Is stress a source of gravity?

In summary: Electric field effects. In addition, the electric field creates an additional quadrupolar moment Qe, orthogonal to Qm and Qs. The net effect is that the amplitude of GW's is proportional to the square of the electric field strength. This is not the case for stress, where the amplitude is proportional to the electric field strength multiplied by the stiffness of the material. This is independent of the material's electric resistance.In summary, the two types of GW's have different amplitudes as a function of the electric field strength.
  • #246
PeterDonis said:
In terms of T_ii terms appearing in the formal expression for the Komar mass integral, yes. In terms of actually contributing, recall that we saw in a series of earlier posts that, if self-gravity can be neglected, the T_ii terms must always cancel in static equilibrium; and if self-gravity cannot be neglected, then whatever "residual" extra contribution remains in the T_ii terms is compensated for by the "redshift factor", which is < 1, multiplying the entire integral; the end result being, in effect, that the negative gravitational potential energy exactly compensates for the positive contribution of T_ii. So in any static equilibrium you can essentially consider the total mass to be the sum of the T_00 contributions alone, with everything else canceling out.
For whatever reason that had never struck home before. Makes the Komar expression a Clayton's really - T_ii contributions yes and no at the same time. Need to chew over that.
It wouldn't be quadrupole; it would be dipole, since by hypothesis the rod only contracts/expands along one dimension. To get a quadrupole variation you would need to have the rod expand/contract along two orthogonal dimensions. Basically what you have described is a time-varying axisymmetric spacetime; I believe there is a general class of EFE solutions that describes these, but I'll have to look it up to be sure.
I cannot see how you figure axially vibrating rod is a dipole source. Conservation of momentum forbids it. It is merely the continuously distributed version of two concentrated masses with a spring in between. And according to this site, that certainly generates quadrupole GW's: http://ned.ipac.caltech.edu/level5/ESSAYS/Boughn/boughn.html - fig.1 and caption. Each rod end has mass dipole-like motion, but they must exactly oppose at any instant to give a net quadrupole source surely. Maybe you were thinking of charge dipole, where opposite motion of unlike charges is a dipole oscillator.
Q-reeus: "As far as I knew F = dp/dt holds perfectly well not just in SR but GR too. Seems not."
No, that force law is still correct, except that it's dp/dtau, not dp/dt (that's true in SR as well); i.e., the derivative is with respect to proper time along the worldline of the object to which the 4-force is being applied, and whose 4-momentum is changing.
Yes understood proper time was to be used and was careless with symbols - I was just focusing on that there is only time derivative of momentum, not extra dynamical terms.
Q-reeus: "If acceleration of matter constitutes in itself a source of added mass..."
It doesn't. T_0i does not appear in the Komar mass integral, so even if you are trying to adopt a model where that integral should be "approximately conserved" in a spacetime that is "approximately stationary", T_0i doesn't come into it.
No, at this stage Komar had been left behind, and my comment was reaction to your statement implying that for the real non-stationary spacetime case d/dtau(T_0i) exactly cancels out T_ii re overall gravitating mass for rod. Since T_ii is supposed to be a periodically sign-reversing source of mass, it can only mean d/dtau(T_0i) is at any instant an equal and opposite source also. That in turn led to question over force law, since active mass should be identical to inertial mass. Given you say there is no change to F = dp/dtau, in what sense then is, or rather can, d/dtau(T_0i) cancel T_ii? As I said earlier, all I could see was d/dtau(T_0i) being part of mass quadrupole moment - with I suppose periodic quadrupolar near-fields as well as GW's the result.
In terms of the true conservation law in GR, covariant derivative of SET = 0, the T_0i terms (more precisely their derivatives) certainly do come into play, since they appear in the covariant derivative. I wrote down the components of that equation in an earlier post, which shows how the covariant derivative constraint relates derivatives of the various components.
Maybe time to go back and pick over your #147 - I did have some questions about meaning of the last four expressions there. :zzz:
 
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  • #247
Q-reeus said:
No, your adding in dissipative processes there is contrary to what I stipulated - idealized case where friction, hysteresis, etc. is assumed absent. That is common practice, even necessary practice, when doing a gedanken experiment. Focus is on inputs and outputs re GW's - or in #162 case, static gravitational field. 'Returned to the input device' is just the T_00 recycling accounted for at the start in #1 - it's a minor part of GW generation.
Oops, my apologies, I do recall that this was an idealized scenario. So, if we remove all dissipative processes we still have energy input, elastic energy, GWs, and any energy returned to the source if you are cycling between strained and unstrained (I don't remember if you were cycling or not). So there are still other places for energy to go besides GW and input energy, so a mismatch between GW and input doesn't imply non-conservation.
 
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  • #248
Q-reeus said:
I cannot see how you figure axially vibrating rod is a dipole source. Conservation of momentum forbids it. It is merely the continuously distributed version of two concentrated masses with a spring in between. And according to this site, that certainly generates quadrupole GW's: http://ned.ipac.caltech.edu/level5/ESSAYS/Boughn/boughn.html - fig.1 and caption.

I see what you're saying, and I may just need to go back and review the mass quadrupole formulas. My visualization of quadrupole oscillations was more like the second illustration on that page, with two pairs of masses oscillating in perpendicular directions. I'm also a bit confused by the statement on that page saying that the system of two masses along a single line can detect incoming GWs; the interferometer GW detectors like LIGO and LISA have two perpendicular arms, not just one. Again, I may just need to go back and dig into the formulas in more detail.

Q-reeus said:
No, at this stage Komar had been left behind, and my comment was reaction to your statement implying that for the real non-stationary spacetime case d/dtau(T_0i) exactly cancels out T_ii re overall gravitating mass for rod.

That's not what the conservation law says. The conservation law, covariant derivative of SET = 0, applies at each individual event in spacetime; it is not an integral law. And at different events, the individual terms in the covariant derivative equations can have different values, as long as they all sum to zero at each event.

Also, the conservation law does not say that d/dt(T_0i) cancels T_ii itself, even at a single event. It only involves derivatives. The i'th component of the law (where i is one of the spatial indices) says that D/Dt(T_0i) + D/Di(T_ii) + D/Dj(T_ji) + D/Dk(T_ki) = 0 (where j and k are the other two spatial indices); but the capital D there is a covariant derivative, not a partial derivative, so the connection coefficients come into play. It's actually qute a bit more complicated than it looks.
 
  • #249
I'm going to go ahead and post what I have on the spherical shell scenario. I don't have a complete analytical expression for the g_tt metric coefficient, but I have expressions for g_rr (I posted that previously but I'll post it again here) and the derivatives of everything, so it's clear enough how things work.

