~duncan/magnetar.html , whose fields are likely to be as many OOM stronger than the strongest we can generate here on Earth is greater than the Earth's own field.

Next, the most energetic gammas may well have a comparable 'curvature effect' to objects whose gravity we can measure. For example, we have 'seen' TeV gammas from the Crab pulsar (or nebula), and expect that GRBs emit PeV or higher gammas (such energetic gammas probably cannot propogate clear across the universe, but as we've 'seen' a couple of nearby AGN in TeV, perhaps a nearby GRB might be 'visible' in PeV gammas). Homework question: if a TeV gamma were converted (magically) into a lump of baryons (yes, it would be magic!), what would the mass of that lump be?

Next (2), it may be that a 'long duration' GRB results in the formation of black hole whose mass is several times that of the Sun. If the progenator star had a magnetic field, it may be that, in the last few microseconds before the BH formed, that field reached an intensity which makes a magnetar's field look like a fridge magnet. Too, the spacetime curvature would be far more extreme than that around the Sun (which has provided the most sensitive tests of some aspects of GR to date).

Finally, as the two neutron stars in a binary merge/collide (they lose energy as gravitational waves; without some external event, a collision is certain), many kinds of extreme environments will likely be created.

Closer to home, you might like to read the reports of the Fundamental Physics Working Group (it's in the middle of the page), from the recent ESA Cosmic Visions workshop. Do you feel that any of the proposed (local) experiments would test a hypothesis related to spacetime curvature resulting from EM?

 

[Gonzolo]

You lost me with GRB and AGN.

But surely, one could calculate the deviation of a beam of light due to an earth-built electric or magnetic field, even if the deviation is in order of femtometers or less.

I finally received my first book on tensors. GR is next. Hopefully, I'll see what the difficulty resides.

 

[Nereid]

You lost me with GRB and AGN.

Gamma ray burst, active galaxy nuclei (http://www.ulo.ucl.ac.uk/~diploma/year_one/heasarc.gsfc.nasa.gov/docs/objects/agn/agntext.html ).

But surely, one could calculate the deviation of a beam of light due to an earth-built electric or magnetic field, even if the deviation is in order of femtometers or less.

You mean, for example, moving a strong permanent magnet close to and away from a LIGO arm?

 

[raodhananjay]

Thanks NEREID . Sorry , you are right , I guess should be asking related questions. Let's start now, I had been inspired by one of the posts , hence I started my queries here. Otherwise although the header " electric .. " of this thread made me curious , I am kind of lost here. Ok Is Somebody ready to do small revision or introduction of things you do here. PLEASE somebody do that for me.

 

[Gonzolo]

Gamma ray burst, active galaxy nuclei (http://www.ulo.ucl.ac.uk/~diploma/year_one/heasarc.gsfc.nasa.gov/docs/objects/agn/agntext.html ).You mean, for example, moving a strong permanent magnet close to and away from a LIGO arm?

Exactly, but perhaps an electrically-generated magnet instead (MRI systems etc.). I would think that superconducting AC would generate the strongest human-made B-fields possible.

 

[Nereid]

Re-reading some of the posts near the start of this thread gave me the idea that perhaps we should be a little more precise about what any experiment is trying to test.

Gonzolo started this thread with a question about whether electric charge could 'curve spacetime', and in the last few posts we've been discussing whether a magnetic field can 'deflect' a beam of light. Would someone please be kind enough to explain why a non-null result in the latter would lead one to conclude 'yes, it can' for the former?

 

[Gonzolo]

I assume that in free space, all that can deflect a beam of light is space-time curvature. So whether the electric charge is moving (B-field) or not (E-field) can be seen as a detail (although certainly not for calculations). The question is for EM in general.

 

[juju]

Hi,

If EM fields curve space/time then time dilation effects should be seen. These would manifest in the changing of particle decay rates.

juju

 

[Calculex]

Re-reading some of the posts near the start of this thread gave me the idea that perhaps we should be a little more precise about what any experiment is trying to test.

Gonzolo started this thread with a question about whether electric charge could 'curve spacetime', and in the last few posts we've been discussing whether a magnetic field can 'deflect' a beam of light. Would someone please be kind enough to explain why a non-null result in the latter would lead one to conclude 'yes, it can' for the former?

