Are the directions of electric fields lines affected by Gravity?

[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:

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

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

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

Convert it to a coordinate basis

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

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.

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

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 boundary 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.