# How does the stress-energy tensor act on gravity?

• I
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How do the components of the stress-energy tensor act on gravity regarding

a) the FRW-universe?
b) a solid ball?

In a FRW-universe ##\rho + 3P## determines the second derivative of the scale factor. So, there are no non-diagonal components. Just theoretically, if the perfect fluid was elastic, how would this show up in the SET?

Now to the solid ball. How does pressure, shear stress and momentum flux influence the spatial curvature?
Does it make sense to say that these components increase or decrease the effective mass of the ball depending on their sign?

I'm not sure if shear stress and momentum flux can be positiv and negative as well. It would be great to have some examples how to achieve that.

Thanks.

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PeterDonis
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How do the components of the stress-energy tensor act on gravity

I'm not sure what you mean by "act on gravity". The SET is the source of gravity via the Einstein Field Equation.

In a FRW-universe ##\rho + 3P## determines the second derivative of the scale factor.

And this is (a part of) the answer to part a) of your question, since the equation that tells you this is the second Friedmann equation, which is the specific form of the Einstein Field Equation (more correctly the appropriate components of it) for an FRW spacetime. (The other part of the answer is the first Friedmann equation, which is how some other relevant components of the EFE look in FRW spacetime.)

How does pressure, shear stress and momentum flux influence the spatial curvature?

In the ball's rest frame, if the ball is in equilibrium and not rotating, there is no shear stress or momentum flux. Just energy density and pressure. In this idealized model, the ball is actually treated as a perfect fluid. In order to solve the EFE for this case, you also need the ball's equation of state, i.e., the relationship between its pressure and its energy density. This depends on the material the ball is made of.

In a more complicated case, such as a rotating ball, you can have momentum flux and shear stress and the equations get a lot more complicated.

Does it make sense to say that these components increase or decrease the effective mass of the ball depending on their sign?

No. In the simplest case of the non-rotating ball at rest, the total mass of the ball ends up being just the integral of its energy density over its volume. If you look at the details of the math, this is because the increase due to the ##3p## in the integrand is exactly cancelled by a decrease due to the correction factor you have to include because of spacetime curvature. See my discussion in this Insights article:

https://www.physicsforums.com/insights/gravity-gravitate-part-2-sequel/

For the more complicated case where the ball is rotating and there can be shear stress (due to different rotation rates at different depths within the ball) and momentum density, the integral for the ball's total mass is more complicated, and I do not have an explicit solution for it, but I would expect the same sort of thing to happen.

pervect
Staff Emeritus
No. In the simplest case of the non-rotating ball at rest, the total mass of the ball ends up being just the integral of its energy density over its volume.

I don't believe this is correct. I recall there's some discussion of this in MTW, but I haven't been able to find the appropriate section to refresh my memory, I'll keep looking. But what I recall is that the mass in the Schwarzschild metric, the mass contained with in a radius r is ##4 \pi r^2 \rho \, dr##, which integrates out to the Schwarzschld mass parameter M, which gives the same mass as the ADM and Bondi masses. However, ##4 \pi r^2 \,dr## isn't the volume element, because dr isn't a distance. The radial distance is ##dr/\sqrt{|g_{rr}|}##, not dr. So the mass is not the integral of the density * the volume.

In the Newtonian limit, for instance, one needs the mass of a gravitational bound sphere to be lower than volume integral of the density - GR reflects the same result in the appropriate weak field limit.

PeterDonis
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However, ##4 \pi r^2 \,dr## isn't the volume element, because ##dr## isn't a distance.

Yes, you're right, I was being sloppy. The correct statement is that the total mass turns out to be equal to the integral of the energy density over the range of ##r## occupied by the body, without taking into account the fact that ##r## is not a radial distance. When you do the actual correct integral that you describe, which has to take into account the actual radial distance (which is ##dr \sqrt{g_{rr}}##, btw; the metric coefficient is not in the denominator), and also has to take into account the normalization of the 4-velocity (since the integrand is a scalar product of an expression involving the SET with the 4-velocity) and the fact that pressure is a source as well as energy density, you find, as I described, that the increase due to including pressure is exactly cancelled by the decrease due to taking proper account of the metric coefficients.

