# Is 'charged black hole' an oxymoron?

1. Jun 7, 2012

### Q-reeus

This is a fork-off from someone else's recent thread that seems destined to languish without response in it's new home. Although the initial query there was from a QED angle, the issue of just how or whether a charged BH makes sense needs tackling from GR angle. The established view evidently is that externally observed net charge is invariant wrt whether infalling charged matter is exterior, at, or 'inside' the EH (event horizon). In other words, from a coordinate perspective and presumably quite generally, charge invariance holds for BH's, as determined by appropriately applying Gauss's law to a static bounding surface enclosing any infalling charge and the BH proper. Below are situations imo casting doubt on that position:

1: Matter of proper mass m gently lowered towards the EH reduces in coordinate measure as m' = fm, with f the usual redshift expression f = √(1-2GM/(rc2)). In keeping with that conservation of energy applies and work is being extracted in the lowering process. Now suppose that m also carries a charge q. it makes no difference to the net reduction in m as all forms of energy reduce the same. Locally there is no variation in the proper charge-to-mass ratio q/m. How can that local invariance of q/m (no free-fall case) not be also reflected as remotely observed - i.e. q' = fq? Certainly the locally invariant q/m will show remotely as a proportionately equally redshifted reduction in the Newtonian gravitational and Coulombic forces of attraction/repulsion between two adjacent such charged masses. The implication is obvious - as vanishes coordinate mass, so vanishes coordinate charge.

2: Gently lowering a charged flywheel (or counter-rotating pair to avoid 'twistup'), while feeding it with power so as to maintain constant externally observed mass. Only advantage of this scenario is that explicitly the locally determined proper q/m ratio steadily declines - a direct consequence of locally observed charge invariance. Which as for example 1: implies the failure of global charge invariance whenever gravity is involved.

3: Now for a 'realistic' BH case, one cannot gently lower mass/charge indefinitely and free-fall is the obvious scenario to consider. Coordinate m is then invariant, as potential energy steadily and conservatively converts to KE. Proper q/m is still invariant but that free-fall determination is surely not the right one to apply here. A hovering observer close to the EH, and defining a fixed location on a bounding Gaussian surface of integration, sees a highly relativistic mass infalling past with a small and continually shrinking q/(γm) ratio, where γ = 1/√(1-(v/c)2) is the usual SR expression locally observed. All-in-all then the quasi-hovering scenarios of slow lowering and that of free-fall dovetails together and imo leads to the same conclusion; asymptotic vanishing of externally observed charge as EH is approached.

While the extreme case of BH implies the asymptotic vanishing of all infalling charge, a finite redshift factor should apply more generally. Hence for a spherical capacitor, where the inner surface is that of a gravitating mass, one expects from the foregoing a net field exists exterior to the outer spherical shell, owing to greater gravitational depression of the charge on the inner surface, despite equal numbers of charges on both surfaces.

The standard position that gravitational redshift of charge does not occur has some ready explanation as to why foregoing is wrong? That would have to be more than simply ab initio enforcing global charge invariance as an axiom I presume? The notion that electric field somehow detaches from source charge and hovers outside a BH EH is perhaps one way of hand-waving an answer. My instant objection would be that for any close distribution of infalling charges, they continue to experience Coulombic fall-off in each other's fields at every stage of infall, something hardly in keeping with a strange delocalization process imo. So, any takers this thread?

2. Jun 7, 2012

### stevendaryl

Staff Emeritus
Yes, there is no problem in GR with having a charged black hole. Black holes are characterized by 3 parameters: (1) The total charge. (2) The total mass. (3) The angular momentum.

Charged black holes with zero angular momentum are described here:
http://en.wikipedia.org/wiki/Reissner-Nordström_black_hole

3. Jun 7, 2012

### Staff: Mentor

Hi Q-reeus, this is an interesting question that I hadn't really thought about before, and I haven't seen much about it in the literature.