First of all, a correction to what I said a while back in post #147 when I posted the four components of the conservation law. When I said tangential stress was completely "uncoupled" from the other components, I was forgetting about the connection coefficient terms in the covariant derivatives. When those are included, tangential stress actually does appear in the modified TOV equation (i.e., hydrostatic equilibrium) for non-isotropic stress (i.e., radial pressure unequal to tangential stress--the tangential stress components still have to be equal by spherical symmetry). I should have remembered that because the presence of the tangential stress in the modified TOV equation is critical for keeping the shell stable (as I said in a previous thread, and as we'll see below).

Once again, we are assuming a static, spherically symmetric spacetime, and so we have five total unknown quantities: the metric coefficients g_tt and g_rr, and the three SET components T_00, T_11, and T_22 = T_33. We'll give easier names to these; in order, they are J(r), K(r), rho(r), p(r), and s(r). As mentioned previously, we have three equations relating these quantities, the three non-trivial components of the EFE (or, equivalently, two EFE components and one non-trivial conservation law, the equation for hydrostatic equilibrium, which is what we'll actually use). So we ought to be able to specify two arbitrary functions and then the rest will be determined. We'll assume in what follows that we know rho(r) and s(r) and are trying to determine the others in terms of them.

(Note: my notation above is a bit different from most of the literature; usually, g_tt and g_rr are written as exponentials, g_tt = exp(Phi) and g_rr = exp(Lambda) for example. This makes some of the equations a bit easier to calculate with for complex problems, but here we're more concerned with the general physical behavior, so I'm keeping J(r) and K(r) as we defined them in a previous thread. This means some of my formulas will look a bit different than the ones in the literature, but they're still describing the same physics.)

The first equation is the 0-0 component of the EFE: it reads:

[tex]G_{00} = \frac{1}{r^{2}} \left( 1 - \frac{1}{K} \right) - \frac{1}{r} \frac{d}{dr} \left( \frac{1}{K} \right) = 8 \pi \rho[/tex]

This simplifies to:

[tex]\frac{1}{r^{2}} \frac{d}{dr} \left[ r \left( 1 - \frac{1}{K} \right) \right] = 8 \pi \rho[/tex]

We define the quantity in brackets as 2m(r), where "m" is a new function of r whose physical interpretation we will see in a moment (of course the name telegraphs it, but bear with me :wink:); we then see that

[tex]\frac{dm}{dr} = 4 \pi \rho r^{2}[/tex]

This means that dm/dr is the "mass added at radius r"; so integrating dm/dr should give us the "total mass inside radius r". We then see that

[tex]K(r) = \left( 1 - \frac{2m(r)}{r} \right)^{-1}[/tex]

which is what I posted before. In other words, purely from the first (0-0) component of the EFE, without looking at anything else, we see that the K factor, the metric coefficient g_rr, depends *only* on the mass inside radius r, and that depends *only* on the function rho(r). It does *not* depend on any other SET components. (Of course, if rho(r) were not one of our "known" functions, but if we instead assumed, say, that we knew p(r) and s(r), then rho(r) and hence m(r) would still depend, indirectly, on p(r) and s(r), since we would be solving for rho in terms of them. But the final result would still be that g_rr "sees" only the mass inside radius r.)

So much of the discussion in the previous thread, about whether stresses were of the "right" magnitude to make the K factor go back to 1 from outside to inside the shell, was really irrelevant. The K factor automatically goes to 1 as m(r) goes to zero, just from the above.

The next equation is the 1-1 component of the EFE, which I'll rewrite using m(r) instead of K(r):

[tex]G_{11} = - \frac{1}{r^{2}} \frac{2m}{r} + \frac{1}{r} \left( 1 - \frac{2m}{r} \right) \frac{1}{J} \frac{dJ}{dr} = 8 \pi p[/tex]

which easily rearranges to

[tex]\frac{dJ}{dr} = 2 J \frac{m + 4 \pi r^{3} p}{r \left( r - 2m \right)}[/tex]

We'll set this aside for a moment.

The third equation is the modified TOV equation for hydrostatic equilibrium. The key change from the standard TOV equation is that the pressure is not isotropic; we allow radial and tangential stress to be different. That adds an extra term to the normal TOV equation; we have

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

Substituting for 1/2J dJ/dr using the second EFE component above, we obtain

[tex]- \frac{dp}{dr} = \left( \rho + p \right) \frac{m + 4 \pi r^{3} p}{r \left( r - 2m \right)} + \frac{2}{r} \left( p - s \right)[/tex]

This allows us to solve for p in terms of known quantities (since we know m(r) from above and we said s(r) was known as well). However, if there is an actual analytical solution for the above, I haven't been able to find one (I don't think there is in the general case). In the special case of rho = constant, MTW give a solution of the standard TOV equation (without the last term on the RHS, i.e., assuming isotropic pressure) for a spherical star, but the equation for p is still pretty messy and I'm not sure exactly how they arrived at it. With the last term on the RHS added in, even their solution for constant rho may no longer work.

But we can still see some things just by looking at the above equation for dp/dr. First of all: in the standard case, where we have a spherical star with matter all the way into r = 0, we can have isotropic pressure because - dp/dr can be positive all the way into the center. In the shell case, however, that won't work; we must have p = 0 at both the outer and inner surfaces of the shell. That means we *need* the last term on the RHS to have a static equilibrium at all, because the (p - s) factor needs to change sign at some point within the shell in order to change the sign of dp/dr. That's why I said in that earlier thread that tangential stresses are key to keeping the shell stable; more precisely, I should have said tangential stresses that go from positive on the inner surface (so p - s can be less than zero) to negative on the outer surface (we'll see why that has to be the case in a moment) are needed to keep the shell stable if there is an interior vacuum region inside.

Why must s be negative on the outer surface? Because, if we cut a "slice" through the center of the shell, and do a force balance on it similar to what is done in the Ehlers paper, we will find that the following must hold (since there is no pressure in the interior vacuum region, so the only force that can balance is the tangential stress integrated over the shell):

[tex]\int_{a}^{b} 2 \pi r s(r) dr = 0[/tex]

where r = a > 0 is the shell's inner radius and r = b > a is the shell's outer radius. This condition requires that, since s is positive on the inner surface (which we've seen it has to be to make dp/dr change sign), it must be negative on the outer surface. Physically, this makes sense because we would expect the shell's material to be compressed tangentially on the inner surface and stretched tangentially on the outer surface.

Now let's go back to the equation for dJ/dr; I'll write it with a changed sign so we are looking at what happens to J as we go *inward* through the shell, from outer to inner surface:

[tex]- \frac{dJ}{dr} = - 2 J \frac{m + 4 \pi r^{3} p}{r \left( r - 2m \right)}[/tex]

Again, I don't have an analytical expression for J itself from this, since I don't have one for p; but just from looking at the above we can see two things: (1) J continues to decrease as we go inward through the shell; but (2) as we approach the inner surface of the shell, - dJ/dr -> 0 smoothly (because m and p both go to zero smoothly). So there is a smooth transition from J decreasing through the shell to J being constant throughout the inner vacuum region.

