Are the directions of electric fields lines affected by Gravity?

[Q-reeus]

The larger point being made isn't terribly hard to grasp in my opinion. Charge does not get "redshifted", no matter how fast something moves, the charge of the body is still the same. I don[t really follow what your difficulty is, I'm afraid I can't put it any more simply than that.

Fair enough, but my difficulty on that point is illustrated by example 3: given here:https://www.physicsforums.com/showpost.php?p=3946413&postcount=1 - and the previous 2 examples prep for that 3: You do not get any sense of an issue there?

 

[Q-reeus]

I will add that it was established as a mathematical proof that GR+Maxwell include charge invariance some time before 1921, as this fact was presented and derived in Pauli's famous relativity exposition in 1921.

Why not just refer to original RN solution that began with the R contribution back in 1916.

Unfortunately, a certain poster here has expressed the point of view that a mathematically based theory can be in conflict with its mathematically based consequences - a concept everyone else finds absurd on its face. The 'method' used is to take specialized results of the theory, or qualitative motivations used pedagogically, as if they are absolute principles and then showing that contradictions result. This whole method is absurd.

More deja vu - this mirrors your approach when I raised the matter of stress as source term in another thread. Same attitude then as now - call it absurd to challenge a 'mathematically exact' solution to the EFE's, but simultaneously avoid any actual detailed evaluation of the specific scenarios presented. In other words - "We have no need to peer through the telescope - we know what must be there - 'holy scripture' (read; GR) tells us what will be seen". Well if you think that a tad harsh or unjust, may I suggest a simple redress - actually present your own analysis of those surely simple enough scenarios in #10, 11 - and enlighten me by showing how it all fits nicely with RN metric. I would then and only then accept that the underlying physical assumption of potential invariant Gauss's law makes sense. If you decline as I expect will be the case, don't expect any respect for your criticism of 'a certain poster' here.

 

[PeterDonis]

MTW confirms that $E = Q/r^2 e_{\hat{r}}$. See pg 841. $\hat{r}$ is the unit length vector in the r direction.

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. The reason I wondered about whether it should be a timelike Killing vector is that I was thinking about an analogue for charge to the Komar mass integral, which uses a timelike Killing vector; but thinking it over, that sort of thing would only work in a region of nonzero charge density. In the vacuum region of R-N spacetime, the charge density is zero, so a "Komar charge" integral over such a region would just be summing up a lot of zeroes; the "source" of the charge is deeper inside the spacetime, not in the vacuum region.

 

[PeterDonis]

https://www.physicsforums.com/showpost.php?p=3560246&postcount=229, and your response was in #232! So, it depends on who says it?

The discussion in that thread evolved quite a bit from the point you refer to; I'm not sure that either DrGreg's statement or my response in #232 is a good reflection of where things ended up. I seriously doubt DrGreg intended anything that would conflict with what I've been saying here about the area of a 2-sphere at radial coordinate r being 4 pi r^2 (since the tangential line element he gives leads direclty to that area when integrated).

either comparing radial vs tangential distances in the one gravity affected setting, or on a before/after basis - gravity switched 'on' vs 'off'. It's the latter case that I was thinking of.

You can't just switch gravity off, so I don't see how the latter case can be physically realized. You *can* take an object from a region with negligible gravity to a region with non-negligible gravity and compare measurements on it before and after; we've gone into this in gory detail in previous threads, and we've seen that what happens depends on how you do the moving. Or you can try to compare two supposedly "identical" objects, one in a region with negligible gravity and one in a region with non-negligible gravity; then you find that what you get depends on how you do the comparing. In short, there is no unique "standard of comparison" that works the way you are suggesting.

Key word there is *local*. In flat spacetime, Poynting theorem holds 'globally' in that steady-state net power flow across any two nesting closed surfaces is conserved. Not so in GR.

You mean "not so in curved spacetime"; yes, because global "conservation of energy" does not hold in a general curved spacetime.

This is naturally not some 'proof' that charge invariance must similarly fail, but is meant to serve as a reminder that 'common sense' results true in flat spacetime can fail when gravity enters the picture.

Yes, which means that instead of just assuming that the flat spacetime result carries over, you have to *show* that it does, in cases where it does. Pervect showed that it does for the case of charge invariance. Nobody is assuming anything; nobody is claiming that Gauss's Law for charge "must" hold in curved spacetime just because it holds in flat spacetime. We're saying it holds because it has been explicitly *shown* that it holds.