Do you have references for the "established view"? I ask because, as I said, I haven't seen much discussion of the charged BH case (Reissner Nordstrom or Kerr-Newman if we add in spin as well). I'm only really familiar with discussions of the Schwarzschild and Kerr (spin but no charge) cases.

Again, do you have a reference? Also, I'm not sure that "no gravitational redshift of charge" is equivalent to "externally observed net charge is invariant". The latter refers purely to what is seen "at infinity", or outside some particular chosen bounding surface. The former refers to the relationship between quantities "at infinity" and quantities at some finite radius r. The two don't have to be the same.

4. Jun 7, 2012

### clamtrox

This seems total gibberish to me.

Why do you say that the mass reduces? According to an observer at constant r?

Why should charge divided by energy should be globally invariant? Just because it's locally constant, the conclusion does not follow...

5. Jun 7, 2012

### Q-reeus

I'm aware of such descriptions and prescriptions, but not aware of anyone tackling the specific objections raised in #1. As per last main para there, I wonder if global charge invariance is mathematically enforced without due regard for all physical consequences. Hopefully that question will be answered here sooner or later.

6. Jun 7, 2012

### Q-reeus

Hi again Peter. Me neither!
Nothing specific for now. Simply take my que from that as per link in #2, it's accepted that a BH can have charge. Given infinite redshift at EH there, it then implies total imperviousness of charge to redshift. Which as I have argued does not really add up.
Agreed the two are technically separate. Given my answer just before, there is for this issue little distinction to be made imo.

7. Jun 7, 2012

### Q-reeus

According to the various scenarios in #1 - reduction is evident for first slow-lowering case, not for second slow-lowering case or free-fall. You do acknowledge that e.g. when a falling mass comes to a crashing halt at a surface, heat energy is radiated away - and that reflects in a reduced net gravitating mass by coordinate measure?

Last edited: Jun 7, 2012
8. Jun 7, 2012

### Staff: Mentor

Hm, ok, Googling hasn't turned up anything useful, and MTW doesn't really talk much about the Reissner-Nordstrom metric, just a couple of homework problems. I'll have to take some time to work this one out.

9. Jun 7, 2012

### stevendaryl

Staff Emeritus
Well, I couldn't really make sense of your objections. You believe that lowering a mass into a black hole causes the mass to reduce? I don't think that's true.

To maintain spherical symmetry, let's suppose instead of a single mass being lowered into the black hole, you instead have lots and lots of little masses arranged in a spherical shell surrounding the black hole. Then they are all lowered into the black hole at the same time, maintaining the spherical symmetry.

Then in the case of an uncharged black hole, the mass doesn't diminish as you lower it. From outside the shell of masses, the gravitational field will look exactly like the gravitational field of total mass Mblack-hole + Mshell. Lowering the shell doesn't change the gravitational field outside the shell at all. (I'm using "gravitational field" loosely here; it's actually the metric that's meaningful, not the gravitational field.)

I don't know for certain about the case of a charged black hole, but I believe it to be the case that the same thing holds for lowering a charged shell. From outside the shell, the mass of the shell gets added to the mass of the black hole, and the charge of the shell gets added to the charge of the black hole. There is no diminishing of either the charge or the mass as you lower the shell toward the event horizon.

In the case of a single mass, instead of a shell, things are more complicated because the mass makes the whole thing not spherically symmetric. However, I think it's still true, that as you lower the mass toward the event horizon, the gravitational and electrical field far from the black hole will approach that of a spherically symmetric black hole with a total mass equal to the original mass and the mass of the particle lowered into the black hole, and with a total charge equal to the original charge, plus the charge of the added particle.