(Note that dK/dr does *not* make a smooth transition at the shell boundaries, even though K itself does. As far as I can tell, this is OK: the "junction conditions" that have been mentioned before do not require that dK/dr be continuous, only that K itself is continuous. They *do* require that dJ/dr be continuous as well as J, which it is; physically, it seems to me this is because dJ/dr contributes to hydrostatic equilibrium whereas dK/dr does not, so a discontinuity in dJ/dr would case a discontinuity in the "acceleration due to gravity".)

So to sum up:

(1) The "K" factor is determined entirely by how much mass is *inside* radius r; so as you descend through the shell, K goes smoothly back to 1 from its value at the outer shell surface. Neither radial nor tangential stress has any effect on K.

(2) Hydrostatic equilibrium for a shell with an interior vacuum region requires that p = 0 at both the outer *and* inner surface, which in turn requires that tangential stress be unequal to radial stress, and that it go from negative on the shell outer surface to positive on the inner surface, and integrate to zero over the shell.

(3) The "J" factor continues to decrease smoothly through the shell, at a rate determined by both the mass and the radial pressure, becoming constant in the interior vacuum region. Tangential stress has no effect on J (except indirectly by its effect on the radial pressure profile).
 
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  • #250
PeterDonis said:
the (p - s) factor needs to change sign at some point within the shell in order to change the sign of dp/dr

Almost forgot to comment specifically on this, since it came up in the previous thread. Looking at the dp/dr equation, you can see that it's not enough just for (p - s) to change sign as we move down through the shell (meaning s > p). It has to also become large enough in magnitude to overbalance the first term, which is always positive. What does it take for that to happen? We just need to set 0 > - dp/dr and rearrange:

[tex]s > p + \frac{1}{2} \left( \rho + p \right)\frac{m + 4 \pi r^{3} p}{r - 2m}[/tex]

For the case of a shell made of ordinary material, we expect r >> m, rho >> p, and m >> r^3 p, so the second term's order of magnitude is rho m/r. If you remember, I calculated in the previous thread that for the case of a steel shell, rho m/r was much less than p. So basically the above condition just reduces to s > p for ordinary materials, and s doesn't have to get much larger than p, so both are still << rho.

So our discussion of how stresses figure into things wasn't *entirely* irrelevant. :wink:
 
  • #251
Peter, I'm going through your derivations in #249, and just a prelimenary query. Not following how you get the simplification of that first G00 expression. When I try it goes like this:

(1/r2)(1-1/K) - (1/r)d/dr(1/K) = 8πρ (that's my version of what we start with)
next perform d/dr operation on K-1, before further grouping of terms ->
(1/r2)(1-1/K)+(1/rK2)dK/dr = 8πρ ->
1/(r2K)(K-1+r/K3dK/dr) = 8πρ

Can't see this coming out to what you get. Past April Fools here but maybe effect lingering for me (the next bit, G11 -> dJ/dr = (...) part looks fine). :yuck:
 
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  • #252
Q-reeus said:
Peter, I'm going through your derivations in #249, and just a prelimenary query. Not following how you get the simplification of that first G00 expression.

It's easier to see if you take it backwards, so to speak; expand the LHS of the second equation to get the LHS of the first:

[tex]\frac{1}{r^{2}} \frac{d}{dr} \left[ r \left( 1 - \frac{1}{K} \right) \right] = \frac{1}{r^{2}} \left[ \left( 1 - \frac{1}{K} \right) + r \frac{d}{dr} \left( 1 - \frac{1}{K} \right) \right] = \frac{1}{r^{2}} \left( 1 - \frac{1}{K} \right) - \frac{1}{r} \frac{d}{dr} \left( \frac{1}{K} \right)[/tex]

In other words, you recognize the first equation as 1/r^2 times a total differential, and just rewrite it that way. You never actually do the d/dr operation on 1/K.
 
  • #253
PeterDonis said:
It's easier to see if you take it backwards, so to speak; expand the LHS of the second equation to get the LHS of the first:

[tex]\frac{1}{r^{2}} \frac{d}{dr} \left[ r \left( 1 - \frac{1}{K} \right) \right] = \frac{1}{r^{2}} \left[ \left( 1 - \frac{1}{K} \right) + r \frac{d}{dr} \left( 1 - \frac{1}{K} \right) \right] = \frac{1}{r^{2}} \left( 1 - \frac{1}{K} \right) - \frac{1}{r} \frac{d}{dr} \left( \frac{1}{K} \right)[/tex]

In other words, you recognize the first equation as 1/r^2 times a total differential, and just rewrite it that way. You never actually do the d/dr operation on 1/K.
Yes, quite so - it looked like an illegal grouping but when I went through and differentiated your expression it worked out the same. April 1st.
 
  • #254
PeterDonis said:
So much of the discussion in the previous thread, about whether stresses were of the "right" magnitude to make the K factor go back to 1 from outside to inside the shell, was really irrelevant. The K factor automatically goes to 1 as m(r) goes to zero, just from the above.
Fine, given what K actually means - not as I had thought (dtau/dr)2.
The next equation is the 1-1 component of the EFE, which I'll rewrite using m(r) instead of K(r):
[tex]G_{11} = - \frac{1}{r^{2}} \frac{2m}{r} + \frac{1}{r} \left( 1 - \frac{2m}{r} \right) \frac{1}{J} \frac{dJ}{dr} = 8 \pi p[/tex]
which easily rearranges to
[tex]\frac{dJ}{dr} = 2 J \frac{m + 4 \pi r^{3} p}{r \left( r - 2m \right)}[/tex]
We'll set this aside for a moment.
Oh no you don't Peter - this is where I feel like having been slipped something in the drink! From that other thread, #23:
Q-reeus: "But even so, let's take the 'huge' atmospheric pressure relevant figure of ~ 10-15. This is the ratio of relevant contributions to that very tiny metric distortion figure of ~ 10-23. So rho contributes a fraction (1-10-15), while p's contribute a fraction 10-15. Isn't that still the only important consideration on this? How can something 10-15 times smaller than the other be overwhelmingly dominant?!"

Because they are "pointing" in different directions (i.e. they affect different components of the tensor). If you were only interested in the time dilation then you would be correct that the pressure is negligible in "ordinary" situations compared to the energy density. However you are specifically interested in the spatial components of the curvature tensor, so the energy density is irrelevant. It is big, but it is in the wrong "direction". Since there is not any momentum flow the only components in the spatial directions are the various normal stress components. You cannot neglect the only source of something you are interested in regardless of how it compares to other things.
(Bold added). And this was reinforced in #24:
Q-reeus: "Well in that case I have very little idea of how it goes. Got the impression from sundry sources (pop sci maybe) that pressure acts simply as an additive term - apart from obvious complications of spatial pressure/density gradients."

