Does an electric charge curve spacetime ?

[Garth]

We are on different wavelengths here.

I'm a bit surprised you don't see why I've been talking about the observer at infinity.

Earlier

I'm surprised this issue hasn't been resolved. It seems to me that the static nature of the charge relative to the observer at infinity means that it should not radiate from the viewpoint of the observer at infinity.

But the observer in question is the one in the local gravitational field - in the laboratory. Indeed with the Earth itself in the Sun's/galactic/intergalactic gravitational fields is not the concept of an observer co-moving at inifinity rather hypothetical?

The reason I brought up the observer at infinity was to answer the second part of your question. Do I really need to go through the whole spiel on asymptotic flatness and energy in GR again? I will if it serves some useful purpose - If I recall correctly you have your own theory with it's own view on energy conservation, but I'd hope you'd be interested in understanding the mainstream view. I believe I'm presenting the mainstream view reasonably fairly, but I'm not, alas, infallible. Anyway, if you want me to clarify this or talk about it more I will, but I'm hoping that pointing out my previous remarks on this topic will be enough.

It is the physical reality that I am trying to understand, whether that is modeled by the mainstream view, or mine, or anybody else's individual theories.
GR is an example of Noether's improper energy theorems and conserves energy-momentum and not in general energy. If a time-like killing vector exists it is possible to define a concept that behaves like energy, the covariant time-component of the 4momentum vector but in many cases it is the contravariant time component that is defined as energy, especially when considering the total energy of a static gravitating body and field as measured by an observer ‘at infinity’. The situation is confusing because energy is not conserved in GR and our natural inclination to want it be so forces an unnatural definition on the theory.

Oh, yes, I guess I haven't mentioned what I see as "the solution". The main solution is that the detection or non-detection of radiation is observer dependent, it's not a physical invariant. It's also not strictly speaking a local pheomenon at all. Google finds (amusingly enough) pmb's webpage with a wide variety of quotes from the literature pointing out the observer dependent nature of the existence of radiation

http://www.geocities.com/physics_world/falling_charge.htm

Thank you for Pete’s link. If I can quote from the sources he quotes;
1. ‘Classical Radiation from a Uniformly Accelerated Charge, Thomas Fulton, Fritz Rohrlich, Annals of Physics: 9, 499-517 (1960)’
"An electron which falls freely in a uniform gravitational field embedded in an inertial frame will radiate, and one which sits at rest on a table in the same field will not radiate; and these two statements do not contradict each other."
2. ‘Radiation from an Accelerated Charge and the Principle of Equivalence, A. Kovetz and G.E. Tauber, Am. J. Phys., Vol. 37(4), April 1969’
“A nonvanishing energy flux is found only if the charge is freely falling and the observer supported, or vice versa”
So 2 is saying that an inertial observer i.e. freely falling – with no forces acting on her – will observe radiation from the desk bound (supported) charge. Whereas 1. says not, the observer in case 1 is presumeably supported.
But again whence the energy? The fact the GR does not conserve energy does not in itself explain where the energy received by the inertial observer comes from.

The situation is confused, which is why Kirk T. McDonald in “Hawking Unruh radiation and radiation of a uniformly accelerated” (Pete’s link) concludes, “We
now see that the quantum view is richer than anticipated, and that Hawking- Unruh radiation provides at least a partial understanding of particle emission in uniform acceleration or gravitation.”

A partial understanding is better than none – but the case is not closed!

Garth

 

[pervect]

I started to write a rather long and technical response, but it seems to me that I simply wind up repeating myself.

It's really very simple. If one is talking about system energy in GR, one is(or should be) talking about an asymptotically flat space-time to apply the usual formulas.

Let's take a very contrived example. Suppose we have a universe that consists of a single star (with optional planets), and a rocketship.

If the rocket accelerates, the star will move faster and faster. It will have more energy. The rocket's exhaust will have to exist, and it will have some energy, but by taking the limit where the star is a lot bigger than the rocket, it is obvious that the total energy of the rocket exhaust can be ignored, as can the rocket itself. So, in the limit, the energy of the universe from the viewpoint of the rocket is going up.