10. Jun 7, 2012

### Q-reeus

Depends what one means by 'lowering'. Lowering: essentially the case of an almost hovering mass yielding energy to the outside (say a winch) as it slowly descends on a cable - not falls - further into the potential well of BH or otherwise central mass. Energy is extracted to the outside - reflected in system net loss of mass/energy. And it's not hard to show the smaller mass m is where essentially all mass is lost (or gained in the reverse process). [I use here M for BH mass, and m or m(r) for the lowered/free-falling mass or masses]
Sure it does - as per above. Provided one stops before reaching the EH, there is a net gain in the sense that M+m(r) > M, but that gain would diminish to zero in the physically impossible case of lowering m(r) all the way, i.e. m(r) -> 0 as r -> rs. But that would require a locally infinitely strong cable - free-fall must take over at some point. But up to that point, there is a net loss of mass/energy from the system M+m(r).
I considered just that case in the other thread (charged and uncharged dust shell). For free-fall yes we just have an invariant M+m (m defined either in coordinate terms or at the beginning radius). I maintain though the correct perspective for evaluating charge in that free-fall case is that of a hovering observer - as discussed in example 3: in #1 (lack or not of spherical symmetry is not important to this).
If one slowly lowers then as per above, coordinate shell mass diminishes by redshift factor, while locally, q/m is invariant. Now just join the dots so to speak - does coordinate charge logically diminish also or not?
That for reasons covered above and in #1 cannot be true unless conservation of energy fails (energy extracted to the outside with no loss of net system mass/energy). Remember too we don't need the extreme of a BH to figure out that local invariance of q/m leads to nonlocal variance of same. Or so I conclude. Theorists surely have thought this stuff through but it escapes me how they try and reconcile it all. :uhh: Must go. :zzz:

11. Jun 7, 2012

### Ich

I think that's exactly the point. Energy is leaving the system, therefore, by Gauss' law, its total mass as seen from the outside decreases.
OTOH, there's definitely no charge leaving the system. According to Gauss, the sum of the field lines going out through some surface doesn't change, no matter how you position the charge inside. Big difference here to the mass case.
I'd say clamtrox identified the weakest point in your argumentation:
Locally it must be the same, no question, but why also as seen from the outside?

12. Jun 7, 2012

### stevendaryl

Staff Emeritus
For problems involving gravity, you can't compute the mass of a collection of objects
by adding up the mass of the component objects.

No, the charge does not change.

No, that's not true. If you extract energy from a system of masses, the total mass will decrease by the same amount. The total mass of a collection of particles is not, as I said, equal to the sum of the masses of the individual particles. Charge, however, does add linearly in that way.

I think the paradox you're talking about has nothing specifically to do with black holes. You can get a similar paradox using charged plates. Suppose you have two positively charged flat plates that are touching each other. (Held together by clamps). This composite object will have a certain total charge, and a certain total mass. Now, allow the charged plates to do work by allowing them to separate because of mutual repulsion. Now, the total charge will be unchanged, but the total mass will be smaller.

Any time a system does work, its total mass changes. I don't see why black holes are any different in this regard.

13. Jun 7, 2012

### jartsa

A physicist that is lowered down in a closed laboratory does not see relative strenghts of different forces change.

A physicist that is not lowered down, but dropped down in a closed laboratory, does not see relative strenghts of different forces change.

A physicist howering at constant altitude says that bigger force is needed to accelerate an object that is falling fast, than a similar object that is moving slowly at the same altitude. This physicist says that at least one force (inertia ) changes in at leat one of free fall and slow lowering, which leads this physicist to think that probably all forces are equally reduced in a lab that is lowered down.

Or does a falling object gain mass?

14. Jun 7, 2012

### Staff: Mentor

I'm still cogitating about this, but I wanted to go ahead and post some general info about the Reissner-Nordstrom metric, which is the unique GR solution for a nonrotating charged BH. I'll also comment on a couple of things in the OP that need clarification/correction.

As noted, the Reissner-Nordstrom spacetime is the unique solution to the Einstein Field Equation for a static, spherically symmetric spacetime where the only stress-energy present is due to a static radial electric field produced by a "point charge" at the origin. (Note that there *is* nonzero stress-energy present; this is *not* a vacuum solution, as the Schwarzschild spacetime is.)

