On Pressure/Gravity/The Expanding Universe

[Riogho]

Now, It is has been derived from GR that the pressure within a substance contributes to it's gravity just as it's mass density does. But I had a thought

Suppose you have negative pressure. A simple example I can come up with would be when a solid object is stretched to support a hanging weight.

Would this cause a gravitational repulsion?

And if the negative pressure was great enough, to counter the gravity substantiated by the mass density would it actually COUNTER gravity?

And could this be why the universe is actually accelerating it's expansion, because we are full of negative pressure in areas that there is very little mass? (the universe is expanding between known points of mass i.e. galaxies).

Now that I think about it, if this is correct, would dark energy play into this? I can't see how dark energy would have +p... And since it's gravitational effects are slim-to-none, that would give it a gravitational repulsion acting on space itself...

Hmm..

Sorry for the blabbering, I'm just writing as I think.

and looking for answers :P

 

[ZapperZ]

Now, It is has been derived from GR that the pressure within a substance contributes to it's gravity just as it's mass density does.

Er... say what? If I heat a gas in a seal container to increase its pressure, can you show how this relates to the gas "gravity"?

Zz.

 

[Riogho]

http://en.wikipedia.org/wiki/Friedmann-Lemaître-Robertson-Walker_metric

Specifically this section:

The second equation states that both the energy density and the pressure causes the universe expansion rate to decrease, i.e. both cause a decceleration in the expansion of the universe. This is a consequence of gravity, with pressure playing a similar role to that of energy (or mass) density, according to the principles of general relativity. The cosmological constant, on the other hand, causes an acceleration in the expansion of the universe.

 

[CaptainQuasar]

Wikipedia probably isn't the best source to quote for a claim like this.

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[ZapperZ]

Oh dear...

You need to make sure you understand the physics, and not simply buy the "name" being attached to such things. This isn't a thermodynamic "pressure". And you should be careful when something is being used as an analogy.

BTW, and anyone who has been here for a while can tell you this, Wikipedia and I do not mix, i.e. I do not consider it as a valid source of reference.

Zz.

 

[Riogho]

Any chance you could give me a quick round-about of what this new 'pressure' is then?

I'd rather not wait 5 years to take a class on it >.<

 

[ZapperZ]

There is a reason why, in many upper level physics courses, one needs to have the prerequisite knowledge first before taking such classes. Many of these complicated formulation require an already-established foundation and knowledge to be able to comprehend the material.

It is unreasonable to expect that you be taught about such concepts, because inevitably, one will have to keep on backtracking to be able to explain the explanation. For example, would you know what is meant by the "flow of the stress-energy tensor" when I use it to explain to you what the pressure is in GR? What about energy density?

http://xxx.lanl.gov/PS_cache/gr-qc/pdf/0103/0103044v5.pdf

You cannot learn or understand physics by bits and pieces. Physics isn't a bunch of disconnected information. This is why "learning" physics via reading Wikipedia is a fallacy, neglecting the fact that the accuracy of the information you get out of it is suspect. If you build your idea based on such information, you will also sink by it.

Zz.

 

[Riogho]

I see.

Well... perhaps I should sign up for a physics class then.

 

[CaptainQuasar]

Don't totally discount teaching yourself about it either, it's just that Wikipedia can be of limited use for a thorough education.

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[shalayka]

If you took an elastic and stretched it, it would contain more energy than it did when there is no stress (the extra energy equals the amount you put in by working to stretch it). So, I suppose the answer to your question is no. The energy you put in would contribute toward the attractive gravitational field, as usual.

http://hyperphysics.phy-astr.gsu.edu/Hbase/pespr.html

 

[shalayka]

Er... say what? If I heat a gas in a seal container to increase its pressure, can you show how this relates to the gas "gravity"?

Zz.

I'm not sure what you're asking (or if you are asking seriously, since you were the one who initially steered me into the direction of GR when I needed some guidance about a year ago).

The difference between 'dust' and a perfect fluid in GR is that the perfect fluid has pressure. Even if one models both dust and a perfect fluid of identical rest energy, the perfect fluid's gravitational field will be stronger.

As mentioned by the OP, gravitational collapse is inevitable in dense enough material. Since the internal pressure contributes to its own gravitational field, it is a self-defeating vicious circle. More pressure leads to more gravitation, which leads to greater density, which leads to more pressure, leading to more gravitation, and so on. Schutz's book has an entire chapter dedicated to this topic, and it rocks!

 

[pervect]

Er... say what? If I heat a gas in a seal container to increase its pressure, can you show how this relates to the gas "gravity"?

Zz.

John Baez:
http://groups.google.com/group/sci....baez+pressure+causes+gravity#04751e088b0f5c5e

"Wait a minute!" said G. Wiz. "Your formula for R^0_0 agrees with the
stuff in the course outline, and you are right that it means the
gravitational effect of pressure is to cause things to contract. In
layman's lingo: like energy, pressure causes an attractive gravitational
force."

However, this does NOT mean that if you seal gas in a container and heat it, that you get extra gravity (other than from the added heat energy). You have pressure in the gas, and tension in the walls. The net result is no overall effect from the pressure terms, but they do cause a redistribution of force.

 

[shalayka]

I'm not sure why the pressure components wouldn't contribute. According to Schutz's book, momentum flux does not require a particle to cross the boundary.

Higher random velocities is the definition of pressure in this context. From what I understand, if you add heat to a gas, its particles will move with greater random velocities.

If the fluid's dust, the particles have no individual random movement, and that's where pressure doesn't count.

