# Question about pressure in GR

1. Aug 1, 2008

### bunburryist

On
http://math.ucr.edu/home/baez/einstein/node3.html
it says

"Given a small ball of freely falling test particles initially at rest with respect to each other, the rate at which it begins to shrink is proportional to its volume times: the energy density at the center of the ball plus three times the pressure at that point.

What is meant by "pressure" here? Why is it three times the pressure?

2. Aug 1, 2008

### MeJennifer

Re: Question about "pressure" in GR

Think about a massive sphere, the pressure inside this sphere decreases from maximum at the center to zero on the outside.

In general relativity, pressure, just like a mass and energy, contributes positively to spacetime curvature.

3. Aug 1, 2008

### bunburryist

Re: Question about "pressure" in GR

The article refers to "A small ball of freely falling test particles." Is this meant to describe particles separated from one another in space but arranged in the shape of a ball (this is what it sounds like to me), or particles that are in contact with one another in the shape of a ball. I don't understand how they can be in contact with one another if they are freely falling. Either they are in contact, and so can "press" against one another (in which case they can't be "freely" falling), or else they are freely falling, in which case there is no physical pressure in the sense of contact. What am I missing here? I don't understand what is pressuring what, so to speak.

4. Aug 1, 2008

### Mentz114

Re: Question about "pressure" in GR

bunburryist:
You seem to misunderstand that 'freely falling' in this context has nothing to do with whether they are touching or not. One marble can fall freely, and if two are glued together, they can also fall freely together.

I seem to recall that J. Baez likens his particles to coffee-grounds. Which reminds me ...

M

5. Aug 1, 2008

### bunburryist

Re: Question about "pressure" in GR

I know - I'm being real dense here, I know, but I want to be sure I have this right.

So it's not that the particles are simply falling together (like a collapsing cloud), but rather that they, as a group, are falling toward another larger object - right?

If that's right, it would also make an earlier statement from Baez-

To state Einstein's equation in simple English, we need to consider a round ball of test particles that are all initially at rest relative to each other. As we have seen, this is a sensible notion only in the limit where the ball is very small. If we start with such a ball of particles, it will, to second order in time, become an ellipsoid as time passes.

- make more sense.

Do they form an ellipsoid because the part of the ball closest to the larger mass will accelerate slightly faster than the far side, stretching the ball into an ellipsoid?

Either I'm starting to understand this or I'm nowhere close! I appreciate your help very much, MeJennifer.

6. Aug 1, 2008

### George Jones

Staff Emeritus
Re: Question about "pressure" in GR

They are not in contact. Pressure is force/area and force is rate of change momentum, so pressure is the amount of momentum flow per of unit area per unit time.

Think of a rigid box that contains an ideal gas. Remove a small patch of one wall of the container. Some molecules escape. At the instant that the patch is removed, the flow of momentum (a component of momentum perpendicular to the wall) per unit area per unit time of the escaping molecules is equal to the pressure of the gas.

The situation in the paper by Baez and Bunn is the reverse of the above paragraph. Imagine placing a small patch of a rigid (no such thing in relativity) wall and fixing it so that it remains at rest in the local inertial frame. Some of the particles bounce off the patch. The pressure on the patch equals the flow of momentum of the particles at this location when the patch isn't there.

Three perpendicular patches are needed to cover all possible directions in space, i.e., spatial momentum is a sptatial vector.

7. Aug 1, 2008

### MeJennifer

Re: Question about "pressure" in GR

No, that is not related to the initial example you quoted however it is related to the second quotation you made.

8. Aug 1, 2008

### cesiumfrog

Re: Question about "pressure" in GR

Why is pressure pertinent to gravity?

Theoretically/historically, are pressure terms shown necessary in order to for a covariant/relativistic description of mass to match the rank of the Einstein tensor?

Are there thought experiments demonstrating why pressure cannot be treated the same as other forms of potential energy (e.g., chemical)?

9. Aug 1, 2008

### Antenna Guy

Re: Question about "pressure" in GR

I can't say that I know for certain, but I think the factor of 3 arises from how an integral over a spherical volume of uniform density can be constructed. For example:

Let's say that the sphere of particles has a uniform density $\rho$, and the mass is given by:

$$M=\rho\frac{4}{3}\pi R^3$$

Now let's divide this sphere up into concentric shells such that the mass of any given shell is given by:

$$m=\rho\frac{4}{3}\pi(r_2^3 - r_1^3)$$

After factoring the difference of cubes we have:

$$m=\rho \frac{4}{3}\pi(r_2^2 +r_1 r_2+r_1^2)(r_2 - r_1)$$

In the limit where $r_1$ approaches $r_2$ this reduces to:

$$m=\rho 4 \pi r^2 \delta r$$

Rearranging terms gives:

$$4 \pi \rho \delta r = \frac{m}{r^2}$$

Integrating the LHS (which appears to have the units of differential pressure) from zero to R yields:

$$4 \pi \rho R$$

Note that the usual "3" in the denominator (of a spherical volume integral) is nowhere to be found.