For determining the inward "pull of gravity", you are correct; the relevant quantity is ρ+3p, and the pressure p would be negligible in the case we're discussing. However, the "pull of gravity" comes under the heading of "time components" of the curvature. We're talking about "space components", where the energy density does not come into play.
(Bold added). Further reinforced in #30, #42 and later. Mark it well; crystal clear from those quotes - stress and only stress was to play in determining the J transition. So mass density rho should not be there in any direct way - only implicitly as the cause of stress. Is that what I'm looking at in #249 & #250 this thread? Far from it. The m - and I believe that should read m(r) in above eqn's reads 4/3πr3ρ - ρ being rest-energy density (when SI units are used we have that the denominator and numerator m's look very different too). So what has gone on here? I need an explanation as to why ρ now figures, and as straight calculations will show, figures overwhelmingly, in what was to be a strictly stress only thing, as above quotes show. Will leave off further commentary on the rest of #249 till this bit of what looks like an indefinitely extended April 1st is cleared up.
 
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  • #255
Q-reeus said:
Mark it well; crystal clear from those quotes - stress and only stress was to play in determining the J transition.

Those quotes say nothing of the kind. J is the "redshift factor" or "time dilation factor". All the statements you quoted talk about stress contributing to "space components", which have nothing to do with J. Also, the quotes refer to components of the curvature tensor, which is *not* the same as K--or J, for that matter. K and J are components of the *metric*, not the Riemann curvature tensor. I didn't compute the curvature tensor components to see what they depend on, since the original discussion in the other thread was about the J and K factors. But if you want to see what the curvature tensor looks like, I can take a look at that too.

You would have a better case if you complained that much of what we said in that previous thread appeared to indicate that stress should play a role in determining the *K* factor, when apparently it doesn't from my equations. But of course that's why I specifically said that it appears, from the actual math, that much of what we said about stress in that previous thread was irrelevant. (Similar remarks would apply to the curvature tensor components if it turned out that stress didn't appear in those in quite the way we thought in the other thread.) The reason I went to the trouble of working the actual computation from the EFE and posting what I did in #249 was to make sure we were looking at the actual math, what GR actually says, instead of talking off the top of our heads about what we thought, intuitively, that GR might say.

Q-reeus said:
The m - and I believe that should read m(r) in above eqn's reads 4/3πr3ρ - ρ being rest-energy density

Only if the density rho is constant. If it varies you have to integrate dm/dr = 4 pi r^2 rho (with variable rho) through the entire range of r (from the inner shell surface to the r you're interested in) to find m(r).

Q-reeus said:
(when SI units are used we have that the denominator and numerator m's look very different too).

Not sure where you're getting this from. m(r) refers to the same quantity wherever it appears. I was writing the formulas in "geometric units" where G = c = 1, but m(r) still refers to the same quantity if the units are switched (provided they're switched the same way everywhere).

Q-reeus said:
So what has gone on here? I need an explanation as to why ρ now figures, and as straight calculations will show, figures overwhelmingly, in what was to be a strictly stress only thing, as above quotes show. Will leave off further commentary on the rest of #249 till this bit of what looks like an indefinitely extended April 1st is cleared up.

Nothing to do with April 1st (it was only March 31st in my time zone when I posted #249). See above for the "explanation".
 
  • #256
PeterDonis said:
Q-reeus: "Mark it well; crystal clear from those quotes - stress and only stress was to play in determining the J transition."
Those quotes say nothing of the kind. J is the "redshift factor" or "time dilation factor". All the statements you quoted talk about stress contributing to "space components", which have nothing to do with J. Also, the quotes refer to components of the curvature tensor, which is *not* the same as K--or J, for that matter. K and J are components of the *metric*, not the Riemann curvature tensor. I didn't compute the curvature tensor components to see what they depend on, since the original discussion in the other thread was about the J and K factors. But if you want to see what the curvature tensor looks like, I can take a look at that too.
How this has come about eludes me. Had figured, rather loosely, that given K turned out ρ only dependent, that only left J as 'the' one to show this strict stress dependence talked about back there. The quotes made it clear that what I wanted to know would depend only on stress components in shell SET region. And what I wanted was explanation for how in particular the transverse, but also the radial spatial metric components ( not directly but expressed as rdθ/dtau, dr/dtau), went from their Schwrazschild values at r>=rb to flat Minkowski for r<=ra. And that the transition supposedly depended only on shell stresses. My shock from an earlier thread again was being told the transverse component jumped in transitioning through the shell, something I could not figure, given it's independence on gravitational potential in exterior Schwarzscild region. Hence a determination to see what if anything in shell region allowed such a jump. Somehow this all got shunted in a different direction. So we finish up with a ρ only dependent K factor = g_rr, which I had originally thought everywhere equalled (dtau/dr)2, and a J factor = g_tt that has mixed dependence on ρ and p.

Well perhaps we can get it clear at last whether what I had asked for in #1 (rdθ/dtau, dr/dtau) does only depend on stress in transitioning through the SET region. That's what I was being told. As I recall things bogged down over coordinate dependence, but these two quantities are well enough defined so surely computable? Their only use here was to see if a physically real stress-as-source really could pull it off. Evidently now neither K or J are relevant to that. Thinking clearly now I should not have tied stress-only to J transition in last para of #254 since I myself always expected redshift factor to experience a very mild shell transition - nearly all the 'work' is done going from infinity down to r = rb, and only a small reduction throught to r<=ra is expected. In particular it's this apparent transverse spatial metric transition jump that has me bugged.
Q-reeus: "The m - and I believe that should read m(r) in above eqn's reads 4/3πr3ρ - ρ being rest-energy density"
Only if the density rho is constant. If it varies you have to integrate dm/dr = 4 pi r^2 rho (with variable rho) through the entire range of r (from the inner shell surface to the r you're interested in) to find m(r).
Sure but for a solid shell we expect negligible variation - anything beyond ~ 0.1% strain in most engineering materials means catastrophic failure.
Not sure where you're getting this from. m(r) refers to the same quantity wherever it appears. I was writing the formulas in "geometric units" where G = c = 1, but m(r) still refers to the same quantity if the units are switched (provided they're switched the same way everywhere).
Probably just a case of looking very different to SI units TOV shown here http://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_equation I was more concerned that -dp/dr expression appears to relate to a solid sphere than to a shell. Owing to the above matters it is not worth at this stage pursuing whether that sign reversal thing is possible given parameters for a small shell - it won't. Something is basically out and I don't know where. That last bit 2/r(p-s) came in how?
Nothing to do with April 1st (it was only March 31st in my time zone when I posted #249). See above for the "explanation".
OK - just feels like being Groundhog day = April 1st. :confused:
 
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  • #257
Q-reeus said:
And what I wanted was explanation for how in particular the transverse, but also the radial spatial metric components ( not directly but expressed as rdθ/dtau, dr/dtau), went from their Schwrazschild values at r>=rb to flat Minkowski for r<=ra.

