Are the directions of electric fields lines affected by Gravity?

[PeterDonis]

It was back there in those 'proof' entries I referred to earlier that field line concept were a key part of that 'proof'.

No, that's not correct; the field line concept was just a way of trying to help you visualize what the proof was saying. The proof itself used the underlying math, not any intuitive reasoning based on "field lines".

So when you e.g. write p(r)=−ρ(r) as in #42, I'm supposed to understand that that p is just the radial (-ve sign) component rather than net value of p (+ve sign) at r?

I'm not sure what you mean by the "net value of p". If the only stress-energy present is that due to the static electric field, then the radial pressure *is* negative--it's a tension. (The sign convention for the SET that I'm using, which is the standard one AFAIK in GR, is for tensile stresses to be negative and compressive stresses to be positive.) That's the point of p(r) = - rho(r); the energy density is positive, and the radial pressure is negative (tension).

If there are other sources of stress-energy present in a region of the spacetime (as in the case of a shell around a charged gravitating body), then the "net" radial pressure p(r) in such a region will be the sum of minus the EM field energy density, plus the radial pressure due to the other sources (which will be positive). Similarly, the net tangential stress will be the sum of the tangential stress due to the EM field (which is equal to rho, the EM field energy density, and is positive so it's a compression) and that due to the other sources (which will vary in sign, being tension at the shell's outer radius and compression at the shell's inner radius--I didn't go into this here because I went into it in detail in a previous thread; the exact numbers will be slightly different because the equations for p(r) and s(r) are for the "net" stress including all contributions).

Mass has no effect on form of E field.

Sure it does; the effect of M (and of spacetime curvature generally) is in the radial unit vector e^\hat{r} (what I called the unit normal n^a. But when you look at the force on an object that happens to be static (i.e., the only nonzero component of its 4-velocity is the "t" component), you find that the curvature contributions cancel out. And since the static force is what you need to evaluate in order to do the charge integral, you find that the charge is invariant for any value of the radius.

Look at my #17 again, the second version (where I said what would have to be the case for Q(r) = Q for all r to hold). There is a contribution of 1 / \sqrt{g_{rr}} from the unit normal vector n^a, and a contribution of 1 / \sqrt{g_{tt}} from the 4-velocity u^b, and those two contributions cancel since g_tt = 1/g_rr for R-N spacetime (it's one of the special cases where that's true). So when you look at the final number it looks like there is no "mass effect", because the metric coefficient factors cancel each other.

It's important to note, though, that the cancellation only happens in special cases; it doesn't necessarily happen in *all* cases. That's one of the reasons I'm taking so much time to delve into the math: to try to nail down exactly *what* the special conditions are for the curvature contributions to cancel, so it "looks like" there is no "mass effect". One such condition is obvious from the above: we must have g_tt = 1/g_rr (and I went into the requirements for that in a previous post). Since, for example, that condition will be violated inside a spherical shell, the effect of the shell on charge conservation needs to be evaluated separately. (I think that it still holds, because the effects of the change in mass with radius offset the effects of the metric coefficients not exactly cancelling; but I am still checking the math to be sure.)

 

[Q-reeus]

No, that's not correct; the field line concept was just a way of trying to help you visualize what the proof was saying. The proof itself used the underlying math, not any intuitive reasoning based on "field lines".

From back there: "The easiest way to define the apropriate sense of integration is to use the idea of counting field lines." Context was clear enough re Gauss's law, and as said before, I have no problem with that - when gravity is not around.

I'm not sure what you mean by the "net value of p". If the only stress-energy present is that due to the static electric field, then the radial pressure *is* negative--it's a tension. (The sign convention for the SET that I'm using, which is the standard one AFAIK in GR, is for tensile stresses to be negative and compressive stresses to be positive.) That's the point of p(r) = - rho(r); the energy density is positive, and the radial pressure is negative (tension).

Sorry - my last comments on that got the signs mixed up somehow. Putting it down to sleep deprivation. Your notation did leave it ambiguous as to which p was meant, but no big deal.

If there are other sources of stress-energy present in a region of the spacetime (as in the case of a shell around a charged gravitating body), then the "net" radial pressure p(r) in such a region will be the sum of minus the EM field energy density, plus the radial pressure due to the other sources (which will be positive). Similarly, the net tangential stress will be the sum of the tangential stress due to the EM field (which is equal to rho, the EM field energy density, and is positive so it's a compression) and that due to the other sources (which will vary in sign, being tension at the shell's outer radius and compression at the shell's inner radius--I didn't go into this here because I went into it in detail in a previous thread; the exact numbers will be slightly different because the equations for p(r) and s(r) are for the "net" stress including all contributions).

This all goes back to a certain monetary challenge, and me thinks radially supported perfect fluid is implied above. For a solid self-supporting thin shell, transverse hoop stresses - of the same sign throughout the shell's radial depth, will be typically completely dominate, unless perhaps net radial forces from charge and mass cancel or near cancel. But I'd rather not get stuck in that groove again!

Sure it does; the effect of M (and of spacetime curvature generally) is in the radial unit vector e^\hat{r} (what I called the unit normal n^a. But when you look at the force on an object that happens to be static (i.e., the only nonzero component of its 4-velocity is the "t" component), you find that the curvature contributions cancel out. And since the static force is what you need to evaluate in order to do the charge integral, you find that the charge is invariant for any value of the radius.

Reads a lot like "mass has no net effect on charge" to me.

Look at my #17 again, the second version (where I said what would have to be the case for Q(r) = Q for all r to hold). There is a contribution of 1 / \sqrt{g_{rr}} from the unit normal vector n^a, and a contribution of 1 / \sqrt{g_{tt}} from the 4-velocity u^b, and those two contributions cancel since g_tt = 1/g_rr for R-N spacetime (it's one of the special cases where that's true). So when you look at the final number it looks like there is no "mass effect", because the metric coefficient factors cancel each other.

Again, if they cancel, then they cancel, right?

It's important to note, though, that the cancellation only happens in special cases; it doesn't necessarily happen in *all* cases. That's one of the reasons I'm taking so much time to delve into the math: to try to nail down exactly *what* the special conditions are for the curvature contributions to cancel, so it "looks like" there is no "mass effect". One such condition is obvious from the above: we must have g_tt = 1/g_rr (and I went into the requirements for that in a previous post). Since, for example, that condition will be violated inside a spherical shell, the effect of the shell on charge conservation needs to be evaluated separately. (I think that it still holds, because the effects of the change in mass with radius offset the effects of the metric coefficients not exactly cancelling; but I am still checking the math to be sure.)

OK then keep checking by all means. But consider too attempting your own answer to crunch problems in #10 sometime soon - that's where it all hits the fan imo. And if parallel plate cap is for some reason a problem geometry, maybe e.g. case of two almost touching concentric charged shells undergoing a differential radial relative displacement. Plain as day to me from #10 scenario there cannot be a proper energy accounting without introducing an 'active' vs 'passive' charge notion - which as shown there, being ad hoc will fail to deliver consistency. But please don't just take my word on that! :zzz:

 

[PeterDonis]

From back there: "The easiest way to define the apropriate sense of integration is to use the idea of counting field lines."

"The idea of counting field lines", again, is a way of describing the underlying math. If you understood the underlying math that would be obvious. Pervect was not giving you a line by line proof; he was merely trying to describe, in a way that would make some kind of intuitive sense, what the proof is saying. If you want a more concrete look at the underlying math, see my post #17 or below.

me thinks radially supported perfect fluid is implied above.

A shell that is supporting its own weight with an "empty" region inside ("empty" here meaning "nothing but the static EM field present") can't be composed of a perfect fluid; the stresses can't be isotropic (though they can be diagonal). That's why I coined the term "quasi-perfect fluid" in one of those previous threads--to have a shorthand way of describing the case where the tangential stresses are equal, but are not equal to the radial stresses. See further comments below.

For a solid self-supporting thin shell, transverse hoop stresses - of the same sign throughout the shell's radial depth

For a shell that is in static equilibrium with vacuum inside and outside, the tangential stress *has* to change sign inside the shell. I went through this in the other thread, but here's a quick recap: the radial pressure due to the shell must be positive inside the shell and zero at both boundaries. That means the sign of dp/dr has to change somewhere inside the shell; dp/dr will be negative at the outer radius and positive at the inner radius. If you look at the formula for dp/dr for this case, it's obvious that the *only* way that can happen is if the sign of s (the tangential stress) changes; every other term has to be negative (since p is positive inside the shell).

For the case where the EM field is present, things are more interesting; I'm working the math on that one now. However, it seems obvious that the *total* stress can't be isotropic, because the EM field stress certainly isn't (radial stress is opposite in sign to tangential stress). I think it's highly unlikely that the shell stress alone (total stress - EM field stress, which is known) will turn out to be isotropic.

Reads a lot like "mass has no net effect on charge" to me.

Again, if they cancel, then they cancel, right?

But they only cancel because there is a "mass effect" involved--two of them. And..

OK then keep checking by all means.

...still checking, but after thinking about it some more I think I was wrong to expect the cancellation to still hold within the shell. Here's why: look at the equation for the charge again, in this form:

Q(r) = \frac{Q}{\sqrt{g_{rr} | g_{tt} |}} = Q\sqrt{\frac{1}{J} \left( 1 - \frac{2m}{r} \right)}

Outside the shell we have J = 1 - 2m / r, so the sqrt factor is just 1 and we have Q(r) = Q. But inside the shell, we have J = f \left( 1 - 2m / r \right), where f < 1! That means that inside the shell, we have

Q(r) = \frac{Q}{\sqrt{f}}

So I now expect that when I work the math, I will find that there *is* a "mass effect" due to the shell--the charge "visible" outside the shell is *smaller* than the charge "visible" inside the shell.

But consider too attempting your own answer to crunch problems in #10 sometime soon

As I said before, I'm not sure how that's even relevant to the question of charge invariance. In any case, I need to finish up the R-N spacetime analysis before doing anything else on this topic.

 

[pervect]

A shell that is supporting its own weight with an "empty" region inside ("empty" here meaning "nothing but the static EM field present") can't be composed of a perfect fluid; the stresses can't be isotropic (though they can be diagonal). That's why I coined the term "quasi-perfect fluid" in one of those previous threads--to have a shorthand way of describing the case where the tangential stresses are equal, but are not equal to the radial stresses. See further comments below.

You can imagine an alternative - a structural core, essentially a pressure vessel, that supports a perfect fluid exterior.

The structural core can be analyzed as a pressure vessel, with only tangential compressive stresses and no radial stresses.

