Are the directions of electric fields lines affected by Gravity?

[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

$$4 \pi j^t = \frac{d}{dr} \left( g^{rr} g^{tt} E \right) + \left( \frac{J'}{2J} + \frac{r m' - m}{r \left( r - 2m \right) } + \frac{2}{r} \right) \left( g^{rr} g^{tt} E \right)$$

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.

 

[PeterDonis]

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

On re-reading, it may be confusing to state things in terms of coordinates and frames as I did. So in case it's needed to improve clarity, let me re-state everything in terms of how it would actually be measured in the scenario we are considering.

The charge on the capacitor inside the shell is q. Since this charge is only "visible" between the plates, if we are going to measure it using Gauss's Law (which is the point at issue), we have to, as I said previously, use a closed surface that encloses just one plate, measure the electric field normal to each small segment of that surface, and integrate over the entire surface. The electric field is nonzero only between the plates, and (in the idealized case) is exactly uniform and exactly normal to the plates, so the measured charge will be q = E * A, where E is the electric field and A is the plate area (in "proper" units). Note that, as before, I am leaving out constant factors of epsilon_0 or 4 pi or whatever that depend only on the units we're using.

One could, I suppose, try to concoct a way of "remotely" measuring the charge, by "remotely" measuring either the electric field or the area. But how would you do that directly? You could measure the electric field indirectly, by measuring energy (we'll discuss that below), but that requires you to "interpret" what the energy measurement says about the electric field. The whole point here is to get down to the direct observables, eliminating all indirect "interpretation" steps. So one way of putting the fact that the charge is q "regardless of which coordinates you use" would be to say that there is only one way to *measure* q, which is locally; there is no way for an observer at infinity to directly measure q on the capacitor inside the shell.

The energy stored in the capacitor, as measured anywhere inside the shell, is W = E * d * q = E^2 * d * A, where d is the plate separation (again in "proper" units). As our direct measurement of energy, we will adopt Q-reeus' method of putting a known charge on the plates (measured using Gauss's Law as above), and then measuring the work required to separate the plates by a proper distance d, by means of a linkage between the "source" of the work and the capacitor itself. If we do this using a "source" of work anywhere inside the shell, we will get W as above.

The energy stored in the capacitor, as measured at infinity, is W sqrt(J). This is easy to see just by comparing the measurement at infinity with the measurement inside the shell, given above. The measurement at infinity requires a linkage extending from infinity to the capacitor inside the shell, and any work done through that linkage will be "redshifted" by a factor of sqrt(J). More precisely, it will be redshifted by a factor sqrt(ratio of g_tt inside shell to g_tt at infinity), but since g_tt at infinity is 1, the ratio is just sqrt(g_tt inside shell) = sqrt(J). (Btw, this analysis also shows us that the reason why the energy measurement gives W anywhere inside the shell, is that g_tt is *constant* inside the shell--if it varied there, the energy measurement would vary too, depending on where the "source" of the work was relative to the capacitor. Only in the limiting case of a truly local inertial frame, where g_tt can be considered constant throughout the "patch" of spacetime under consideration, will we always get W as the energy measurement.)

The voltage on the capacitor, measured anywhere inside the shell, is, as is obvious from the above, V = E * d. The obvious direct measurement of voltage would simply be to put a voltmeter across the capacitor plates, which amounts to measuring the work required to move a test charge from one plate to the other against the electric field. Note that this is *not* the same as the total energy stored in the capacitor; it is better viewed as a "cross check" of sorts. In other words, once all three measurements are taken (charge, energy, and voltage), we must have W = V * q, even though all three were measured using independent methods.

As with charge above, however, I do not see any way to *directly* measure the voltage on the capacitor "at infinity"; the only way I see to do it is indirectly, by measuring energy at infinity and then deducing, since W = V * q and q is the same, that V must "redshift" the same way W does. However, this does open up a question: since both V and q can't be measured directly "at infinity", could we adopt an interpretation of the energy "redshifting" that has q "redshifting" *instead* of V? In other words, we would say that V(at infinity) = V and q(at infinity) = q sqrt(J), so W(at infinity) = V(at infinity) * q(at infinity) = W sqrt(J) still holds.