Gauss's law to gravity is a Sacred Cow tenet incorporated into GR as axiom - manifest as RN metric

The R-N metric is a solution to the EFE; that's been proved. The EFE doesn't incorporate Gauss's Law as an axiom. The fact that Gauss's Law for *charge* holds in R-N spacetime is easily shown; pervect did it. Gauss's Law for charge does not have to be assumed or incorporated into GR as an axiom; it is a *theorem*.

Gauss's Law for *gravity* does *not* hold in curved spacetime; as pervect noted, the corresponding integral for gravity is *not* conserved.

 

[PeterDonis]

In the post I hope to make soon regarding the R-N metric and what it says about charge

I want to tie up this loose end (which, as noted earlier in this thread, has actually spanned multiple threads), but it will probably take at least two posts. This first one is partly to review some general properties of static, spherically symmetric spacetimes, and then to make a few comments about the R-N spacetime's geometric properties. I'll defer the specific issues relating to charge to a subsequent post.

(Note: I posted much of the following in a previous thread, here:

https://www.physicsforums.com/showpost.php?p=3843463&postcount=249

However, there I was concentrating specifically on the case of a shell of matter with vacuum both outside and inside, so I didn't comment on some more general properties that are of interest.)

The most general line element for a static, spherically symmetric spacetime can be written as follows (my notation is slightly different from what you'll find in most textbooks, for example MTW):

$$ds^2 = - J(r) dt^2 + \frac{1}{1 - \frac{2 m(r)}{r}} dr^2 + r^2 d\Omega^2$$

where J(r) and m(r) are functions of the radius. The function J(r) can be called the "redshift factor", and will be less than or equal to 1; here we will only consider cases where it is positive (i.e., regions outside any horizon that might be present). The function m(r) can be called the "mass inside radius r", and can be defined via its radial derivative as follows:

$$\frac{dm}{dr} = 4 \pi r^2 \rho (r)$$

where $\rho (r)$ is the energy density seen locally by a static observer (which is the t-t component of the SET, $\rho = T^t_t$). This equation for m comes directly from the t-t component of the EFE.

The redshift factor J(r) is governed by the following equation, which comes directly from the r-r component of the EFE:

$$\frac{1}{2J} \frac{dJ}{dr} = \frac{m(r) + 4 \pi r^3 p(r)}{r \left( r - 2 m(r) \right)}$$

where the LHS is written this way because it turns out to be more convenient. Here we see one additional function of r, the radial pressure p(r), which is the r-r component of the SET, $p = T^r_r$. Note that we are *not* assuming isotropic pressure; that is, whatever stress-energy is present need *not* be a perfect fluid. However, the SET must be diagonal (in the chart in which the above line element is expressed), and the tangential components must be equal; we'll see that in a moment.

The radial pressure p(r) is governed by a generalized form of the Tolman-Oppenheimer-Volkoff (TOV) equation which does not assume isotropic pressure; it turns out that that just adds one additional term to the standard TOV equation. The equation can be derived from the tangential component of the EFE, but it turns out to be easier to evaluate the r component of the covariant divergence of the SET, which is equivalent but involves a lot less algebra. The result is:

$$\frac{dp}{dr} = - \left( \rho(r) + p(r) \right) \frac{1}{2J} \frac{dJ}{dr} - \frac{2}{r} \left( p(r) - s(r) \right)$$

where we now can see the convenience of writing the J equation as we did above, and where s(r) is the tangential stress, which is the tangential component of the SET, $s = T^\theta_\theta = T^\phi_\phi$.

I emphasize that all this applies to *any* static, spherically symmetric spacetime; it includes *all* of the cases we have discussed in various threads, including not just vacuum regions, not just the exterior of R-N spacetime, but also interior regions of spherically symmetric bodies such as planets or stars, and interior regions of spherically symmetric shells with vacuum inside (it also applies to the inner vacuum region itself, of course). All we need to do in any specific case is to find appropriate expressions for any two of the five unknowns, rho, p, s, J, m. Then, since we have three equations relating all these unknowns, we can determine the other three from the two we have expressions for.

