
#1
Oct1711, 10:14 AM

P: 1,115

This is really a continuation from another thread but will start here from scratch. Consider the case of a static thin spherical mass shell  outer radius r_{b}, inner radius r_{a}, and (r_{b}r_{a})/r_{a}<< 1, and with gravitational radius r_{s}<< r(shell). According to majority opinion at least, in GR the exterior spacetime for any r >= r_{b} is that given by SM (Schwarzschild metric) as expressed by the SC's (Schwarzschild coordinates) http://en.wikipedia.org/wiki/Schwarz...zschild_metric. In the empty interior region r=< r_{a}, a flat MM (Minkowski metric) applies. Owing probably to it's purely scalar nature, there seems little controversy as to the temporal component transition, as expressed as frequency redshift scale factor S_{f} = f'/f (f' the gravitationally depressed value as seen 'at infinity'). In terms of the potential V = (1r_{s}/r)^{1/2}, with r_{s} = 2GMc^{2}, one simply has S_{f} = V, which for any r=< r_{a} has it's r parameter value 'frozen' at r = r_{a}. There is a smooth transition from r_{b} to r_{a} that depends simply on V only. So far, so broadly reasonable.
What of the spatial metric components? In terms of corresponding scale factors S_{r}, S_{t} that operate on the radial and tangent spatial components respectively, it is readily found from the SC's that S_{r} = V, S_{t} = 1, everywhere in the exterior SM region. S_{r} = V is physically reasonable and identical to frequency redshift factor S_{f} here. Whether according to GR S_{f} and S_{r} diverge for some reason in the transition to the interior MM region is not clear to me, but that would be 'interesting'. An apparently consensus view that S_{t} dives relatively steeply from S_{t} = 1 at r = r_{b}, to some scaled fraction of V at r=< r_{a}. [A previous attempt here at PF at the shell transition problem found something different; an invariant S_{t} = 1 for all r, but an S_{r} that jumped back from V to 1 in going from r_{b} to r_{a}. That gave flat interior spatial metric with unity scale factors S_{r}, S_{t} (i.e. 'at infinity' values) but a redshift factor S_{f} = V]. Seems though the consensus view is for an equally depressed S_{r}, S_{t} in the interior MM region, the exact value of S_{r}, S_{t}, and S_{f} relative to V is unclear from recent entries on that matter, where parameters A, B, were used as equivalent to S_{f}, (S_{r}, S_{t}) respectively. I wish to focus on the question of physical justification for S_{t} diving from unity to some fraction of V. Given that an invariant S_{t} = 1 for all r >= r_{b} (SM regime) stipulates complete independence on V or any combination of it's spatial derivatives ∂V/∂r, ∂^{2}V/∂r^{2}, etc. In the shell wall region, where shrinkage of S_{t} apparently occurs, there is owing to a nonzero stressenergy tensor T a different relative weighting of V and all it's derivatives to that applying for r => r_{b}, but otherwise, only the weighting factors are different. What permits variance of S_{t} in one case, but not the other? An answer was that in the shell wall where T is nonzero, the Einstein tensor G (http://en.wikipedia.org/wiki/Einstein_tensor) operates and this is the explanation. That seems unsatisfying, and should be justifiable at a basic, 'bare kernel' level. By that is meant identify the 'primitives' from which everything in G can be derived, and show what particular combo leads to a physically justifiable variance of S_{t}, applying nowhere but the shell wall region. So what are the 'primitives'. I would say just V and it's spatial derivatives, which owing to the spherical symmetry, are of themselves purely radial vector quantities (but obviously not all their combinations as per div, curl etc). One caveat here is to nail down the relevant source of V  taken simply as total mass M exterior to r_{b}. Has been pointed out that for a stable shell there must be pressure p present in addition to just rest matter density ρ, ie T = ρ + p, = T_{00} + T_{11}+T_{22}+T_{33}. My assumption is that for a mechanically stable thin shell of normal material, ρ >>> p, and so to a very good approximation, just use ρ. If one feels the p terms should be included regardless, then I would further suggest they will act here just as a tiny addition to ρ. That is, the contribution of T_{11} for instance in some element of stressed matter introduces no 'directionality' per se to the potential, yes? Can't think of any physical quantity  relevant to this case anyway  that could not be expressed as some function of the above primitives. But recall, all these primitives exist in the region exterior to r_{b}, where S_{t} is strictly = 1!. Only remotely relevant quantity I can think of that *may* be zero in the exterior region but obviously nonzero in the wall region is the three divergence nabla^{2}V. And that could account for a varying S_{t}? Can't imagine how. So something real, real special has to be pulled out of the hat imo. So special I consider it impossible, but open to be shown otherwise. Seems to me the anomaly is intractable in GR and a cure requires a theory where isotropic contraction of spatial and temporal components apply. Then and only then the transition issue naturally resolves. But that's my opinion. So, any GR pro willing to give this a go, let's get on with the show! 