The outer fluid layer will have an isotropic non-negative pressure that increases with depth (zero at the surface).

For a shell that is in static equilibrium with vacuum inside and outside, the tangential stress *has* to change sign inside the shell.

I don't quite see why this would have to be- perhaps it's some artifact of your particular idealization. It's certainly not true in the pressure vessel + fluid exterior case, where nothing is under tension.

 

[PeterDonis]

You can imagine an alternative - a structural core, essentially a pressure vessel, that supports a perfect fluid exterior.

Yes, this case would be different. I was specifically analyzing the case of a thin shell with either (1) *vacuum* inside and outside, or (2) a static, Reissner-Nordstrom electric field inside and outside. Those cases are what are relevant to the scenarios Q-reeus has proposed (in this and other threads).

I don't quite see why this would have to be- perhaps it's some artifact of your particular idealization.

I actually didn't give the full argument in my last post; I only referred to the dp/dr equation, but that in itself is not enough to show that s (the tangential stress) must change sign. Let me go into a bit more detail.

Thin shell with fluid inside:

A pressure vessel (thin shell) with fluid inside is supported against its own gravity by the pressure of the fluid inside; and since the fluid inside goes all the way down to r = 0, it can be a perfect fluid with isotropic pressure (just like an idealized planet or star). So the radial pressure in the shell itself can be negligible; the fluid inside can provide all the radial force required.

Also, the force balance that determines the tangential hoop stress in the pressure vessel (cut a plane through the center of the vessel and balance forces in both directions across the plane) includes both the fluid inside and the hoop stress in the vessel; so the hoop stress can be the same sign (tension) throughout the entire shell (while the stress in the fluid inside is compressive everywhere).

Thin shell with vacuum inside:

A thin shell with vacuum inside must support itself against its own gravity in order to be in static equilibrium, and the force balance across a plane cut through the center of the shell (meaning, through the vacuum region inside) *only* has the hoop stress in the shell to balance. That means two things:

(1) The radial pressure inside the shell *cannot* be negligible: it has to be enough to hold the shell up against its own gravity. (Of course for a typical real shell this amount of force *is* negligible, but we're talking here about a relativistic case where in principle it might not be.) Also, since there is vacuum inside and outside the shell, the radial pressure must go to zero at the shell's inner and outer surfaces. That means dp/dr must change sign, which means, if you look at the dp/dr equation, that s, the tangential stress, must be *positive* in part of the shell (because that's the only way there can be any term in dp/dr at all that is positive--all the other terms are negative). It can't be negative everywhere, as it is in the case of the pressure vessel with fluid inside.

(2) Because the force balance across a plane cut through the center of the shell *only* includes the shell's tangential stress, that tangential stress *must* change sign inside the shell for the net force across the plane to be zero. So the tangential stress can't be positive everywhere; it must be negative (tension) somewhere.

Looking at dp/dr again, we see that it should be positive (which requires a positive tangential stress) towards the shell's inner surface, and negative towards the shell's outer surface. So we expect the tangential stress to be positive (compression) towards the shell's inner surface, and negative (tension) towards the shell's outer surface.

 

[PAllen]

So I now expect that when I work the math, I will find that there *is* a "mass effect" due to the shell--the charge "visible" outside the shell is *smaller* than the charge "visible" inside the shell.
.

This will be interesting. I was checking over my references and found very few of that had as much detail as I thought I remembered (mostly - develop in detail in SR, and gloss over generalization of electrodynamics to GR. I misremembered the SR treatment being appropriately generalized to GR). Even MTW had but a few pages with no substantive results for GR. However, Synge (1960) has a sizable chapter on electrodynamics in GR, developed with failry general assumptions (He claims they apply to any distribution of charged and neutral fluids, combined with vacuum regions, under assumptions of classical physics - no active field other than EM. For proofs, he simplifies the type of fluid and considers varying amounts of charge of only one sign - but claims the results generalize ).

One key conclusion is a general form of Gauss's Law that says that the integral of E through a surface is the same as a volume integral for which only volume elements with nonzero charge density contribute. Thus comparing two nested surface integrals, for which all charge density is in the inner one, the surface integrals must come out the same. Presence of uncharged matter between the integration surfaces cannot have any effect - because the volume element is multiplied by charge density. [Edit: I know this book is out of print and hard to find - I could not find anything on the internet with a comparable level of treatment. If you can lay hands on it, the key result is eq. 93 on p.366 of my edition. I don't think copyright would allow me to scan in the 12 pages developing this result - but would be happy to if someone knowledgeable ruled it ok].

 

[PeterDonis]

One key conclusion is a general form of Gauss's Law that says that the integral of E through a surface is the same as a volume integral for which only volume elements with nonzero charge density contribute. Thus comparing two nested surface integrals, for which all charge density is in the inner one, the surface integrals must come out the same. Presence of uncharged matter between the integration surfaces cannot have any effect

Yes, you're right, which means that the change in the surface integral of the charge from outside to inside the shell indicates that the shell *cannot* be neutral. I am verifying that by explicitly computing Maxwell's Equations inside the shell; it looks like the covariant divergence of the EM field tensor is indeed nonzero there, indicating the presence of nonzero charge density. I'll post that later when I've finished checking some things.

More precisely, what the computations appear to indicate is that if you take a charged massive body (i.e., something that can be described by an external R-N metric, gravity plus a static E field) and put a shell of matter around it, assuming everything stays spherically symmetric, one of two things must happen:

(1) If we stipulate that the EM field tensor within the shell is the same as the EM field tensor in the inner and outer non-shell regions--i.e., we assume the matter of the shell has no effect on the field (for example, its permittivity and permeability are the same as those for free space)--then the shell *cannot* be neutral; there must be charge density inside the shell, which compensates for the shell's effect on spacetime curvature to keep the EM field tensor the same. This is the case that I've been working out.

(2) If we stipulate that the shell is neutral, meaning that the Gauss's Law integral in the inner region is the same as the Gauss's Law integral in the outer region, then the EM field tensor within the shell *cannot* be the same as it is in the inner or outer regions. It can still be purely radial (assuming spherical symmetry--as far as I can tell that assumption can still be retained), but the E field will *not* be Q/r^2, as it is in the inner and outer regions; it will be something different. The computations I'm doing for the other case should also give at least an idea of what the E field within the shell would look like for this case.

A quick note on terminology: we have three spacetime regions, one from r_outer to infinity (the "outer" region), one from r_inner to r_outer ("within" the shell), and one inside r_inner (the "inner" region). Hopefully that helps clear up any possible ambiguity in the terms.

 

[Q-reeus]

A shell that is supporting its own weight with an "empty" region inside ("empty" here meaning "nothing but the static EM field present") can't be composed of a perfect fluid; the stresses can't be isotropic (though they can be diagonal). That's why I coined the term "quasi-perfect fluid" in one of those previous threads--to have a shorthand way of describing the case where the tangential stresses are equal, but are not equal to the radial stresses. See further comments below.

I tried in the past but had no luck finding an authoritative explicit expression for radial vs transverse stress distributions for a self-gravitating and self-supporting spherical shell. While the following specifically deals with cylinders, there is a clear parallel with spherical shells: www.me.ust.hk/~meqpsun/Notes/Chapter2.pdf (Ponder the last graphical results shown), or similar here: http://www.scribd.com/doc/85488099/144/Saint-Venant%E2%80%99s-Principle#page=320 (need to click on pointer thing to get to doc p309-311, which is 320-322 on pointer). Key feature is just how well Saint Venant's principle applies to the load sharing behavior. Even for very thick-walled pressurized or spinning cylinder cases, no hint of sign reversal for radial or transverse stresses, and clearly transverse stresses completely dominate for t/r << 1. Spherical pressure vessel stress-strain relations are very similar: http://solidmechanics.org/text/Chapter4_2/Chapter4_2.htm. I expect even greater uniformity of hoop stresses when body forces (self-gravitation) are involved. Exception being a shell with surface charge electrostatically near balancing gravitational forces. Not particularly fussed with your idea of how it goes stress-wise - as long as we are dealing with a realistic scenario where shell material energy densities are many orders of magnitude greater than stress contributions (weak gravity regime), won't make any substantive difference to what is important here.

Your tentative finding of different behavior in the 'within shell' region is sort of interesting, and I note further input in #55, 57 on that. I say 'sort of' because it stems from the RN position that in vacuo at least charge invariance holds. Hate to keep riding the issue but imo you would be making a big-short cut by tackling that issue in #10. Still, let's see where your approach goes. :zzz:

 

[pervect]

Thin shell with vacuum inside:

A thin shell with vacuum inside must support itself against its own gravity in order to be in static equilibrium, and the force balance across a plane cut through the center of the shell (meaning, through the vacuum region inside) *only* has the hoop stress in the shell to balance. That means two things:

(1) The radial pressure inside the shell *cannot* be negligible: it has to be enough to hold the shell up against its own gravity.

I'm not sure I agree with this. I suspect that you'll find a solution with the radial stress zero everywhere, if you look for one, that will be mathematically and logically consistent.

If we divide the shell into layers, the innermost shell can only be held up by hoop-stress, I think we're agreed on that. So it's possible for a shell to just have hoop-stress.

Now, just imagine a series of self-supporting shells, stacked one inside the other. Each shell has enough hoop-stress to hold up its own weight, and no more.

Basically there are several possible ways for the stress to balance out. One extreme is the perfect fluid over a shell that supports the fluid and everything above it. This solution has radial pressure that increases with depth until you reach the supporting shell.

Another extreme is the stacked network of self-supporting shells. There wouldn't be any radial pressure in this scheme that I can see.

 

[PAllen]

I suspect that you'll find a solution with the radial stress zero everywhere, if you look for one, that will be mathematically and logically consistent.

If we divide the shell into layers, the innermost shell can only be held up by hoop-stress, I think we're agreed on that. So it's possible for a shell to just have hoop-stress.

Now, just imagine a series of self-supporting shells, stacked one inside the other. Each shell has enough hoop-stress to hold up its own weight, and no more.

I'm confused. Consider the Newtonian approximation. To a first approximation, the innermost shell has no weight to support. Each shell further out has more weight to support, as there is more mass inside of it (that outside has no effect). How can this not produce a radial stress component?

 

[pervect]

I'm confused. Consider the Newtonian approximation. To a first approximation, the innermost shell has no weight to support. Each shell further out has more weight to support, as there is more mass inside of it (that outside has no effect). How can this not produce a radial stress component?

It produces either a radial stress component, or a tangential one (with no radial stress component), or possibly some combination of the two.