One other general question we can ask that might be of interest () is, under what circumstances will the redshift factor, J(r), take the following form?

$$J(r) = f \left( 1 - \frac{2 m(r)}{r} \right)$$

Notice first of all that there is a constant factor f in front. That is there because, as you can see from the above, we do not have an equation for J or its derivative in isolation; we only have an equation for the *ratio* of dJ/dr to J. That means that, whatever expression we derive for J from the above equations, we can *always* put some constant factor f in front of it and still satisfy the equations. In order to determine that constant factor, we have to look at boundary conditions: for example, in any exterior vacuum region, where J -> 1 as r -> infinity, we must have f = 1; that is, if we have some expression for J that goes to 1 as r goes to infinity, we *cannot* put any constant factor f in front of it and still have a valid solution except f = 1, which is trivial. But there will be cases, as we will see, where we *can* find some f that is not 1 but which satisfies the boundary conditions.

To see what the above condition on J implies, we can simply take its derivative and divide by 2J to obtain:

$$\frac{1}{2J} \frac{dJ}{dr} = \left( \frac{m(r)}{r^2} - \frac{1}{r} \frac{dm}{dr} \right) \frac{1}{1 - \frac{2 m(r)}{r}}$$

which quickly simplifies to

$$\frac{1}{2J} \frac{dJ}{dr} = \frac{m(r) - 4 \pi r^3 \rho (r)}{r \left( r - 2 m(r) \right)}$$

Comparison with the equation for 1/2J dJ/dr above makes it clear that we must have $p(r) = - \rho (r)$ in order for J(r) to take the special form given above. There are two cases of interest where this condition might be satisfied in a static, spherically symmetric spacetime. Obviously it will be satisfied in any vacuum region, where rho = p = 0. But it is also satisfied, as it happens, by the SET of a static electric field, as in R-N spacetime (I'll give more detail on this in a subsequent post). So in those two cases, we expect to see a relationship between the "redshift factor" and the radial metric coefficient. (The condition itself is also satisfied by "dark energy", such as a cosmological constant, but in cases where that is present the spacetime will not be static, so we won't consider those cases here.)

But the condition p = - rho is obviously *not* going to be satisfied by any kind of normal matter. So in cases like the interior of a shell or the interior of a planet, we do *not* expect to see a relationship between the redshift factor and the radial metric coefficient.

The above should be all the general machinery we will need; in my next post I'll consider special cases corresponding to (at least some of) the scenarios that have been proposed by Q-reeus.

 

[PeterDonis]

The above should be all the general machinery we will need

Actually, there is another thing; we'll need expressions for the proper acceleration of a radially moving test object in a static, spherically symmetric spacetime. Unfortunately these expressions are rather messy and I haven't been able to simplify them very much, but we'll need them for reference if nothing else.

The 4-velocity $u^a$ of a radially moving object has two components, $u^t$ and $u^r$. However, it will make things look somewhat simpler if we adopt the notation that $u^r = w u^t$, where $w = dr / dt$. Also, since the 4-velocity is a timelike unit vector, we must have $g_{ab} u^{a} u^{b} = -1$. This allows us to derive an expression for $u^t$ in terms of w and other known quantities:

$$\left( u^t \right)^2 = \frac{r - 2m}{r \left( J - w^2 \right)}$$

The general formula for the 4-acceleration is

$$A^b = u^a \nabla_a u^b = u^a \partial_a u^b + u^a \Gamma^b_{ac} u^c$$

We will have two components to $A^b$, the t component and the r component (since the motion is purely radial). Expanding out the above formula for each case, we get

$$A^t = \left( u^t \partial_t + u^r \partial_r \right) u^t + 2 \Gamma^t_{tr} u^t u^r$$

$$A^r = \left( u^t \partial_t + u^r \partial_r \right) u^r + \Gamma^r_{tt} \left( u^t \right)^2 + \Gamma^r_{rt} u^t u^r + \Gamma^r_{rr} \left( u^r \right)^2$$

Substituting for everything we can substitute for, to minimize the number of distinct quantities, we obtain:

$$A^t = u^t \left( \partial_t + w \partial_r \right) u^t + \frac{w \left( r - 2m \right)}{r J \left( J - w^2 \right)} \frac{dJ}{dr}$$

$$A^r = u^t \left( \partial_t + w \partial_r \right) u^r + \left( \frac{\left( r - 2m \right)^2}{r^2} + \frac{w \left( r - 2m \right)}{r} \right) \frac{1}{2 J \left( J - w^2 \right)} \frac{dJ}{dr} + \left( r \frac{dm}{dr} - m \right)\frac{w^2}{r^2 \left( J - w^2 \right)}$$