The transverse metric coefficients don't change at all. All the change in the K factor does is bring the radial metric coefficient "back into line" as the mass m(r) decreases to zero. Look at the metric I wrote down again; K(r) only multipilies dr^2; the tangential part of the metric is the standard metric on a 2-sphere, and doesn't change at all as we descend through the shell.

Bear in mind, also, that I used standard Schwarzschild coordinates, so that the factor K(r) only multiplies dr^2 in the metric. As was discussed in that previous thread, different coordinate charts will make the behavior of the individual metric coefficients look different. If we used isotropic coordinates, there would be a "spatial" factor L(r) (as I wrote it in that previous thread) that would multiply the *entire* spatial metric, not just the radial portion. But in those coordinates, a 2-sphere at radial coordinate R does *not* have area 4 pi R^2; and in those coordinates the EFE components look different. I stuck to standard Schwarzschild coordinates, where the K factor only multiplies the dr^2 term in the metric, because that's the easiest to match up to a "global" physical interpretation--K(r) in the metric corresponds directly to the observable "non-Euclideanness" of a spacelike slice of constant Schwarzschild time t, as I defined it in that previous thread.

Q-reeus said:
My shock from an earlier thread again was being told the transverse component jumped in transitioning through the shell, something I could not figure, given it's independence on gravitational potential in exterior Schwarzscild region.

Once again, the transverse component, in the coordinates I was using, does not change at all. What changes is the K factor, and that only multiplies the *radial* component. Again, that matches up well with what we actually observe physically; the spacetime is spherically symmetric, so the transverse components *can't* "change"; any given spacelike slice can always be viewed as a series of nested 2-spheres, each of which has the standard 2-sphere metric with no distortion. The Schwarzschild "r" coordinate simply *labels* each 2-sphere such that its area A = 4 pi r^2, to make the labeling match up easily with the metric on each 2-sphere.

Q-reeus said:
Hence a determination to see what if anything in shell region allowed such a jump. Somehow this all got shunted in a different direction.

No, you just forgot all the stuff I took such pains to explain in the previous thread about coordinate-dependent quantities vs. real physical observables, some of which I just recapped above.

Q-reeus said:
So we finish up with a ρ only dependent K factor = g_rr, which I had originally thought everywhere equalled (dtau/dr)2,

Since g_rr multiplies dr^2 in the metric, if you're looking at a small line element where only dr is nonzero, that line element will be spacelike, and g_rr would be interpreted, physically, as (ds/dr)^2 (or, equivalently, sqrt(g_rr) = ds/dr), where s is the physical length along the line element. Spacelike intervals usually aren't labeled with tau, since that is usually used to refer to proper time. This is true everywhere; changing the K factor's numerical value doesn't change what K means physically; it always refers to the "conversion factor" from a radial coordinate differential to an actual radial physical distance.

Q-reeus said:
and a J factor = g_tt that has mixed dependence on ρ and p.

Yes, and this was talked about before, when we said that for ordinary objects, p is so much smaller than rho that its effect on J can be ignored; but the effect is there.

Q-reeus said:
Well perhaps we can get it clear at last whether what I had asked for in #1 (rdθ/dtau, dr/dtau) does only depend on stress in transitioning through the SET region. That's what I was being told. As I recall things bogged down over coordinate dependence, but these two quantities are well enough defined so surely computable?

We went over all this in the other thread. The physical observables are J and K; nothing else varies since the spacetime is spherically symmetric (meaning that each 2-sphere at a given radius has the standard 2-sphere metric, which doesn't change). I have now shown you the exact physical dependence of J and K on the relevant SET components. So in the terms you just used above, dtheta/dtau, or dtheta/ds as I would call it, is unchanged (so r dtheta/dtau varies only as r), and dr/dtau, or dr/ds as I would call it, varies as sqrt(1/K). And dt/dtau (here I agree with using tau since t is timelike here) varies as sqrt(1/J).

Q-reeus said:
In particular it's this apparent transverse spatial metric transition jump that has me bugged.

I can see why you're bugged since there is *no* transverse spatial metric jump, so you'll have a hard time figuring out what it depends on.

Q-reeus said:
Probably just a case of looking very different to SI units TOV shown here http://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_equation

To actually convert that equation to "geometric" units, you would first have to multiply it, both sides, by 1/c^2. If this unit correction is made, the M in the second bracket is multiplied by G/c^2, just as it is in the last bracket. So there's really no difference in the two M's when you normalize the units properly.

Q-reeus said:
I was more concerned that -dp/dr expression appears to relate to a solid sphere than to a shell.

I was, too, which is why I spent considerable time verifying that the only change in the TOV equation when going from a solid sphere to a shell with vacuum inside is to add the (p - s) term that appears in my earlier post. The rest of dp/dr stays the same as the standard TOV equation, because it comes directly from the equation for dJ/dr, which is derived from the 1-1 component of the EFE and is the same for a shell as for a solid sphere, provided m(r) is computed correctly (i.e., for a shell m(r) = 0 at some positive radial coordinate a, instead of at r = 0). (And, obviously, we have to set the right boundary conditions on radial pressure p as well, p = 0 at the inner surface of the shell as well as at the outer.)

Q-reeus said:
Owing to the above matters it is not worth at this stage pursuing whether that sign reversal thing is possible given parameters for a small shell - it won't. Something is basically out and I don't know where. That last bit 2/r(p-s) came in how?

From the connection coefficients that have to be included to properly compute the covariant derivative of the pressure with respect to r in the "1" component of the conservation law (covariant derivative of SET = 0). In the standard TOV equation those particular terms from the connection coefficients cancel because the pressure is isotropic, but they're there.
 
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  • #258
PeterDonis said:
Q-reeus: "And what I wanted was explanation for how in particular the transverse, but also the radial spatial metric components ( not directly but expressed as rdθ/dtau, dr/dtau), went from their Schwrazschild values at r>=rb to flat Minkowski for r<=ra."

The transverse metric coefficients don't change at all.
Well I've just done a lot of tracing back and rereading to find where that notion came from, and here it is:
https://www.physicsforums.com/showpost.php?p=3560020&postcount=224

https://www.physicsforums.com/showpost.php?p=3560127&postcount=227

As I say that was a shock to me. It has different physical consequences to what I now see your position is, which is that radial length coefficient only changes in shell transition, not transverse. Thus for flat interior metric we must have that interior length scale is identical to that at infinity. Which means, because interior clock rate is rescaled as per J factor (or sqrt thereof), interior observer sees shell size as smaller than coordinate observer would. Whereas your position back then amounts to interior observer seeing shell as larger to some extent - depending on just how the spatial components were meant to 'share' the change in transition. Anyway that provided the impetus for the later thread we have been recently referring to. All for naught.