#2
Oct1711, 10:19 AM

Sci Advisor
P: 8,004




#3
Oct1711, 11:26 AM

Mentor
P: 16,472

Don't know if you saw this in the other thread:
Here is a arxiv paper you may like. It uses an analytical model for the shells, so it is not the usual "step function" you would normally consider, but it describes things like the radial and transverse pressures: http://arxiv.org/abs/0911.4822 



#4
Oct1711, 12:54 PM

P: 1,115

How does GR handle metric transition for a spherical mass shell? 



#5
Oct1711, 12:56 PM

P: 1,115





#6
Oct1711, 01:44 PM

Physics
Sci Advisor
PF Gold
P: 5,505





#7
Oct1711, 02:00 PM

P: 1,115





#8
Oct1711, 02:53 PM

Physics
Sci Advisor
PF Gold
P: 5,505

(Quick back of the envelope calculation to verify that it's reasonable: metric coefficient correction for the Earth is of order 10^6, the ratio of r to M. For a typical material like steel, pressure components are of order 10^9 in SI unitsgigapascalsand energy density is of order 10^20 in SI unitsone kilogram per cubic meter is 9 x 10^16, water is 1000 kg/m^3, and steel is several times the density of water. So the ratio of pressure to energy density in a typical structural material is of order 10^11, five orders of magnitude smaller than the metric correction for r a million times M. Looks like we're OK.) http://en.wikipedia.org/wiki/Cylinder_stresses This talks about cylinders, not spheres, but the general idea is the same for spheres, including the sign of the stress. The Wiki page isn't very careful about signs, but the convention for the stressenergy tensor is clear: pressure (compressive stress) is positive and tension (tensile stressstretching) is negative. 



#9
Oct1711, 03:32 PM

Physics
Sci Advisor
PF Gold
P: 5,505

Start somewhere in the EV, well above the outer surface of the shell, but still far enough from "infinity" that the metric differs significantly from the Minkowski metric. Take two concentric 2spheres, both centered exactly on the "shell" (but much larger), one with physical area A, the other with slightly larger physical area A + dA. By "physical area" I mean the area measured by covering the 2sphere with very small identical objects and counting the objects (and by "identical objects" I mean that if you bring any two of them to the same location they will be exactly the same shape and size). Pack the volume between these two spheres with the same identical little objects (which we take to be exactly spherical so they have the same dimension in all directions). Euclidean geometry would lead us to predict that the volume dV between the two spheres that we will find from this packing is given by: [tex]dV = \frac{1}{4} \sqrt{\frac{A}{\pi}} dA[/tex] However, when we actually do the packing, we will find the physical volume dV between the two spheres is *larger* than this, by some factor K; and K will get *larger* as we descend in region EV, getting closer to the outer surface of region NV. Finally, at the outer surface of region NV, K will have reached some value K_o > 1. As we continue to descend through NV, we find the opposite effect now taking place; the factor K begins to get smaller, and when we reach the inner surface of NV, we find that it is now 1, the same value it has at "infinity". In other words, K_i = 1. Once we are inside region IV, the K factor does not change; spacetime is flat. So since K is 1 at the inner surface of NV, it is 1 throughout IV. If we take any pair of spheres centered in IV, with areas A and A + dA, the volume between them, as shown by packing with identical little objects, will be exactly as given by the above formula. I should also include what happens to the "rate of time flow" as we descend through the three regions. (This "rate of time flow" would be observable as gravitational redshift/blueshift, so substitute that in wherever I say "time flow" if that makes it easier to see what I mean.) In region EV, time flow, relative to its rate "at infinity", is slowed by the same factor K that volume between two spheres is increased from its Euclidean value. So at the outer surface of NV, time flow, relative to infinity, is multiplied by a factor J_o = 1 / K_o. However, as we descend through region NV, time flow continues to slow, so when we reach the inner surface of NV, time flow relative to infinity is multiplied by a factor J_i < J_o. Therefore, we can see that J_i is *not* equal to 1 / K_i; the relationship between J and K that held in region EV no longer holds in region NV. And since the rate of time flow is the same throughout NV, the relationship between J and K that held in EV doesn't hold in IV either. I won't comment at this point on how any of the above relates to the descriptions in terms of coordinate systems; I'll save that for another post. But I think the above summarizes the physical observations that would apply in the scenario. 



#10
Oct1711, 03:44 PM

Mentor
P: 16,472

I feel like you are deliberately ignoring a clear physical source of the effect you are interested in for no reason other than your apriori assumption that it cannot possibly be large enough. 



