I How big are the non-mass parts of the stress-energy tensor?

ohwilleke

Gold Member
Summary
I am trying to get a sense of the magnitude of the mass contribution to the gravitational effects via the stress-energy tensor relative to other components (EM flux, kinetic energy, pressure, angular momentum) in a typical system like a solar system or galaxy.
In Newtonian gravity, non-rest mass contributions to gravitational effects are ignored and for many purposes (e.g. low precision solar system astronomy, N-body approximations of galaxy or galaxy cluster dynamics), the other contributions that enter Einstein's field equations through the stress-energy tensor can be ignored without changing the final results very much.

But, the other contributions such as electro-magnetic flux, linear momentum, pressure, and angular momentum do enter into Einstein's field equations and if you are being sufficiently precise, you need to consider them. For example, you need to consider them when calculating frame dragging effects or gravito-magnetic effects.

As a practical matter, in a typical system, like determining the gravitational pull on objects not moving at relativistic speeds in the solar system, or in a galaxy or galaxy cluster, not in the immediate vicinity of (or within) a black hole or neutron star, what percentage of gravitational effects in classical GR can be attributed to rest mass and how much can be attributed to other non-rest mass gravitational source?

Maybe specific examples would be easier to talk about. For example, the precession of Mercury is a system affected by non-mass contributions to the stress-energy tensor. What proportion of the total gravitational pull on Mercury is due to these non-mass contributions and what is simply due to mass (treated in a GR way rather than with Newton's gravity equation)? Or, what difference do non-mass contributions make to the path of the solar system around the Milky Way?

In lieu of a percentage, a sense of the order of magnitude contribution of mass v. non-mass components (and ideally of the respective non-mass components relative to each other), would be fine and helpful.

My sense is that it is usually pretty small, but that's a pretty vague understanding.

Are we talking one part in ten, one part in a thousand, or one part in millions or trillions?

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Nugatory

Mentor
For example, the precession of Mercury is a system affected by non-mass contributions to the stress-energy tensor.
Metcury’s orbit is calculated from the Schwarzschild solution, which is a vacuum solution; the entire stress-energy tensor is zero.

The general relativistic deviations from the predictions of Newtonian gravity are not related to non-mass terms of the SE tensor. They’re inherent in the Schwarzschild solution.

• ohwilleke

PeterDonis

Mentor
you need to consider them when calculating frame dragging effects or gravito-magnetic effects
No. As @Nugatory says, these effects have nothing to do with "non-mass terms" in the stress-energy tensor. They are purely spacetime geometry effects in well-known vacuum solutions.

What you need to look at for significant "non-mass contributions" to the stress-energy tensor are models and predictions regarding the internal structure of dense objects like stars and planets. As a simple example, consider a planet like the Earth, but non-rotating, idealized as a static perfect fluid. Then the only nonzero SET components are energy density and pressure. Using rough figures for the Earth's core, we have $\rho = 1.2 \times 10^{21}$ and $p = 3.6 \times 10^{11}$ in SI units. So the ratio $p / \rho$ is roughly $3 \times 10^{-10}$. As objects get more dense, this ratio goes up--you might try estimating it for a neutron star, for example.

Note, btw, that for a static object, these internal contributions do not appear in the externally measured mass of the object: roughly speaking, the (negative) gravitational binding energy cancels out the (positive) contribution from pressure, so the externally measured mass is just the energy (mass) density integrated over the object's volume.

• Paul Colby and ohwilleke

pervect

Staff Emeritus
Summary: I am trying to get a sense of the magnitude of the mass contribution to the gravitational effects via the stress-energy tensor relative to other components (EM flux, kinetic energy, pressure, angular momentum) in a typical system like a solar system or galaxy.

In Newtonian gravity, non-rest mass contributions to gravitational effects are ignored and for many purposes (e.g. low precision solar system astronomy, N-body approximations of galaxy or galaxy cluster dynamics), the other contributions that enter Einstein's field equations through the stress-energy tensor can be ignored without changing the final results very much.