The metric in Boyer-Lindquist coordinates (the generalization of Schwarzschild coordinates) is:

$$ds^2 = - \left( 1 - \frac{2M}{r} + \frac{Q^{2}}{r^{2}} \right) dt^{2} + \frac{1}{1 - \frac{2M}{r} + \frac{Q^{2}}{r^{2}}} dr^{2} + r^{2} d\Omega^{2}$$

where M is the mass of the hole and Q is its charge (in geometric units, i.e., Q, like M, has units of length). Note that M is determined the same as it is for Schwarzschild spacetime--put an electrically neutral object in orbit about the hole at a large distance, measure its orbital parameters, and apply Kepler's Third Law--and Q is determined similarly, but using a charged object and the equivalent of Kepler's Third Law for electromagnetism.

I'll note in passing that the above metric actually describes three different geometries, depending on the relationship of Q and M. We'll only be considering the Q < M case here; Q = M is a sort of "degenerate" case of that, where the two horizons (we'll see below that there are two for Q < M) merge into one, and Q > M has no horizon, but a "naked singularity" at r = 0.

So we assume Q < M, and we find that the above metric describes a spacetime with two horizons, which we'll call r+ and r-. They are at radial coordinates given by:

$$r = M +/- \sqrt{M^{2} - Q^{2}}$$

which is just the obvious solution to the quadratic in r that you get when you set the expression in the metric coefficients above to zero. (You can see, btw, that if Q = M the two solutions are degenerate at r = M, and if Q > M there are no real solutions.)

Outside r+, the spacetime works much the same as the exterior region of Schwarzschild spacetime, just with some extra terms in Q. For example, the proper acceleration of a "hovering" observer, who stays at the same radial coordinate r, is:

$$a = \left( \frac{M}{r^{2}} - \frac{Q^{2}}{r^{3}} \right) \frac{1}{\sqrt{1 - \frac{2M}{r} + \frac{Q^{2}}{r^{2}}}}$$

You can see that this will be *less* than the corresponding quantity in Schwarzschild spacetime at the same r coordinate, because both of the terms in Q make it smaller. However, it will still diverge at the outer horizon, r+, because the "redshift factor" still goes to infinity there. It should also be noted that the formula for the "redshift factor" has the extra term in Q in it; the OP does not include this.

It's important to be clear about exactly what the proper acceleration above represents. It represents the reading on an accelerometer carried by a "hovering" observer, regardless of *how* that observer maintains his altitude. (For a charged object, as we'll see, it is possible for the object to "hover", if the sign of its charge is the same as that of the hole's charge, purely due to the electrical repulsive force; but the object, if it "hovers", will feel the proper acceleration given above.) For an electrically neutral object, since there are no other forces acting on it, this proper acceleration represents the rocket thrust that must be applied to make it hover. This may seem confusing because the hole's charge, Q, is in the formula; but the reason it is there is that there is stress-energy associated with that charge, as noted above, and that stress-energy affects the spacetime curvature, which affects the motion of *all* objects. In other words, the presence of the charge affects the "force of gravity" as well as the "electromagnetic force", because of the stress-energy associated with the charge. This means that the "mass" M of the hole does *not* represent the entire "source" of gravity associated with the hole, as it does for Schwarzschild spacetime; there is an additional effect due to "stress" associated with the charge Q, which is not accounted for in M.

(One might wonder how the above can be true given that, if we measure the orbit of an electrically neutral object, we attribute the orbital parameters only to M. Bear in mind that we stipulated above that the orbit has to be at large r, where the term in Q in the metric is negligible compared to the term in M. Orbits close to the hole *will* be affected by both M and Q, even for a neutral object; at least, it looks to me like they will, but I have not done the detailed calculations to verify that.)

Next, we need the force on a charged object due to the charge of the hole. Actually, we'll need the "force per unit mass" on a charged test object, which is the proper acceleration induced by the hole's charge. This is:

$$a_{e} = \frac{q}{m} \frac{Q}{r^{2}} \frac{1}{\sqrt{1 - \frac{2M}{r} + \frac{Q^{2}}{r^{2}}}}$$

where q/m is the charge/mass ratio of the test object. This should be easily recognizable as just the static Coulomb force times the same "redshift factor" that appears above.