I'm not sure how to explain it any clearer than using the math. Note that the physics concept of pressure is perhaps subtly different from the engineering concept.

Consider peter's metric:

<br /> g_{\mu\nu} = \left[ \begin {array}{cccc} -J \left( r \right) &amp;0&amp;0&amp;0<br /> \\0&amp;{\frac {r}{r-2\,m \left( r \right) }}&amp;0&amp;0<br /> \\0&amp;0&amp;{r}^{2}&amp;0\\0&amp;0&amp;0&amp;{r}^{2}<br /> \left( \sin \left( \theta \right) \right) ^{2}\end {array} \right] <br />

Define T_{\hat{a}\hat{b}} in an orthonormal basis:

<br /> <br /> T_{\hat{a}\hat{b}} = \left[ \begin {array}{cccc} \rho \left( r \right) &amp;0&amp;0&amp;0\\0&amp;P \left( r \right) &amp;0&amp;0\\0&amp;0&amp;S<br /> \left( r \right) &amp;0\\0&amp;0&amp;0&amp;S \left( r \right) <br /> \end {array} \right] <br />

Convert it to a coordinate basis

<br /> T_{ab} = \left[ \begin {array}{cccc} J \left( r \right) \rho \left( r \right) &amp;0&amp;0&amp;0\\0&amp;{\frac {rP \left( r \right) }{r-2\,m<br /> \left( r \right) }}&amp;0&amp;0\\0&amp;0&amp;{r}^{2}S \left( r<br /> \right) &amp;0\\0&amp;0&amp;0&amp;{r}^{2} \left( \sin \left( \theta<br /> \right) \right) ^{2}S \left( r \right) \end {array} \right] <br />

Take the covariant derivative \nabla^a T_{ab}

There is only one term, in the r direction. In Newtonian terms, this represents the force balance equation.

<br /> \frac{\left(dJ/dr\right)}{2J} \, \left(\rho+P\right) + dP/dr + \frac{2}{r} \left( P - S \right) = 0<br />

If S=0, then we have the usual differential equation for P(r). But S is a free choice, there are no constraints it has to follow. In particular, there's nothing to prevent us from choosing S such that P(r) = 0. This particular choice represents a stable solution which has only tangential pressure and no radial pressure - in the physics sense of the term , i.e. in the sense used by the stress-energy tensor.

 

[PeterDonis]

I say 'sort of' because it stems from the RN position that in vacuo at least charge invariance holds.

Everything I've posted so far does not "assume" anything about the R-N metric. I am assuming that the EFE and Maxwell's Equations are valid because without those I can't compute anything. In so far as I am using the R-N metric for anything (which I'm not for much at this point, most of what I've posted is general and applies to any static, spherically symmetric spacetime), I have *derived* the R-N metric from the EFE and Maxwell's Equations. The only other assumption is a particular form for the EM field tensor, but I'm even allowing for the possibility that that might vary, as you can see from recent posts.

 

[PeterDonis]

In particular, there's nothing to prevent us from choosing S such that P(r) = 0.

Isn't there? Let's see what the equation looks like when we do that. If P(r) = 0, then dP/dr = 0 as well, so we must have

\frac{\left(dJ/dr\right)}{2J} \, \rho = \frac{2 S}{r}

The LHS is positive, so the RHS must be as well, meaning that S would be *positive* (a compression, not a tension) throughout the shell. That's impossible if the shell is in static equilibrium with vacuum inside and outside, for the reasons I gave in a previous post; the argument I gave there does not depend on what value the radial pressure takes.

 

[pervect]

Yes (I called the e_{\hat{r}} vector n^{a} in the equation I wrote down, but it's the same thing).

My question was about the *timelike* vector u^{b} in my equation; should it be a timelike *unit* vector, or the timelike *Killing* vector of the spacetime? As PAllen commented, if the electric field E is going to be measured by local Lorentz observers, then u^{b} must be a timelike *unit* vector.

I agree, u should be a timelike unit vector.

 

[pervect]

Isn't there? Let's see what the equation looks like when we do that. If P(r) = 0, then dP/dr = 0 as well, so we must have

\frac{\left(dJ/dr\right)}{2J} \, \left(\rho+P\right) = \frac{2 S}{r}

The LHS is positive, so the RHS must be as well, meaning that S would be *positive* (a compression, not a tension) throughout the shell. That's impossible if the shell is in static equilibrium with vacuum inside and outside, for the reasons I gave in a previous post; the argument I gave there does not depend on what value the radial pressure takes.

I'd expect S to be positive - but the only needed boundary condition should be that the radial component , which we've been calling P, equals zero.

Taking the divergence of T_ab and setting it to zero is the easiest way of getting the necessary equations - but you already did that, you don't agree about the significance. Let's try a free body diagram - see attached.

The free body is that of some wedge, which is pulled down by gravity, and held up, by the force S and possibly some pressure P

Let's do the simple Newtonian analysis.

Then there is some force rho * volume pulling it down. THe volume will will to first order be equal to dV = (R dU)^2 * dr, so the force will be rho*g*dV

Now, what is the vertical force due to S? S is a force / unit area, the total area that it acts on is 4 R dU dr, because there are four squares of height dr and width R du. The vertical component of this force is the total force time the sine of the angle, the angle being dU / 2. For small dU, the sine(dU) = dU. So we have a total vertical force due to S of just 2 R dU^2 dr, or (2/R) dV

If we had a pressure difference, it would give an additional force of (P+dP)(RdU+dr)^2 - P(R^2 dU^2), which gives components dP R^2 dU^2 and 2 P R dU dr, or equivalently (dP/dr)*dV and (2/R)*P*dV

So we can write the total force balance equation as

(rho*g + (2/R)*S + (dP/dr) + (2/R)*P ) = 0.

We can see this is just what we worked out via the stress-energy tensor approach, modulo a few different sign choices - and a few additions due to relativity.

We can easily see that for the proper value of S, the wedge is under no net force, even with P=0.

We can also appreciate why P=0 is the correct boundary condition when there is a vacuum above the wedge .

 

[Q-reeus]

Everything I've posted so far does not "assume" anything about the R-N metric. I am assuming that the EFE and Maxwell's Equations are valid because without those I can't compute anything.

As I harped on in the past here, it's not a question of either separately being valid, but how they are combined. The assumed ab initio fundamental guiding principle is the universal validity of that bit about counting of flux lines through any given bounding enclosed surface (i.e. Gauss's law), and #10 in particular strongly suggests there are self-consistency issues in following that guiding principle (when gravity enters the equation) - the one which you (and everyone else in GR community it seems), work from as 'a given'. that Gauss's law applies *locally* in arbitrarily curved spacetime I do not question - but *global* validity is another matter.

In so far as I am using the R-N metric for anything (which I'm not for much at this point, most of what I've posted is general and applies to any static, spherically symmetric spacetime), I have *derived* the R-N metric from the EFE and Maxwell's Equations. The only other assumption is a particular form for the EM field tensor, but I'm even allowing for the possibility that that might vary, as you can see from recent posts.

See previous comments. When your current line of attack runs it's course, please consider taking up my suggestion!

 

[PAllen]

I'd expect S to be positive - but the only needed boundary condition should be that the radial component , which we've been calling P, equals zero.

Taking the divergence of T_ab and setting it to zero is the easiest way of getting the necessary equations - but you already did that, you don't agree about the significance. Let's try a free body diagram - see attached.

The free body is that of some wedge, which is pulled down by gravity, and held up, by the force S and possibly some pressure P

Let's do the simple Newtonian analysis.

Then there is some force rho * volume pulling it down. THe volume will will to first order be equal to dV = (R dU)^2 * dr, so the force will be rho*g*dV

Now, what is the vertical force due to S? S is a force / unit area, the total area that it acts on is 4 R dU dr, because there are four squares of height dr and width R du. The vertical component of this force is the total force time the sine of the angle, the angle being dU / 2. For small dU, the sine(dU) = dU. So we have a total vertical force due to S of just 2 R dU^2 dr, or (2/R) dV

If we had a pressure difference, it would give an additional force of (P+dP)(RdU+dr)^2 - P(R^2 dU^2), which gives components dP R^2 dU^2 and 2 P R dU dr, or equivalently (dP/dr)*dV and (2/R)*P*dV

So we can write the total force balance equation as

(rho*g + (2/R)*S + (dP/dr) + (2/R)*P ) = 0.

We can see this is just what we worked out via the stress-energy tensor approach, modulo a few different sign choices - and a few additions due to relativity.

We can easily see that for the proper value of S, the wedge is under no net force, even with P=0.

We can also appreciate why P=0 is the correct boundary condition when there is a vacuum above the wedge .

Hey, you beat me to it. Based on what your prior math post suggested, I had worked out exactly the analysis above - that a pure tangential pressure that is an increasing function of r is sufficient for shell equilibrium.

 

[PAllen]

As I harped on in the past here, it's not a question of either separately being valid, but how they are combined. The assumed ab initio fundamental guiding principle is the universal validity of that bit about counting of flux lines through any given bounding enclosed surface (i.e. Gauss's law), and #10 in particular strongly suggests there are self-consistency issues in following that guiding principle (when gravity enters the equation) - the one which you (and everyone else in GR community it seems), work from as 'a given'. that Gauss's law applies *locally* in arbitrarily curved spacetime I do not question - but *global* validity is another matter.

See previous comments. When your current line of attack runs it's course, please consider taking up my suggestion!

Can you try to restate in as simple and clear a form as possible, this supposed inconsistency. I just read through and gave reference to a proof of Gauss's law globally assuming only Maxwell + EFE + differential geometry. Which part of this do you claim is internally inconsistent?

 

[Q-reeus]

Can you try to restate in as simple and clear a form as possible, this supposed inconsistency. I just read through and gave reference to a proof of Gauss's law globally assuming only Maxwell + EFE + differential geometry. Which part of this do you claim is internally inconsistent?

I'll do my best in offering, in words basically, why there must be something conceptually wrong. it repeats earlier input, but I will itemize. Given we have settled on a thin spherical mass shell as an appropriate 'test chamber' (owing to flat spacetime applying within), what are the generally accepted, basic effects such a test chamber has on a perturbatively small EM systen enclosed within - as determined both locally, and remotely? Straight away it can be said there is no effect locally - not even tidal effects since flat spacetime prevails within the shell interior. Here's an itemized check-list of what I believe there is, separately, agreed upon 'remote' effects:

1: Frequency and therefore energy redshift. Perform any operation whatsoever locally (within shell interior) that results in some energy release/exchange to the outside, and the usual redshift formulae apply - factor of √-g_tt for frequency and net energy release, and factor of -g_tt for radiated power. Or for an observer closer in than 'infinity', substitute the appropriate ratio of √-g_tt, -g_tt factors applying to radii r₁, r₂, where r₁ = shell mean radius, and r₂ = observer's radius.