As I said, these are messy expressions and I haven't been able to simplify them. But we can at least check one special case: a static observer, for whom w = 0 (i.e., u^r = 0). For this case, all the partial derivatives are zero, so A^t is zero, and we have only

$$A^r = \left( 1 - \frac{2m}{r} \right)^2 \frac{1}{2 J^2} \frac{dJ}{dr}$$

Finally, note that the actual measured proper acceleration will be the magnitude of the 4-vector A^b, which in general is

$$A = \sqrt{ g_{ab} A^a A^b }$$

For the special case of a static observer, we have

$$A = \sqrt{ g_{rr} } A^r = \left( 1 - \frac{2m}{r} \right)^{3/2} \frac{1}{2 J^2} \frac{dJ}{dr}$$

For the further special case where J = 1 - 2m / r (I discussed last post when this special case applies), we obtain

$$A = \left( \frac{m}{r^2} - \frac{1}{r} \frac{dm}{dr} \right) \frac{1}{\sqrt{1 - \frac{2m}{r}}}$$

One other note: I think there is another special case of interest and I will probably need to add a post about that before going into details about R-N spacetime and charge. This is the case where A^t = 0, which corresponds to the case where the "total energy" of the test object is conserved.

 

[PeterDonis]

I will probably need to add a post about that before going into details about R-N spacetime and charge.

Still considering some items relating to the acceleration equations I posted, but I'm going to go ahead and post a few things about some special cases, including R-N spacetime.

First special case: Schwarzschild spacetime. This is what you get when you set $\rho = p = s = 0$, i.e., complete vacuum, no stress-energy anywhere. This immediately gives $dm / dr = 0$, so the mass $m(r) = M$ is constant. And, as noted previously, this is a case where we know $J(r) = f ( 1 - 2 m / r )$, and for the exterior vacuum region we know from the boundary condition at infinity that f = 1, so we recover the standard Schwarzschild metric.

Second special case: Schwarzschild exterior, a thin shell of matter, and flat Minkowski interior. (Some of what follows recaps things I have posted in previous threads.) Here we have $\rho = p = s = 0$ everywhere except inside a thin shell lying between radial coordinates $r_{i}$ (inner) and $r_{o}$ (outer). For simplicity we assume that inside the shell we have constant density $\rho$. Then we have $m(r) = 0$ in the interior vacuum region, $m(r) = M$ in the exterior vacuum region (where M is the shell's total mass), and $m(r) = 4 / 3 \pi \rho ( r^3 - r_i^3 )$ in the region occupied by the shell.

First, what exactly do we mean by saying the interior vacuum region is a "flat Minkowski" region? The fact that m(r) is zero in the interior vacuum region confirms that it is spatially flat (g_rr = 1); and since the pressure p is also zero we know that dJ / dr is zero, so J(r) is constant. However, J(r) is *not* equal to 1 in this region; at least, not with the standard time coordinate we have been using. Instead, J(r) will be equal to its value at r_i, the inner radius of the shell, everywhere in the interior vacuum region. In other words, this is a case where the factor f that I talked about before is not equal to 1. (In fact, we can see that it must be *less* than 1, because the value of J(r) at r_i must be less than 1, since dJ/dr is positive everywhere so J at any finite radius r must be less than its value at infinity, which is 1.)

Of course we can re-scale the time coordinate in the interior vacuum region to make the metric explicitly take the Minkowski form: we just apply the transformation $t' = \sqrt{J(r_i)} dt$; this is the sense in which the interior vacuum region is indeed a "flat Minkowski" region. Locally, in fact, the transformed coordinates will be the "natural" ones for static observers to use. However, there is still a "gravitational redshift" in this region relative to infinity; for example, if static observers in the interior vacuum region emit light outward, it will be redshifted when it is received by observers at infinity.

How does it happen that the interior region ends up being flat, even though there is still a redshift relative to infinity? As noted before, in the region occupied by the shell, the condition $p = - \rho$ does not hold, so the relationship between g_tt and g_rr that holds in the exterior vacuum region is broken; and when that relationship is "re-established" in the interior vacuum region, it is re-established with a *different* factor f than the one that applied in the exterior region (which, of course, is f = 1).