Just to be clear on the physical interpretation of the above: if q and Q are the same sign, the force per unit mass is directed radially outward, as expected; the hole repels the object. If the signs are opposite, the hole attracts the object (this is in addition to the "attraction" due to gravity). It's important to note that this is a proper acceleration--an accelerometer attached to the object would measure this acceleration if no other forces were acting (i.e., no rocket engines or other sources of proper acceleration).

We can put the two acceleration formulas together to find the actual rocket thrust required to make a charged object "hover":

$$a_{rocket} = a - a_{e} = \left( \frac{M}{r^{2}} - \frac{Q^{2}}{r^{3}} - \frac{q}{m} \frac{Q}{r^{2}} \right) \frac{1}{\sqrt{1 - \frac{2M}{r} + \frac{Q^{2}}{r^{2}}}}$$

So an object with the same sign of charge as the hole could potentially hover solely due to the repulsive electric force, without requiring any rocket thrust at all; we can make $a_{rocket}$ vanish by choosing q/m appropriately given a radius r. Conversely, an oppositely charged object will require *more* rocket thrust than a neutral object to hover. In both cases, the key point is that a charged object will *not* free fall in this spacetime if "left to itself" (i.e., no rockets attached or other forces acting on it); unlike a neutral object, a charged object will feel nonzero proper acceleration on its "natural" trajectory. So when the OP says that the "natural" case to consider is free fall, that is not correct for a charged object; the "natural" case there is the trajectory where a_e is the proper acceleration of the object's worldline. This also means that the object's energy at infinity will *not* be constant, unlike that of a neutral object.

I'll save deeper analysis for another post, but I wanted to get the basics down while they're fresh in my mind, and for reference so we all have a common basis to work from.

15. Jun 7, 2012

### pervect

Staff Emeritus
Charged black holes are not an oxymoron at all. If we limit ourselves to static geometries (Schwarzschild and Reissner-Nordstom are the cannoical examples), we can describe the results as follows in language that is very similar to the Gauss law formulation.

I'll start out by mentioning that the surface area of a enclosing sphere of r=constant, as measured by local rulers, is well defined and is equal to 4 pi r^2 in the Schwarzschld and Reisser-Nordstrom geometry.

The gravitational force in a static geometry is well defined, as there is a well-defined notion of "staying in place". The electric force is always well defined, because we have uncharged test particles that we can use as a reference, the relative acceleration between a charged and an uncharged particle when both are standing still at the same point is entirely due to the electric field.

We can note that the product of the area of the enclosing sphere multiplied by the electric force is a constant, and that said constant is proportional to the enclosed charge, Q, of the black hole. This is one illustration that there's no problem with defining the electric charge of a black hole.

Aside from working this out for yourself, I'm not sure if there's any quick proof unless you use the notion of differential forms, in which case the proof is trivial. (Finding it may not be trivial.)

The area of the enclosing sphere multiplied by the gravitational force is NOT a constant, but the area of the enclosing sphere multiplied by the "force at infinity" IS a constant. This is discussed somewhat in depth in the Wiki article http://en.wikipedia.org/w/index.php?title=Komar_mass&oldid=465638523.

And that's all there is to it. I suspect from the way that the topic keeps popping up that some confusion remains (especially about the second point) but I cant write a description of the physics any more simply than the above.

I also have to caution again against applying or trying to apply the above ideas to a non-stationary metric. It Won't Work!

16. Jun 7, 2012

### Staff: Mentor

By "electric force", do you mean the "redshifted electric force"? From the formula I posted above, there should be a "redshift factor" in the electric force felt by a charged object at a finite radius r; but that means you would have to take the "force at infinity" to make area * electric force constant.

