I'm asking for a qualitative explanation if there is one.

My own answer doesn't work. I would have guessed it's because when a gas has pressure the kinetic energy adds to the rest mass of a given quantity of the gas, so the pressure contribution would be equal to whatever energy density it contributes. But that can't be right. If one had an ideal monatomic gas where the atoms are randomly moving around at non-relativistic speeds, the kinetic energy per volume of the atoms is 1.5 times greater than the pressure, but in chapter 4 of Schutz's book "A First Course in General Relativity" (or any other GR text) he says that rho plus pressure (in units where c=1) plays the role of inertial mass density. In my incorrect view the equation would be rho plus 1.5 pressure

Why is my answer wrong? I'm guessing part of the problem is that the kinetic energy of the atoms is already part of the mass density term--that is, a hot gas of one mole of helium atoms would have a higher mass than a cold gas composed of helium. Then the pressure is tacked onto that, which seems like counting it twice to me, but clearly I'm confused.

  • $\begingroup$ Lubos Motl's answer is, not surprisingly, completely correct. I'm not sure that it'll satisfy your desire for a qualitative explanation, though. Unfortunately, I don't have anything better at the moment! $\endgroup$ – Ted Bunn Jan 23 '11 at 23:35

Einstein's equations of general relativity say that $$R_{\mu\nu}-\frac{1}{2}(R-2\Lambda) g_{\mu\nu} = \frac{8\pi G_N}{c^4}T_{\mu\nu}$$ Ignore the cosmological term proportional to $\Lambda$ for a while.

The left hand side is the Einstein tensor and the right hand side is proportional to the stress energy tensor. For low velocities, the dominant component of the equation is the $\mu\nu=00$ (time-time) component, and it effectively reduces to $\Delta \phi_{grav}=4\pi G\rho $, the Poisson equation for Newton's gravity, which implies all the inverse squared distance law, and so on.

However, in relativity, the energy is just 1 component of a 4-vector, the energy-momentum vector, and the density of anything is just one (the time-like) of 4 components of a vector which also includes the flux as the 3 spatial components.

In particular, the mass or energy density $\rho$ becomes just the component $T_{00}$ of a whole symmetric tensor that has $4\times 3/2\times 1$ components (in four dimensions). Relativity implies that all of them are equally important because they can transform into each other by the Lorentz transformations.

In particular, the pressure appears as the doubly spatial components of the stress-energy tensor. Typically, $T_{xx}=T_{yy}=T_{zz}=p$, the pressure. For solids, this pressure is why the tensor contains the word "stress" - stress is a kind of pressure. For all materials, you may imagine that the pressure it the flux of the $p_x$ component of the momentum in the $x$-direction - that's why a gas or liquid will push a wall behind it. In relativity, all these components of the stress-energy tensor have to contribute to the corresponding components of the Einstein tensor (the curvature).

Now I may return to the cosmological constant term. It is effectively the same thing as a stress energy tensor with $p=-\rho$, a negative pressure: you could put it on the right-hand side. Such a form of uniform matter density with a negative pressure deforms the Minkowski space into de Sitter space that is still "maximally symmetric": dust with no pressure wouldn't be able to do so.

I am convinced that any valid - however qualitative - explanation why pressure curves the spacetime in general relativity has to boil down to Einstein's equations in one way or another.

Cheers LM

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Part of the argument can be reasoned without using any "heavy handed" general-relativity, but it is a long road.

Part 1: Relativistic objects are "pulled toward" gravitational sources more than predicted by newtonian mechanics. Light is pulled twice as much. This needs general relativity, see here.

Part 2 Pressure results from exchange of moving particles. In an ideal gas, the moving particles cause pressure. In a solid of liquid, it's more complex. There is degeneracy pressure and attractive forces that allow object to exist in tension. This is due the the exchange of negative energy virtual photons.

A system under pressure (such as a hot gas) contains moving particles that have "extra" gravitational attraction to our test mass. Conversely, our test mass feels an extra attraction to these particles. Although the kinetic energy of the particles contributes to the total mass and thus the gravity, the pressure creates extra gravity beyond that of the mass and kinetic energy alone.

An example: Consider a thin hollow spherical mirror filled with photons (again, we are working with a weak gravity source). We place a test mass (that does not interact with light) just inside the mirror. Due to the shell theorem we need only consider the tiny amount of mass/energy that closer to the center of the sphere than our test mass. The Newtonian calculation would be G(our_mass)(energy_of_photons)/r^2, but the actual force is twice that due to the pressure. If we are outside the mirror the newtonian formula again applies. Although there is pressure inside, there is tension in the walls of the mirror, in effect it is a a balloon inflated with photons! The pressure and tension terms cancel themselves out. When you are outside of a spherically-symmetric object, only total mass matters and the internal pressures will always cancel.

When pressure is due to gravitational compaction you can't escape hard-core relativity if you want to account for it: On earth, the pressure contribution to gravity in the core is only 1e-9 of the density (mass) contribution. This "tiny" amount of pressure isn't canceled out by tension as in the case of our mirror balloon. However, to understand why it gets canceled out we would need to invoke full-beast-mode-general-relativity because relativistic effects are also 1e-9 as strong as Newtonian gravity for Earth (no it's not a coincidence that both are 1e-9). Pressure is only important, in comparison to density, when P ~ (density)c^2, and that c^2 makes even core-of-Earth pressures look small.

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I am going to give a largely physical argument.

The contribution of pressure to the curvature occurs if the motion of the molecules is relativistic. If you have a gas with a certain pressure it means the molecules of the gas are moving rapidly. The momentum of a molecule $p~=~\gamma mv$ will collide off of a wall and impart some momentum to the wall. Given an area $A~=~x^2$ of a wall in a box, if it is massive enough the molecules bounce back with momentum $p’~=~-\gamma mv$. For $N$ molecules impacting the wall in a time $\delta t$ this is a net momentum transfer $\delta p~=~2N\gamma mv$. We assume the $\delta t$ is the time it takes each atom to travel between walls of the box $\delta t~=~x/v$. In this way within that time interval nearly all the molecules impact the walls. So there is a force $$ F~=~\frac{\delta F}{\delta t}~\simeq~2N\gamma mv^2/x $$ The pressure $P~=~F/A$ $=~2N\gamma mv^2/x^3$. Now for the molecules highly relativistic $v~\simeq~c$ and this is an energy density. So the pressure acts as a source of the gravity field --- it has mass-energy.

For cosmologies where galaxies are treated as particles the pressure is usually set to zero. For the interior of a collapsing star the pressure becomes quite large and contributes to the source of the gravity field.

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    $\begingroup$ Either I don't understand this argument, or it's not right. The pressure shows up in the Einstein equation independently of the energy density term: Pressure is $T_{xx},T_{yy},T_{zz}$, while energy density is $T_{tt}$. The argument you give suggests that pressure in a sense shows up in the $T_{tt}$ bit (i.e., energy density), but the question is asking why the other terms gravitate. $\endgroup$ – Ted Bunn Jan 23 '11 at 23:34
  • $\begingroup$ The question asked for a qualitative explanation, so I did not go into tensors. The argument is that pressure density performs a type of "work," which is a form of mass or energy which can act as a source of gravity. Evidently this has not gone over very well. $\endgroup$ – Lawrence B. Crowell Jan 24 '11 at 0:54

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