Understanding Metrics & Forces in Extreme Situations

[pervect]

Rather than argue about Newtonian theory, let's try a different aproach.

Consider the GR case, as outlined in http://math.ucr.edu/home/baez/einstein/

more specifically
http://math.ucr.edu/home/baez/einstein/node3.html

In any event, we may summarize Einstein's equation as follows:

{\ddot V\over V} \Bigr\vert _{t = 0} = -{1\over 2} (\rho + P_x + P_y + P_z).<br />

This equation says that positive energy density and positive pressure curve spacetime in a way that makes a freely falling ball of point particles tend to shrink. Since E = mc^2 and we are working in units where c = 1, ordinary mass density counts as a form of energy density. Thus a massive object will make a swarm of freely falling particles at rest around it start to shrink. In short: gravity attracts.

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 the pressure in the x direction at that point, plus the pressure in the y direction, plus the pressure in the z direction.

So, the second derivative of the rate of change of the volume of a sphere of coffee grounds is proportional to the volume of the coffee grounds multiplied by the density of matter - plus, a pressure term.

Note that, as Baez & Bunn explain, this depends only on the amount of matter contained within the ball. External gravity via external matter does not cause the volume to shrink. If we put the coffee grounds near the Earth, there will be the tension and compression terms you noted. The result will be that the ball will grow longer in one direction, shorter in the other two, and the second derivative of the volume will be zero.

Now, skip on a bit to:

http://math.ucr.edu/home/baez/einstein/node7.html

They apply the above result to a sphere of coffee grounds, placed at rest relative to each other.

The result is that the ball shrinks, due to gravity, and that the accleration required is just the volume of the ball * (rho+3P).

Thus, with no cosmological constant, we see that the expansion deaccelerates due to gravity.

If we insist that our ball hold its shape, by making it rigid, and put strain gauges on it, we will see that the ball is in compression. Gravity tries to make the ball shrink, and the forces that make the ball a rigid body oppose gravity.

I thought that this was easy to explain from a Newtonian viewpoint, but we got tied up in a lot of difficulties. So perhaps the GR argument as presented by Baez & Bunn will actually make the explanation simpler.

I do agree that the force is independent of direction. I'm replacing the ball by a very-lightweight and open "rigid bar". And the goal is to find the tension (or compression) in the bar. The endpoints of the bar will correspond with two of the particles in Baez's ball of coffee ground. The fact that the ball of coffee grounds shrinks means that the bar would also shrink, unless a force opposes it.

Thus the B&B argument shows that a bar, in free space, would experience compressive forces. This is not due to the expansion of space, but is due to the matter contained within space, so that one has to be careful that the bar experiment doesn't change the density of the local matter distribution.

 

[MeJennifer]

Rather than argue about Newtonian theory, let's try a different aproach.

Consider the GR case, as outlined in http://math.ucr.edu/home/baez/einstein/

more specifically
http://math.ucr.edu/home/baez/einstein/node3.html



So, the second derivative of the rate of change of the volume of a sphere of coffee grounds is proportional to the volume of the coffee grounds multiplied by the density of matter - plus, a pressure term.

Note that, as Baez & Bunn explain, this depends only on the amount of matter contained within the ball. External gravity via external matter does not cause the volume to shrink. If we put the coffee grounds near the Earth, there will be the tension and compression terms you noted. The result will be that the ball will grow longer in one direction, shorter in the other two, and the second derivative of the volume will be zero.

Now, skip on a bit to:

http://math.ucr.edu/home/baez/einstein/node7.html

They apply the above result to a sphere of coffee grounds, placed at rest relative to each other.

The result is that the ball shrinks, due to gravity, and that the accleration required is just the volume of the ball * (rho+3P).

Thus, with no cosmological constant, we see that the expansion deaccelerates due to gravity.

If we insist that our ball hold its shape, by making it rigid, and put strain gauges on it, we will see that the ball is in compression. Gravity tries to make the ball shrink, and the forces that make the ball a rigid body oppose gravity.