17. Jun 8, 2012

### Q-reeus

And I disagree. Just saying 'energy leaves the picture' can easily be construed as akin to say sucking out part of the water in a filled tank = 'system mass loss'. Bad analogy. In actual scenario there is no particle count change involved - n1 atoms, n2 charges before, n1 atoms, n2 charges after. Gravity depresses resting mass (read; 'gravitational charge') of each of n1 atoms, but I am asked to believe it has absolutely no equivalent effect on each of n2 electric charges. Is it really so - that is the crux of the matter.
Not sure to whom that is directed. My response: evidently having difficulty getting message in #1 across.

18. Jun 8, 2012

### Q-reeus

Depending on exactly what that means, I probably agree - assuming it means assembly of matter originally dispersed, then naturally a more compact *and static* arrangement exhibits a reduced assembled mass. Which is entirely in accord with what I have been saying all along - energy is given off during assembly, and lowering a mass is part of an assembly process in this context. Assuming my above interpretation of what you meant is correct, actually it is you who previously claimed otherwise - e.g. from #9:
stevendaryl - I take you as sincere in your position, but this kind of thing has me scratching head.
I will put that down to a complete misunderstanding of what I was saying - which was refuting your own previous claim as per e.g. quote from #9 above!
That's your's and others position here (apart from perhaps one), but I am to be convinced it amounts to more than dogmatic assertion.
Agreed - I said as much back in #1, and used spherical capacitor as example. You drew a different conclusion though.
It's a case of yes and no. Yes clearly if the work extracted from charged plates is channelled exterior to the system there is net loss to that system. But in BH case involving purely free-fall, no there is no net change as everything stays inside. None of that is paradoxical. let's see if I can't turn a variant of your charged plates example around on you. Just consider slow lowering of a charged parallel-plate capacitor down to some floor level in a potential well. If we agree all forms of energy redshift the same, what happens to the field strength between those capacitor plates? Only one guess.

19. Jun 8, 2012

### Q-reeus

Depends on your frame of reference jartsa. Drop a mass from a high tower, then catch it in a spoked wheel. It spins around with KE - that means total mass/energy is greater than the rest mass observed up in the tower before the drop. But from pov of far away observer, the system earth+tower+mass has not changed in gravitating mass at all - provided no frictional loss results in heat escaping.

20. Jun 8, 2012

### Q-reeus

Peter - splendid job of explaining various contributions to Reissner-Nordstrom metric - thanks. But first I comment on your comments on my OP.
And the reason is two-fold:
a: My scenario was that of a Schwarzschild BH (but more generally an uncharged central mass M) perturbed slightly by a vastly smaller mass/charge in the process of being lowered/dropped in. Which is not a pre-existing R-N system with central charge already in place. So no conflict there.
b: Being highly sceptical of the premise of finite exterior charge for a BH, my use of an R-N model would be rather inconsistent.
Again, as I did not model things using a pre-existing 'charged' R-N BH, it boils down to whether free-fall is an appropriate description for infall of an assumed small charged mass toward a large neutral mass M. If one ignores the typically vanishingly small radiative back reaction (dependent on the q/m ratio for starters), I would say yes given the context of what we are about here. And while agreeing your energy argument may be strictly correct, any relatively infinitesimal radiative energy loss is surely not really germaine to the topic. Also isn't there continuing dispute about whether any radiative loss at all applies to charge in radial free-fall?

Anyway, back to your interesting R-N dissection. I concentrate on just the results for proper acceleration a of a neutral test mass, and ae for a charged test mass (agree with pretty much all your other pertinent observations as far as what R-N metric implies.):
Which, along with the later expression you give for ae, is really baffling to me. While axial stress component for E field of Q is negative, the two transverse components are positive. So I would expect a net positive stress adds to a positive energy density, yet together appears as negative contribution in R-N expression?