I personally don't see any reason to adopt this interpretation, but I can't say on a quick look that it's actually inconsistent--although I haven't checked it with the math, it's quite possible that there would be an inconsistency (or more precisely, that there would be no way to consistently formulate the "q redshifting instead of V" interpretation mathematically). However, one thing I do want to note is that this is *not* what Q-reeus is proposing! Q-reeus is saying that q(at infinity) = q sqrt(J) *and* V(at infinity) = V sqrt(J) (because he says the E field "redshifts" *and* charge "redshifts"). I don't see how to fit this in with the fact that W = V * q should hold.

 

[Q-reeus]

One could, I suppose, try to concoct a way of "remotely" measuring the charge, by "remotely" measuring either the electric field or the area. But how would you do that directly? You could measure the electric field indirectly, by measuring energy (we'll discuss that below), but that requires you to "interpret" what the energy measurement says about the electric field. The whole point here is to get down to the direct observables, eliminating all indirect "interpretation" steps. So one way of putting the fact that the charge is q "regardless of which coordinates you use" would be to say that there is only one way to *measure* q, which is locally; there is no way for an observer at infinity to directly measure q on the capacitor inside the shell.

This is true, but no more a problem imo than remotely measuring the surface temp of a star via it's received spectrum here on earth. Or of determining it's mass via Keplerian orbital dynamics of say a circling planet. We are free to accurately infer so long as there is an agreed set of physical relations to work from.

As with charge above, however, I do not see any way to *directly* measure the voltage on the capacitor "at infinity"; the only way I see to do it is indirectly, by measuring energy at infinity and then deducing, since W = V * q and q is the same, that V must "redshift" the same way W does. However, this does open up a question: since both V and q can't be measured directly "at infinity", could we adopt an interpretation of the energy "redshifting" that has q "redshifting" *instead* of V? In other words, we would say that V(at infinity) = V and q(at infinity) = q sqrt(J), so W(at infinity) = V(at infinity) * q(at infinity) = W sqrt(J) still holds.

It's possible to have it both ways here to an extent if, as I have looked at in #10, one does the split between 'active' charge q_a and 'passive' charge q_p. Work W will redshift as required, and Gauss's law holds good. One serious fly in the ointment is described in the second last para in #$10. Unfortunately, imo can't really have cake and eat it too! If there is some other solution apart from my own fix in #10, I'm 'all ears'.

I personally don't see any reason to adopt this interpretation, but I can't say on a quick look that it's actually inconsistent--although I haven't checked it with the math, it's quite possible that there would be an inconsistency (or more precisely, that there would be no way to consistently formulate the "q redshifting instead of V" interpretation mathematically). However, one thing I do want to note is that this is *not* what Q-reeus is proposing! Q-reeus is saying that q(at infinity) = q sqrt(J) *and* V(at infinity) = V sqrt(J) (because he says the E field "redshifts" *and* charge "redshifts").

Yes, but recall I have it as owing to modification of vacuum permittivity and permeability. When that is done, there is automatically an even split between reduced q and reduced V E, with the latter's reduction projecting out to a distant observer.[Edit: E owing to a charged sphere say - not of course the 'ideal very thin capacitor' discussed above] No suggestion any of that is locally observed within the shell of course! Using loose talk here, the idea of vacuum as more than nothing and having physical structure which includes dielectric and magnetic susceptibilities is in line with, to get it back to ME's, Maxwell displacement current concept. Or QED concept of vacuum polarization/breakdown, which there is considerable effort in testing with e.g. high power lasers. So to me it makes sense that gravity will alter these vacuum quantities in the way suggested.

I don't see how to fit this in with the fact that W = V * q should hold.

There is it seems an interesting situation here. :zzz:

 

[PeterDonis]

This is true, but no more a problem imo than remotely measuring the surface temp of a star via it's received spectrum here on earth.