2: A physically meaningful remote linkage ratio of 1:1 as locally determined at the two locales. Example: using idealized light and stiff connecting rods and bell cranks, we find that an observer within the shell will concur that when distant observer's rod is radially moved x units, x units are observed within shell also. This is independent of the outside observer's potential at radius r₂ - need not be at infinity.

3: Owing to finding that √g_rr within shell is unity - i.e. identical to coordinate value, we can meaningfully extend 2: above. If our remote linkage connects to say two parallel capacitor plates within the sphere, it is perfectly proper to infer that remotely pushing a rod x units will change the plate separation distance by x units, not just locally, but as determined in coordinate measure. Which amounts to this: the sole effect of shell mass M is to alter coordinate clock-rate and thus relative energy of whatever lies within the shell 'test chamber'. Spatial displacements are not effected locally or as remotely determined.

4: According to RN logical foundations, Gauss's law holds exactly, which in turn means E field of charge has no dependence on √-g_tt, and local values for electric field strength E match with coordinate values.

5: Now combine 2-4 above. Say, via linkages, a remote operator alters separation of charged capacitor plates enclosed within the shell (plates of effective area A, field strength E between plates) by some differential displacement dx. Then the change in field energy is dW = 1/2ε₀E²Adx - as determined both locally and remotely. Assuming that is, both dx and E have identical local and remotely inferred values, and that F = qE holds 'normally'. We left out item 1: from this consideration. As things stand, there cannot be a match with 1: which requires the remotely observed energy change obey the experimentally verified redshift requirement that dW = √-g_tt1/2ε₀E²Adx. So there is an evident conflict. I anticipated what seems at first sight the obvious fix that manages to preserve Gauss's law globally but also energy redshift - introduce a redshift of 'passive' charge such that F = q_pE = √-g_ttqE applies to coordinate value , 'active' charge |E| = q_a/(4πε₀r²) remaining unaffected. That would fix, here, the mechanical energy balance, but in order to fix things in terms of coordinate computed field energy, one must slap on that redshift factor of √-g_tt, and just how that could be justified other than on an ad hoc book-keeping basis is questionable.

The problem is it does not work under all situations - as shown back in #10. Newton's third law fails if the 'active'/'passive' charge fix is consistently adopted. Which makes that fix untenable imo. There may be some better way than suggested in #10 - assume modification of e₀, u₀ by factor1/√-g_tt, but if so it alludes me. Sorry if this is not what you consider an answer, but that's my line of thought.

There are further angles on this issue and #11 looks at one, but I'm not perfectly comfortable now that bit is completely sound. So anyway, my reasoning is itemized and I welcome anyone pointing to any weak links in that chain.

 

[PeterDonis]

I'd expect S to be positive - but the only needed boundary condition should be that the radial component , which we've been calling P, equals zero.

No, it isn't--at least, there is another constraint, though "boundary condition" may not be the right term for it. See below.

Taking the divergence of T_ab and setting it to zero is the easiest way of getting the necessary equations - but you already did that, you don't agree about the significance.

I'm not disputing anything about that equation--as you say, I already derived it. I'm saying there's *another* constraint equation on the tangential stress, s(r), which you aren't taking into account. See below.

We can also appreciate why P=0 is the correct boundary condition when there is a vacuum above the wedge .

I'm talking about a case where there is vacuum *below* the wedge. We have a thin, spherically symmetric shell with vacuum both outside *and* inside. The *only* nonzero stress-energy present at all is that within the shell.

Consider a plane through the center of the sphere (i.e., through the point r = 0). The net force on either hemisphere of the shell, normal to that plane, must be zero, or the shell will not be in static equilibrium. If there is vacuum inside and outside the shell, then the *only* force normal to that plane is the shell tangential stress, s(r). That means we must have

\int_{r_i}^{r_o} 2 \pi r s(r) dr = 0

for the shell to be in static equilibrium, where r_i and r_o are the shell inner and outer radius. Unless s(r) is zero everywhere, the only way that integral can be zero is if s(r) changes sign somewhere between r_i and r_o; it *cannot* be positive everywhere.

 

[PeterDonis]

Hey, you beat me to it. Based on what your prior math post suggested, I had worked out exactly the analysis above - that a pure tangential pressure that is an increasing function of r is sufficient for shell equilibrium.

See my response to pervect. We went all over this in a previous thread; I don't think pervect was involved in it, but you were.

 

[Q-reeus]

Consider a plane through the center of the sphere (i.e., through the point r = 0). The net force on either hemisphere of the shell, normal to that plane, must be zero, or the shell will not be in static equilibrium. If there is vacuum inside and outside the shell, then the *only* force normal to that plane is the shell tangential stress, s(r). That means we must have

\int_{r_i}^{r_o} 2 \pi r s(r) dr = 0

for the shell to be in static equilibrium, where r_i and r_o are the shell inner and outer radius. Unless s(r) is zero everywhere, the only way that integral can be zero is if s(r) changes sign somewhere between r_i and r_o; it *cannot* be positive everywhere.

Peter - I could never figure before your reasoning in getting that hoop stresses necessarily changed sign. Now it is evident. Sorry, but imo you are missing a basic consideration. That force balance eq'n is wrong - should not be 0 on the RHS, but ~ -(1/π)ρ2πr²δrg, where g is the mean value of radial self-gravitation acting over the shell thickness of δr and density ρ (I think that factor 1/π is correct for integration over a hemisphere). There are two equal and opposite net forces to consider. Hoop stresses are not opposing nothing - they oppose the 'weight' of self-gravity! Hence hoop stresses of the same sign throughout is perfectly ok and indeed expected.

 

[pervect]

No, it isn't--at least, there is another constraint, though "boundary condition" may not be the right term for it. See below.



I'm talking about a case where there is vacuum *below* the wedge. We have a thin, spherically symmetric shell with vacuum both outside *and* inside. The *only* nonzero stress-energy present at all is that within the shell.

Consider a plane through the center of the sphere (i.e., through the point r = 0). The net force on either hemisphere of the shell, normal to that plane, must be zero, or the shell will not be in static equilibrium. If there is vacuum inside and outside the shell, then the *only* force normal to that plane is the shell tangential stress, s(r). That means we must have

\int_{r_i}^{r_o} 2 \pi r s(r) dr = 0

for the shell to be in static equilibrium, where r_i and r_o are the shell inner and outer radius. Unless s(r) is zero everywhere, the only way that integral can be zero is if s(r) changes sign somewhere between r_i and r_o; it *cannot* be positive everywhere.

If you're integrating over the entire half-sphere, the force doesn't have to be zero. Consider a point on the half-sphere, right in the middle. In a Newtonian sense, there will be a force pushing down on this half-sphere at this point, due to gravity, that will contribute to this integral.

In fact, every piece of the half-sphere will contribute ato this integral due to gravity.

The tangential stress component simply supports the half-sphere against the gravitational forces.

Setting the radial pressure to zero means that the amount of radial angular momentum transported in the r direction is zero. This condition basically makes each part of the shell self-supporting against gravity and totally independent of the presence or absence of the other shell sections.

Another way of convicing yourself is to analyze the Newtonian forces on a wedge, as per my previous post. Assume P=0. You'll see that the presence of the S terms is both necessary (necessary, because P=0), and sufficient to counteract the downward force due to "gravity" on the wedge.

 

[pervect]

Hey, you beat me to it. Based on what your prior math post suggested, I had worked out exactly the analysis above - that a pure tangential pressure that is an increasing function of r is sufficient for shell equilibrium.

Yes - I agree!

To finish up Peter's analysis , the metric for a hollow sphere of total mass M with an inner boundry of R1 and an outer boundary of R2 is then just

for R1<r<R2
m(r) = \frac{M \left( r^3 - R1^3\right)}{R2^3 - R1^3}
J(r) = K \, exp \, \left( \int_{x=R1}^{x=r}\frac{2\,m(x)}{x\left(x-2\,m(x)\right)}\right)

K is an arbitrary constant, it should be chosen so that J(R2) = (1-2M/R2) if one wants to make the metric appear Schwarzschild in the exterior region. Alternatively, one can chose K=1, and have J(r)=1 in the interior region.

I haven't been able to integrate the integral analytically.

for r<R1
m(r)=0
J(r) = J(R1)

for r>R2
m(r)=M
J(r)=J(R2)\frac{1-2\,M/r}{1-2\,M/R2}

and finally the metric given m(r) and J(r)

ds^2 = -J(r) dt^2 + \frac{dr^2}{1-2\,m(r)/r} + r^2 d\theta^2 + r^2 sin^2 \theta d\phi^2

I'm not sure how much time I'll have to argue with Peter - I don't see any problem with the above solution, however, and I thought I'd post it.
If you throw it into GRTensor or an equivalent program, it gives a constant density solution with zero radial pressure, and a rather complicated formula for the tangential stress S(r) that vanishes at r=R1.

It's not the only possible solution of course- it's just one of the simpler ones. We can make P(r) nonzero as long as it vanishes at the r=R1 and r=R2, but it's simplest if it just vanishes everywhere.

 

[PeterDonis]

If you're integrating over the entire half-sphere, the force doesn't have to be zero.

The *total* force does, but you're right, I had left out a term in the integral.

Consider a point on the half-sphere, right in the middle. In a Newtonian sense, there will be a force pushing down on this half-sphere at this point, due to gravity, that will contribute to this integral.

In fact, every piece of the half-sphere will contribute to this integral due to gravity.

The tangential stress component simply supports the half-sphere against the gravitational forces.

Ok, yes, I see this. The total force will be the sum of two terms that cancel each other: the s(r) term that I wrote down, and a "weight of the shell" term which I can't write down an explicit formula for right now, I'll have to think about it some more.

Another way of convicing yourself is to analyze the Newtonian forces on a wedge, as per my previous post.

Yes, I already saw this part; I just didn't see how to reconcile it with the part above. I agree now with your entire analysis of the hollow shell with vacuum inside and outside; now I need to look at the case of a shell surrounding a charged gravitating body again.

 

[PeterDonis]

A (hopefully) quick general post on Maxwell's Equations for a static, spherically symmetric electric field. [Edit: corrected the E factors in the equations towards the end.]