Regardless though it is all based on there actually being a finite external electric field. We have this R-N expression, and yes one then derives specific interesting results as you have done in #14. My question is, what is the physical basis for that Q lying at the EH can project a finite E to the BH exterior? Still wondering if it comes down to, as I suspected, enforcing global charge invariance as axiom, and out comes R-N. Might be thought a jaundiced outlook, but I get back to scenarios given in #1. Here's another situation worth pondering. A balloon is tethered on a string to the centre of a large mat. Both are electrically charged with the same sign of charge - just sufficient to have the balloon float against earth's gravity. For a distant observer, the mass of earth and balloon are redshifted wrt to that locally observed (not equally of course - much of the earth's mass is already partially redshifted at the balloon location, but that's not important here). Now the balloon is still floating from that observer's pov, so what does that tell us about the value of balloon/mat charge and E fields in coordinate measure? Sleep on it maybe.

21. Jun 8, 2012

### Q-reeus

It's clear that statement of global charge invariance is the standard position, but does it amount to more than an axiom.
Does use of said differential forms free one from reliance on the axiom of global charge invariance?
Komar expression explicitly acknowledges the redshift of 'gravitational charge'. Said my piece in responding to PeterDonis on why a similar redshift of electric charge should imo be true. As done back in #1 really. What might really convince eitherway is a numerical GR simulation of infalling test charge E-field. Some chance.

22. Jun 8, 2012

### stevendaryl

Staff Emeritus
You call something "dogmatic assertion" because someone did not present a textbook deriving it from first principles? What exactly are you expecting from a discussion forum? You can't just demand that people convince you--only you can convince yourself, one way or the other. The best you can do is ask questions.

I was mistaken in my first response--allowing a dropping mass to do work does change the total mass/energy. As far as the question of whether charge is unchanged by dropping it into a black hole, I'm not exactly sure what you would take as a convincing argument. Charge is a scalar; since charges come in discrete multiples of the charge on an electron, it's just a matter of counting. The charge of something can't change except by emitting or absorbing a charged particle.

I guess I don't understand what it is that you think is paradoxical. Draw a surface around the outside of the black hole a small distance from the event horizon. Inside this surface is a certain charge Q and a certain mass M. If you drop a charge through the surface, Q goes up and so does M. If you extract energy from the system, then M goes down (but Q doesn't). I'm not sure what you think is oxymoronic about any of this.

The field strength doesn't change. The charge doesn't change.

23. Jun 8, 2012

### stevendaryl

Staff Emeritus
It's not at all clear what your point is. People are telling you what the theory says; conservation of charge is an inherent aspect of our current understanding of electromagnetism. Our current understanding could very well be wrong, of course. For example, if the photon has a very tiny mass, then charge is not conserved.

So it's not clear whether you are asking theoretical questions: what does the theory predict? Or are you asking what's really true? Nobody knows what's really true, they just have the current best guess.

24. Jun 8, 2012

### Q-reeus

Well so far that's what it amounts to - in so many words I am simply told it is so - take it or leave it. In #1 I gave reasons, right or wrong, for doubting it.
Whoa on a little - have I made such demands? You perhaps just misread my style. Hoped for more than expected was that each scenario presented has a ready explanation according with global charge invariance. Numbered for easy reference, but needn't have bothered.
Well that is an argument of sorts but you are probably aware that many here hold that otherwise sacred conservation of energy/momentum fails in GR. Why insist charge remains aloof necessarily?
Again, you simply assert there will be no change in externally observed charge. I'm not convinced and am not demanding you do convince me - just pointing out there is nothing in that statement to change my position of doubt.
Probably not such a good example to give as coordinate perspective depends on orientation of the plates for one thing. However a physical fact is that if a remote observer discharges the capacitor, the extracted energy remotely received is for sure redshifted. And that energy resided in the field. Same redshift will apply to the force remotely applied if prising the charged plates apart - and F = qE. Something to think over.

25. Jun 8, 2012

### Q-reeus

Conservation of charge is actually a different thing - that charge always comes in pairs with equal and opposite sign.
There is afaik absolutely no experimental/observational evidence that charge invariance globally holds. So it gets down as I have tried to test for internal consistency via gedanken experiments. And yes thought maybe someone has a neat first-principles theoretical explanation why GR forbids any violation. Might be hoping too much.