I thought that this was easy to explain from a Newtonian viewpoint, but we got tied up in a lot of difficulties. So perhaps the GR argument as presented by Baez & Bunn will actually make the explanation simpler.

I do agree that the force is independent of direction. I'm replacing the ball by a very-lightweight and open "rigid bar". And the goal is to find the tension (or compression) in the bar. The endpoints of the bar will correspond with two of the particles in Baez's ball of coffee ground. The fact that the ball of coffee grounds shrinks means that the bar would also shrink, unless a force opposes it.

Thus the B&B argument shows that a bar, in free space, would experience compressive forces. This is not due to the expansion of space, but is due to the matter contained within space, so that one has to be careful that the bar experiment doesn't change the density of the local matter distribution.

While I don't disagree with anything here I fail to see how this explains anything. I am sure the problem is mine and not yours.

Clearly the effect represented by the cosmological [ initially and incorrectly written "gravitational". meJennifer] constant is treated differently from {\ddot V\over V} \Bigr\vert _{t = 0} = -{1\over 2} (\rho + P_x + P_y + P_z)..
I cannot see any explanation other than , "we'll just make some constant to even it out". The same goes for the numerous "expansion factors", to me they just seem ad hoc additions to match experimental data.
By the way there is *nothing* wrong with that, but what would be wrong is to assert that this all logically follows from GR.

Perhaps I have a fundamental misunderstanding here, I love to be educated here. I simply fail to see why a simple constant for "dark matter" would suffice while for regular matter we would need a more complicated method.

 

[pervect]

While I don't disagree with anything here I fail to see how this explains anything. I am sure the problem is mine and not yours.

Clearly the effect represented by the gravitational constant is treated differently from {\ddot V\over V} \Bigr\vert _{t = 0} = -{1\over 2} (\rho + P_x + P_y + P_z)..
I cannot see any explanation other than , "we'll just make some constant to even it out". The same goes for the numerous "expansion factors", to me they just seem ad hoc additions to match experimental data.
By the way there is *nothing* wrong with that, but what would be wrong is to assert that this all logically follows from GR.

Perhaps I have a fundamental misunderstanding here, I love to be educated here. I simply fail to see why a simple constant for "dark matter" would suffice while for regular matter we would need a more complicated method.

What makes you say that?

The only real difference with the above expression is that in GR, pressure causes gravity. In Newtonian physics, this isn't the case. This doesn't matter much until we get to "dark energy" and its negative pressure. It doesn't affect "dark matter" at all.

Note that we can apply Gauss's law to show that something is defintely fishy about the previous argument that was being presented about the gravitational field in an infinite and/or very large universe.

If we draw a sphere somewhere in this universe, and integrate the Newtonian gravitational field perpendicular to the surface of the sphere, we must get the enclosed mass, by Gauss's law.

The argument that the field is everywhere zero can't be corrrect - for then the surface intergal would be everywhere zero, and the enclosed mass would be zero. But we just got through saying that we wanted a universe that did have matter in it.

So let's go back to the finite case, of a very large (but not infinte) spherical planet, and forget about infinites for a bit.

By applying Gauss's law, we can conclude that the field always points towards the center of the sphere. In the constant density case, we find that the field is proportional to the distance away from center.

This is because F = GM/r^2 and M = rho * r^3, therefore F = g M rho r

Note that this sort of Hooke's law force is what we'd get for a constant density Earth.

Note also that this sort of Hooke's law force implies a tidal force in the radial direction, because the force increases with distance.

 

[oldman]

Rather than argue about Newtonian theory, let's try a different approach.

I quite agree. Thanks for the URL's.

From a Newtonian perspective all I could conclude, from the symmetry of the situation, is that the only possible manifestation of gravity in the "cosmic fluid" would be pressure, rather than the mistaken idea of tidal forces I was hung up on. But I can't calculate what that pressure would be. For this, it seems, GR is needed.

The conclusion you come to:

Thus, with no cosmological constant, we see that the expansion deaccelerates due to gravity.