And in that later thread, as per previous quotes, I was led to believe there is this one-to-one match up between metric coefficients and SET components. Thus to explain g_tt shell transition change, T_00 only is involved, and for g_rr, T_rr only is involved, and so on. It is now evident that also is not so; K = g_rr has zero dependence on T_rr and purely depends on T_00, while J = g_tt has mixed dependence on T_00, T_rr, T_ss. From all this you may appreciate why I feel somewhat bamboozled - one might say somewhat misled.

So anyway it now seems obvious stress in static shell does not have the metric altering properties I came to understand it was suppposed to have. Which sort of makes any further discussion of that scenario pointless re this thread. Unless that is there is some sort of 'clean' (not mixing with other T terms) 1:1 match up involving radial and transverse stresses to curvature/metric terms that can be said to have well defined physical meaning?
Since g_rr multiplies dr^2 in the metric, if you're looking at a small line element where only dr is nonzero, that line element will be spacelike, and g_rr would be interpreted, physically, as (ds/dr)^2 (or, equivalently, sqrt(g_rr) = ds/dr), where s is the physical length along the line element. Spacelike intervals usually aren't labeled with tau, since that is usually used to refer to proper time. This is true everywhere; changing the K factor's numerical value doesn't change what K means physically; it always refers to the "conversion factor" from a radial coordinate differential to an actual radial physical distance.
I was using the Wikipedia notation, which is a tau for total differential line element interval. Strictly one should use either ∂tau/∂r or ds/dr I suppose to avoid any confusion.
We went over all this in the other thread. The physical observables are J and K; nothing else varies since the spacetime is spherically symmetric (meaning that each 2-sphere at a given radius has the standard 2-sphere metric, which doesn't change). I have now shown you the exact physical dependence of J and K on the relevant SET components. So in the terms you just used above, dtheta/dtau, or dtheta/ds as I would call it, is unchanged (so r dtheta/dtau varies only as r), and dr/dtau, or dr/ds as I would call it, varies as sqrt(1/K). And dt/dtau (here I agree with using tau since t is timelike here) varies as sqrt(1/J).
So when all is said and done, this is just what I had thought back then, that dr/ds = K-1/2. Well I maintain then the shell transition behavour of K is a purely mathematical consequence of adopting SC geometry and without sensible physical justification, since in my view, J = 1/K should hold in all regions involving shell. But that's my view and gets us away from this thread topic.
 
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  • #259
Q-reeus said:
Well I've just done a lot of tracing back and rereading to find where that notion came from, and here it is:
https://www.physicsforums.com/showpost.php?p=3560020&postcount=224

https://www.physicsforums.com/showpost.php?p=3560127&postcount=227

Yes, I see how those posts could be confusing. That's yet another thread that touched on this topic, and I'll have to go back through it to see how that whole discussion developed. That thread was quite a bit before the others. See further comments below on some specific items of confusion.

Q-reeus said:
As I say that was a shock to me. It has different physical consequences to what I now see your position is, which is that radial length coefficient only changes in shell transition, not transverse.

As long as you define "radial length coefficient" properly. As I've said before, it's important to pay careful attention to the distinction between coordinate-dependent quantities and actual physical observables. The "J" and "K" factors as I defined them in a previous thread are actual physical observables: "J" is the "gravitational redshift" factor, which can be observed by exchanging light signals with someone far away from the gravitating body, and "K" is the "non-Euclideanness" of space, which can be observed by counting how many small identical objects can be packed between two 2-spheres of area A and A + dA, relative to the Euclidean prediction.

It just so happens that in Schwarzschild coordinates, J = g_tt and K = g_rr; but that is a special property of that particular coordinate system, that its metric coefficients just happen to match up so nicely with physical observables. Since calling K the "radial length coefficient" depends on the fact that K = g_rr, that terminology is only really appropriate if you are using Schwarzschild coordinates; i.e., it is coordinate-dependent. Whereas the physical definition of K that I gave above is *not* coordinate-dependent; it may be harder to *express* K in terms of coordinate differentials and metric coefficients in other coordinates, but the physical meaning of K is still the same, and can be described without referring to any coordinates at all, as I did above.

Q-reeus said:
Thus for flat interior metric we must have that interior length scale is identical to that at infinity.

What do you mean by "interior length scale"? This looks to me like a coordinate-dependent quantity. The physical observable, K, being 1 in the interior vacuum region just means that, in that region, exactly as many small identical objects can be packed between two 2-spheres of areas A and A + dA as Euclidean geometry predicts. If that's all you mean by "interior length scale", then yes, it's the same in the interior as at infinity. But if you mean something else, you'll have to specify what you mean. The term "length scale" does not have a unique, well-defined meaning in GR, at least not if you are trying to use it to compared lengths in different parts of spacetime; it depends on the coordinates.

Q-reeus said:
Which means, because interior clock rate is rescaled as per J factor (or sqrt thereof), interior observer sees shell size as smaller than coordinate observer would.

You are correct that J < 1 in the interior vacuum region, so if one could look at light emitted from within that region that was somehow transmitted through the shell and received far away, that light would appear redshifted. That's the physical meaning of J.

Equating this to a different "clock rate" in the interior is OK, but relating that in turn to "shell size" depends on how you do the relating.

Q-reeus said:
Whereas your position back then amounts to interior observer seeing shell as larger to some extent - depending on just how the spatial components were meant to 'share' the change in transition. Anyway that provided the impetus for the later thread we have been recently referring to. All for naught.

In other words, "the size of the shell to an interior observer, compared to its size at infinity" is coordinate-dependent; with one choice of interpretation of coordinates, the interior observer sees the shell as "larger", but with another choice, it is seen as "smaller". None of this has anything to do with physical observables; the physical observable is that K = 1 in the interior vacuum region, and J < 1.

Perhaps I should adopt a policy of refusing to talk at all about coordinate-dependent quantities with you, since it seems to confuse you so much, and just stick exclusively to physical observables. But you in turn would have to agree to accept "sorry, that's coordinate-dependent, so there isn't a well-defined answer" as the answer to a *lot* of the questions you have been asking. You haven't shown much of a desire to do that.

Q-reeus said:
And in that later thread, as per previous quotes, I was led to believe there is this one-to-one match up between metric coefficients and SET components. Thus to explain g_tt shell transition change, T_00 only is involved, and for g_rr, T_rr only is involved, and so on. It is now evident that also is not so; K = g_rr has zero dependence on T_rr and purely depends on T_00, while J = g_tt has mixed dependence on T_00, T_rr, T_ss. From all this you may appreciate why I feel somewhat bamboozled - one might say somewhat misled.

Sorry if you were confused. Now that I've shown you the math, perhaps it will be clearer. But you have always maintained before that you were allergic to math. :confused: Now that I know it actually makes sense to you, I'll feel less of an urge to try to translate into ordinary English, which as I've said many times, is fraught with inaccuracy.