I have a similar problem with the question. Curvature of space-time necessarily means that the frames of reference of all inertial observers would we affected. Since only charged masses would be affected by the presence of another electric charge, electrically neutral masses would measure time and space without being affected. How can this be a curvature of space-time?

Besides, Maxwell's equations state that the speed of propagation of an electromagnetic wave depends only on the values for \epsilon_0 and \mu_0 (the permittivity and permeability constants for space). These are experimentally derived and and neither of these constants depend on the magnitude of the respective electric or magnetic fields.

Calculex

 

[raodhananjay]

As light is an EMW I think (feel) the magnetic fields (extremely strong of course) may be able to deflect the light.

 

[pervect]

I have a similar problem with the question. Curvature of space-time necessarily means that the frames of reference of all inertial observers would we affected. Since only charged masses would be affected by the presence of another electric charge, electrically neutral masses would measure time and space without being affected. How can this be a curvature of space-time?

Uncharged masses *are*affected by the presence of electric charge (according to GR). This is why the Reissner-Norsdstrom metric for a charged black hole is different from the Schwarzschild metric for an uncharged black hole. Note the additional Q^2/r^2 terms in the Reissner-Nordstrom metric.

http://reissner-nordstrom-black-hole.wikiverse.org/

http://en.wikipedia.org/wiki/Schwarzschild_metric

This happens because electromagnetic fields are a form of energy, and thus contribute to the gravitatioanl field.

In GR, electromagnetic fields contribute to the stress energy tensor (the RHS of Einstein's field equations), which implies they contribute to curvature (the LHS of Einstein's field equations)

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

Now, the Einstein field equations are

Gmu,nu = 8pi Tmu,nu

Here Gmu,nu is the Einstein curvature tensor, which encodes information about the curvature of spacetime, and Tmu,nu is the so-called stress-energy tensor, which we will meet again below. Tmu,nu represents the energy due to matter and electromagnetic fields, but includes NO contribution from "gravitational energy".

 

[juju]

Hi all,

There is a problem when dealing with the EM energy of the field created by a charged particle,

If it is only potential energy activated by another charge, then there would be no effect on neutral particles and probably no curvature in the GR sense,

However, if it results from a scaler energy field, then even if the force vectors cancel from the fields of two or more charges, the scaler energy fields from all the particles still exist and would add, producing more curvature than would be calculated just from a resultant EM vector field.

juju

 

[Andrew Mason]

Uncharged masses *are*affected by the presence of electric charge (according to GR). ...

This happens because electromagnetic fields are a form of energy, and thus contribute to the gravitational field.

A charge in an electric field has energy which is the product of its charge and the electric potential of the field where the charge is located. The units of an electric field are Energy/charge (eg. volts or joules/coulomb). So is it correct to say that an electric field represents stored energy in the absence of another mass having a charge? I should think that there would be no energy represented by the field itself.

AM

 

[Stingray]

Yes, the field itself has a local energy density (at least in SR where such things are well-defined). Think of radiation for example.

Anyways, GR states that geometry couples to the stress-energy tensor, not "energy density," whatever that may be. Electromagnetic fields have a well-defined stress-energy tensor, so they do generate gravitational fields. They are usually very small, but were significant in the early universe. I think that the CMB experiments have verified this.

As for measuring the gravity of EM fields directly, I think that's hopeless for now. For order of magnitude purposes, it is ok to think of the mass as being due to an energy density. Then the effective mass is ~E/c^2. I don't think we can measure gravitational effects for things less than a few kilograms (I may be wrong!), which implies an energy of ~10^17 Joules. This must also be localized to within less than 10 cm or so. So the required energy density for a "tabletop" experiment is about 10^20 J/m^3. The electric field strength required is about 10^15 V/m. The requisite magnetic field strength is about 10^7 Tesla. Very large fields can be achieved for with special lasers, but they are so fleeting that I doubt any gravitational effect could be measured.