If we insist that our ball hold its shape, by making it rigid, and put strain gauges on it, we will see that the ball is in compression. Gravity tries to make the ball shrink, and the forces that make the ball a rigid body oppose gravity.

I agree, even though I can't see it from a Newtonian perspective. GR rules.

When it comes to estimating these forces, or rather pressure (which I failed to calculate in a Newtonian way), I have one question about your earlier comments. It probably arises because I had muddied the waters by talking of tidal forces instead of pressure.

The pressure isn't given by the H^2 you mentioned earlier (as found from the components of the Riemann tensor), is it?

If it were, I would worry that the dynamics have something to do with determining this pressure, which seems ridiculous. Of course right now, when H is about 10 ^ - 19 per sec, nobody would give two hoots about such a tiny quantity as H^2 --- it would be a mere scientific curiosity. But during inflation, when I estimate that H is huge, H^2 has a startlingly large value which might have physical consequences if the pressure you mention were to depend on H.

I expect that this quibble arose from crossed wires, though.

 

[pervect]

I will agree with one thing about the Newtonian analysis, though. The finite spherical case I was talking about has a preferred direction, towards the center of the sphere.

I suspect that the Newtonian limit for an infinite universe may not really be well defined, that it may depend on how exatly one approaches infinity (infinte sphere, infinite plane, etc). I still don't believe the "no force" solution, as I explained earlier, because it conflicts with Gauss law.

The infinite GR case won't have such a preferred direction and doesn't run into this issue though. In the neighborhood of a point, everything nearby the point follows a geodesic will accelerate towards that point - given that there is no a positive matter density and no "dark energy" or "qunitessence".

 

[oldman]

I suspect that the Newtonian limit for an infinite universe may not really be well defined, that it may depend on how exactly one approaches infinity (infinite sphere, infinite plane, etc).

This may well be so.

I still don't believe the "no force" solution, as I explained earlier, because it conflicts with Gauss law.

Here we disagree. Your explanation was, I assume:

Note that we can apply Gauss's law to show that something is definitely fishy about the previous argument that was being presented about the gravitational field in an infinite and/or very large universe.

If we draw a sphere somewhere in this universe, and integrate the Newtonian gravitational field perpendicular to the surface of the sphere, we must get the enclosed mass, by Gauss's law.

The argument that the field is everywhere zero can't be correct - for then the surface intergal would be everywhere zero, and the enclosed mass would be zero. But we just got through saying that we wanted a universe that did have matter in it.

So let's go back to the finite case, of a very large (but not infinite) spherical planet, and forget about infinities for a bit.

By applying Gauss's law, we can conclude that the field always points towards the center of the sphere. In the constant density case, we find that the field is proportional to the distance away from center.

This is because F = GM/r^2 and M = rho * r^3, therefore F = g M rho r

Note that this sort of Hooke's law force is what we'd get for a constant density Earth.

Note also that this sort of Hooke's law force implies a tidal force in the radial direction, because the force increases with distance.

There is conflict embedded here that I don't understand.

You can show that there is no field at any point in the "cosmic fluid" by integrating the field inside spherical shells centred on the point from zero radius out to infinity. And inside each shell Gauss's law shows that the field due to that shell is zero. So the integration must also give zero. But, as you point out, then using Gauss's law to find the mass inside any sphere will then give zero, which is wrong. I don't understand this.

A simpler and in my view more believable argument is just to say that the gravitational field is zero throughout the cosmic fluid because the fluid is everywhere always isotropic, i.e highly symmetic.

I suspect that this conflict is somehow connected with the applicability of Gauss's law to "compact" domains. The cosmic fluid isn't compact; it may be infinite. I then agree that:

The infinite GR case won't have such a preferred direction and doesn't run into this issue though. In the neighborhood of a point, everything nearby the point follows a geodesic will accelerate towards that point - given that there is no a positive matter density and no "dark energy" or "qunitessence".

Whatever the resolution of the Gauss's law argument, my main question remains: how do you quantify the above acceleration in an inflating universe? as H^2?