The only nonzero components of the EM field tensor are F_{rt} = - F_{tr} = E(r) (we *define* E this way in the general case; we aren't yet specifying that E has any particular functional form, except that it obviously can only be a function of r).

One issue that always makes me nervous when writing down tensor components is which "version" of the tensor is the "canonical" one that just looks like what we expect from flat spacetime (in the R-N case, it would be F_rt = E = Q/r^2), and which versions of the tensor have extra factors of metric coefficients thrown in. In this case, I believe that the "canonical" version of the EM field tensor is the 2-form version (both indexes lower) that I wrote down above; thus E(r) should be the "pure" E field we expect from flat spacetime, the field a local Lorentz observer would see, with no factors of metric coefficients in it. This is how it appears to be in MTW, for example; also, when I wrote down the Gauss's Law integral earlier, it came out very simply as a contraction of F_ab with two vectors, where the final value had no metric coefficient factors in it. Everything seems to work out OK this way, so we'll go with it.

But we should note one thing at the outset: the standard equation for the Lorentz force, written in the global Schwarzschild coordinates, *does* have a metric coefficient factor in it. That equation is

A^a = e F^a_b u^b

where A^a is the acceleration induced on a particle with charge e per unit mass, and u^b is the particle's 4-velocity. For the case of a (possibly momentarily) static particle, the only nonzero component is

A^r = e F^r_t u^t = e F_{rt} g^{rr} u^t = e E \frac{g^{rr}}{\sqrt{| g_{tt} |}} = e E \frac{1}{g_{rr} \sqrt{| g_{tt} |}} = e E \frac{r - 2m}{r \sqrt{J}}

For the standard R-N case, we have J = 1 - 2m/r, and the actual measured acceleration will be the magnitude of the A vector, or

A = \sqrt{g_{rr}} A^r = e E \frac{r - 2m}{r \sqrt{J}} \sqrt{\frac{r}{r - 2m}} = e E \sqrt{\frac{r - 2m}{r J}} = eE

so all the metric coefficient factors cancel and we get the expected result (the same as we would have calculated in a local Lorentz frame where all the metric coefficient factors are unity).

Now we look at Maxwell's equations; specifically, the Maxwell Equation with source. (The other maxwell equation, dF = 0, is easy to check and I won't give it explicitly here.) This equation is

\nabla_a F^{ab} = 4 \pi j^b = \partial_a F^{ab} + \Gamma^a_{ac} F^{cb} + \Gamma^b_{ac} F^{ac}

where j^b is the charge-current 4-vector. Since F is an antisymmetric tensor, the last term above will always be zero (terms in the summation will always occur in pairs that cancel each other because the Christoffel symbols are symmetric in the lower two indices).

The only nonzero component of this equation, given the EM field tensor above, turns out to be

\nabla_a F^{at} = 4 \pi j^t = \partial_a F^{at} + \Gamma^a_{ac} F^{ct}

Since F^rt is the only nonzero component of F with a "t" as the second index, the above becomes

4 \pi j^t = \partial_r F^{rt} + \Gamma^a_{ar} F^{rt} = \partial_r F^{rt} + \left( \Gamma^t_{tr} + \Gamma^r_{rr} + \Gamma^\theta_{\theta r} + \Gamma^\phi_{\phi r} \right) F^{rt}

which gives

4 \pi j^t = \frac{d}{dr} \left( g^{rr} g^{tt} E \right) + \left( \frac{J&#039;}{2J} + \frac{r m&#039; - m}{r \left( r - 2m \right) } + \frac{2}{r} \right) \left( g^{rr} g^{tt} E \right)

where I have used primes for radial derivatives for brevity, and where the metric coefficient factors next to E are there because we are looking at the bivector F^rt instead of the 2-form F_rt, so we have to raise both indexes. Substituting for J'/2J and m', we find that the m's cancel and we have

4 \pi j^t = \frac{d}{dr} \left( \frac{r - 2m}{rJ} E \right) + \left( \frac{2}{r} + \frac{4 \pi r^3 \left( \rho + p \right)}{r \left(r - 2m \right)} \right) \left( \frac{r - 2m}{rJ} E \right)

I'll go into the specific implications of this for the scenarios we've been discussing in a follow-on post.

 

[pervect]

y can only be a function of r).

One issue that always makes me nervous when writing down tensor components is which "version" of the tensor is the "canonical" one that just looks like what we expect from flat spacetime (in the R-N case, it would be F_rt = E = Q/r^2), and which versions of the tensor have extra factors of metric coefficients thrown in.

Styles may vary, but using MTW as a guide, I assume that anything written as F_{ab} is in a coordinate basis, or a holonomic basis, see http://en.wikipedia.org/w/index.php?title=Holonomic&oldid=505441575, in which the basis vectors are the partial derivatives of the coordinates i.e. \partial / \partial r, etc. Because of this, the basis vectors are not usually unit length, which implies that you have "extra factors of the metric coefficients thrown in".

If one instead chooses a basis of unit vectors, i.e. \hat{t}, \hat{r}, \hat{\theta}, \hat{\phi} (this isn't exactly MTW's notation but is fairly common usage), and use them as a nonholonomic basis for the tensor, then the tensor is written as F_{\hat{a}\hat{b}}. The "hats" tell you that it's a nonholonomic basis, and also mean that you won't get any extra metric coefficients.

To work with nonholonomic basis explicitly, you're supposed to use the "Ricci rotation coefficients", usually written as \omega_{\alpha \mu \nu}, rather than the more usual Christoffel symbols, but I let GRTensor deal with all that. Wald does go through the math on pg 50 as to how the Ricci rotation coefficients are defined.

Some authors will write out the basis vectors specifically for you whenever they use a nonholonomic basis (which is needed to make the metric coefficeints reliably vanish). This is the most time consuming, but the most clear.

I always have the sneaking feeling that I lose about 90% of the readers whenever I use either approach, however (the hats, or writing down the basis vectors).

 

[Q-reeus]

Still waiting feedback on #69 - I was asked!

 

[PeterDonis]

Styles may vary, but using MTW as a guide, I assume that anything written as F_{ab} is in a coordinate basis, or a holonomic basis

This is the convention I usually use as well, and the one I have been using in all my posts in this thread.

the basis vectors are not usually unit length, which implies that you have "extra factors of the metric coefficients thrown in".

Only for some "versions" of the tensor; that's the issue that makes me nervous. For example, take the SET you wrote down earlier (post #61). In the coordinate basis, the 2-form version, T_{ab}, has metric coefficient factors in each component. Now compute the "mixed" components T^a_b; when you raise an index on each component, you multiply it by the corresponding inverse metric coefficient, so the mixed components end up having *no* metric coefficient factors in them. That's why I wrote down the EFE using the "mixed" components (MTW do the same thing in a number of their examples)--it looks simpler that way. (The only quirk is that T^t_t = - \rho, the minus sign is there because the (-+++) metric sign convention is being used.)

The reason all the above works is that if we express the actual physical observables, \rho and so forth, as contractions of the SET with 4-vectors, we write expressions like this:

\rho = u^a T_{ab} u^b

In a local inertial frame, with a basis of unit vectors (the "hat" vectors), the 4-velocity components are just (1, 0, 0, 0), so we obviously have to have, for example, T_{\hat{0} \hat{0}} = \rho. In the coordinate basis, though, the 4-velocity (assuming a "static" observer, the simplest case) is ( 1 / \sqrt{| g_{tt} |}, 0, 0, 0), so we must have T_{00} = g_{tt} \rho = J(r) \rho as you wrote down. So if we want to sanity check what the coordinate transformation equations are telling us, we can always link things back to the definitions of observables in terms of scalars--contractions of vectors and tensors that leave no indexes free.

Similar reasoning is what led me to the conclusion that, in the Schwarzschild coordinate basis, the 2-form "version" of the EM field tensor, F_{ab}, is the one that has no metric coefficients appearing in it. This seems to be consistent with what I find in MTW, but I don't know that they justify it in terms of the way the coordinate transformation from a local inertial frame to the coordinate basis works. I'm justifying it by the definition of a physical observable, the Gauss's Law integral for the charge, which only comes out right in the coordinate basis (for the case of R-N spacetime) if F_{rt} = Q / r^2, with no metric coefficients appearing, because the full expression (once we do the angular part of the integral) is

Q = r^2 F_{ab} n^a u^b = r^2 F_{rt} n^r u^t

and n^r contributes a factor 1 / \sqrt{g_{rr}} and u^t contributes a factor 1 / \sqrt{| g_{tt} |}, which cancel (for R-N spacetime). But the same expression holds in a local inertial frame, with unit basis vectors, so we must also have F_{\hat{r} \hat{t}} = Q / r^2. That makes me nervous; it makes me think that there may be metric coefficient factors I have left out, that just happen to cancel when g_rr = 1 / g_tt, as is true in R-N spacetime but will not be true, for example, inside a shell around a charged gravitating body. I don't see any metric coefficient factors written in MTW when describing the EM field tensor in curved spacetime, or in other papers I have found on R-N spacetime, but that may just mean they are assuming that any such factors will cancel.

 

[PeterDonis]

Say, via linkages, a remote operator alters separation of charged capacitor plates enclosed within the shell (plates of effective area A, field strength E between plates) by some differential displacement dx. Then the change in field energy is dW = 1/2ε0E2Adx - as determined both locally and remotely.

Locally, yes. Remotely, no. You left out a key point: how much work does the remote operator have to do to alter the plate separation by dx (where dx is a proper distance, not a coordinate distance), compared to the local case? Because of the potential difference, these two cases will require a different amount of work to be done by the operator. That different amount of work will exactly compensate for the redshift of the field energy.

Edit: I see that you agree that the work *should* be "redshifted" like the energy is, but you think this somehow contradicts Gauss's Law. It doesn't. You talk about Gauss's Law in R-N spacetime, but your scenario isn't set in R-N spacetime; there is no global E field, only local E fields between the plates of each capacitor. The Gauss's Law integral for the capacitor inside the shell, when viewed in coordinate terms, *will* include a "redshifted" E field between the plates; but that E field is only local, between the plates; it doesn't extend through the entire spacetime, so there's no reason why that capacitor's E field has to be the same, in coordinate terms, as the E field of the capacitor at infinity.

 

[Q-reeus]

Q-reeus: "Say, via linkages, a remote operator alters separation of charged capacitor plates enclosed within the shell (plates of effective area A, field strength E between plates) by some differential displacement dx. Then the change in field energy is dW = 1/2ε0E2Adx - as determined both locally and remotely."