I won't say any more about this special case; the above should be enough for here. (In other threads I've talked about the behavior of p and s in the region occupied by the shell.)

Third special case: R-N spacetime. Here the parameters are as follows:

$$\rho (r) = \frac{1}{8 \pi} \frac{Q^2}{r^4}$$

$$p (r) = - \rho (r)$$

$$s (r) = \rho (r)$$

$$m (r) = M - \frac{Q^2}{2r}$$

Since the condition $p = - \rho$ holds, we know that $J = 1 - 2m(r) / r$, so we have a complete solution; when written out, of course, we have

$$J (r) = 1 - \frac{2M}{r} + \frac{Q^2}{r^2}$$

By plugging into other formulas in my past few posts, we can recover the other formulas for R-N spacetime that I posted way back in the early part of this thread.

Some quick observations:

We can now see explicitly that the "mass inside radius r" goes down as r decreases, and that it goes to M as r goes to infinity. (We are limiting ourselves to the region outside the horizon and to Q < M, so m(r) will always be positive; in other words, we are not, strictly speaking, considering an R-N black hole--at least not yet--but rather a charged gravitating body like a planet or star with an appreciable static E field.) This explains, as I noted in a much earlier post, why the "redshift factor" J(r) does not decrease as fast as it does in Schwarzschild spacetime; as you descend, more and more of the mass-energy in the spacetime is above you rather than below you, so it doesn't contribute to the effects of "gravity" that you see. You can also read this off from the equation for dJ / dr, where the negative pressure (p = - rho) makes dJ / dr *smaller*.

We can also see, though, that the "mass inside radius r" behaves *differently* in this respect than the "charge inside radius r", which, as shown in previous posts dealing with the Gauss's Law integral, is *constant*, equal to Q at any radius r. So we have at least one key difference in the behavior of charge vs. the behavior of mass-energy.

 

[PeterDonis]

Actually, there is another thing; we'll need expressions for the proper acceleration of a radially moving test object in a static, spherically symmetric spacetime. Unfortunately these expressions are rather messy and I haven't been able to simplify them very much, but we'll need them for reference if nothing else.

On further review, I'm not sure all of these expressions are right. (I made at least one transcription error, but I also need to check against some other things.) Unfortunately I'm out of the "edit window" now so I can't edit or delete the post, but I'll go over things again and post an update.

Also, there is one more special case I want to make some comments about: the case of a charged gravitating body with a spherical shell some distance outside it. This is basically just a combination of the second and third special cases in my previous post:

We have an exterior vacuum region with the standard R-N geometry, where the mass M that appears in the metric is M_body + M_shell, the mass of the central charged body plus the mass of the shell. So in this region we have $m(r) = M_{body} + M_{shell} - Q^2 / 2r$, where Q is the charge of the central body (we assume the shell is uncharged), and $J(r) = 1 - 2m(r) / r$.

We have an interior vacuum region, between the central body and the shell, where the geometry is also R-N, but with two changes from the exterior: the mass M in the metric is now just M_body, the mass of the central charged body, and there is a "rescaled" time coordinate; in other words, the g_tt metric coefficient in this region, written using the "standard" time coordinate, has a factor f in front of it that is less than 1. So in this region we have $m(r) = M_{body} - Q^2 / 2r$, and $J(r) = f \left( 1 - 2m(r) / r \right)$.

In the shell region, if we assume the shell has constant density, we have $m(r) = M_{body} + 4 / 3 \pi \rho_{shell} \left( r^3 - r_i^3 \right) - Q^2 / 2r$. We can also see that dJ / dr will still be positive inside the shell, so the value of J at r_i, the shell inner radius (which determines the factor f in the interior vacuum region, see above), will be less than its value at r_o, the shell outer radius. The difference is that there is also a contribution to dJ / dr from the energy and pressure of the static electric field, so the change in J across the shell will be different than in the case of a shell with vacuum inside and outside, assuming identical parameters for the shell (same density and thickness).

 

[Q-reeus]

The discussion in that thread evolved quite a bit from the point you refer to; I'm not sure that either DrGreg's statement or my response in #232 is a good reflection of where things ended up.