 

[oldman]

I claimed in my previous post (#36) that:

You can show that there is no field at any point in the "cosmic fluid" by integrating the field inside spherical shells centred on the point from zero radius out to infinity. And inside each shell Gauss's law shows that the field due to that shell is zero. So the integration must also give zero. But, as [Pervect, post#33] points out, then using Gauss's law to find the mass inside any sphere will then give zero, which is wrong. I don't understand this.

A simpler and in my view more believable argument is just to say that the gravitational field is zero throughout the cosmic fluid because the fluid is everywhere always isotropic, i.e highly symmetic.

This post is a note to show that the view I've taken in this last paragraph is, I think, consistent with Einstein's law of gravity. The argument (which may be quite wrong!) runs like this:

Accept for a moment that the Newtonian gravitational field is indeed zero everywhere in an infinite cosmic fluid, as I've argued above. This means that there is no spatial gradient of field anywhere, no net force on any element of the fluid and no tidal stresses of the tear-apart or squash-together kind caused by force-gradients.

But a non-directional compressive or dilational stress, namely pressure is permitted. It must be the same everywhere --- it's not allowed to have gradients, because of high symmetry and the forces gradients would create, which are taboo.

Now pressure is the flow of momentum in a fluid, due to the motion of its elements. And momentum implies kinetic energy, which is equivalent to mass. The pressure that is allowed in the cosmic fluid is therefore the contribution to the mass/energy density of the fluid of the motion of its elements, in this case provided that such motion preserves the high symmetry of the cosmic fluid.

Expansion (or contraction) is just this kind of motion. It seems to me that a uniform pressure in the cosmic fluid will arise because of its expansion (or contraction), in exactly the way laid down by Einstein's law of gravity, which is, from

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

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 the pressure in the x direction at that point, plus the pressure in the y direction, plus the pressure in the z direction.

Of course this doesn't reconcile the paradox raised by Pervect (see above) with the no-field view. I still don't know how this is to be settled.

 

[pervect]

A few technical notes:

The method of measurement of tidal force is that basically one has a rigid bar, of fixed length. The ends of the bar are thus NOT following geodesics. One measures the stress, or strain, in the bar required to keep the ends of the bar at a constant distance.

For the flat FRW metric

ds^2 = -dt^2 + a^2(t) (dx^2 + dy^2 + dz^2)

the components of the tidal acceleration in the x direction can be computed via the geodesic deviation equation

http://math.ucr.edu/home/baez/gr/geodesic.deviation.html

Note that this computes the tidal force for an observer following a geodesic, hence the name. It won't compute the correct tidal force for an accelerating or rotating observer. For instance, a non-accelerated observer in flat Minkowskian space-time won't experience any tidal forces, but an accelerated observer in flat Minkowskian space time will experience tidal forces according to GR. See for instance Bell's spaceship paradox. This effect is not one that one would expect from Newtonian physics.

The tensor used to calculate the tidal force doesn't have an input for the acceleration of the observer, it only computes the tidal forces for a geodesic observer.

When the center of the bar is following a geodesic, and the bar is short, the equation can also be used to find the stress in the bar as below.

The equation is:

a_tidal = R^\hat{x}{}_{\hat{t}\hat{x}\hat{t}}

Note that the four-velocities in Baez's equation are unity, and serve to pick out a particular component of the Riemann. I've not included them above, because they are unity. The "hats" indicate that I'm using a frame-field basis. This can be thought of as a local coordinate system that "wears the hats" that is locally Minkowskian.

Note that the y and z components are identical by symmetry.

The value for a general expansion is \frac{-\frac{d^2 a}{d t^2}}{a} where a is the scale factor.

While I computed this via the Riemann tensor, it is possible to take a simpler more physical approach.

We are trying to find the separation between geodesics. We know that comoving objects in the "cosmic flow" follow geodesics. We know that the distance as a function of time for comving objects goes as (coordinate-difference) * a(t). Thus we can write

accel = K d^2 a / dt^2

where K is a constant equal to the difference in comoving coordinates.

Tidal acceleration is just acceleration per unit length. Since length = K * a, the tidal acceleration is just

(d^2 a / dt^2 ) / a

This is our previous result, except for the sign, which I've been sloppy about keeping tract of.