This is not a GR issue. The same thing happens in Newtonian mechanics. The only issue is, that we can no longer simply say "the rocket is not in an inertial frame" the way we used to in Newtonian mechanics. So we have to do something more complicated.

There's a perfectly good defintion for energy in GR with asymptotically flat space-times. You can also define energy in GR when you have only a time-like killing vector without asymptotic flatness, to some extent, but you have a calibration problem witht the formulas I've seen. (It's easy enough to come up with a conserved quantity when you have a time-like killing vector, but it's hard to scale this quantity correctly without an asymptotically flat space-time to serve as a reference. A conserved quantity remains conserved when multiplied by an arbitrary constant - setting the value of the constant in the usual manner requires asymptotic flatness).

In short, to define system energy, it is sufficient to have an asymptotically flat space-time (and it's close to necessary, as well as being sufficient).

If you really do want to understand reality, you should take a look at how asymptotic flatness defines energy in GR. You come up with oddball defintions of energy, then complain that they don't act the way you want. At least use one of the accepted notions of energy (such as Bondi mass, or ADM mass) - then complain that *they* doesn't work the way you want .

As far as practical issues go, they aren't as bad as you think. Compare and contrast

acceleration of Earth's sun due to galactic rotation (250 million year rotation period, 26,000 light years from center): 10^-10 m/s^2

moon's orbital accelration around earth: .002 m/s^2
earth-moon's acceleration around sun: .006 m/s^2

You're welcome to work out the galaxies acceleration due to the local group, if you can figure out how - I think you'll find that it's smaller than the sun's acceleration due to galactic rotation, so there's a general downward trend.

The trend is generally downward for the same reason that the night sky is black. (Of course, the reason the night sky is black isn't particularly obvious).

 

[Garth]

If the rocket ship in your example turns off its engine and free falls towards the star then in its frame of reference the star/universe's energy continues to increase.

In GR energy is not conserved, in general, even for inertial observers.
That is what I have been saying all along! It is one of the starting points of SCC.

However the issue with the supported charge and free falling observer is whether the radiation actually exists and can be measured. Therefore, can energy be taken out of the system? Is any work being done, either by the support of the charge or the deviation of the observer from its geodesic - and if not whence the energy? In other words, is this a free lunch?

I am glad to see you seem to have the situation sorted, others do not!

Garth

 

[pervect]

I'm going to take another shot at the more technical approach. But I've said it all before. I can still hope that it will "make sense" this time.

If a time-like killing vector exists it is possible to define a concept that behaves like energy, the covariant time-component of the 4momentum vector but in many cases it is the contravariant time component that is defined as energy

It's true that E₀ is a constant of motion for particles following a geodesic. But this is a red herring. It does not lead to good defintion of the energy of a system. This should be no surprise, it's not even coordinate independent.

I'm assuming that is what we are interested in, yes? The energy of a system, not a constant of motion of a particle following a geodesic?

Several good and equivalent defintions for the energy M of a system with a timelike killing vector k^a are:

a surface intergal
$$M = -\frac{1}{8 \pi} \int_S \epsilon_{abcd} \nabla^c k^d$$
here $$\epsilon_{abcd}$$ is the Levi-civita tensor normalized to be a volume element of the space-time, this is the surface intergal of a two form.

a volume intergal
$$M = \frac{1}{4 \pi} \int_{\Sigma} R_{ab} n^a k^b dV$$

here R_ab is the Ricci, and n^a is a unit future perpendicular to the volume element dV.

$$M = 2 \int_{\Sigma} (T_{ab} - \frac{1}{2} T g_{ab}) n^a k^b dV$$

here T_ab is the stress energy tensor, g_ab is the metric tensor, and n^a remains the unit future perpendicular to the volume element dV.

Note that this last expression illustrates why you can't just intergrate T_ab overe a volume of space and expect to come up with a system energy . (Except as an approximation that's only valid in the weak field case).

These can all be found in Wald, pg 288-291

Now that we've talked about the easy case of a system with a timelike killing vector, let's talk about the harder case - what if you don't have a timelike killing vector.