You can measure the star's surface temp "remotely" by measuring the spectral lines in the radiation from it (and adjusting for the redshift due to the star's mass, if it's large enough to matter). In other words, you are making a direct measurement of something that traveled to you from the star. What corresponding measurement allows you to "remotely" measure the charge on the capacitor inside the shell? What can "travel to you" from the charge bearing information about it? Saying "energy" or "voltage" won't work because you would have to assume a relationship between those things and charge, and that's precisely the point at issue.

Yes, but recall I have it as owing to modification of vacuum permittivity and permeability.

But that's not a matter of interpretation--that's a change in the physics. You can independently measure the vacuum permittivity and permeability, for example by making measurements with a magnet inside the shell and also measuring the speed of light. GR predicts that all those measurements will yield the same vacuum values as they do at infinity--again, this is obvious by the equivalence principle. So if an experiment were actually done in a vacuum inside a spherical massive shell, and it was found that the electromagnetic constants were different, we would need to rework our entire structure of physical theories. (Which means that this particular subtopic is probably verging on being too speculative to take further here.)

(There is one possible way it could be somewhat simpler than that--see below.)

QED concept of vacuum polarization/breakdown, which there is considerable effort in testing with e.g. high power lasers.

It will definitely be interesting to see what those efforts turn up. Bringing up QED and vacuum polarization does raise a good point, though. If any kind of QED effect like this is involved, we would have to change the model we have been using; the "vacuum" inside the shell in the capacitor scenario will not actually be vacuum in the GR sense. A "vacuum" in the GR sense means an SET of *zero*, period. A "vacuum" in the QED sense means an SET which, in the simplest case, amounts in GR terms to a small positive cosmological constant--i.e., *not* zero.

Such a case could in principle be modeled in GR as we have it, without requiring a wholesale change in the theory, but I don't know that I'll have time any time soon to try to extend what I've already done in this thread to cover such a model.

 

[Q-reeus]

You can measure the star's surface temp "remotely" by measuring the spectral lines in the radiation from it (and adjusting for the redshift due to the star's mass, if it's large enough to matter). In other words, you are making a direct measurement of something that traveled to you from the star. What corresponding measurement allows you to "remotely" measure the charge on the capacitor inside the shell? What can "travel to you" from the charge bearing information about it? Saying "energy" or "voltage" won't work because you would have to assume a relationship between those things and charge, and that's precisely the point at issue.

Ideally one could directly check for any 'anomalous' E field owing to grav. potential of a central mass of a spherical capacitor. If my idea is correct, there will in fact be a net E external to the outer shell, owing to a relatively depressed effective charge on the inner shell surface. I could try some figures, but gut instinct says way too feeble for any terrestrial experiment to detect. So it gets back to argument based on self-consistency criteria. Issue remains that, using those ideal 1:1 remote-linkage rods, a reduced force is experienced 'out here' when moving those cap plates located 'down there' within the shell, against a locally determined field E. It needs explaining somehow. A fully self-consistent one at that.

Q-reeus: "Yes, but recall I have it as owing to modification of vacuum permittivity and permeability."
But that's not a matter of interpretation--that's a change in the physics. You can independently measure the vacuum permittivity and permeability, for example by making measurements with a magnet inside the shell and also measuring the speed of light. GR predicts that all those measurements will yield the same vacuum values as they do at infinity--again, this is obvious by the equivalence principle. So if an experiment were actually done in a vacuum inside a spherical massive shell, and it was found that the electromagnetic constants were different, we would need to rework our entire structure of physical theories. (Which means that this particular subtopic is probably verging on being too speculative to take further here.)