It's also worth noting that there is a separate way of deriving the integral ##\int 4 \pi r^2 \rho(r) dr## for the mass of a spherically symmetric body that does not require going the roundabout route that I just described (which, as I noted in my Insights article, is called the Komar mass of the system). Solving the Einstein Field Equation for the spherically symmetric case in Schwarzschild coordinates actually leads naturally to the expression ##1 / (1 - 2m(r) / r)## for the metric coefficient ##g_{rr}##, where ##m(r)## is the integral I just described. The reason only ##\rho## appears in the integral is that the expression arises from solving the ##t-t## component of the EFE, for which only ##\rho## appears on the RHS. So for this particular case, you don't actually need to solve the Komar mass integral; but it's still useful to see how both methods arrive at the same answer, particularly since the Komar mass is applicable much more generally (it works for any stationary spacetime), so it's good to see how it works in a simple case.

atyy
I recall there's some discussion of this in MTW, but I haven't been able to find the appropriate section to refresh my memory, I'll keep looking.

I think it's in Schutz, not MTW.

pervect
Staff Emeritus
As far as the stress energy tensor acting on gravity, I also find the question rather vague, but Baez's paper, "The Meaning of Einstein's Equation", might shed some light on the signifinace of ##\rho + 3P##.

Gold Member

I'm not sure what you mean by "act on gravity". The SET is the source of gravity via the Einstein Field Equation.
In the FRW case I'm interested in the effect of shear stress and momentum flux on the value ##\ddot{a}##. This looks strange if not silly. The only purpose is to get a feeling how the components of the SET act. Intuitively I would expect that an elastic perfect fluid would tend to decelerate the expansion which means that it acts like positive pressure. But would tension due to elasticity be represented by ##T^{ii}## with ##i<>0## or by ##T^{ik}## with ##i<>k##?

Before we go to the rotating ball, maybe better to a rotating disc I'd like to understand how the flux of momentum density creates pressure or shear stress respectively. As I understand it pressure results if momentum density flows in the same direction across the surface (e.g. x-momentum in x-direction). And shear stress results if the flux is parallel to the surface. If correct, I would expect that tension due to elasticity creates pressure, because the force is perpendicular to the surface. If so, then rotation of a universe with matter density seems to be the only possibility to create shear stress. How would it act on ##\ddot{a}##? Sorry for such weird ideas. I'm not sure at all, so please correct.

pervect
Staff Emeritus
Rotation is a property of a congruence of worldlines, but not so much (not at all, as far as I know) of the stress-energy tensor, which is what I think you're interested in.

Basically, we can decompose the stress energy tensor into the following format

$$\begin{bmatrix} Energy (1x1) & Momentum (1x3) \\ Momentum(3x1) & Pressure (3x3) \end{bmatrix}$$

Proper choices of the basis vectors always allows us to diagonalize the 3x3 pressure matrix. So the off diagonal terms are all about the choice of basis vectors, rather than having any physical significance, it's a statement about our choice of coordinates.

I believe that the name for the 3x3 part of the stress-energy tensor that I call pressure is the "Cauchy stress tensor" in classical mechanics. I could be wrong about the name, but it doesn't really affect the mathematical argument. The Cauchy stress tensor is convenient because discussions of how to diagonalize it exist in Wiki and in the literature.

The orientation of coordinate axes that diagonalize the Cauchy stress tensor are called the "principal stresses". If our coordinate axes are aligned with the principal stresses, the off-diagonal terms of the tensor vanish. If they are not aligned, they do not in general vanish.

wiki said:
A coordinate system with axes oriented to the principal directions implies that the normal stresses are the principal stresses and the stress tensor is represented by a diagonal matrix:

I suppose I should say that in one sense, a spatial rotation of the coordinate axes does turn a diagonalized pressure tensor into a non-diagonal one, and vica-versa.

Also, if you have all diagonal elements the same (isotropic pressure), symmetry will suppress the genreation of off-diagonal elements created by a spatial rotation.

But this is different sense of 'rotation' from the idea of a rotating ball of fluid.

