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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.

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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! –  Ted Bunn Jan 23 '11 at 23:35
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2 Answers

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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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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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. –  Ted Bunn Jan 23 '11 at 23:34
    
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. –  Lawrence B. Crowell Jan 24 '11 at 0:54
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