First, pardon please delayed response - only partly owing to I assume a lengthy PF server outage. And I must say bravo Peter for a huge and thoughtful campaign of five postings in a row - last three quite lengthy and detailed! Now, re above; understood. Issue there likely only relevant for thick shells or solids where extended radial differential effects could conceivably induce 'appreciable' internal stress/strains, but a side-issue here.

The R-N metric is a solution to the EFE; that's been proved. The EFE doesn't incorporate Gauss's Law as an axiom. The fact that Gauss's Law for *charge* holds in R-N spacetime is easily shown; pervect did it. Gauss's Law for charge does not have to be assumed or incorporated into GR as an axiom; it is a *theorem*.

Read through #16 and #32 again. Not sure proof is the right word. If deciding ab initio that field lines concept is inviolable re E field - lines must begin and end on charge globally, regardless of spacetime curvature or not, what follows - RN solution - will inevitably be made to conform to that position on how nature works. (Realize your next postings do put up a decent argument for that case, but here I am responding as though unaware of those)

Gauss's Law for *gravity* does *not* hold in curved spacetime; as pervect noted, the corresponding integral for gravity is *not* conserved.

Which is interesting and I find it hard to avoid seeing it as a tacit admission gravity really does gravitate; 'lines of g' do indeed terminate in mid-air! Alternately, given the striking parallels between gravitating mass and charge in the almost-Newtonian weak-gravity regime, it begs why and on what basis the assumed departure re 'Gauss's almost law for g', if gravity is rather taken as not gravitating! This is referring here to the restricted case where mass/charge is static in a gravitational potential - thus excluding from consideration the SR velocity/energy dependence for gravitating mass but not charge (though very important when looking at free-infall case).

 

[Q-reeus]

I want to tie up this loose end (which, as noted earlier in this thread, has actually spanned multiple threads), but it will probably take at least two posts. This first one is partly to review some general properties of static, spherically symmetric spacetimes, and then to make a few comments about the R-N spacetime's geometric properties. I'll defer the specific issues relating to charge to a subsequent post.

(Note: I posted much of the following in a previous thread, here:

https://www.physicsforums.com/showpost.php?p=3843463&postcount=249

However, there I was concentrating specifically on the case of a shell of matter with vacuum both outside and inside, so I didn't comment on some more general properties that are of interest.)

The most general line element for a static, spherically symmetric spacetime can be written as follows (my notation is slightly different from what you'll find in most textbooks, for example MTW):

$$ds^2 = - J(r) dt^2 + \frac{1}{1 - \frac{2 m(r)}{r}} dr^2 + r^2 d\Omega^2$$...and much following.

There is much that's familiar and also I agree sound and well presented in this and some of the following two entries, but one quite simple and crucial relation has me perplexed. It's in this para:

Comparison with the equation for 1/2J dJ/dr above makes it clear that we must have p(r)=−ρ(r) in order for J(r) to take the special form given above. There are two cases of interest where this condition might be satisfied in a static, spherically symmetric spacetime. Obviously it will be satisfied in any vacuum region, where rho = p = 0. But it is also satisfied, as it happens, by the SET of a static electric field, as in R-N spacetime (I'll give more detail on this in a subsequent post).

The stress components for a magnetostatic field (and E field is perfectly analogous) are given in eq'n (3) here: http://fieldp.com/myblog/2010/theory-and-applications-of-the-maxwell-stress-tensor/ Take x as field direction, then we easily see that for diagonals, |σ_xx| = B²/μ₀, |σ_yy| = |σ_zz| = -1/2|σ_xx|. Convention there is for tension to be positive, but it doesn't really matter as will be seen. When it comes to total stress/pressure contribution, correct me if wrong, but one simply does a scalar sum of the principal stresses - whether stress is owing to a material mechanically/hydrostatically stressed, or that of EM field. And clearly, at least in vacuo, that comes to zero for any EM field! So what's going on here? Is there some justification for excising two components only in EM case but not otherwise? All the rest of your findings is impacted strongly on the resolution of this seemingly straightforward point. Your derivation for that last expression in #42 was though quite helpful, and I will be cataloging these entries as handy reference material - with care. I have no problem with that particular one-way coupling - as already stated in another thread. But two-way coupling still makes more sense to me. Particularly when it gets down to the thorny matter of resolving the difficulties shown up in say #10.