Q-reeus said:
So anyway it now seems obvious stress in static shell does not have the metric altering properties I came to understand it was suppposed to have.

The metric, strictly speaking, is also coordinate-dependent; as I said above, it just happens to be a special property of Schwarzschild coordinates in spherically symmetric spacetimes that the "J" and "K" physical observables happen to match up exactly with g_tt and g_rr. In most cases that doesn't happen. It's much better to focus on the actual physical observables whenever possible.

Q-reeus said:
Which sort of makes any further discussion of that scenario pointless re this thread. Unless that is there is some sort of 'clean' (not mixing with other T terms) 1:1 match up involving radial and transverse stresses to curvature/metric terms that can be said to have well defined physical meaning?

I'll take a look at the curvature tensor components, which will show how physical observables associated with tidal gravity are affected. But again, in general, no, I would not expect there to be a "clean" match up between SET components and metric coefficients; first, since both, strictly speaking, are coordinate-dependent; but second, even looking at physical observables I would not expect there to be a "clean" match up in general.

Q-reeus said:
I was using the Wikipedia notation, which is a tau for total differential line element interval. Strictly one should use either ∂tau/∂r or ds/dr I suppose to avoid any confusion.

Wikipedia's notation is somewhat inconsistent; I've seen both dtau and ds used, and not always with attention to the metric sign convention either.

Q-reeus said:
So when all is said and done, this is just what I had thought back then, that dr/ds = K-1/2.

*If* you are using Schwarzschild coordinates. But that relationship is coordinate dependent; it is *not* true if you use a different radial coordinate. The physical meaning of the K factor is what I defined above.

Q-reeus said:
Well I maintain then the shell transition behavour of K is a purely mathematical consequence of adopting SC geometry and without sensible physical justification

No, the shell transition behavior of K, as I defined it *physically*, is directly observable; you can measure it wherever you like, in principle, by testing how many small identical objects can be packed between 2-spheres of area A and A + dA, and comparing the result to the Euclidean value. This can be done, in principle, inside the shell, and the result will change as you go down through the shell from outer to inner surface. There's nothing "mathematical" about it. It's a physical observable.

The only "mathematical consequence" here is how the physical observable, K, shows up in the metric, since that depends on the coordinates. But that doesn't change the actual physics; it only changes how the physics is represented in the math. It seems to me that you are creating a lot of confusion for yourself by failing to grasp this basic point.

Q-reeus said:
since in my view, J = 1/K should hold in all regions involving shell. But that's my view and gets us away from this thread topic.

Well, it would be interesting if you could give some actual physical reason why J = 1/K (with J and K defined as physical observables, as I did above) should hold everywhere, instead of just in the exterior vacuum region.
 
  • #260
PeterDonis said:
Q-reeus: "Thus for flat interior metric we must have that interior length scale is identical to that at infinity."
What do you mean by "interior length scale"?
As referenced to coordinate measure - 'at infinity', or if you like, referenced to if shell gravity were 'switched off'. Since we agree dr/ds = K-1/2 here, a little man descending through a porthole in the shell will appear to grow in length but not girth, wrt coordinate observer. Of course locally the man observes nothing strange about himself but still something 'weird' will be noticed because whereas it took him so many steps to circumnavigate the shell at r=rb, he finds that for r<=ra, it takes less than expected to circumnavigate at r=ra, or to cross a diameter, if based on the steps taken before. The shell has shrunk on descending, by his local measure.

On the other hand, adopting the position as per first link:
"The faraway observer, looking at how the packing of the little objects changes as you descend through the non-vacuum region, could interpret what he sees this way: radial lengths continue to "contract" through this region, but now *tangential* lengths start to contract as well (they did not in the exterior vacuum region), and the tangential lengths contract *faster* than the radial lengths, so that when the inner surface of the non-vacuum region is reached, the packing of the little objects is now isotropic again; the relationship between radial and tangential packing is now Euclidean, but *all* lengths are now "contracted" compared to lengths at infinity."
In that case we obviously have the shell to locally appear 'less shrunk' or not shrunk at all on descent through the porthole, depending on exactly how the transition works. So it's certainly more than just this counting tiny spheres packed between two concentric spheres - shell diameter/circumference will vary or not as locally measured. The current stance has that locally determined shell circumference/diameter shrinks on descent.
Whichever way one wants to look at it - tiny sphere packing fraction between concentric shells, or changing/not changing shell circumference/diameter, there are real, locally measurable differences on either measure involved, consistent with what I said last post.
Q-reeus: "Which means, because interior clock rate is rescaled as per J factor (or sqrt thereof), interior observer sees shell size as smaller than coordinate observer would."
You are correct that J < 1 in the interior vacuum region, so if one could look at light emitted from within that region that was somehow transmitted through the shell and received far away, that light would appear redshifted. That's the physical meaning of J.
Equating this to a different "clock rate" in the interior is OK, but relating that in turn to "shell size" depends on how you do the relating.
As per my above remarks - varying shell size and varying 'sphere packing fraction' are two sides to the same coin imo. Both will be locally observed, not coordinate artifacts.
Q-reeus: "Whereas your position back then amounts to interior observer seeing shell as larger to some extent - depending on just how the spatial components were meant to 'share' the change in transition."

In other words, "the size of the shell to an interior observer, compared to its size at infinity" is coordinate-dependent; with one choice of interpretation of coordinates, the interior observer sees the shell as "larger", but with another choice, it is seen as "smaller". None of this has anything to do with physical observables; the physical observable is that K = 1 in the interior vacuum region, and J < 1.
Yes it does - shell size varying is a locally measured consequence - again as per above remarks.
Perhaps I should adopt a policy of refusing to talk at all about coordinate-dependent quantities with you, since it seems to confuse you so much, and just stick exclusively to physical observables. But you in turn would have to agree to accept "sorry, that's coordinate-dependent, so there isn't a well-defined answer" as the answer to a *lot* of the questions you have been asking. You haven't shown much of a desire to do that.
No need for such an embargo I would hope, just a matter of each having clear and consistent positions. :rolleyes:
Well, it would be interesting if you could give some actual physical reason why J = 1/K (with J and K defined as physical observables, as I did above) should hold everywhere, instead of just in the exterior vacuum region.
Gets down to what is correct metric in the first place - I believe the exponential metric adopted by a different theory is more consistent but then we are into a different topic and I'd rather not have a hijacking develop. My hope was to find 'interesting' situation re stress as source revealed for static shell - and it's now clearer why that hasn't panned out. Best then to leave this bit behind here. if you want, we could reactivate that other thread, but I'd rather just tidy up this one first if possible. :zzz:
 
  • #261
Q-reeus said:
As referenced to coordinate measure - 'at infinity', or if you like, referenced to if shell gravity were 'switched off'. Since we agree dr/ds = K-1/2 here, a little man descending through a porthole in the shell will appear to grow in length but not girth, wrt coordinate observer.

