Does an electric charge curve spacetime ?

[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.

 

[juju]

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..

This is only true if it is assumed that the field structure close to the convergence point of the field (the point particle) is the same as the field structure further away (the normal field structure.)

juju

 

[Garth]

This may be of interest:
"Space-time Curvature of Classical Electromagnetism"
http://arxiv.org/abs/gr-qc/0410043

Garth

 

[selfAdjoint]

Weyl's conformal tensor?

 

[pervect]

Weyl's conformal tensor?

It's the usual Weyl tensor, the traceless part of the Riemann. Someone explained to me once why it's sometimes called a "conformal tensor", but alas I don't recall the details.

http://mathworld.wolfram.com/RiemannTensor.html

reviews the breakdown of the Riemann into the Ricci and the Weyl tensors.

Unfortunately this is about as far as I've gotten with the paper so far, I don't understand the main point of it yet.

[edit]

The key term I'm missing the meaning of is
"the same support as the matter", it probably has something to do with homology which I don't know much about.
[end edit]

 

[Stingray]

The key term I'm missing the meaning of is
"the same support as the matter", it probably has something to do with homology which I don't know much about.

If you're asking what support means, that's the portion of a function's domain where it is nonzero. The sentence is saying that whatever it is discussing is nonzero only where there is matter.

The Weyl tensor is also called the conformal tensor because it is invariant under conformal transformations (multiplying the metric by a scalar function).

 

[pmb_phy]

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.

The energy of a electric field is proportional to the integral of the square of the E-field over the region that the field doesn't vanish. Following the derivation which leads to that conclusion leads one to believe that what you say is true. However if it were true then this proves to be troublesome. It would seem to imply that the field had no momentum. But if the field has no momentum then the principle of momentum conservation would be violated.

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.

There are various ways to express EM energy, not all of which include the integration mentioned above. As such there are different ways to look at the energy involved. Shadowitz covers this in his EM text.

To be quite literal - Look at how the association of energy<-> field came to be. One starts with a single point charge and assigns a total potential energy of this isolated system to be zero. One then brings in a charged particle from infinity to a nearby point. The work done will equal the change in the potential energy of the system. It makes no sense to say "The potential energy is at this place..etc". Bring in more charges - there will be more potential energy. Now let the charges become infinitesimal and let the number of them approach infinity. The discrete charge distribution then becomes a continuous charge distribution. Then the potential energy can then be expressed as being proportional to the integral of the square of the E-field over all space. If one then applies this relation to the field of a point charge then the integral diverges. What this means physically is that you tried to assemble a point charge. The work required to do that would be infinite. But now recall what this energy means. We *defined* the energy of a point charge to be zero. Thus it would seem reasonable to hold that the energy associated with the E-field of a point charge is zero.

However I have been unable to justify this assertion since when I attempted to do so it led to other problems (I can't recall exactly but I do remember trouble in doing so).

Pete

 

[4newton]

No Difference

All observation indicates that there is no basic difference between forces, charge, mass, magnetic. All these forces act on the mechanisms that produce the forces in the same way. They all act over infinite distance. The only difference is the strength of the forces and their geometry.

 

[Nereid]

Summary, so far

Q: "Does an electric charge 'curve spacetime'?"
A: GR predicts this, however the effect is way, way smaller than anything we can measure (or even detect) with 'lab experiments' here on Earth.

So, remaining questions? My €0.02's worth:
- what are the experimental limits on 'spacetime curvature' by electric (or magnetic) charges?
- what astronomical observations or space-based experiements, in principle, could be made/done to measure the GR predicted effect?
- how might the effect be indirectly detected (other than by experiments and observations which test GR in general)?

 

[pervect]

I'd add "anything we can detect directly" to your summary. Part of the mass of gold and aluminum atoms is due to their electromagnetic field, so the fact that there is no difference in the Eotovos experiments for these two elements tells us *something* about gravitation and energy.

One could also try and question whether or not gold and aluminium have the predicted mass change due to their electromagnetic energy. I'm not aware of the experimental details here, but it looks to me like there is enough precision in the measurements where the total electronic binding energy should be measuarable - though it may be tricky to separate from the nuclear binding energy.