 

[PeterDonis]

And I must say bravo Peter for a huge and thoughtful campaign of five postings in a row - last three quite lengthy and detailed!

Thanks! And more to come...

If deciding ab initio that field lines concept is inviolable re E field - lines must begin and end on charge globally, regardless of spacetime curvature or not

The "field lines concept" is not fundamental; it's just one way of visualizing what the math says. The fundamental concept is the math--in this case, Maxwell's Equations in curved spacetime. Those equations, combined with the EFE, are all that you need to "assume" to derive the R-N metric. The stuff I've been posting all comes out of that derivation. If you're taking the position that Maxwell's Equations or the EFE might not be right, then there's not much point in continuing the discussion. (Even if we wanted to have a discussion along those lines, it's not enough just to say those equations might not be right--you have to put something in their place that we can use to make predictions. What?)

Which is interesting and I find it hard to avoid seeing it as a tacit admission gravity really does gravitate; 'lines of g' do indeed terminate in mid-air!

This is one reason why the "field lines" concept is not fundamental; this interpretation only works in spacetimes where a conserved "energy" can be defined, i.e., in stationary spacetimes. The analogous concept for EM works, to the extent it works, in *any* spacetime, stationary or not.

In a stationary spacetime, as has been discussed ad nauseam in other threads, yes, you can interpret Gauss's Law for gravity not holding as meaning that "gravity gravitates". But you have to be very careful what inferences you draw from that; as those other threads make clear, *lots* of people have real trouble being that careful. This is why I tend not to favor such interpretations, and to prefer concentrating on statements that hold in any spacetime, like local energy conservation--the covariant divergence of the SET is always zero.

 

[PeterDonis]

Take x as field direction, then we easily see that for diagonals, |σ_xx| = B²/μ₀, |σ_yy| = |σ_zz| = -1/2|σ_xx|.

That's right; the radial stress component (in the field direction) is *minus* the tangential components (orthogonal to the field direction). And, when you add in the time component (energy density), you find that the radial stress is *opposite* in sign (but equal in magnitude) to the energy density. I.e., we have radial tension and tangential compression. Also, at least for a static electric field, the tangential stresses are equal in magnitude to the radial stress; there isn't a factor of 1/2. (I've verified that while doing the computations for the R-N metric, since the stress components have to match the components of the Einstein tensor, and they all have the same magnitude--no factors of 1/2.) I'm not sure where you are getting the factor of 1/2 for the magnetic case here.

When it comes to total stress/pressure contribution, correct me if wrong, but one simply does a scalar sum of the principal stresses - whether stress is owing to a material mechanically/hydrostatically stressed, or that of EM field. And clearly, at least in vacuo, that comes to zero for any EM field!

No, it doesn't. At least, not if by "total stress/pressure contribution" you mean what appears in the equation for "attractive gravity"--the initial inward acceleration of a small ball of test particles immersed in whatever "material" we are talking about. That is indeed the sum of *all* the diagonal SET components, or $\rho + \sigma_x + \sigma_y + \sigma_z$, which in this case comes to $2 \rho$--i.e., *not* zero.

Perhaps it's that "in vacuo" that's the issue--an EM field *always* has a positive energy density, so there is never a true "vacuum" when an EM field is present. In "natural" units the energy density is just $\left( E^2 + B^2 \right) / 8 \pi$.

I have no problem with that particular one-way coupling - as already stated in another thread. But two-way coupling still makes more sense to me. Particularly when it gets down to the thorny matter of resolving the difficulties shown up in say #10.

I'm afraid you lost me here--what is the "one-way" vs. "two-way" coupling referring to?

 

[PeterDonis]

Take x as field direction, then we easily see that for diagonals, |σ_xx| = B²/μ₀, |σ_yy| = |σ_zz| = -1/2|σ_xx|.

On taking a slightly longer look at the page you linked to, it seems to me that for a B field purely in the x direction (i.e., B_x = B, all other components zero), we would have

$$\mu_0 \sigma_{xx} = B^2 - 1/2 B^2 = 1/2 B^2$$

$$\mu_0 \sigma_{yy} = \mu_0 \sigma_{zz} = 0 - 1/2 B^2 = - 1/2 B^2$$

So the factor of 1/2 is present in *all three* components, and their magnitudes are all the same. And since converting to "natural" units just means replacing $\mu_0$ with $4 \pi$, we have that the magnitude of each stress component in "natural" units is $B^2 / 8 \pi$, as I said in my last post.