But, the other contributions such as electro-magnetic flux, linear momentum, pressure, and angular momentum do enter into Einstein's field equations and if you are being sufficiently precise, you need to consider them. For example, you need to consider them when calculating frame dragging effects or gravito-magnetic effects.
The momentum terms would be needed for the frame dragging effects, but not the pressure terms.

Here is a reasonably clear thought experiment that should give you an idea of the magnitude of the effects. Suppose you have some sort of spherically symmetric and very strong shell, and inside it in the very center you have some spherical mass.

You measure the gravitational field (i.e. the gravitational acceleration, with a gravimeter), just inside the inner edge of the shell. The symmetry of the shell means that the shell doesn't affect the reading of the gravimeter.

Now you explode the spherical mass in the center of the shell and convert it into pure energy. And the shell is so strong that it traps all the radiation and other particles that are emitted. Actually, what you want is for the explosion to generate a photon gas of pure radiation, but a matter-antimatter explosion would be a reasonable approximation thereof, though there might be small, insignificant differences as you would have massive particles moving slower than "c" in the debris of the explosion, for instance pions.

The gravimeter reading as a result of the explosion just inside the shell will not stay the same, it will double. This is a result of a paradox called Tollman's paradox, though if you try to google it, you'll get too many hits from one of his other paradoxes to find this particular one.

Note that the field as measured by the gravimeter outside the shell won't change as the result of the explosion. This counter-intuitive effect is due to the tension in the shell needed to hold it together under the enormous pressures.

If the average matter density in is $\rho = m / V$, where m is the mass of the spherical mass at the center, and V is the volume of the sphere, we can write the stress-energy tensor inside the pressure vessel before the explosion as

$$T^{\mu\nu} = \begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

After the explosion, the stress energy tensor will be

$$T^{\mu\nu} = \begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & \frac{\rho}{3}c^2 & 0 & 0 \\ 0 & 0 & \frac{\rho}{3}c^2 & 0 \\ 0 & 0 & 0 & \frac{\rho}{3}c^2 \end{bmatrix}$$

If $\rho$ is 1 kg/m^3, $\frac{\rho}{3}c^2$ is about $3*10^{16}$ pascals. And that pressure doubles the gravimeter reading. Before the explosion, the reading would be $GM/r^2 = G \rho V / r^2 = \frac{4}{3} \pi G \rho r$. After the explosion, the reading would be $\frac{8}{3} \pi G \rho r$, G being the gravitational constant, $\rho$ being the average density, V being the volume $\frac{4}{3} \pi r^3$, and r being the radius of the spherical pressure vessel.

Note that the reading of a gravimeter outside the spherical pressure vessel would be affected by the gravity of the shell as well as the interior. And the reading outside the shell wouldn't change as a result of the explosion.

There was a paper on this, but I don't recall the correct keywords - Tolman's paradox brings up too many hits on a different paradox by the same scientist.

• ohwilleke

PeterDonis

Mentor
The momentum terms would be needed for the frame dragging effects, but not the pressure terms.
More precisely, in order to have frame dragging, you need a spacetime that is not static, which means the interior of the gravitating massive body will have momentum terms in its SET when evaluated in global coordinates in which the center of mass of the body is at rest. The SET in the vacuum region outside the body where frame dragging is measured is zero.

The gravimeter reading as a result of the explosion just inside the shell will not stay the same, it will double.
I don't think it's that simple. See below.

we can write the stress-energy tensor inside the pressure vessel before the explosion as
No, only inside the spherically symmetric mass at the center. The SET just inside the shell is zero before the explosion since there is vacuum there. So the value of $\rho$ is given by $M / V$ for the volume $V$ of the object at the center only, not the entire interior volume of the shell. And the gravimeter reading just inside the shell is the simple Schwarzschild proper acceleration for a static observer.

After the explosion, the stress energy tensor will be
...spread throughout the entire interior of the shell, not just concentrated at the center. So $\rho$ will be different from the $\rho$ before the explosion, because the non-vacuum region is spread over a larger volume. And the gravimeter reading will now not be given by the Schwarzschild formula because the gravimeter is no longer sitting in vacuum, it's in the interior of the object.

• ohwilleke

pervect

Staff Emeritus
No, only inside the spherically symmetric mass at the center.
You read what I wrote, not what I meant. But it's a fair point, you can't be expected to read my mind and un-muddle my posts.