 

[Nereid]

I'd add "anything we can detect directly" to your summary. Part of the mass of gold and aluminum atoms is due to their electromagnetic field, so the fact that there is no difference in the Eotovos experiments for these two elements tells us *something* about gravitation and energy.

One could also try and question whether or not gold and aluminium have the predicted mass change due to their electromagnetic energy. I'm not aware of the experimental details here, but it looks to me like there is enough precision in the measurements where the total electronic binding energy should be measuarable - though it may be tricky to separate from the nuclear binding energy.

Thanks Pervect, I'd forgotten your earlier comment on this. IIRC, these experiments (there've been several, over the decades) have just the results predicted from GR (to parts per thousand? or better??), but, as you say, the interpretation is more about nuclear binding energy than electronic ... hmm, worth hunting down some of the papers?

 

[pervect]

I suppose the other thing we ought to mention is Mallet's idea for detecting the gravitational effect of light. The problem is that I'm not sure whether or not his idea will work. It appears (see the Mallet thread) that Mallet's analysis has some major problems, I expect the time machine results to disappear when these are corrected, but it's unclear to me whether or not the rest of his results will likewise vanish.

 

[Andrew Mason]

One could also try and question whether or not gold and aluminium have the predicted mass change due to their electromagnetic energy. I'm not aware of the experimental details here, but it looks to me like there is enough precision in the measurements where the total electronic binding energy should be measuarable - though it may be tricky to separate from the nuclear binding energy.

Is this not just a matter of proving the invariance of the speed of light? Einstein proved mathematically that the principle of relativity implies that the absorption/release of energy by a body increases/reduces its mass by E/c²

So, in a real sense, Michelson Morley confirms that electronic binding energy contributes to the mass of the atom.

But, getting back to the question, the issue here is whether a single electric charge curves space time. A body with electric charge of mass m and charge q and a non charged body of mass m will, according to GR, curve space time in exactly the same way.

Now, to the extent that the charge q contributes to the body's energy, and therefore its mass, the charge q will contribute part of that mass, m. Hence, it would contribute to the curvature of space-time. But my point is that I don't see how a single charge contributes energy. You need two charges separated by a distance in order to have electrical energy.

Andrew Mason

 

[pervect]

Is this not just a matter of proving the invariance of the speed of light? Einstein proved mathematically that the principle of relativity implies that the absorption/release of energy by a body increases/reduces its mass by E/c²

So, in a real sense, Michelson Morley confirms that electronic binding energy contributes to the mass of the atom.

I would be very, very, very surprised if electronic binding energy didn't obey the standard formula E=mc^2.

That said, I think one needs a stronger statement than the constancy of the speed of light to conclude that m=E/c^2. One needs at a minimum to assume that *all* physical laws, not just the speed of light, are the same for moving observers and stationary observers. I'm not positive if even that's sufficient - I think one has to assume additionally that the laws of physics are given by an action principle, and that the laws of physics do not change with time, or with location in space. The last two assumptions are more-or-less necessary for the laws of physics to be the same for differently moving observers, but we might as well be explicit about all the assumptions up front.

But, getting back to the question, the issue here is whether a single electric charge curves space time. A body with electric charge of mass m and charge q and a non charged body of mass m will, according to GR, curve space time in exactly the same way.

Look up the metric for a charged black hole. Look up the metric for an uncharged black hole. [I believe I've posted links to them elsewhere in the thread BTW.]. If what you said above were correct, then the metric of a charged black hole would have to be the same as the metric of an uncharged black hole. But they metrics are not the same.

 

[Chronos]

This is all consistent with the equivalence principle. A charge, or magnetic field has the same gravitational effect as the mass equivalent of the field strength, which is obviously a very minute effect and probably unmeasureable by any technology I can imagine.

 

[Andrew Mason]

That said, I think one needs a stronger statement than the constancy of the speed of light to conclude that m=E/c^2. One needs at a minimum to assume that *all* physical laws, not just the speed of light, are the same for moving observers and stationary observers.