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PeterDonis
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Intuitively I would expect that an elastic perfect fluid would tend to decelerate the expansion

A perfect fluid with positive energy density and positive pressure does decelerate the expansion. More generally, a perfect fluid for which ##\rho + 3p > 0## decelerates the expansion.

would tension due to elasticity be represented by ##T^{ii}## with ##i<>0##

That is pressure; or, in the terminology some sources use, positive values are called "pressure" and negative values are called "tension".

or by ##T^{ik}## with ##i<>k##?

This is shear stress (for ##i > 0## and ##k > 0##). A perfect fluid by definition has zero shear stress, so these components are zero.

As I understand it pressure results if momentum density flows in the same direction across the surface (e.g. x-momentum in x-direction).

Not quite. Pressure can be present even if the fluid is entirely at rest, which means momentum density is zero. So pressure does not require any net flow of momentum.

Gold Member
Rotation is a property of a congruence of worldlines, but not so much (not at all, as far as I know) of the stress-energy tensor, which is what I think you're interested in.

Basically, we can decompose the stress energy tensor into the following format

$$\begin{bmatrix} Energy (1x1) & Momentum (1x3) \\ Momentum(3x1) & Pressure (3x3) \end{bmatrix}$$
Thanks. So, considering a rotating disc ##T^{12}##, ##T^{13}## and ##T^{23}## (shear stress) and ##T^{21}##, ##T^{31}## and ##T^{32}## (momentum flux) are all zero, correct?

In a rotating disc we have radial and circumferential stresses, as mentioned here. If these manifest themselves as pressure in the SET does it mean that theses stresses are tension or tensile loading resp.?

Gold Member
That is pressure; or, in the terminology some sources use, positive values are called "pressure" and negative values are called "tension".

Not quite. Pressure can be present even if the fluid is entirely at rest, which means momentum density is zero. So pressure does not require any net flow of momentum.
Ok, understand.

If I understand @perfect correctly, the tangential and radial stresses in a rotating rigid body are pressure. And because pressure results from tension due to rotation in this case it should be represented as negative pressure in the SET. Now one can compare the spatial curvature due to a body with and without its rotation. The curvature is determined by ##\rho + 3P##, so the rotational effect can be compensated by ##\rho##, right? Then a rotating body should cause the same curvature like this body (geometrically the same) at rest but with accordingly decreased ##\rho## .

What happens if the negative pressure of the rotating body exceeds the matter density? Is the curvature hyperbolic then, of flat in case of cancellation?

pervect
Staff Emeritus
Thanks. So, considering a rotating disc ##T^{12}##, ##T^{13}## and ##T^{23}## (shear stress) and ##T^{21}##, ##T^{31}## and ##T^{32}## (momentum flux) are all zero, correct?
It depends on the coordinates you use. I believe that's correct for cylindrical coordinates, but don't shoot me if I'm wrong. I don't believe it's correct for Cartesian coordinates, though.

In a rotating disc we have radial and circumferential stresses, as mentioned here. If these manifest themselves as pressure in the SET does it mean that theses stresses are tension or tensile loading resp.?
I believe all the stresses in a rotating disk would be tensions, and your reference says the same. As other posters have mentioned, tensions are just negative pressures.

So to recap, in a rotating disk, I would expect that ##T^{rr}## is some negative number, ##T^{r\theta}## and ##T^{\theta r}## are zero, and ##T^{\theta\theta}## is some negative number.

If you use or convert to Cartesian coordinates, though, I'd exepect ##T^{xy}## to be nonzero.

PeterDonis
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Now one can compare the spatial curvature due to a body with and without its rotation.

No, you can't, because the structure of the body has to change from the rotating to the non-rotating state. There is no way to just change the rotation while keeping everything else the same.

The curvature is determined by ##\rho + 3P##,

The thing that ##\rho + 3P## produces, when it does, isn't spatial curvature, it's spacetime curvature. More precisely, it's one aspect of it. But it's the wrong aspect for what you're trying to analyze.

If you read the Baez article, you will see that he is analyzing the case of a small ball of test particles that changes volume because of the gravity produced by stress-energy. The amount of volume change is proportional to ##\rho + 3P## as calculated for the stress-energy that is present.

