Stress-energy tensors in GR

  • #26
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You are correct, it is not feasible to have anisotropic pressure and no shear because anisotropic pressure is shear. The only difference is a spatial rotation.

This is assuming that the spatial-spatial part of the stress-energy tensor is the same as the stress tensor from mechanics.
Yes, I think this is routinely assumed.
Of course in realistic physical condition neither staticity nor perfect spherical symmetry are expected, and I guess one would have one of the transverse pressure overcome the other and the sphere would rotate ever faster as the mass increased, limited by light speed.

I don't know, anyway it would be interesting if anyone could confirm if a static spherically symmetric SET not corresponding to a perfect fluid is in fact admissible.
 
  • #27
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Yes, I think this is routinely assumed.
Of course in realistic physical condition neither staticity nor perfect spherical symmetry are expected
I agree.

, and I guess one would have one of the transverse pressure overcome the other and the sphere would rotate ever faster as the mass increased, limited by light speed.
Huh? What are you talking about?

I don't know, anyway it would be interesting if anyone could confirm if a static spherically symmetric SET not corresponding to a perfect fluid is in fact admissible.
Why wouldn't it be admissible? As far as I know this is a very standard metric.
 
  • #28
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Huh? What are you talking about?
Just trying to explore the consequences of relaxing staticity, like allowing the sphere to rotate(helped by its pressure anisotropy).


Why wouldn't it be admissible? As far as I know this is a very standard metric.
The metric is perfectly admisible mathematically, I'm obviously referring to the physicality of the imperfect fluid SET for the reasons discussed in previous posts.
 
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  • #29
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it is not feasible to have anisotropic pressure and no shear because anisotropic pressure is shear.
And also it is not feasible to have shear and at the same time assume a spherically symmetric diagonal SET, as they do in the paper, thus my doubts.
 
  • #30
PeterDonis
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And also it is not feasible to have shear and at the same time assume a spherically symmetric diagonal SET
Sure it is, because the SET will only be diagonal in one particular coordinate chart--the standard Schwarzschild chart, with the center of the gravitating body at the origin. More precisely, the SET will only be diagonal in the family of coordinate charts including all possible choices of "zero points" for the angular coordinates. In other words, the spherical symmetry of the spacetime ensures that, *if* we choose a chart centered on the gravitating body (i.e., the gravitating body's center of mass is at the spatial origin), we can rotate the angular coordinates however we like and the SET will still be diagonal. Spherical symmetry only requires that the tangential stress be the same in all tangential directions; it does not require that the tangential stress be the same as the radial stress.

But if the tangential stress is *not* the same as the radial stress, then if we transform to any chart whose spatial origin is *not* at the center of mass of the gravitating body, there will be off-diagonal terms in the spatial part of the SET, indicating shear stress. One obvious way to construct such a chart is to spatially rotate the standard Schwarzschild chart about any point *other* than the origin, i.e., a "spatial rotation", as DaleSpam said (with the proviso that the axis of rotation can't pass through the CoM of the gravitating body). So perhaps a better way to say what DaleSpam was saying is that "shear stress" and "anisotropic pressure" are just two different ways of describing the same physics, using two different coordinate charts.

Note that changing charts does not change the fact that the *spacetime* is spherically symmetric. Spherical symmetry just means there are a set of 3 spacelike Killing vector fields with the appropriate commutation relations. It doesn't require the stress to be isotropic.
 
  • #31
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Sure it is, because the SET will only be diagonal in one particular coordinate chart--the standard Schwarzschild chart, with the center of the gravitating body at the origin. More precisely, the SET will only be diagonal in the family of coordinate charts including all possible choices of "zero points" for the angular coordinates. In other words, the spherical symmetry of the spacetime ensures that, *if* we choose a chart centered on the gravitating body (i.e., the gravitating body's center of mass is at the spatial origin), we can rotate the angular coordinates however we like and the SET will still be diagonal. Spherical symmetry only requires that the tangential stress be the same in all tangential directions; it does not require that the tangential stress be the same as the radial stress.

But if the tangential stress is *not* the same as the radial stress, then if we transform to any chart whose spatial origin is *not* at the center of mass of the gravitating body, there will be off-diagonal terms in the spatial part of the SET, indicating shear stress. One obvious way to construct such a chart is to spatially rotate the standard Schwarzschild chart about any point *other* than the origin, i.e., a "spatial rotation", as DaleSpam said (with the proviso that the axis of rotation can't pass through the CoM of the gravitating body). So perhaps a better way to say what DaleSpam was saying is that "shear stress" and "anisotropic pressure" are just two different ways of describing the same physics, using two different coordinate charts.

Note that changing charts does not change the fact that the *spacetime* is spherically symmetric. Spherical symmetry just means there are a set of 3 spacelike Killing vector fields with the appropriate commutation relations. It doesn't require the stress to be isotropic.
Ok, you mean as long as the tangential stresses are exactly equal in the sphere one can keep the tensor diagonal, right? I was automatically thinking about a situation where the three normal stresses are different.
 
  • #32
PeterDonis
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Ok, you mean as long as the tangential stresses are exactly equal in the sphere one can keep the tensor diagonal, right?
Yes; there can only be two independent principal stresses (tangential and radial), not three. If all three are different, then you're right, the SET and the spacetime can't be spherically symmetric.
 
  • #33
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Yes; there can only be two independent principal stresses (tangential and radial), not three. If all three are different, then you're right, the SET and the spacetime can't be spherically symmetric.
Right, but I'm back to my previous concern, why is this imperfect fluid solution not preferred over the perfect fluid one then?
 
  • #34
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Just trying to explore the consequences of relaxing staticity, like allowing the sphere to rotate(helped by its pressure anisotropy).
Ahh, ok. I haven't explored non spherical metrics much (not even the standard ones like Kerr), so I cannot comment much. I haven't heard of any solutions resembling your comments, but I don't know enough to claim that they don't exist.

The metric is perfectly admisible mathematically, I'm obviously referring to the physicality of the imperfect fluid SET for the reasons discussed in previous posts.
As long as λ and η don't violate the energy conditions then it is physically admissible.
 
  • #35
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And also it is not feasible to have shear and at the same time assume a spherically symmetric diagonal SET, as they do in the paper, thus my doubts.
Sure it is. But I see that I am late responding. I will just "second" Peter Donis' comments.
 
  • #36
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Right, but I'm back to my previous concern, why is this imperfect fluid solution not preferred over the perfect fluid one then?
It is more complicated. It has two free functions instead of just one.
 
  • #37
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It is more complicated. It has two free functions instead of just one.
Compared to the physical advantage that it provides this doesn't seem very relevant.
 

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