Einstein made only one basic assumption: that the laws of electro-magnetism were the same to all inertial observers. I am not aware of any other assumption that is necessary to his conclusion "m=L/c²"

Look up the metric for a charged black hole. Look up the metric for an uncharged black hole. [I believe I've posted links to them elsewhere in the thread BTW.]. If what you said above were correct, then the metric of a charged black hole would have to be the same as the metric of an uncharged black hole. But they metrics are not the same.

I am not suggesting the analysis of black holes might lead to such a result. I don't pretend to understand the math involved in GR well enough to question it. I am just saying that I don't understand the physical basis for charge alone having energy. I was hoping someone might explain how it does.

AM

 

[pervect]

I am not suggesting the analysis of black holes might lead to such a result. I don't pretend to understand the math involved in GR well enough to question it. I am just saying that I don't understand the physical basis for charge alone having energy. I was hoping someone might explain how it does.

AM

The point isn't that charges have energy - the point is that fields have energy. Or, if you prefer, fields act just exactly as if they have energy.

The simplest discussion I could find in any of my textbooks was not particularly simple, and was based on Maxwell's equaitons.

Look for "Poynting's theorem" in an E&M textbook.

It starts with the vector calculus identity

\nabla \cdot (A \times B) = B \cdot (\nabla \times A) - A \cdot (\nabla \times B)

now we let E = A and B = H. Then we get

\nabla \cdot (E \times H) = B \cdot (\nabla \times E) - E \cdot (\nabla \times H)

and we substitute

\nabla \times E = -\frac{\partial B}{\partial t}
\nabla \times H = J + \frac{\partial D}{\partial t}

We wind up with

- \nabla \cdot (E \times H) = E \cdot J + (E \cdot \frac{\partial D}{\partial t} + H \cdot \frac{\partial B}{\partial t})

This can be expressed in intergal form

-\oint (E \times H) \cdot n \, da = \int_v E \cdot J \, dv + \int_v (E <br /> \cdot \frac{\partial D}{\partial t} + H \cdot \frac{\partial B}{\partial t} ) \, dv

The right hand side is the rate at which work is being done on the charges. the power.

Power = Voltage * current = (E * dl) . (J * da) = (E . J) dv is the most obvious example.

We add to this two other terms - a similar voltage*current term, but with the displacement current, and finally the rate at which magnetic fields do work.

The left hand side is the Poynting vector, the surface integral of which gives the amount of energy being transferred into the volume.

The right hand side can be re-written in the usual isotropic media, where D = eo E and B = u0 H as

\frac{d}{dt} \int_v (\frac{1}{2} \epsilon E^2 + \frac{1}{2} \mu H^2) \, dv

And this term can therefore be interpreted as the total energy stored in the volume V.

Which is why the total energy is proportional to e E^2 + u H^2.

Which can also be seen to be the right answer from the bare result presented at:

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

 

[Andrew Mason]

The point isn't that charges have energy - the point is that fields have energy. Or, if you prefer, fields act just exactly as if they have energy.

The simplest discussion I could find in any of my textbooks was not particularly simple, and was based on Maxwell's equaitons.

It looks to me that you are describing time dependent electromagnetic fields. Time dependent EM fields, of course, propagate as EM radiation. So it is apparent that time dependent electro-magnetic fields represent energy. The Poynting vector represents energy flow rate through a surface.

But I thought we were talking about non-time dependent electric field of a point charge.

If so, we are talking about a situation in which current densities are 0, charge density, \rho \rightarrow \infty. So \frac{\partial B}{\partial t} = 0 and \frac{\partial E}{\partial t} = 0 and \frac{\partial D}{\partial t} = 0

In a real situation, the charge density is not infinite, so there is some charge distribution in a finite volume. This represents energy because we have charges separated by distances. The energy of such a charge distribution will, therefore, contribute mass to the charges.

AM

 

[pervect]

It looks to me that you are describing time dependent electromagnetic fields. Time dependent EM fields, of course, propagate as EM radiation. So it is apparent that time dependent electro-magnetic fields represent energy. The Poynting vector represents energy flow rate through a surface.

But I thought we were talking about non-time dependent electric field of a point charge.

The Poynting vector does indeed describe the rate at which energy flows, that's on the left hand side.