For the particular case of a De-sitter universe, a(t) = exp(Ht) where H is Hubble's constant. We can then see that for this case, the above expression reduces to H^2.

In the general case, the tidal force can be derived from the decleration parameter q and the Hubble constant

http://scienceworld.wolfram.com/physics/DecelerationParameter.htmlAs far as your remarks about pressure in the fluid go, I'll have to study them at more length to make sure we aren't "crossing wires".

 

[oldman]

A few technical notes...

It's good to see actual machinery in motion. I guess that most of my puzzles have been resolved, thanks to you, and that we're now on the same page. It's been quite a long road, and I'm grateful that your persistence has prevailed.

I have one remaining question about the interpretation of the word "pressure" in general relativity (GR). It is often made clear that in GR pressure contributes to gravitating mass. You have shown that in the "cosmic fluid" the component of the Riemann? tensor is, in the x direction:

a_tidal = R^\hat{x}{}_{\hat{t}\hat{x}\hat{t}} {and that} ...the y and z components are identical by symmetry

By taking into account these three equal diagonal components you can calculate "the pressure", which in the case of an inflating cosmic fluid turns out to be H^2, and you seem to regard this as a symmetrical "tidal force" . I hope I've got this right.

In ordinary physics, by "pressure" one usually means phenomena like hydrostatic pressure, gas pressure, degeneracy pressure, or radiation pressure, which one understands intuitively to be a driver of compression (or dilation).

In GR, where pressure gravitates, I suspect that the interpretation of pressure is a bit wider than this. In GR, pressure seems to be a convenient (and quite appropriate) label for the flux of momentum, and through this for the mass equivalent of the energy of motion of constituent masses. Formally, of course, this label has the same derivation as "ordinary" pressure. In a gas, for example, GR pressure can be regarded as the mass equivalent of the kinetic energy of the random motion of gas molecules, via m = this energy/c^2.

Is this interpretation correct?

If it is, then is not the pressure you calculate for an inflating cosmic fluid nothing more than the mass equivalent of the energy of the motion of expansion of the cosmic fluid elements? In which case I can't quite make out why it would induce strains in a "rigid" bar that could be measured with strain gauges, as if the bar were at the bottom of a sea.

But perhaps I have got entirely the wrong idea about "pressure" as it is used and calculated in GR.

 

[pervect]

The pressure that is the source term in GR is not related to the tidal forces that we just calculated.

For instance, if we have a "dust" universe, the pressure, as used in Einstein's equations is zero - the tidal forces, however, we just got through calculating and won't be zero in a dust universe.

While your remark that what we've calculated is essentially a pressure is true, you shouldn't confuse it with the sort of pressure that's the source term in Einstein's equations, the pressure that's part of the stress-energy tensor.

To compute the pressure as it is usually definied, we look at the total force on the wall of a box of fixed volume, when the box is of infinitesimal size.

Now, you are probably thinking "well, the universe is expanding, so that there is a force on the box walls due to the geodesic motion of the particles, therefore there is a pressure on the walls".

But if we work out the pressure due to cosmological expansion, it depends on the size of the box, and goes to zero when the box has an infinitesimal volume. At small volumes, all particles essentially have the same velocity, so there is no force on the walls of a small box. In other words, the force on our box walls is proportional to the length of the side of a box, and when this length goes to zero, the force is zero. Therfore the pressure, which is the limit of the force / area as the size goes to zero, is also zero.

You can think of it this way.

If you have take the particle swarm model of a fluid, and all the particles at any point have the same velocity, the pressure in the fluid is zero.

I.e. if all particles at one particular spot have the same velocity in magnitude and direction, the fluid is pressureless.

However, if you have random, thermal motion, like that in an ideal gas, so that at anyone spot the "swarm of particles" are all moving in different directions, that represents a fluid with pressure.

The cosmological "pressure" for a dust universe can thus be seen to be zero, because all the individual particles are moving in one direction, the direction of the Hubble flow.