Edit: I see that you agree that the work *should* be "redshifted" like the energy is, but you think this somehow contradicts Gauss's Law. It doesn't. You talk about Gauss's Law in R-N spacetime, but your scenario isn't set in R-N spacetime; there is no global E field, only local E fields between the plates of each capacitor. The Gauss's Law integral for the capacitor inside the shell, when viewed in coordinate terms, *will* include a "redshifted" E field between the plates; but that E field is only local, between the plates; it doesn't extend through the entire spacetime, so there's no reason why that capacitor's E field has to be the same, in coordinate terms, as the E field of the capacitor at infinity.

How do you get that E can be redshifted between capacitor plates in coordinate measure and still be compatible with global validity of Gauss's law? Change the setup then to that I suggested last para. in #52:

And if parallel plate cap is for some reason a problem geometry, maybe e.g. case of two almost touching concentric charged shells undergoing a differential radial relative displacement.

This is a mere rearrangement from parallel plates to concentric shells, but makes it particularly clear one either accepts:
(a); Gauss's law holds globally - absolutely enforcing zero redshift of E field between shells in coordinate measure, or
(b); One concedes, based on the energy redshift that will actually be so, either E field between shells redshifts in coordinate measure ('active' charge redshift) - and therefore must be redshifted as received at any remote location, or F = qE fails ('passive' charge redshift). This is all about the logic (or not) of 'global conservation of flux lines'.

I have covered these possibilities in #10, #69, and shown in #10 that 'active'/'passive' split is no real answer. To repeat - if as it seems from your comment above, you concede a redshifted E in coordinate measure, full consistency with that demands Gauss's law fails globally. It is logically inconsistent to have diminished flux line count between parallel plates cap in coordinate measure, whilst a mere rearrangement of the charge distribution (shells) magically re-establishes zero diminution of flux lines.

 

[PeterDonis]

How do you get that E can be redshifted between capacitor plates in coordinate measure and still be compatible with global validity of Gauss's law?

"Global validity of Gauss's Law" doesn't mean what you think it means. It does mean that the integral of E over the area of a 2-sphere in R-N spacetime is independent of radius, which means that E itself is Q/r^2 in that case, with no extra metric coefficient factors. It does *not* mean "E field components are never affected by spacetime curvature in any scenario whatsoever". You appear to be taking a conclusion that is specific to R-N spacetime and treating it as a universal law. It isn't.

[Edit: I should make clear that even the limited statement I made above, about the Gauss's Law in R-N spacetime, may change when a shell of matter is introduced. Read my previous posts on that, including the one on the implications of Maxwell's Equations: if the matter in the shell is assumed to not affect the field, then it can't be neutral, it must be charged; and if the matter in the shell is stipulated to not be charged, then it must affect the field--the E field inside the shell *cannot* be just Q/r^2.]

 

[Q-reeus]

"Global validity of Gauss's Law" doesn't mean what you think it means. It does mean that the integral of E over the area of a 2-sphere in R-N spacetime is independent of radius, which means that E itself is Q/r^2 in that case, with no extra metric coefficient factors.

Actually that's just what I thought it meant. And if true, redshift of E cannot occur in coordinate measure. Think about it in terms of field line count. Say one line for every elemental charge (electron). Gauss's law holding = line count through any bounding enclosed surface is invariant. Apply that to interior region of shell - it follows that line density (field strength E) is independent of gravitational potential for any EM structure within. There is no way around that as long as one sticks to lines having to always begin and end on charge (Gauss's law!). Concentric charged shells config. makes that plainly obvious surely. Hence the energy dilemma must arise - and, once this is firmly grasped, the one seemingly obvious compromise cure, as already covered, will be found to lead to another paradox. I can see only one true remedy - relax Gauss's law via potential dependent vacuum permittivity, permeability.

It does *not* mean "E field components are never affected by spacetime curvature in any scenario whatsoever".

I'm inclined to think 'totally warp free' probably does logically follow. Strange indeed if indifference to √-g_tt is not matched by similar indifference to √g_rr, √g_θθ etc. But then I see any indifference to metric components as a fantasy. Witness gravitational redshift, bending of light.

[Edit: I should make clear that even the limited statement I made above, about the Gauss's Law in R-N spacetime, may change when a shell of matter is introduced. Read my previous posts on that, including the one on the implications of Maxwell's Equations: if the matter in the shell is assumed to not affect the field, then it can't be neutral, it must be charged; and if the matter in the shell is stipulated to not be charged, then it must affect the field--the E field inside the shell *cannot* be just Q/r^2.]

Good luck with that one. If so, and you wish to retain Gauss's law as working principle, imo something is terribly amiss! Unless it's purely a fairly inconsequential artifact of mass density creating a coordinate measured jog in the r part of q/r2. Just my layman's musing on that. :zzz:

 

[PeterDonis]

Apply that to interior region of shell - it follows that line density (field strength E) is independent of gravitational potential for any EM structure within.

In your capacitor scenario, the E fields of the capacitors are confined to between their plates; a Gauss's Law integral over a surface enclosing an entire capacitor gives zero. So there's nothing to "redshift" when you go outside the shell and try to "look inside".

There is no way around that as long as one sticks to lines having to always begin and end on charge (Gauss's law!). Concentric charged shells config. makes that plainly obvious surely.

In this case the only nonzero E field is between the shells; the net charge of both shells together is zero. So a Gauss's Law integral will only give a nonzero result if the surface of integration only encloses one shell. A spherical surface between the shells will enclose only the inner shell, and that integral *will* be invariant regardless of which surface (between the shells) you choose. Can you think of any other closed surface that encloses just one shell, without intersecting one of them (which makes the Gauss's Law integral invalid)?

 

[pervect]

How do you get that E can be redshifted between capacitor plates in coordinate measure and still be compatible with global validity of Gauss's law? Change the setup then to that I suggested last para. in #52:

This is a mere rearrangement from parallel plates to concentric shells, but makes it particularly clear one either accepts:
(a); Gauss's law holds globally - absolutely enforcing zero redshift of E field between shells in coordinate measure, or
(b); One concedes, based on the energy redshift that will actually be so, either E field between shells redshifts in coordinate measure ('active' charge redshift) - and therefore must be redshifted as received at any remote location, or F = qE fails ('passive' charge redshift). This is all about the logic (or not) of 'global conservation of flux lines'.

I have covered these possibilities in #10, #69, and shown in #10 that 'active'/'passive' split is no real answer. To repeat - if as it seems from your comment above, you concede a redshifted E in coordinate measure, full consistency with that demands Gauss's law fails globally. It is logically inconsistent to have diminished flux line count between parallel plates cap in coordinate measure, whilst a mere rearrangement of the charge distribution (shells) magically re-establishes zero diminution of flux lines.

Gauss's law holds globally if, and only if, one uses coordinate independent methods. You seem to me to either incapable or unwilling of learning coordinate independent methods (aka tensors), however - this observation is based on both past experience, and also just the fact that this thread exists.

So, let's try a different approach. If you write it in the coordinate-dependent style that you've adopted as your entire approach to physics, what we call "Gauss's law" is a different law in every different coordinate system. The good news is that all these different laws, which share the same name, are related mathematically by the methods that we use to transform from one coordinate system to another. The bad news is that unless you learn how to do these transformations, which as far as I know requires learning tensors, you'd need to learn each particular version of Gauss' law "by rote" for every coordinate system you might want to use. And the same applies for any other physical law, really.

 

[Q-reeus]

In your capacitor scenario, the E fields of the capacitors are confined to between their plates; a Gauss's Law integral over a surface enclosing an entire capacitor gives zero. So there's nothing to "redshift" when you go outside the shell and try to "look inside".

This is misplaced analysis. The fact that 1 + (-1) = 0 doesn't mean there are no 1's there. The relevant integral then is to take the enclosing surface over just one plate. Do you then find that line count, and from that line density, varies with potential?

In this case the only nonzero E field is between the shells; the net charge of both shells together is zero. So a Gauss's Law integral will only give a nonzero result if the surface of integration only encloses one shell.

And similarly as per parallel cap scenario, that's the obvious way of looking at it. again, we are not interested in proving 1 + (-1) = 0, we know that trivial fact. In my book, global validity of Gauss's law = invariance of E field wrt potential, no if's, but's, or maybe's. Additionally assume F = qE, apply to any quasi-static scenario such as I have presented in #10 etc. and we have an instant recipe for clash with the known fact of energy redshift. This is just continual repetition, and I'd like to think we can move forward somehow.

A spherical surface between the shells will enclose only the inner shell, and that integral *will* be invariant regardless of which surface (between the shells) you choose.

Which is entirely in accord with my argument! Have you not understood, after so long now, what I have been driving at? Gauss's law boils down to that field lines must begin and end on charge. No? And that single principle *if taken as fact* automatically enforces the impossibility of E field strength linkage to gravitational potential (obviously in coordinate measure). Can you give me a single example showing otherwise, in line with what you wrote in #80 which claimed redshift of E will occur - whilst also observing Gauss's law? Cannot be.

Can you think of any other closed surface that encloses just one shell, without intersecting one of them (which makes the Gauss's Law integral invalid)?

No; and see last comments. Can you in turn provide that example refuting what I just wrote above - please. And just to reiterate. I introduced spherical shells, rods and levers attached to capacitors etc. owing to it making things physically obvious imo. Coordinate values for length were there unaffected by potential and it implied a direct correspondence with coordinate field values based on forces transmitted. But really this argument hinges just on validity of the one principle - do field lines globally always begin and end on charge?

I see from entry #85 there seems to be some question mark placed over the equivalence of that to Gauss's law in general setting. Which, unless I have missed something in the translation, would greatly surprise me - especially given the content of #16.

 

[Q-reeus]

Gauss's law holds globally if, and only if, one uses coordinate independent methods.

Does this mean that using Schwarzschild coords, it actually fails? If so how and to what extent? Be specific on this please.

You seem to me to either incapable or unwilling of learning coordinate independent methods (aka tensors), however - this observation is based on both past experience, and also just the fact that this thread exists.

On that last bit - just check who the OP was - not me. Or maybe you meant 'continues to run'? I appreciate that my low brow approach rankles the likes of yourself and other participants here with degrees and Ph.D's. But whether labelled 'intuition' or otherwise, note that though frozen out of the subsequent discussion (and I get the message there), I did manage to get it right from the start re a certain issue running on here for some length. Anyway let's get to the point of what you are trying to say below, because any specific impact it's supposed to have on this topic is not clear to me.