Well, the idea is simple. If you don't have a killing vector that's timelike everywhere, maybe you have a killing vector that's asymptotically timlike, at infinity. We can then apply the first definition directly. If turns out that if you have an asymptotically flat space-time that's a vacuum at infinity, you do have killing vectors which are asymptotically timelike. To formulate this rigorously requires a defintion of conformal infinity. In this case, we also have to specify "which infinity" (it's null infinity, in the jargon of conformal infinity). It turns out

Fortunately ... the asymptotic symmetry group of null infinity has a preferred 4 parameter subgroup of translations, so the notion of "an asymptotic time translation" is well defined.

So, if I may conclude with a few words

asymptotic flatness, asymptotic flatness, and asymptotic flatness, are the three keys to energy conservation in standard GR.

 

[Garth]

Agreed - which goes to show how slippery a concept energy is in GR.

However for ordinary energy 'book keeping', in experiments in the laboratory etc. we do not have the luxury of being able to integrate over the whole gravitational system out to an asymptotically flat null infinity.
We do experiments within a limited volume in which we keep an energy account and quite naturally expect energy to be conserved in such a closed system. It is in these cases that it is possible to question whether the approach of GR is adequate or not, and the detailed study of a supported electric charge in a gravitational field may be helpful in answering that question.

Garth

 

[Garth]

Hi Garth,

I was thinking of a situation where each pair of charges only curved the space between the charges. Thus, in the aggregate these would cancel, but still exist locally.

juju

Hi juju - your question and ideas have become a little lost in the discussion here!

You are taking the same approach as Einstein did when he tried to formulate a unified field theory. He wanted to explain the electromagnetic force in the same way as he explained the gravitational force by space-time curvature only he couldn't get it to work.
There is of course repulsion as well as attraction to explain together with the electric and magnetic fields. As pervect said earlier there is another theory, the Kaluza-Klein theory, which uses an extra dimension, and this helped towards developing string theory.
So yours are good fertile ideas.
Garth

 

[selfAdjoint]

You are taking the same approach as Einstein did when he tried to formulate a unified field theory. He wanted to expalin the electromagnetic force in the same way as he expalined the gravitational force by space-time curvature only he couldn't get it to work.

See http://www.einstein-schrodinger.com/ . He got it to work. He makes some striking assumptions (a very large cosmological constant, etc.) but it's all legitimate EUFT.

 

[Garth]

See http://www.einstein-schrodinger.com/ . He got it to work. He makes some striking assumptions (a very large cosmological constant, etc.) but it's all legitimate EUFT.

Thank you for that very interesting link, I shall download E & S's original papers and study them. Do you have any idea why, "This was supposedly disproven way back in 1953, but there are a few stubborn souls such as myself who still think it is correct, and who work to prove it", it was supposedly disproven?

A large cosmological constant might tie in with present questions about DE for example.
Garth

 

[juju]

Hi Garth,

Here's are some link to information about other theories that unify gravity and electromagnetics within the general framework of GR.

http://www.americanantigravity.com/davidmaker.shtml

http://www.compukol.com/mendel/

http://www.aias.us/

Also, I understand that Roger Penrose also believes that the path to a unified field theory lies through GR, rather than QM.

Thanx

juju

 

[selfAdjoint]

Thank you for that very interesting link, I shall download E & S's original papers and study them. Do you have any idea why, "This was supposedly disproven way back in 1953, but there are a few stubborn souls such as myself who still think it is correct, and who work to prove it", it was supposedly disproven?

A large cosmological constant might tie in with present questions about DE for example.
Garth

Neither Einstein not Schroedinger, I believe, were able to derive Maxwell's equations as a limiting case of their theories. But also consider Hlavaty's Geometry of Einstein's Unified Theory circa 1958, which is cited in the papers I linked to. Hlavaty claimed to have derived the "Maxwell" (i.e. Faraday) tensor, which turned out to be not at all obvious.

 

[pmb_phy]

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? Both are $$F = k/r^2$$ in elementary physics.