But the case you are trying to analyze now is that of a rotating disk whose volume is constant; it is spinning but not expanding or contracting. So ##\rho + 3P## is irrelevant to this case, at least for understanding the tensions produced by the rotation. (It might still be relevant for a large rotating body like a planet, where you have to include hydrostatic equilibrium in the analysis in order to explain how the planet holds itself up against its own gravity; but for the rotating disk case you are considering, for example as described in the link you gave, the disk's self-gravity is negligible.)

In short, you appear to be mixing up different cases that don't really belong together. You need to decide what particular thing you are trying to understand, and focus on the particular case that illustrates that particular thing.

Gold Member
No, you can't, because the structure of the body has to change from the rotating to the non-rotating state. There is no way to just change the rotation while keeping everything else the same.
Ah ok. Can I - then deviating from the SET discussion - compare the mass with and without rotation? I'm thinking of the mass increase of a stretched spring. If I remember correctly @Dale has argued here with the energy-momentum relation some time ago.

The thing that ##\rho + 3P## produces, when it does, isn't spatial curvature, it's spacetime curvature. More precisely, it's one aspect of it. But it's the wrong aspect for what you're trying to analyze.

If you read the Baez article, you will see that he is analyzing the case of a small ball of test particles that changes volume because of the gravity produced by stress-energy. The amount of volume change is proportional to ##\rho + 3P## as calculated for the stress-energy that is present.
I should have mentioned that I don't intend to analyze the rotating disc in the FRW context, but rather as a mass which curves space. But then I end up in the Schwarzschild spacetime and eventually the Kerr metric which makes no sense.

Is the following correct? The SET for a disc at rest is determined by ##\rho_{rest}## and the same disc rotating by ##\rho_{rot} + 3P##, whereby P is negativ. Because of the tension forces exerted on the rotating disc I would expect ##\rho_{rest} > \rho_{rot}## because the volume should be slightly increased. But kinetic energy increases the value of ##\rho##. Perhaps this effect is more important.

So far we had tension which is represented as negativ pressure in the SET.
Are the components corresponding to shear stress always positive?
Is the mass increased by exerting shear forces, analogous to the stretched spring?

I know Baez' article and like it very much.

PeterDonis
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Can I - then deviating from the SET discussion - compare the mass with and without rotation?

How are you going to do the comparison? How do I know a given rotating object is "the same except for rotation" as some non-rotating object, so that I can compare them?

I don't intend to analyze the rotating disc in the FRW context, but rather as a mass which curves space.

Curves spacetime, not space. I keep emphasizing this because it's a very important point. You can't just think in terms of curved space; you have to include time and you have to treat space and time as interrelated, not separate things.

Is the following correct?

Before we can even get to that point, you need to answer the question I posed at the top of this post.

Gold Member
It depends on the coordinates you use. I believe that's correct for cylindrical coordinates, but don't shoot me if I'm wrong. I don't believe it's correct for Cartesian coordinates, though.
Ok, thanks.

I believe all the stresses in a rotating disk would be tensions, and your reference says the same. As other posters have mentioned, tensions are just negative pressures.

So to recap, in a rotating disk, I would expect that ##T^{rr}## is some negative number, ##T^{r\theta}## and ##T^{\theta r}## are zero, and ##T^{\theta\theta}## is some negative number.
Ok, thank for helping!

Gold Member
How are you going to do the comparison? How do I know a given rotating object is "the same except for rotation" as some non-rotating object, so that I can compare them?
In the discussion stretched spring and change in massit didn't play a role that the spring isn't the same after stretching. Why can't one treat the rotating disc analogously?

Curves spacetime, not space. I keep emphasizing this because it's a very important point. You can't just think in terms of curved space; you have to include time and you have to treat space and time as interrelated, not separate things.
Why can't I say a mass curves space, e.g. as measured by a non-euclidean circumference to radius ratio, well knowing that the mass curves spacetime too?

PeterDonis
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In the discussion stretched spring and change in massit didn't play a role that the spring isn't the same after stretching.