You may have missed reading this passage in #102: "No suggestion any of that is locally observed within the shell of course!"
In principle one might detect a local, exceedingly small 'tidal polarization', similar to detecting 'tidal clock-rate' variation. [exterior to the shell interior of course]

It will definitely be interesting to see what those efforts turn up. Bringing up QED and vacuum polarization does raise a good point, though. If any kind of QED effect like this is involved, we would have to change the model we have been using; the "vacuum" inside the shell in the capacitor scenario will not actually be vacuum in the GR sense. A "vacuum" in the GR sense means an SET of *zero*, period. A "vacuum" in the QED sense means an SET which, in the simplest case, amounts in GR terms to a small positive cosmological constant--i.e., *not* zero.

Just to clarify comment in #102 re vac. pol. - the usual usage of vacuum polarization in QED context refers to highly non-linear effects. I merely drew on it as indicative of the view there is physically real structure to the vacuum. Have mentioned it briefly before, but worth bringing up again - if we posit physically real vacuum polarizability (classical context a la Maxwell), it could be seen as preserving Gauss's law 'in reality' though not 'in effect'. That's because vacuum permittivity and permeability modification by factor 1/sqrt(J) is clearly introducing non-linearity to those quantities (need I remind - not as locally observed!). Apply an E field to a non-linear dielectric medium and effective volume bound charge density ρ_p results according to ρ_p = -div P, which in the spherically symmetric cases we have been considering comes down to ~ d/dr (1/sqrt(J)), which is non-zero. If this ρ_p is physically real, lines always begin and end on charge, but some of that charge is of the fleeting vacuum kind. Take it or leave it.

Such a case could in principle be modeled in GR as we have it, without requiring a wholesale change in the theory, but I don't know that I'll have time any time soon to try to extend what I've already done in this thread to cover such a model.

I get the drift. Things have been going nowhere much and won't lose any sleep if above turns out to be my closing statement. Cheers, and thanks for sticking around awhile longer than others.

 

[PeterDonis]

a reduced force is experienced 'out here' when moving those cap plates located 'down there' within the shell, against a locally determined field E. It needs explaining somehow. A fully self-consistent one at that.

I've already given the self-consistent explanation, here and in other threads, but to recap briefly: the reduced force at infinity is due to the effect of the spacetime in between infinity and the region within the shell. The force transmitted down the linkage is "blueshifted" because of the change in potential, and energy transmitted back up is "redshifted" for the same reason. So if you measured the force exerted at the *bottom* end of the linkage, right where it hooks to the capacitor, it would *not* be "redshifted"--it would be the *same* force that would be measured at the same point if everything were being done locally. Sorry if this part wasn't clear from my previous posts.

You may have missed reading this passage in #102: "No suggestion any of that is locally observed within the shell of course!"

But the permittivity/permeability are local quantities--they appear in the local formulation of Maxwell's Equations. Saying that they "look different" when viewed from infinity makes no sense to me, at least not in the context of standard GR + EM; nothing corresponding to them is "transmitted" anywhere. See further comments below.

Just to clarify comment in #102 re vac. pol. - the usual usage of vacuum polarization in QED context refers to highly non-linear effects. I merely drew on it as indicative of the view there is physically real structure to the vacuum.

Yes, I understand that view, and I agree with it, if we're talking about quantum physics. But the word "vacuum" in quantum physics means something different than it does in classical GR. If we're talking about classical GR, then saying "there is physically real structure" present means there *can't* be a "vacuum" in the GR sense; there *has* to be some nonzero SET corresponding to the "physically real structure", otherwise your model, at the GR level, is incomplete. One could say that the SET is "approximately" zero, but then the model won't include any effects from the "physically real structure of the vacuum".

That's because vacuum permittivity and permeability modification by factor 1/sqrt(J) is clearly introducing non-linearity to those quantities

This part is fine, it just means that, in the Maxwell's Equations portion of the model, we are treating the region inside the shell as a material medium, where the permittivity and permeability can vary from their "vacuum" values. The word "vacuum" is not really appropriate in this case, though, at least not if we're using classical GR + EM, for the reason given above.

(need I remind - not as locally observed!)

This part is *not* fine--Maxwell's Equations are local. If the local values are the normal "vacuum" values, then there's no room in our standard theories to have them "look any different" from infinity. So in this case we're back to "too speculative to discuss further here".