So, let's try a different approach. If you write it in the coordinate-dependent style that you've adopted as your entire approach to physics, what we call "Gauss's law" is a different law in every different coordinate system. The good news is that all these different laws, which share the same name, are related mathematically by the methods that we use to transform from one coordinate system to another. The bad news is that unless you learn how to do these transformations, which as far as I know requires learning tensors, you'd need to learn each particular version of Gauss' law "by rote" for every coordinate system you might want to use. And the same applies for any other physical law, really.

Given what you say here is so as generalization, what is the bottom-line effect? Is there some version of Gauss's law which undermines the line counting argument you gave back in #16 for instance? Just what is bedrock if not that? Does anything in your above passage actually invalidate the substantive conclusions of what I wrote in #10, #69 or subsequently - in particular is there some failing you can identify in the guts of #86?

 

[PeterDonis]

Do you then find that line count, and from that line density, varies with potential?

I explicitly said the charge integral is independent of radius, which I think is what you are asking here. See below.

Gauss's law boils down to that field lines must begin and end on charge.

Yes, obviously.

And that single principle *if taken as fact* automatically enforces the impossibility of E field strength linkage to gravitational potential (obviously in coordinate measure).

That last parenthetical comment destroys your argument, as pervect noted. The quantity that cannot vary is the *invariant* value of the Gauss's Law charge integral over any surface that encloses the same charge. That means you have to construct an integral that does *not* depend on "coordinate measure".

what you wrote in #80 which claimed redshift of E will occur - whilst also observing Gauss's law?

I already answered this, but I'll pose a question to check to see if you understood the full implications of my answer. Consider the capacitor that's inside the shell, so the "redshift factor" J(r) is less than 1. Consider any 2-surface that encloses just one plate of the capacitor; the Gauss's Law integral over that surface will give the charge on that capacitor plate, and will be the same for any surface that encloses just that one plate.

Now consider the second capacitor at infinity, where we verify by some linkage mechanism that the proper distance between the plates is the same as the first capacitor. Consider a 2-surface that encloses just one plate of the second capacitor; the Gauss's Law integral over that surface will give the charge on the second capacitor's plate, and will be the same for any surface that encloses just that one plate.

Assume the charge observed on the second capacitor's plate (the one at infinity) is q. What do you predict will be the charge observed on the first capacitor's plate, and why?

[Edit: Actually there should be two questions here: (1) what do you, Q-reeus, predict based on *your* understanding of physics, your understanding of the "correct" meaning of Gauss's Law, etc., for the charge on the first capacitor? (2) What do you, Q-reeus, think *GR* predicts for the charge on the first capacitor? I am particularly curious to see if the answers to these two questions are different.]

[Edit: I suppose I should clarify that by "charge observed" I mean "number obtained by evaluating the Gauss's Law integral".]

But really this argument hinges just on validity of the one principle - do field lines globally always begin and end on charge?

Of course they do; but that doesn't mean they extend through all of spacetime. Field lines can be confined to a small region, as they are in the case of a capacitor (at least the idealized kind we're discussing here): there are field lines *only* between the plates, not anywhere else. What do you suppose this implies for the answer to the question I posed above?

 

[Q-reeus]

Q-reeus: "Do you then find that line count, and from that line density, varies with potential?"
I explicitly said the charge integral is independent of radius, which I think is what you are asking here.

Follow that logically through, and game's over then, even if certain participants still fail to recognize and/or acknowledge that.

Q-reeus: "And that single principle *if taken as fact* automatically enforces the impossibility of E field strength linkage to gravitational potential (obviously in coordinate measure)."

That last parenthetical comment destroys your argument, as pervect noted. The quantity that cannot vary is the *invariant* value of the Gauss's Law charge integral over any surface that encloses the same charge. That means you have to construct an integral that does *not* depend on "coordinate measure".

What is the actual operational, relevant distinction? Can you apply this distinction to the scenarios I have given and show, explicitly, how it invalidates the core of my argument? Pervect made generalized statements without applying them specifically to the situation at hand. I note he has not responded to my #87, which encourages me to to draw conclusions based on: If someone has answers, they tend to supply them, if not, they tend to go silent. Or was it a case of giving up in disgust? Not really sure.

Q-reeus: "what you wrote in #80 which claimed redshift of E will occur - whilst also observing Gauss's law?"
I already answered this,...

Then refresh my memory please, because I cannot recall where you have in fact done so.

...but I'll pose a question to check to see if you understood the full implications of my answer. Consider the capacitor that's inside the shell, so the "redshift factor" J(r) is less than 1. Consider any 2-surface that encloses just one plate of the capacitor; the Gauss's Law integral over that surface will give the charge on that capacitor plate, and will be the same for any surface that encloses just that one plate.

Agreement!

Now consider the second capacitor at infinity, where we verify by some linkage mechanism that the proper distance between the plates is the same as the first capacitor. Consider a 2-surface that encloses just one plate of the second capacitor; the Gauss's Law integral over that surface will give the charge on the second capacitor's plate, and will be the same for any surface that encloses just that one plate.

Further agreement!

Assume the charge observed on the second capacitor's plate (the one at infinity) is q. What do you predict will be the charge observed on the first capacitor's plate, and why?

[Edit: Actually there should be two questions here: (1) what do you, Q-reeus, predict based on *your* understanding of physics, your understanding of the "correct" meaning of Gauss's Law, etc., for the charge on the first capacitor? (2)

My answer to that is in effect supplied way back here: https://www.physicsforums.com/showpost.php?p=3964623&postcount=248 Unless I am intended to be trapped by smart lawyer tactics here (heaven forbid!), it amounts to 'effective' charge owing to 'effective' dielectric screening (coordinate determined). Can you grasp my angle on that?

What do you, Q-reeus, think *GR* predicts for the charge on the first capacitor?

My understanding follows yours and others input here - it will be just q - i.e invariant wrt potential. Otherwise, as I have maintained from the very beginning, one could not logically have a finite external E for the so-called RN charged BH.

I am particularly curious to see if the answers to these two questions are different.]

Curiosity no more.

[Edit: I suppose I should clarify that by "charge observed" I mean "number obtained by evaluating the Gauss's Law integral".]

No difference.

Q-reeus: "But really this argument hinges just on validity of the one principle - do field lines globally always begin and end on charge?"

Of course they do; but that doesn't mean they extend through all of spacetime. Field lines can be confined to a small region, as they are in the case of a capacitor (at least the idealized kind we're discussing here): there are field lines *only* between the plates, not anywhere else. What do you suppose this implies for the answer to the question I posed above?

Nothing at all. As observed in #86, take relevant surfaces of integration, not those designed to trivialize.

 

[PeterDonis]

it will be just q - i.e invariant wrt potential.
Curiosity no more.

No difference.

Ok, so you agree that the charge on both capacitor 1 (inside the shell where J(r) < 1) and capacitor 2 (at infinity) will be q, and you agree that GR predicts this (and I agree with that as well). Since q is an invariant number, calculated by taking a simple Gauss's Law surface that encloses one plate of each capacitor, all observers must agree on it; i.e., it doesn't matter whether you evaluate the integral, say for capacitor 1, in a local inertial frame or in the global Schwarzschild coordinates. So now I have some further questions:

Assume that the E field between the plates of capacitor 2 (at infinity) is E, and the plate separation (proper distance) is d. Since capacitor 2 is at infinity, it doesn't matter whether we evaluate these quantities in a local inertial frame or in the global Schwarzschild coordinates; in both cases the relevant metric coefficients are just +/- 1 (-1 for g_tt, +1 for g_rr_). By hypothesis, the plate separation of capacitor 1 (inside the shell) is also d, in terms of proper distance.

Now for the questions:

(1a) What is the E field between the plates of capacitor 1, as evaluated in a local inertial frame (where g_tt = -1 and g_rr = 1)?

(1b) What is the energy stored in capacitor 1, as evaluated in a local inertial frame?

(2a) What is the E field between the plates of capacitor 1, as evaluated in the global Schwarzschild coordinates (where g_tt = - J(r), and J(r) < 1; and g_rr = 1)?

(2b) What is the energy stored in capacitor 1, as evaluated in the global Schwarzschild coordinates?

Same "split" of the questions as before, if you think the "correct" answer given by your understanding of the physics is different than the answer GR would predict.

 

[Q-reeus]

Ok, so you agree that the charge on both capacitor 1 (inside the shell where J(r) < 1) and capacitor 2 (at infinity) will be q, and you agree that GR predicts this (and I agree with that as well). Since q is an invariant number, calculated by taking a simple Gauss's Law surface that encloses one plate of each capacitor, all observers must agree on it; i.e., it doesn't matter whether you evaluate the integral, say for capacitor 1, in a local inertial frame or in the global Schwarzschild coordinates. So now I have some further questions:

Firstly, recall our relative stances here. I agree with the above *only insofar as it represents the GR (your) perspective*. Take note! My own perspective was given in that prior posting's link to #248 in that other thread. Which means imo q *in effect* is reduced in coordinate measure - as consistently maintained. Which means - forget about 'line counting'. Let's not confuse the two! No clever lawyer tactics!

Assume that the E field between the plates of capacitor 2 (at infinity) is E, and the plate separation (proper distance) is d. Since capacitor 2 is at infinity, it doesn't matter whether we evaluate these quantities in a local inertial frame or in the global Schwarzschild coordinates; in both cases the relevant metric coefficients are just +/- 1 (-1 for g_tt, +1 for g_rr_). By hypothesis, the plate separation of capacitor 1 (inside the shell) is also d, in terms of proper distance.

Now for the questions:

(1a) What is the E field between the plates of capacitor 1, as evaluated in a local inertial frame (where g_tt = -1 and g_rr = 1)?

E - on my or your (standard RN/GR) perspective.

(1b) What is the energy stored in capacitor 1, as evaluated in a local inertial frame?

Logically can only be that inferred by extrapolation in #69: dW = 1/2ε₀E²Adx, so total = 1/2ε₀E²Ad, where d = plate separation. We all agree on this.

(2a) What is the E field between the plates of capacitor 1, as evaluated in the global Schwarzschild coordinates (where g_tt = - J(r), and J(r) < 1; and g_rr = 1)?

Depends on one's perspective. On your (standard RN/GR) perspective, it has to be E, which follows directly from assuming global validity of Gauss's law, as evidenced in e.g. #34. My own perspective entails a reduced *coordinate evaluated* q' = √-g_ttq, as per #248 in thread previously linked.

(2b) What is the energy stored in capacitor 1, as evaluated in the global Schwarzschild coordinates?