#11
Oct1711, 04:31 PM

P: 1,115





#12
Oct1711, 04:33 PM

P: 1,115





#13
Oct1711, 04:39 PM

P: 1,115





#14
Oct1711, 04:46 PM

Mentor
P: 16,472





#15
Oct1711, 05:38 PM

Physics
Sci Advisor
PF Gold
P: 5,505

But the radial distance, as I've defined it above, is *not* necessarily the same as the radial coordinate r. You can define r to always be the same as the radial distance, but that may not be the easiest definition to work with. More on that in a future post when I talk about coordinates. [tex]J = 1 + 2 \phi[/tex] where [itex]\phi[/itex] is the potential in units where c = 1, and with the usual convention that the potential is zero at "infinity" and negative in a bound system such as this one. But this potential is not what is directly observed; that's J. 



#16
Oct1711, 09:21 PM

Physics
Sci Advisor
PF Gold
P: 5,505

The most familiar radial coordinate for these types of scenarios is the Schwarzschild radial coordinate, which I'll label [itex]r[/itex]; it is defined by [tex]A = 4 \pi r^{2}[/tex] for a 2sphere with physical area A. In terms of this radial coordinate, the metric must take the form: [tex]ds^{2} =  J(r) dt^{2} + K(r) dr^{2} + r^{2} d\Omega^{2}[/tex] where J and K are the gravitational redshift and "packing excess" coefficients I defined in my previous post, which must be functions of r only because of what we said above about the other coordinates. Two things are immediately evident: (1) For this definition of the radial coordinate, the coefficient in front of the tangential part of the metric *must* be [itex]r^{2}[/itex], and nothing else, because we *defined* r that way (so the area of the sphere at r is [itex]4 \pi r^{2}[/itex]). (Btw, this observation also shows why K in the metric above *has* to be the "packing excess" coefficient I defined in my previous post.) (2) In region IV (interior vacuum), K = 1, so the spatial part of the metric with this radial coordinate is exactly in the Minkowski form. The only difference, in terms of this radial coordinate, between IV and the asymptotically flat region "at infinity" is the coefficient J, the gravitational redshift or "time dilation" factor. However, there is another way to define the radial coordinate, the "isotropic" way; the idea here is that we want the relationship between the radial coordinate and actual physical distance to be the same in all directions (as it obviously is not for the Schwarzschild r coordinate). We'll call this isotropic radial coordinate [itex]R[/itex], and the metric in terms of it looks like this: [tex]ds^{2} =  J(R) dt^{2} + L(R) \left( dR^{2} + R^{2} d\Omega^{2} \right)[/tex] I've put a different coefficient, L(R), in front of the spatial part because now, as you can see, it multiplies *all* of the spatial metric, instead of just the radial part. So the area of a sphere at R is [tex]A = 4 \pi R^{2} L[/tex] What does the function L(R) look like? We can stipulate without loss of generality that it must go to 1 at infinity, so that R and r become the same there. We also expect L to get larger as we go deeper into region EV, closer to the outer surface of region NV. But what about in regions NV and IV? I'm still working on that, so I'll follow up with another post. 



#17
Oct1811, 09:34 AM

Physics
Sci Advisor
PF Gold
P: 5,505

I did say in #8 that the key factor is not whether pressure is small compared to energy density, but whether the ratio of pressure to energy density is small compared to the "correction factor" in the metric coefficients. So let's estimate those numbers for a "toy globe". Let's suppose our "toy globe" is a spherical shell made of steel; its total radius is 1 meter, and the shell is 0.1 meter thick (with a hollow interior). We'll assume that the stress components inside the shell are of the same order as atmospheric pressure, or about 10^5 pascals. (They may actually be somewhat larger, but we'll use the lower bound since that will make the pressure/energy density ratio as small as possible.) The mass density of steel is about 8000 kg/m^3. Energy density of the shell = 8000 kg/m^3 x c^2 = 7 x 10^19 J/m^3 Ratio of pressure to energy density = about 10^15 Volume of the shell = 4/3 pi x (1^3  0.9^3) = about 1 m^3 (how convenient!). Mass of the shell = 8000 kg Correction to metric coefficient at shell surface = 2GM / c^2 r = (2 x 6.67 x 10^11 x 8000) / (9 x 10^16 x 1) = about 10^23 So I was wrong in post #8; for a typical object made of ordinary materials, the pressure is *not* negligible; the pressure/energy density ratio is many orders of magnitude *larger* than the correction to the metric coefficients for the object. So there is plenty of "room" for the spatial parts of the stressenergy tensor inside the object to "correct" the metric to flat from the outside to the inside of the shell. I should note, though, that even the above computation is not really "correct", in that I haven't actually tried to compute any components of the field equation; I've just done a quick order of magnitude estimate of the same quantities I estimated in post #8, to show that the relationship I talked about there doesn't hold for an ordinary object. 



#18
Oct1811, 10:04 AM

P: 1,115




Register to reply 
Related Discussions  
Spherical Mass Shell  What Actually Happens?  Special & General Relativity  15  
The virtue particle is mass onshell or offshell?  Quantum Physics  15  
Spherical Shell  Advanced Physics Homework  3  
spherical shell  Advanced Physics Homework  6  
spherical shell  Quantum Physics  1 