 

[pervect]

A charge in an electric field has energy which is the product of its charge and the electric potential of the field where the charge is located. The units of an electric field are Energy/charge (eg. volts or joules/coulomb). So is it correct to say that an electric field represents stored energy in the absence of another mass having a charge? I should think that there would be no energy represented by the field itself.

The field itself has energy, even if there is no charge in it. Consider two electric charges that repel each other. Bring them closer together. It requires work to do so. The energy to do this work is not lost, but it is stored as potential energy. Where is this energy stored? In the electric field.

The expression for the field energy of an electromagnetic field is given at


http://scienceworld.wolfram.com/physics/ElectromagneticField.html

Anyways, GR states that geometry couples to the stress-energy tensor, not "energy density," whatever that may be.

Yes, this is true.

For general information, the energy density (energy per unit volume) is one of the components of the stress-energy tensor.

If you represent a volume of space by a 4-vector orthogonal to the volume, the stress-energy tensor can be thought of as the density of momentum-energy per unit volume. Multiplying the stress energy tensor T_ij by the vector Vⁱ representing a volume element yields the energy momentum 4-vector contained in that volume.

 

[Nereid]

As for measuring the gravity of EM fields directly, I think that's hopeless for now. For order of magnitude purposes, it is ok to think of the mass as being due to an energy density. Then the effective mass is ~E/c^2. I don't think we can measure gravitational effects for things less than a few kilograms (I may be wrong!)

AFAIK, tests (and measurements?) of gravity are now into the 100 micron, 1 mg range (e.g. this interesting summary)

which implies an energy of ~10^17 Joules. This must also be localized to within less than 10 cm or so. So the required energy density for a "tabletop" experiment is about 10^20 J/m^3. The electric field strength required is about 10^15 V/m. The requisite magnetic field strength is about 10^7 Tesla.

10^7 Tesla = 10^11 Gauss, which is far, far above what's achieved here on Earth (compression via explosives); however, it's weak for a neutron star, and mere noise near a http://solomon.as.utexas.edu/~duncan/magnetar.html .

Very large fields can be achieved for with special lasers, but they are so fleeting that I doubt any gravitational effect could be measured.

Aye, that's the real challenge - how to build an experiment to measure a gravitational effect that's both weak and fleeting? Perhaps repetition and nulling (differencing) are the answers?

 

[Andrew Mason]

The field itself has energy, even if there is no charge in it. Consider two electric charges that repel each other. Bring them closer together. It requires work to do so. The energy to do this work is not lost, but it is stored as potential energy. Where is this energy stored? In the electric field.

You have merely pointed out that a charged particle in an electric field has potential energy. My point was that a static electric field which surrounds a charged mass contains no energy UNLESS you bring another charge into it.

When there is a charge in an electric field, that charge has electric potential and one can think of the energy being stored in the field, with its units being that of kQq/r. When there is no charge q in the electric field created by charge Q, I don't see how the field can have any energy.

One way to ask the question is this: when a charge is created, does energy go into creating its electric field? (e.g neutron decaying to proton, electron and antineutrino?). I don't know the answer to that. I am just wondering.

AM

 

[Nereid]

One way to ask the question is this: when a charge is created, does energy go into creating its electric field? (e.g neutron decaying to proton, electron and antineutrino?). I don't know the answer to that. I am just wondering.

It's interesting that 'charge' is created in pairs, in this example, and in all cases (?).

A somewhat similar situation is the 'destruction' of 'mass', e.g. annihilation (e^- + e⁺ -> 2 photons, for example) ... from the GR perspective there is no loss of 'that which creates spacetime curvature', merely conversion of one form (two particles) to another (two different particles, moving at c).

So perhaps a different way of asking this is whether the 'QM' conservation laws (e.g. charge, those 'built in' to the weak and strong forces) are 'built in' to GR?

 

[pervect]

This gets into some interesting but advanced territory. It's fairly well known that the field energy of a charged point particle (usually, this is talked about as the self-energy of the electron) is infinite.

Google finds the archived physicsforum post

https://www.physicsforums.com/archive/t-29348_Electrostatic_Field_Energy_of_Electron.html

as a previous discussion about this point.

This issue requires quantum mechanics to resolve - unfortunately, GR isn't a quantum theory.