If one follows the logic of RN/GR, it is unaltered from the local value. As shown in #10 and various postings subsequently (note carefully here - this entails that one either assumes an undifferentiated q in F = qE, or one attempts a split into 'active/'passive' charge as per #10) . You should have by now no doubt of my view - Schwarzschild = coordinate value is redshifted by factor √(-g_tt), in accordance with experience and the logic of applying potential modified permittivity, permeability, as per #10, and #248 in that other thread referenced above.

Not wishing to be too melodramatic about this saga, I nonetheless feel somewhat like a hapless defendent way back in a Stalinist show trial. Guilty is guaranteed outcome! But hey, this is 2012, and things are very different, right?

 

[Dale]

Not wishing to be too melodramatic about this saga, I nonetheless feel somewhat like a hapless defendent way back in a Stalinist show trial. Guilty is guaranteed outcome! But hey, this is 2012, and things are very different, right?

Then perhaps you should consider following the rules of evidence when presenting your arguments. Even in non-show trials the outcome is guaranteed if one of the lawyers is unfamiliar with large parts of the law and refuses to follow the rules of the court.

 

[Q-reeus]

Then perhaps you should consider following the rules of evidence when presenting your arguments. Even in non-show trials the outcome is guaranteed if one of the lawyers is unfamiliar with large parts of the law and refuses to follow the rules of the court.

Fair comment DS - but then what exactly are the rules here?
[PS: only recently got involved with Classical Physics section. Amazed quite frankly with how long it took you to come to a 'final decision' here: https://www.physicsforums.com/showpost.php?p=4024036&postcount=348]

 

[Dale]

Fair comment DS - but then what exactly are the rules here?

The EFE, Maxwells equations, and the corresponding mathematical frameworks of Riemannian geometry and classical mechanics.

Amazed quite frankly with how long it took you to come to a 'final decision' here

That is a good example of a thread where I did in fact change my mind based on the evidence presented. Regarding the amount of time, I thought it was reasonable, I certainly haven't seen you change your mind over that timeframe.

 

[PeterDonis]

Firstly, recall our relative stances here. I agree with the above *only insofar as it represents the GR (your) perspective*. Take note! My own perspective was given in that prior posting's link to #248 in that other thread. Which means imo q *in effect* is reduced in coordinate measure - as consistently maintained. Which means - forget about 'line counting'. Let's not confuse the two! No clever lawyer tactics!

I'm not sure what you mean by that last comment, but you appear to me to have changed your answer. I asked you specifically what *your* answer was, in addition to asking what you thought the GR answer was. I also asked it with respect to a specific, well-defined question about the number that results from the Gauss's Law integral over a surface enclosing one plate of the capacitor. I didn't say anything about "line counting", or about any other "interpretation" of what the integral "means". I just asked about the actual, numerical value you get when you do the integral.

From your previous post I understood that your answers to both questions were the same: you thought the correct value of the integral was q, and you thought that GR says that the value of the integral is q. Now you appear to be saying that your answers to the two questions are different, but I can't tell for sure what the difference is. I *think* what the above quote means is: your correct value for the integral is q, you think that I think the GR value for the integral is q, but you think the "real" GR value for the integral is q sqrt(J), which is less than q--in other words, you think I have incorrectly stated what the "real" GR value for the integral is, so that I think the GR value is correct, but actually it is wrong. But the above quote could also mean: your value for the integral is q sqrt(J), but the GR value for the integral is q (which is what I think it is), and so you think the GR value for the integral is wrong, period. Which is it?

I can't respond to the rest of your post since I'm not sure now what your answer is on the previous questions about charge.

 

[Q-reeus]

The EFE, Maxwells equations, and the corresponding mathematical frameworks of Riemannian geometry and classical mechanics.

OK but my rule here is: no paradoxes when applied to any specific situation. And you know what I have been arguing on that.

That is a good example of a thread where I did in fact change my mind based on the evidence presented. Regarding the amount of time, I thought it was reasonable, I certainly haven't seen you change your mind over that timeframe.

Fair enough about taking your time - I'm not always above a little tit-for-tat. Maybe wrongly perceived the intent of your comments. And true I haven't changed my mind.

 

[Q-reeus]

I'm not sure what you mean by that last comment,

Sorry; I did get a little testy there, and maybe there is mutual misunderstanding as to what we both were meaning.

but you appear to me to have changed your answer.

No. Will clear that up below.

I asked you specifically what *your* answer was, in addition to asking what you thought the GR answer was. I also asked it with respect to a specific, well-defined question about the number that results from the Gauss's Law integral over a surface enclosing one plate of the capacitor. I didn't say anything about "line counting", or about any other "interpretation" of what the integral "means". I just asked about the actual, numerical value you get when you do the integral.

From your previous post I understood that your answers to both questions were the same: you thought the correct value of the integral was q,

No. I have said the correct value is q *if one accepts global validity of Gauss's law*, and that's not my view of how it actually goes.

and you thought that GR says that the value of the integral is q.

If ever I may have expressed it as 'What GR says', it referred to the RN incorporation of Gauss's law into GR as globally valid. GR per se doesn't afaik say anything itself about the matter. But given the seemingly complete acceptance of RN solution within GR community, one could loosely say 'GR says' by association.

Now you appear to be saying that your answers to the two questions are different, but I can't tell for sure what the difference is. I *think* what the above quote means is: your correct value for the integral is q,

No. Have been at pains to state numerous times my view is we obtain an effective q = sqrt(J)q by way of effective dielectric shielding when factor 1/sqrt(J) operates to increase the vacuum permittivity and permeability. See for instance, among many subsequently, #10 this thread, and 248# and later in that other thread.

you think that I think the GR value for the integral is q,

Correct.

but you think the "real" GR value for the integral is q sqrt(J), which is less than q--in other words, you think I have incorrectly stated what the "real" GR value for the integral is, so that I think the GR value is correct, but actually it is wrong.

No, we both agree (I think) the real GR value is just q - by 'real' meaning the one in accord with global Gauss's law holding and incorporated in RN sol'n.

But the above quote could also mean: your value for the integral is q sqrt(J), but the GR value for the integral is q (which is what I think it is), and so you think the GR value for the integral is wrong, period. Which is it?

That passage is correct as representing my view.

I can't respond to the rest of your post since I'm not sure now what your answer is on the previous questions about charge. Sorry

No need to be. Hope that is all now clear - and again my sorry for the 'lawyer references'. :shy::zzz:

 

[Dale]

OK but my rule here is: no paradoxes when applied to any specific situation.

Agreed. Part of the reason for "the rules of evidence" is to prevent paradoxes.

 

[PeterDonis]

Hope that is all now clear

I think so, but just to recap:

* You believe the charge on the capacitor inside the shell should be q sqrt(J).

* You believe that GR says the charge on the capacitor inside the shell is q.

* You believe the E field inside the capacitor, evaluated in a local inertial frame, is E, and you believe GR agrees with that.

* You believe the E field inside the capacitor, evaluated in global Schwarzschild coordinates, should be E sqrt(J), by consistency with your answer on charge above.

* You believe that GR says the E field inside the capacitor, evaluated in global Schwarzschild coordinates, is E, because that's the only way it can be consistent with GR's answer above on charge.

* You believe that the energy stored in the capacitor, evaluated in a local inertial frame, is W = E^2 d A (modulo some factors of epsilon_0 or 4 pi, depending on which units we're using, and which don't affect any of our discussion here), and you believe GR agrees with that.

* You believe the energy stored in the capacitor, evaluated in global Schwarzschild coordinates, should be W sqrt(J), by consistency with your answers above, and by the fact that the energy has to "redshift" as it climbs out of the gravity well (to speak somewhat loosely).

* You believe that GR says the energy stored in the capacitor, evaluated in global Schwarzschild coordinates, is W, because that's the only way it can be consistent with GR's answers above on charge and the E field.

I will follow up with a separate post on my analysis of the scenario.

my sorry for the 'lawyer references'. :shy::zzz:

No offense taken.

 

[PeterDonis]

I will follow up with a separate post on my analysis of the scenario.

This is the quick version, without math; I'll post the math separately (and may not get a chance to do so for a bit).

First, a key difference between the charge q and the other variables (E field and energy). The charge q is a Lorentz scalar; it is the result of an integral that can be written as an invariant contraction of vectors and tensors with no "free" indices, meaning that its value must be the same in *any* coordinate chart. The E field and energy are *not* scalars; in full generality, they are components of tensors (the EM field tensor and the stress-energy tensor, respectively), though they can be modeled more simply (energy, for example, for a "test object" can be modeled as the time component of the 4-momentum). So there is nothing a priori inconsistent about claiming that the E field and the energy can "redshift" (meaning, can have "metric coefficient factors" in them in global coordinates), while the charge q can't; they are different kinds of things. I believe DaleSpam commented a while back about this same issue.

Second, a bit of stage-setting. Since by hypothesis the plate separation (and area, though we haven't talked about that explicitly) d of the capacitors is constant, in terms of proper distance, we can talk about the voltage on the capacitors instead of the E field, since voltage V is just E * d (again modulo a factor of epsilon_0 or 4 pi depending on the units). So "the E field redshifts" is the equivalent of "the voltage redshifts", and since voltage is just energy per unit charge, this is also the equivalent of saying "the energy redshifts", which is why it's easier to talk about voltage in this scenario, since it makes clear the direct relationship between the field and the energy.

Now, the quick summary of what I think GR actually says about this scenario:

The charge on the capacitor inside the shell is q; since, as above, this is a Lorentz scalar, it has the same value in *any* coordinate chart, including the global Schwarzschild chart. The capacitor charge does not "redshift" in this scenario.

In a local inertial frame, the voltage on the capacitor inside the shell is V (the same as the voltage on the capacitor at infinity), and its energy is just W = V * q, since energy is voltage times charge. This is obvious just from the equivalence principle: the local observer next to the capacitor has no way of knowing, just from local observations, that he is inside a potential well where the "redshift factor" is less than 1. In the local inertial frame, physics looks just like it does in SR, where the voltage is obviously V, with no "redshift". But it can be derived rigorously from the math in a local inertial frame.

In the global Schwarzschild coordinates, the voltage on the capacitor inside the shell is V sqrt(J), and its energy is W sqrt(J) = V sqrt(J) * q, since the charge q is an invariant Lorentz scalar. This tells us that an observer "at infinity" will only be able to extract the "redshifted" energy from the capacitor, so global energy conservation (in the form suitable for a static spacetime) is preserved. This result can be derived by taking the result in a local inertial frame, above, and transforming it to the global Schwarzschild coordinates.