There is no such thing as a "coordinate observer" in general. There is an observer "at infinity" whose measured radial lengths *at infinity* match up with the r coordinate *at infinity*; but that doesn't tell us anything about how those lengths measured *at infinity* compare with lengths measured elsewhere. Once again, you are looking at coordinate-dependent quantities instead of physical observables, and it is leading you astray. Immediately, in this case:

Q-reeus said:
Of course locally the man observes nothing strange about himself

Or about the spacetime immediately around him.

Q-reeus said:
but still something 'weird' will be noticed because whereas it took him so many steps to circumnavigate the shell at r=rb, he finds that for r<=ra, it takes less than expected to circumnavigate at r=ra

Nope, you're getting confused again by looking at coordinates instead of physical observables. On any given 2-sphere with area A, it will take exactly the "right" amount of steps to walk around the circumference of that 2-sphere, because the metric of the 2-sphere itself is unchanged by the K factor. If by "r" in what's quoted above you mean the Schwarzschild r coordinate, then what's quoted above is wrong, period: a shell at Schwarzschild radial coordinate "r" has physical circumference 2 pi r and physical area 4 pi r^2, so it will take *exactly* as many steps to circumnavigate it as its "radius" r indicates. That's how the Schwarzschild r coordinate is *defined*.

What changes as the K factor changes is the number of "steps" need to walk *radially* between two 2-spheres of areas A and A + dA. So this:

Q-reeus said:
or to cross a diameter

is almost correct--but the number of steps it takes to travel radially between two 2-spheres of areas A and A + dA is *larger* when K > 1 (as it is in the exterior vacuum region, getting larger as we descend, and inside the shell, getting smaller again as we descend until it is = 1 again at the shell's inner surface). This is assuming that the "step" is a constant unit of physical distance (which is fine, but I wanted to make that explicit).

Q-reeus said:
The shell has shrunk on descending, by his local measure.

The *areas* of the shells are unaffected by any of this. The K factor doesn't change the metric on a given 2-sphere. So if you are going to say a shell has "shrunk", shrunk relative to what? Relative to what he "expects" the area of a given shell to be, given how far he had to walk radially to get to it from the shell above it? What makes that the "right" measure of "shrinkage"?

Also, you have it backwards again; the distance that the man has to walk radially for the area of the shells he passes to decrease by a given amount is *larger* than it would be if the geometry of the space he is in were Euclidean. So if anything, the spheres are "expanding" relative to their expected Euclidean size. So I don't see how "shrinkage" is an apt term to describe this.

But *why* do you need a term to describe this? What's wrong with just describing what the K factor means, physically, as I did, and then describing how the K factor changes in each region of the spacetime, as I did? That captures all of the actual physics. What more do you need?

Q-reeus said:
On the other hand, adopting the position as per first link:

Which is an "interpretation" that could be made by a faraway observer--but only an "interpretation". It does not say anything about what a local observer sees. It's just one possible way of interpreting what the change in K factor means. In hindsight, it's a rather tortured interpretation; I haven't had time to go back through that old thread yet to see what the context was when I wrote it.

Q-reeus said:
The current stance has that locally determined shell circumference/diameter shrinks on descent.

No, it does *not*. I have repeatedly explained what the K factor means, physically, in this thread, and I have *not* said anything like this.

Q-reeus said:
As per my above remarks - varying shell size and varying 'sphere packing fraction' are two sides to the same coin imo.

In your opinion--meaning your interpretation. But the physical observable is *only* the K factor, which I think is what you are referring to by "sphere packing fraction"--your continual switching of terminology does not help with clarity. It would be nice if you would just stick with one term, like "K factor", to refer to the physical observable, as I have tried to do. The K factor is *not* an observation of "varying shell size"--it's only an observation of "varying amount of radial distance between shells compared to the Euclidean expected value". Everything else is your interpretation, and your interpretation seems to be adding to your confusion instead of helping to resolve it.

Q-reeus said:
Gets down to what is correct metric in the first place - I believe the exponential metric adopted by a different theory is more consistent but then we are into a different topic and I'd rather not have a hijacking develop.

Do you have a reference for this "exponential metric"?

Q-reeus said:
if you want, we could reactivate that other thread

Which other thread? There have been at least three linked to in this one.
 
  • #262
PeterDonis said:
It's just one possible way of interpreting what the change in K factor means. In hindsight, it's a rather tortured interpretation; I haven't had time to go back through that old thread yet to see what the context was when I wrote it.

Having gone back and looked at the context in that old thread, I think there were two crucial points that I was missing back then, because I hadn't actually gone and done the explicit calculation as I did in this thread:

(1) I was still thinking in terms of isotropic pressure then; but as I've shown in this thread, having tangential stress be different from radial pressure (and in fact having the tangential stress change sign, being positive at the inner surface of the shell and negative at the outer surface) is crucial to the shell being stable. Some of the things I said in that older thread really were only applicable to the case of isotropic pressure.

(2) I was still guessing that everything could be described in terms of a "potential" then; but the calculations in this thread make clear that that can't be done. The J and K factors simply have different dependencies as you descend through the shell; K depends only on m(r) (and hence only on T_00), while J depends on m and p, and p depends indirectly on s, via the extended TOV equation for dp/dr), so J depends ultimately on all the diagonal SET components.

Btw, the fact that K depends only on m makes it clear why the "anisotropy" that was talked about in the other thread goes away as you descend through the shell--the presence of the "anisotropy" is equivalent to K being greater than 1, and K can be greater than 1 at a given radius only if there is nonzero mass inside that radius. So as m goes to zero, the "anisotropy" has to go to zero as well. That seems to me to be a much simpler way of looking at it than the speculations I was making in that other thread.
 
  • #263
Peter - if I were to take you to task on practically every comment made, just in #261, #262, things would most likely finish up pretty nasty. That I don't think either of us need. Sufficient to say sadly we disagree on many issues, interpretations, and points raised. I do thank you for continued participation and useful input when so many others just lurked (not all). If others want to add some fresh input here I will respond if asked to. But probably won't be needing to. Need a rest anyway. Cheers.
 
  • #264
Q-reeus said:
Sufficient to say sadly we disagree on many issues, interpretations, and points raised.

It appears so. I will say that this thread has been instructive for me by prompting me to actually look at the EFE for the shell case; as I said in #262, there were points that did not match up with my previous intuitive guesses. But that just illustrates what I've said before: if we really want to be sure we understand what GR says, we have to look at what it actually says, and that means looking at the actual math. I probably should have done that in previous threads.

Q-reeus said:
I do thank you for continued participation and useful input when so many others just lurked (not all).

You're welcome.
 
  • #265
Of course stress is a source of gravity. I know when I am stressed out I always feel weighted down.
 

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