If there is a configuration of charges which has an associated energy distribution then the mass associated with that energy will curve spacetime. I am not conviced that a single charge can do that since the energy of a charge distribution is defined as the potential energy of the distribution and that energy is the energy required to assemble the distribution. One does not assemble a point charge. As a matter of fact if one follows the derivation of the ennergy in the electric field then one sees that it starts by assembling point charges by bringing them from infiity to a finite distance to each other. One starts with a single point charge and a starting energy of zero. One then ends up with a relationship for the energy in terms of a sumation. One then takes the limit and goes to an integral and one can then show that there is a one to one relationship between the total potetial energy (i.e. energy required to assemble the distribution) and the E field. But one assumes that if there are finite discrete charges then the initial energy of the configuration is zero. So there is no energy in the field of a point charge - At least in my opinion ... today ... as of 7:41am. However I had some percocet not to long ago so it might be the drugs.

I can justify it though. If one takes a dumbell consisting of two identical point charges held together by a rod then the momentum of the rod at low speed will be p = Mv where

M = 2m_e + m_rod + m_em

The first term represents the bare mass of the charges, the second the bare mass of the rod and the last is the mass of the EM field as measured in the rest frame. If the mass of the EM field includes the energy of the point charges then the mometum is zero. If the mass of the EM field includes only the mutual potential energy then the mass is finite. This makes sense for many reasons. If this actually true then EM field of a point charge carries no momentum either. For details see Griffiths and Owen's paper Mass renormalization in classical electrodynamics, Am. J. Phys.

A postulate could be that an electron in an elevator (made of electons, or a negatively charged inside surface) cannot tell the difference whether :

1. the elevator is stopped and that there is a large + charge underneath or :

2. the elevator is accelerating upwards

Where does this lead? What happens if you take magnetism into account?

I have no idea what your saying here but I will say this. 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

 

[Gonzolo]

I have no idea what your saying here but I will say this...Pete

Just trying to make an electric analogy with a mass in an elevator, which cannot tell whether it is accelerating or near a planet. I have pretty much given up on this (idea and thread) for the time being, due to some of the points that were mentioned + the fact I believe that there is a strict asymetry between masses and charges : m -> $$\gamma m$$, while q -> q (invariant?). So the analogy has many limits.

 

[pmb_phy]

Just trying to make an electric analogy with a mass in an elevator, which cannot tell whether it is accelerating or near a planet. I have pretty much given up on this (idea and thread) for the time being, due to some of the points that were mentioned + the fact I believe that there is a strict asymetry between masses and charges : m -> $$\gamma m$$, while q -> q (invariant?). So the analogy has many limits.

As I've explained, you can distinguish whether you're accelerating or near a planet (i.e. Schwarzschild spacetime, curved spacetime etc.). Just weigh the charge. The weight is spacetime curvature dependant.

Pete

 

[pervect]

Agreed - which goes to show how slippery a concept energy is in GR.

However for ordinary energy 'book keeping', in experiments in the laboratory etc. we do not have the luxury of being able to integrate over the whole gravitational system out to an asymptotically flat null infinity.
We do experiments within a limited volume in which we keep an energy account and quite naturally expect energy to be conserved in such a closed system. It is in these cases that it is possible to question whether the approach of GR is adequate or not, and the detailed study of a supported electric charge in a gravitational field may be helpful in answering that question.

Garth

From a practical or experimental point of view, I think the problem is more or less the opposite - finding strong enough fields / high enough curvatures that these effects manifest.

For instance, the weak-field approximation of integrating T_ab to get the total energy is "good enough" to handle anything in the solar system, via the PPN approximation even though it's not the actual correct expression for the total energy of a system. Really strong fields (like those near a black hole), would be needed to find any departure even in theory.

Furthermore, there are experimental problems in finding, for instance, T_ab in the Sun experimentally to carry out the intergration to verify the above statement. It's much more practical to compute the mass of the sun by observing satellite orbits and studying it's far gravitational field than it is to actually measure its stress-energy tensor, or even its conventional density, everywhere in its interior and integrate.

 

[nrqed]

Whether or not radiation exists depends on the coordinates, it's not a physical invariant. Photon number is conserved by the Lorentz boost, but not by arbitrary coordinate changes. It's possible for the observer accelerating and co-moving with the electron not to see any radiation, while another observer accelerating with respect to it will see radiation.

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

 

[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

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

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?