 

[CaptainQuasar]

Coming from someone who doesn't understand any of the nuances here, my understanding has always been that it's mass which produces gravity. But it does seem attractive to a lay ideation that since per Einstein and GR energy and mass are the same thing, the energy that shalayka points out to exist in a pressurized environment would also contribute to gravity? But that's a complete shot in the dark on my part.

It also seems notable that within a given body like a star or planet increased pressure compacts the same given amount of mass into a smaller volume, thus reducing the possible distance between the given mass and other matter and increasing the r² component of the force of gravity.

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[pervect]

I'm not sure why the pressure components wouldn't contribute. According to Schutz's book, momentum flux does not require a particle to cross the boundary.

Higher random velocities is the definition of pressure in this context. From what I understand, if you add heat to a gas, its particles will move with greater random velocities.

If the fluid's dust, the particles have no individual random movement, and that's where pressure doesn't count.

The pressure terms in the gas do contribute. So do the tension terms in the shell enclosing the gas.

For more details, see for instance http://en.wikipedia.org/w/index.php?title=Mass_in_general_relativity&oldid=186756158

(sorry for using the wikipedia as a reference, but I believe the treatment is accurate. I may be biased, because I wrote much of it :-)).

Imagine that we have a solid pressure vessel enclosing an ideal gas. We heat the gas up with an external source of energy, adding an amount of energy E to the system. Does the mass of our system increase by E/c2? Does the mass of the gas increase by E/c2?

The question is somewhat ambiguous as stated. Interpreting the question as a question about the Komar mass, the answers to the questions are yes, and no, respectively. Because the pressure vessel generates a static space-time, the Komar mass exists, and can be found by treating the ideal gas as an ideal fluid. Using the formula for the Komar mass of a small system in a nearly Minkowskian space-time, one finds that the mass of the system in geometrized units is equal to E + ∫ 3 P dV, where E is the total energy of the system, and P is the pressure.

The integral ∫ P dV over the entire volume of the system is equal to zero, however. The contribution of the positive pressure in the fluid is exactly canceled out by the contribution of the negative pressure (tension) in the shell. This cancellation is not accidental, it is a consequence of the relativistic virial theorem (Carlip 1999).

If we restrict our region of integration to the fluid itself, however, the integral is not zero and the pressure contributes to the mass. Because the integral of the pressure is positive, we find that the Komar mass of the fluid increases by more than E/c2.

The significance of the pressure terms in the Komar formula can best be understood by a thought experiment. If we assume a spherical pressure vessel, the pressure vessel itself will not contribute to the gravitational acceleration measured by an accelerometer inside the shell. The Komar mass formula tells us that the surface acceleration we measure just inside the pressure vessel, at the outer edge of the hot gas will be equal to G\left(E + 3 P V \right) / r^2 c^2

This surface acceleration will be higher than expected because of the pressure terms. In a fully relativistic gas, (this includes a "box of light" as a special case), the contribution of the pressure term 3 P V will be equal to the energy term E, and the acceleration at the surface will be doubled from the value for a non-relativistic gas.

One might also ask about the answers to this question if one assumed that one were asking about the mass as it is defined in special relativity rather than the Komar mass. If one assumes that the space-time is nearly Minkowskian, the special relativistic mass exists. In this case, the answer to the first question is still yes, but the second question cannot be answered without even more data. Because the system consisting only of the gas is not an isolated system, its mass is not invariant, and thus depends on the choice of observational frame. A specific choice of observational frame (such as the rest frame of the system) must be specified in order to answer the second question. If the rest frame of the object is chosen, and special relativistic mass rather than Komar mass is assumed, the answer to the second question becomes yes. This problem illustrates some of the difficulties one faces when talking about the mass of non-isolated systems.

 

[CaptainQuasar]

So pervect, in the case of a body like Jupiter, is the force of gravity near it greater than what can be accounted for in its conventional mass - by approximately E + ∫ 3 P dV, because it doesn't have a shell containing it?

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[pervect]

So pervect, in the case of a body like Jupiter, is the force of gravity near it greater than what can be accounted for in its conventional mass - by approximately E + ∫ 3 P dV, because it doesn't have a shell containing it?

⚛​

No, because of an effect that I didn't mention. In flat space-time, the Komar mass of an object (which must be an isolated system) is given by (rho+3P/c^2)*volume. I've included the factors of 'c' here to show what the expression looks like in standard, non-geometric units. The space-time around Jupiter, however, is curved. Suppose we divide Jupiter up into a lot of small chunks. In each individual chunk, we adopt a coordinate system so that that chunk is in flat space-time. We do this because it's easy to explain and envision, not because it's a good way to actually carry out the calculation.

Using this approach, the total Komar mass of Jupiter is equal to the sum over chunks of

K(ch) * volume(ch) * (rho(ch) + 3P(ch)/c^2)

where K is the "redshift factor", computed from the metric, and is less than 1, and K(ch) indicates that K is a function of which particular chunk we are dealing with.

See for example http://en.wikipedia.org/w/index.php?title=Komar_mass&oldid=170982833 (sorry for using a wikipedia article as a reference, especially one in which I had a hand in writing.) This also explains what I mean by "redshift factor", which you can think of as the time dilation factor of the metric. A full technical treatment can be found in Wald, "General Relativity", but will likely be much harder to follow.)

If we take all the chunks and separate them out to infinity, the process requires work. For Jupiter this will be roughly equal to the Newtonian binding energy of the planet. Thus Jupiter weighs less than it would if we divided it up into chunks, took each chunk well away from the planet, measured the mass of each chunk, and added the measurements together.