I suppose the best way of describing what I should have written is to say we have spherical shell full of both matter and antimatter, with some sort of membrane separating the two components so they don't mix. The density in the sphere is constant initially, and equal to $\rho$, it fills the entire volume of the sphere. The pressure in the sphere is also intitially zero. It's convenient to maintain spherical symmetry, so the membrane should be spherical and divide the bigger sphere into two separate areas.

Then we remove the membrane. The matter and antimatter mix. When everything reaches equilibrium, we'll have a sphere with the same density $\rho$, but a dramatically increased pressure P.

This is sort of a dry way of saying that if we explode something much more powerful than an atomic bomb in a closed box and don't let anything - no particles or radiation - escape, the interior of the box will have a high pressure.

We approximate the resulting state in the hellish interior of this spherical box as a photon gas.

We also assume the sphere is reasonably small, so that we can used linearized theory and Newtonian formula to calculate the gravimeter reading, rather than using the GR formula. This implies that the mass is small enough that the time dilation term, which is of second order, is negligible. This is OK, because we are not interested in the GR effects that arise for a large amount of matter, we are interested in the effects of pressure.

The point is that the Komar mass density will be $\rho+3P/c^2$, and since P = $\rho c^2/3$, the Komar mass density will double when we allow the contents to explode.

Searching through old PF posts for papers, I find "Pressure as a source of gravity", https://arxiv.org/abs/gr-qc/0510041

Ehlers et al said:
The active mass density in Einstein's theory of gravitation in the analog of Poisson's equation in a local inertial system is proportional to ρ+3p/c^2. Here ρ is the density of energy and p its pressure for a perfect fluid. By using exact solutions of Einstein's field equations in the static case we study whether the pressure term contributes towards the mass.
I'm not particularly fond of the term "active mass density" that the paper uses, the $\rho+3P/c^2$ result is easily recognizable as the Komarr mass density.

The above is from the abstract, another quote from the body of the article talks about the closely related Tolman paradox. The cited papers on Tolman's paradox would also be quite relevant - many of them are paywalled, though.

It was Richard Tolman  who studied universes filled with radiation whobegan to wonder about the consequences of the 3p-term for gravitational theory.The following scenario is known as Tolman’s Paradox: A static spherical boxhas been filled with a gravitating substance of a given mass. If this substanceundergoes an internal transformation (e.g. matter and anti-matter turning into radiation) raising the pressure, the active mass in the box would change be-cause of the 3p-term since the energy is conserved. However, such an internaltransformation should not affect the mass measured outside the box, say by anorbiting particle obeying Kepler’s third law. In a spherically symmetric fieldthe particle should be oblivious to all spherically symmetric changes inside itsorbit, a consequence of the vacuum equations known as Birkhoff’s Theorem
The remaining point is to justify using the Komarr mass as the source of gravitation for small enough systems. While the paper has it's own justification, I think of it as a consequence of the Komar mass being the surface integral of the acceleration (force/unit mass) "at infinity", and the fact that with no time dilation the accelerations are the same locally as they are at infinity.

My approach might be a bit quirky, if so, use the published approach.

MathematicalPhysicist

Gold Member
@pervect how can you have a membrane that separates between matter and anti-matter that won't let them mix? I mean the membrane will be made of matter or anti-matter, so this scenario is impossible even theoretically, unless there's a third type of matter which is not anti-matter and not matter.

Perhaps dark matter?! PeterDonis

Mentor
The density in the sphere is constant initially, and equal to $\rho$, it fills the entire volume of the sphere.
Then we remove the membrane. The matter and antimatter mix. When everything reaches equilibrium, we'll have a sphere with the same density $\rho$, but a dramatically increased pressure P.
Ah, ok, that clarifies the scenario. Yes, in this case $\rho$ will be the same before and after.