Regards,

Bill

10. Aug 1, 2008

### bunburryist

Re: Question about "pressure" in GR

Is this right -

If we were to put the walls inside a balloon and compress the balloon, the pressure on the walls would increase, and, of course, the pressure would be there whether or not there was a wall. Only in the case of the particles in the example, rather than our compressing the balloon, and so the air inside, thus increasing the pressure, the particles are moving into more and more "compressed" space-time (I'm sure this is a sloppy way of putting this) and so are moving closer together, compressing their momentum into a smaller and smaller area of space-time, thereby increasing the overall pressure.

As for their momentum, the only place I can see it coming from would be as they move closer together as they move into more compressed space-time, just as two balls falling toward the earth, and, moving closer to one another, would acquire a momentum relative to each other as they converge.

From the way I read the text the particles are not moving relative to one another at the beginning, but are motionless. So, if I understand, the pressure in the text is not so much a mechanical pressure of things bouncing off of, or pressing against other things (although that would be just as relevant as the motionless particles in the example), but rather one of increasing momentum as they move closer and closer together.

11. Aug 2, 2008

### George Jones

Staff Emeritus
Re: Question about "pressure" in GR

So there are more particles per unit area moving across a surface, and thus a greater flow of momentum per unit area.
Yes, the pressure at the start is zero, but become non-zero as the particles move and have momentum with respect to the local inertial frame. The only initial contribution to equation (2) is from the mass/energy density.

12. Aug 2, 2008

### MeJennifer

Re: Question about "pressure" in GR

I would say they have momentum with respect to each other as there really is no local inertial frame.

13. Aug 3, 2008

### JimJast

Re: Question about "pressure" in GR

Landau once said that it is difficult to explain something that one does not understand himself. It seems very true in the case of many people explaining GR "in plain English". That's one thing. The other thing is that "a small ball of freely falling test particles initially at rest with respect to each other" will never shrink because of virial theorem.

It seems that it is better to start learning physics Feynman's style, from knowing how the nature works. Not mathematician's style, ignoring the true behavior of nature and in the process inventing never observed phenomena like non conservation of energy, expanding space, or shrinking of small balls of free falling particles. However I must admit that mathematician's style might have its good points regarding two legs of science: experimental (eg. Feynman's) and theoretical (eg. Baez's).

14. Aug 3, 2008

### George Jones

Staff Emeritus
Re: Question about "pressure" in GR

In a vacuum, this is true to second order in time, but not because of the virial theorem. If the particles are test particles that don't influence the curvature of spacetime, then the mass/enregy density is zero as well as the initial pressure, and equation (2) of Baez and Bunn gives that

$$\frac{d^2 V}{dt^2} = 0.$$

15. Aug 3, 2008

### JimJast

Re: Question about "pressure" in GR

That's about my point: in this sense there are no "test particles" in physics. They are purely mathematical invention, never observed in the real world which is not only relativistic but also quantum, unlike mathematical, idealized world. Now, in which one do we live in and which mechanisms should we rather learn (and teach) first?

Once I have a fight with the head of "Gravitation and Cosmology Dept." in our U, when I insisted that student's should be taught general relativity at the first year instead of Newtonian magic which they tend to absorb as real stuff. And the head maintained that they are to stupid for that and that it was OK to teach them Newtonian stuff for a few hundred years with good results, so it is OK now. And I believe that it is why so many people believe that the universe is expanding since they are isolated for the real knowledge and they are taught only a mathematical abstractions that don't work the same as the nature does.

16. Aug 4, 2008

### cosmik debris

Re: Question about "pressure" in GR

In the Stress Energy Tensor the diagonal consists of 4 components. The time (00) component is the energy density. The other three are spatial momentum flow components in the x, y, and z directions. Momentum flow through a unit area is a pressure. If the matter is homogeneous and isotropic the pressures are the same in all three directions, add them together or multiply by three and add the energy density. The whole of this tensor contributes to the Einstein curvature. G = T.