The right hand side has two terms. The first is just voltage*current (the volume intergal of E . J dv. The second term is

\frac{d}{dt} \int_v (\frac{1}{2} \epsilon E^2 + \frac{1}{2} \mu H^2) \, dv

When there is no energy flow, this term is zero. When there is energy flow, energy is being either stored or released from storage.

One can (and does) interpret this term as the energy stored in the electromagnetic field.

Consider a capacitor being charged, for instance, and look at the above equations to see where it is going.

We know that power is flowing into it (V * I). There is no actual current flow in the capacitor through the dielectric - J is zero - the capacitor is not dissipating any energy (like a resistor does).

But while J is zero, there is still a displacement current \frac{\partial D}{\partial t} through the dielectric. This current represents energy that is being stored in the capacitor. You can think of this energy as being stored in the electric field of the capacitor with a magnitude of .5*e0*E^2 per unit volume between the plates of the capacitor. (Plus possibly some small amount of energy stored in the fringing fields).

It's fairly clear that a capacitor does store energy. The above formula gives the right answer for the amount of energy a capacitor stores.

A charged capacitor is a "static" system, but it still has stored energy, waiting to be released - just connect a load across it, and the presence of the energy will make itself manifest.

The other term involving H^2 (often rewritten in terms of B^2 which is proportional to H^2) gives magnetic energy storage. This describes the energy stored in an inductor.

Note that as I mentioend before, there are issues with point charges. At zero radius, the energy in the electric field is infinite using classical formulas. The classical electron radius is the radius at which the energy in the electric field of the electron is equal to it's mass. It's about 2.8*10^-15 meters.

see for instance

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

GR doesn't deal well with point charges, just as classical electrodynamics doesn't. If one is looking at charges at distances much greater than the classical electron radius, GR and classical electrodynamics will give a good answer to the problems one formulates. One can approximate electrons as having the classical electron radius, rather than being point particles.

If one is looking at distances closer to an electron than the classical electron radius, one probably needs a different theory - quantum electrodynamics, or quantum gravity, depending on whether one is looking at the electric or magnetic field.

 

[Andrew Mason]

When there is no energy flow, this term is zero. When there is energy flow, energy is being either stored or released from storage.

One can (and does) interpret this term as the energy stored in the electromagnetic field.

There is no question about that. A charged capacitor represents a charge distribution - charges separated by distances. The energy is stored in the electric field between the charges. But the field surrounding a charged capacitor should be essentially 0 (there is a small electric dipole effect because the + and - charges are separated by a distance). We are talking about a large charge (+ or -) sitting in space all by itself.

It appears from all of this, that:

1. A quantity of charge Q distributed within a finite volume of space V contains energy, E.

2. This energy, E contributes to the mass within that volume of space, V by the factor m_Q=E/c²

3. E is contained in the electric field within V

4. This mass contributes to the curvature of space time external to V (ie it is caused by the total mass contained in V, which includes m_Q).

5. There is zero energy in, and 0 mass contributed by, and 0 contribution to the curvature of space-time by, the electric field in the space outside of V.

The classical electron radius is the radius at which the energy in the electric field of the electron is equal to it's mass. It's about 2.8*10^-15 meters.

If one is looking at distances closer to an electron than the classical electron radius, one probably needs a different theory - quantum electrodynamics, or quantum gravity, depending on whether one is looking at the electric or magnetic field.

Isn't the calculation of the energy of the electric field of the electron based on the classical model of a volume of space (the size of an electron) having point charges distributed throughout that volume?

AM

 

[pervect]

There's one other point I should mention. The stress-energy tensor, the right hand side of Einstein's equation, is actually a 4x4 matrix. In many situations, only one term of this tensor, the energy density per unit volume, T₀₀, is important. This is not the case for the electromagnetic field.

For the case of an electrostatic field, additional diagonal terms T₁₁, T₂₂, T₃₃ can be as large as the energy density term, T₀₀.

You'll see negative terms (tension) along the diirection of the field lines, but positive terms (pressure) in the two directions perpendicular to the field lines. The off diagonal terms will be zero for an electrostatic field (they are given by the Poynting flux).

These pressure and tension terms also cause gravity.