We also assume the sphere is reasonably small, so that we can used linearized theory and Newtonian formula to calculate the gravimeter reading, rather than using the GR formula. This implies that the mass is small enough that the time dilation term, which is of second order, is negligible.
This is obviously workable for the "before" state. However, I don't think it is for the "after" state, because the Newtonian formula does not include the effects of pressure, and the whole point of the scenario is that the effects of pressure are equal to those of energy density in the "after" state. I think you have to use the full TOV equation in the "after" state to calculate hydrostatic equilibrium, and that implies a different "acceleration profile" than the Newtonian formula.

the Komar mass density will double when we allow the contents to explode.
More precisely, the Komar mass density not counting the shell will double. But the overall integrated Komar mass of the system, i.e., the externally measured mass, will remain unchanged. The shell's contribution changes in such a way as to offset the change inside the shell.

pervect

Staff Emeritus
@pervect how can you have a membrane that separates between matter and anti-matter that won't let them mix? I mean the membrane will be made of matter or anti-matter, so this scenario is impossible even theoretically, unless there's a third type of matter which is not anti-matter and not matter.

Perhaps dark matter?! You wave your hands a lot. It'd be much better to have the matter and antimatter in separate containers, and some device to merge them. But it complicates the analysis.

The reason I revised the setup it was to make the math simple, when I had (correct) complaints that the math didn't match the setup.

The whole thought experiment isn't very physical. A membrane or field to separate matter from antimatter isn't that much more unrealistic than a container than can contain the resulting explosion without letting anything escape.

It doesn't obviously violate any laws of physics to assume you can do that (though if you dig into it far enough, perhaps it will). So we can think of it as a thought experiment to illustrate just what's needed to have a gravitationally signficant pressure, without a lot of math.

And provide the math on the side for those who can utilize it.

pervect

Staff Emeritus
Ah, ok, that clarifies the scenario. Yes, in this case $\rho$ will be the same before and after.

....

More precisely, the Komar mass density not counting the shell will double. But the overall integrated Komar mass of the system, i.e., the externally measured mass, will remain unchanged. The shell's contribution changes in such a way as to offset the change inside the shell.
Correct, the way I got around the issue of the shell is to have the gravimeter placed just inside the shell, and appeal to spherical symmetry of the shell to say that the shell had no effect on the reading of the gravimeter.

My focus is on the gravimeter inside the shell. The gravimeter outside the shell is a bit of a distraction to the argument, but I couldn't resist mentioning it. Unfortunately, mentioning it may confuse things :(. But when we focus on the gravimeter INSIDE the shell, the reading changes as a result of the explosion, and the reason it changes is the pressure term.

The paper I cited comes to the same conclusion I believe. I found the paper's reasoning a bit opaque, perhaps my reasoning may also appear opaque. My reasoning is ultimately based on some discussion in Wald, that motivates the Komar mass as the integral of the "force at infinity" over a spherical spatial boundry.

Given a familiarity with Wald's argument, we only need to argue that the force at infinity is just equal to the force for a small mass, one when there is no significant time dilation. Formally, the ratio of the force at infinity to the local force is just the redshift factor $\sqrt{| \xi^a \xi_a | }$, where $\xi$ is the Killing vector associated with the time-translation symmetry. See Wald , "General Relativity", pg 288, and it's reference to a "string" and a problem in chapter 6.

PeterDonis

Mentor
My focus is on the gravimeter inside the shell.
Yes, but inside the shell is not vacuum so you can't use the Schwarzschild formula (or its Newtonian approximation) for the gravimeter's reading. You have to actually calculate the gradient of the potential (or, equivalently, the gradient of the redshift factor) from the Einstein Field Equation and the known stress-energy tensor.

In the "before" state this is not really an issue because the potential difference across the shell can be assumed to be negligible, so the gravimeter reading just inside will be basically the same as the gravimeter reading just outside, and that reading we do know from the Schwarzschild formula (or its Newtonian approximation).

But in the "after" state, you cannot assume that the potential difference across the shell is negligible any more, because the shell is now under significant tension in order to hold in the photon gas. So you have to actually calculate the potential and its gradient just inside the shell in order to get the gravimeter reading there; it will no longer be basically the same as the reading outside, and the Schwarzschild formula or its Newtonian approximation won't work.

MathematicalPhysicist

Gold Member
You wave your hands a lot. It'd be much better to have the matter and antimatter in separate containers, and some device to merge them. But it complicates the analysis.