 

[Q-reeus]

46
The "field lines concept" is not fundamental; it's just one way of visualizing what the math says.

Quite the point I have been emphasizing. It was back there in those 'proof' entries I referred to earlier that field line concept were a key part of that 'proof'.

The fundamental concept is the math--in this case, Maxwell's Equations in curved spacetime. Those equations, combined with the EFE, are all that you need to "assume" to derive the R-N metric. The stuff I've been posting all comes out of that derivation. If you're taking the position that Maxwell's Equations or the EFE might not be right, then there's not much point in continuing the discussion.

As said numbers of times, it's over how the two are married. Sooner or later, I [STRIKE]expect[/STRIKE] hope the specific issues in #10, 11, will be faced square-on. They haven't gone away.

(Even if we wanted to have a discussion along those lines, it's not enough just to say those equations might not be right--you have to put something in their place that we can use to make predictions. What?)

Well what did I suggest in #10 - originating actually in thread linked to in #5. It's there - not in sophisticated math, but as key concept.

This is one reason why the "field lines" concept is not fundamental;

Again, can only agree - except I extend that observation to include E, B fields.

 

[Q-reeus]

Q-reeus: "Take x as field direction, then we easily see that for diagonals, |σxx| = B2/μ0, |σyy| = |σzz| = -1/2|σxx|."

That's right; the radial stress component (in the field direction) is *minus* the tangential components (orthogonal to the field direction). And, when you add in the time component (energy density), you find that the radial stress is *opposite* in sign (but equal in magnitude) to the energy density. I.e., we have radial tension and tangential compression. Also, at least for a static electric field, the tangential stresses are equal in magnitude to the radial stress; there isn't a factor of 1/2. (I've verified that while doing the computations for the R-N metric, since the stress components have to match the components of the Einstein tensor, and they all have the same magnitude--no factors of 1/2.) I'm not sure where you are getting the factor of 1/2 for the magnetic case here.

I must admit to getting that wrong - there is no factor 1/2 as you say. [edit out mistaken passage] should have remembered what I argued main para here:https://www.physicsforums.com/showpost.php?p=3840301&postcount=237 .

So when you e.g. write p(r)=−ρ(r) as in #42, I'm supposed to understand that that p is just the radial (-ve sign) component rather than net contribution of p (+ve sign) at r? Dear dear - more precision then! :tongue2:

No, it doesn't. At least, not if by "total stress/pressure contribution" you mean what appears in the equation for "attractive gravity"--the initial inward acceleration of a small ball of test particles immersed in whatever "material" we are talking about. That is indeed the sum of *all* the diagonal SET components, or ρ+σx+σy+σz, which in this case comes to 2ρ--i.e., *not* zero.

Sure, as essentially sorted out above.

Q-reeus: "I have no problem with that particular one-way coupling - as already stated in another thread. But two-way coupling still makes more sense to me. Particularly when it gets down to the thorny matter of resolving the difficulties shown up in say #10."

I'm afraid you lost me here--what is the "one-way" vs. "two-way" coupling referring to?

Raised before in earlier thread - charge via it's E field energy density distorts Schwarzschild geometry (witness your J expression in #42), but as per #34, Mass has no effect on form of E field. Still don't buy it.

 

[PeterDonis]

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

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

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

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

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

Mass has no effect on form of E field.

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

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

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

 

[Q-reeus]

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

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

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

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

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

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

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

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

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

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

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

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

 

[PeterDonis]

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

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

me thinks radially supported perfect fluid is implied above.

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

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

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

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

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

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

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

OK then keep checking by all means.

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

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

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

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

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

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

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

 

[pervect]

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

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

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

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

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

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

 

[PeterDonis]

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

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

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

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

Thin shell with fluid inside:

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

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

Thin shell with vacuum inside:

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

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

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

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

 

[PAllen]

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

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

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

 

[PeterDonis]

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

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

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

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

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

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

 

[Q-reeus]

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

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

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

 

[pervect]

Thin shell with vacuum inside:

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

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

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

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

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

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

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

 

[PAllen]

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

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

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

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

 

[pervect]

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

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

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

Consider peter's metric:

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

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

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

Convert it to a coordinate basis

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

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.