That's because an explicit process was described for going from the unstretched spring to the stretched spring. So one valid answer to the question I posed would be that "we know that the rotating disk is the same as the non-rotating disk because we obtained the rotating disk by spinning up the non-rotating disk". Then we could analyze the spinning up process to see exactly what changes and how it affects whatever quantities you're interested in.

Why can't I say a mass curves space

Because "space" depends on your choice of coordinates. Spacetime does not.

Dale
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How do the components of the stress-energy tensor act on gravity regarding

a) the FRW-universe?
b) a solid ball?
The only purpose is to get a feeling how the components of the SET act. Intuitively I would expect
I think there is a substantial danger in this approach of misgeneralizing. Consider if you were to take the same approach to understanding Maxwell’s equations. By restricting yourself to spherical symmetry you would conclude that currents have no effect on the field, even though they are clearly a source term. You would also never discover magnetic fields, or EM waves, Poynting vectors, field momentum, etc. About the only thing you would get right is Coulomb’s law. The rest of electromagnetism would be hidden by the excessive symmetry of the source.

Unfortunately, with less symmetry the calculations become more complicated. So I don’t have a better alternative.

PeterDonis
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I think there is a substantial danger in this approach of misgeneralizing.
And I confess that "act on gravity" is vague as @perfect has mentioned. But the statement 'the components of the stress-energy tensor are sources of gravity' one reads sometimes is also vague and may have influenced the wording of my question. But I hope to make progress with the help of all of you. The problem is my own limit as a layman in physics.

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Gold Member
So one valid answer to the question I posed would be that "we know that the rotating disk is the same as the non-rotating disk because we obtained the rotating disk by spinning up the non-rotating disk".
I was implicitly assuming that, because otherwise such a comparison makes indeed no sense. But sorry, I should have mentioned it.
Because "space" depends on your choice of coordinates. Spacetime does not.
When I said "space" I was thinking of the Schwarzschild spacetime, but explicitly of the curvature factor ##(1 - 2M/r)##. As this factor is determined by M it should be invariant. But I think the whole idea is wrong since the SET doesn't yield M. On the other side, a mass under shear stress or under tension should at the end create a different curvature factor than the same mass with no stresses. In other words M should grow differently if I through m or the same m under such stresses into the black hole. So I'm almost lost. I do appreciate your help very much.

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Dale
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'the components of the stress-energy tensor are sources of gravity' one reads sometimes is also vague
Let me be less vague then. The EFE governs gravity in GR. If you divide the equations up into terms describing gravitational field and terms describing the source of the field then we find that the only source is the stress energy tensor.

The problem is my own limit as a layman in physics
I just don’t think this approach is capable of solving the problem for the reasons that I gave above. This approach wouldn’t even give you an understanding of EM, and EM is much simpler.

But I think the whole idea is wrong since the SET doesn't yield M. On the other side, a mass under shear stress or under tension should at the end create a different curvature factor than the same mass with no stresses.
This is exactly what I am talking about. The equivalent thing happens in EM with a spherical source distribution. You could say a spherical charge distribution with a radial current should create a different field than the same charge with no current, but it doesn’t. The fact that it doesn’t does not imply that current is not a source of the EM field, just that the geometry is too simple.

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PeterDonis
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When I said "space" I was thinking of the Schwarzschild spacetime, but explicitly of the curvature factor ##(1 - 2M/r)##.

Which depends on your choice of coordinates. The factor you describe is there in Schwarzschild coordinates (in the denominator of ##g_{rr}##), but it is not in other coordinates; isotropic coordinates have a factor ##\left( 1 + M / 2R \right)^4## in front of the entire spatial part of the metric, and Painleve coordinates have no spatial curvature at all; "space" in those coordinates is flat.

PeterDonis
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As this factor is determined by M it should be invariant.

No, because it also depends on ##r##, which is a coordinate.

PeterDonis
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a mass under shear stress or under tension should at the end create a different curvature factor than the same mass with no stresses

And the way to approach this, as I've already said, is to start with the mass with no stresses, figure out the spacetime curvature it produces, then apply whatever change you are interested in that produces the stresses (for example, set the mass rotating), figure out what that involves and how it changes the spacetime curvature.