The reason I revised the setup it was to make the math simple, when I had (correct) complaints that the math didn't match the setup.

The whole thought experiment isn't very physical. A membrane or field to separate matter from antimatter isn't that much more unrealistic than a container than can contain the resulting explosion without letting anything escape.

It doesn't obviously violate any laws of physics to assume you can do that (though if you dig into it far enough, perhaps it will). So we can think of it as a thought experiment to illustrate just what's needed to have a gravitationally signficant pressure, without a lot of math.

And provide the math on the side for those who can utilize it.
I am just trying to understand from what the membrane that separates between anti-matter and matter is made of?

I mean if anti-particle meets a particle they annihilate each other, so your scenario is impossible if there isn't a third type of substance which doesn't interact with both of them.
Unless of course, there are particles and anti-particles that don't annihilate each other, for example anti-proton and electron; but I don't know how you can know which particles and anti-particles are there, and actually differentiate between them.

PeterDonis

Mentor
I am just trying to understand from what the membrane that separates between anti-matter and matter is made of?
This is a thought experiment, so the membrane can be made of unobtainium. I mean if anti-particle meets a particle they annihilate each other, so your scenario is impossible if there isn't a third type of substance which doesn't interact with both of them.
Such a substance could be any particle of a different species. For example, if the matter and antimatter is electrons and positrons, the membrane could be made of neutrons--say a thin sheet of neutronium.

Unless of course, there are particles and anti-particles that don't annihilate each other, for example anti-proton and electron
Of course there are. See above.

I don't know how you can know which particles and anti-particles are there
Um, by picking which ones to use? I don't understand what the problem is. Do you think we can't differentiate protons or neutrons from electrons and positrons?

pervect

Staff Emeritus
Yes, but inside the shell is not vacuum so you can't use the Schwarzschild formula (or its Newtonian approximation) for the gravimeter's reading. You have to actually calculate the gradient of the potential (or, equivalently, the gradient of the redshift factor) from the Einstein Field Equation and the known stress-energy tensor.
You can use the _interior_ schwarzschild solution.

Wald gives for the interior metric

$$-f(r)dt^2 + h(r)dr^2 + r^2d\theta^2 + r^2\sin^2\theta \,d\phi^2$$

with $f = e^{2\Phi}$ and $m(r)=\int_0^r 4\,\pi\,\rho r^2 dr = \frac{4}{3}\pi\,r^3$ as

$$\frac{d\Phi}{dr} = \frac{m(r)+4\pi\,r^3\,P}{ r(r-2m(r))}$$

Now when P=0 this is just

$$\frac{d\Phi}{dr} = \frac{m(r)} { r(r-2m(r))}$$

with $m(r) = \frac{4}{3}\pi\,r^3\,\rho$

When P is equal to $\frac{\rho}{3}$ (using geometric units where c=1) the numerator becomes

$$\frac{4}{3}\pi\,r^3\,\rho + 4 \pi \,r^3 \frac{\rho}{3} = \frac{8}{3}\pi\,r^3\,\rho = 2m(r)$$

We don't need to wade through the details of the interior solution to see that when $P = \frac{\rho}{3}$ we simply double the potential $\Phi$ that we had with $P=0$. Wald mentions that $\Phi$ can be regarded as the potential.

The other approaches work too, perhaps you'll find this approach more convincing.

It might also be useful to re-write $d\Phi/dr$ as
$$\frac{d\Phi}{dr} = \frac{\frac{m(r)}{r^2}+4\pi\,r\,P}{ 1-\frac{2m(r)}{r}}$$

and consider the case where $2m(r)/r << 1$, so the denominator can be taken as unity.

Since $m(r) \propto \rho r^3$ , $m(r)/r^2 \propto \rho r$. Since $P \propto \rho$, the second term is also proportional to $\rho r$

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pervect

Staff Emeritus
I am just trying to understand from what the membrane that separates between anti-matter and matter is made of?
It's not really made of anything, it's a thought experiment. But there is something a bit more physical you could make.

You could have an outer solid hollow sphere of matter, and an inner sphere of anti-matter. Both spheres have equal volumes, and the same density. They are both charged, so they repel each other. And there is a small gap between them, so they don't touch.

Now the modelling of the gap wouldn't be perfect, and the charge would affect the geometry a bit, but you can probably ignore both of these issues.

PeterDonis

Mentor
with $m(r)=\frac{4}{3} \pi r^3 \rho$
Ah, I see, you are assuming constant density throughout. Somehow I missed that before. In that highly idealized case (which is unrealistic but has the virtue of being analytically solvable), yes, setting $P = \rho / 3$ just doubles the gradient of the potential by doubling the numerator.

• ohwilleke

pervect

Staff Emeritus
Ah, I see, you are assuming constant density throughout. Somehow I missed that before. In that highly idealized case (which is unrealistic but has the virtue of being analytically solvable), yes, setting $P = \rho / 3$ just doubles the gradient of the potential by doubling the numerator.
I did a bit more work, the summary of what I found was

$$\nabla^2 \Phi \approx 4\pi\left( \rho(r) + 3P(r) + r \frac{\partial P}{\partial r} \right)$$
This is under the condition that we neglect time dilation, i.e. that r << 2m(r)

It's conceptually an easy calclation, knowing $d\Phi / dr$ to get $\nabla^2 \Phi$.

Why are we looking at $\nabla^2 \Phi$?

This relates to (1.4) in https://arxiv.org/pdf/gr-qc/0510041.pdf, which basically suggests that

$$\nabla^2 \Phi = 4\,\pi\,G\,(\rho + 3P)$$

They did not assume G=1, while I did assume G=1.

So if this is true, we get the same solution for $\Phi$ high $\rho$ and low P as we do for low $\rho$ and hi P, i.e. all we need to know is $\rho+3P$ .

The reference above uses $\Delta V$ rather than $\nabla^2 \Phi$ , I think that's just notational differences.

We note there is an apparent discrepancy in the term involving the change in pressure in what I derive as compared to the text.

I believe that this apparently extra assumption can be reasoned away, but I'll simply note that it is not necessary that $\rho(r)$ be constant to get my earlier result, though it is necessary for P(r) to be constant.

It seems quite natural to me that a small enough ball of a photon gas should have a constant interior pressure. Analyzing a small ball is what the r << 2m(r) assumption is all about.

PeterDonis

Mentor
This is under the condition that we neglect time dilation, i.e. that r << 2m(r)
I think you mean r >> 2m(r). That is what equates to $g_{tt} \approx 1$.

it is not necessary that $\rho(r)$ be constant to get my earlier result, though it is necessary for $P(r)$ to be constant.
For a photon gas, since $P = \rho / 3$, if one is constant the other must be.

pervect

Staff Emeritus
I think you mean r >> 2m(r). That is what equates to $g_{tt} \approx 1$.
No, it's definitely r << 2m(r)

For a photon gas, since $P = \rho / 3$, if one is constant the other must be.
That's a good point - for the photon gas, both would have to be constant.

PeterDonis

Mentor
No, it's definitely r << 2m(r)
It can't be. If $r << 2m(r)$ in a static, spherically symmetric spacetime, which is what we are discussing, then $g_{rr}$ changes sign, which means that either the metric is no longer Lorentzian (two negative eigenvalues instead of one) or, if $J(r)$ also changes sign, there is a horizon and you are inside it, which means the spacetime is no longer entirely static. See the line element given in my Insights article:

PeterDonis

Mentor
it's definitely r << 2m(r)
Even just looking at your own posts, this can't be right. At one point, you (correctly) say:

and consider the case where $2m(r)/r<<1$
This is the same as saying $r >> 2m(r)$.

pervect

Staff Emeritus
It can't be. If $r << 2m(r)$ in a static, spherically symmetric spacetime, which is what we are discussing, then $g_{rr}$ changes sign, which means that either the metric is no longer Lorentzian (two negative eigenvalues instead of one) or, if $J(r)$ also changes sign, there is a horizon and you are inside it, which means the spacetime is no longer entirely static. See the line element given in my Insights article:
Ooops. Yeah, we want a small r, but 2m (2Gm/c^2) , the Schwarzschid radius, has to be even smaller.

"How big are the non-mass parts of the stress-energy tensor?"

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