A universe with radiation represents a universe with pressure, a universe with just comoving dust has no pressure.

 

[oldman]

The pressure that is the source term in GR is not related to the tidal forces that we just calculated.

The argument you have given to justify this clear statement doesn't convince me, I'm afraid. We seem to be talking past each other, as here and in your post #38 you keep mentioning a "tidal force" that is being calculated, whereas I keep on banging on and on about "pressure".

I think that what you are calculating is neither a "force" nor "tidal", but the accelerated motion of fluid elements.

From the geodesic deviation equation you can calculate accelerations, which can be associated with proportional "forces" onlywhen the fluid elements have mass . The dust particles you mention in your post #40 don't have mass, so unsurprisingly there is indeed no pressure in a dusty fluid. You can't use a dusty fluid as an example to justify the statement quoted above. When fluid elements have mass, their accelerated motion must contribute to momentum flux and therefore act as a source term.

Then the adjective "tidal" is inappropriate in the cosmic fluid, as I realized some time back in this thread. As Peacock puts it in Cosmological Physics (p.43): in a tidal field "any tendency for test particles to approach one another along one direction means that they will be separated along the other direction". The pressure we are discussing is therefore not a tidal phenomenon.

Lastly, your argument that in the cosmic fluid:

At small volumes, all particles essentially have the same velocity, so there is no force on the walls of a small box... the pressure due to cosmological expansion, it depends on the size of the box, and goes to zero when the box has an infinitesimal volume. ... If you have take the particle swarm model of a fluid, and all the particles at any point have the same velocity, the pressure in the fluid is zero.

I.e. if all particles at one particular spot have the same velocity in magnitude and direction, the fluid is pressureless.

I can't accept this. All volumes of the cosmic fluid, no matter how small, are expanding and partake of the Hubble flow, and neighbouring particles never have the same velocity, no matter how close they are. "Essentially" is a bit of a weasel word!

I believe that I am not the person who is guilty of confusing ordinary pressure --- the kind that in a gas produces forces on the walls of a box with "the sort of pressure that's the source term in Einstein's equations, the pressure that's part of the stress-energy tensor." But I surely don't want to turn an amicable disagreement into an acrimonious argument. I hope I'm not doing this!

 

[pervect]

Let me provide a reference for what I'm trying to say:

http://en.wikipedia.org/wiki/Fluid_solution

In general relativity, a fluid solution is an exact solution of the Einstein field equation in which the gravitational field is produced entirely by the mass, momentum, and stress density of a fluid.

...

Several special cases of fluid solutions are noteworthy:

* A perfect fluid has vanishing viscous shear and vanishing heat flux:
* A dust is a pressureless perfect fluid:
* A radiation fluid is a perfect fluid with μ = 3p:

Dust solutions by defintion do not have a pressure by the standard defintion of the term, as per the Wikipedia defintion.

If you want to call what we have been talking about a "pressure" rather than a "tidal force" or "tidal acceleration", that's OK with me as long as I know what you're talking about.

But what you must not do is to confuse the concept we have been discussing with the standard defintion of the pressure of a fluid, the one that is used in the GR defintion, the P in rho + 3P.

And if you call the concept that we have been discussing just "pressure" without any other modifier, I don't see any way to avoid confusion, having two different ideas with the same name.

To try and be really specific:

A solution to the FRW equations where rho > 0 and P = 0 is called a 'dust solution' (per the wikipedia article).

[add] It's a red herring to talk about whether any partricular grain of dust has mass. The important thing is that in the specific solution I'm talking about, the average density of matter per unit volume is non-zero.

Because rho > 0, the volume of a sphere of comoving coffee grounds will contract due to gravity in the above solution. This leads to the effects we've been discussing.

Because P is defined to be equal to zero by the standard defintion, the rate of contraction of the volume of the sphere of comvoing coffee grounds cannot be considered to be a measure of pressure of the fluid, since the former is non-zero and the later is zero in the example case of a FRW dust solution.

 

[oldman]

If you want to call what we have been talking about a "pressure" rather than a "tidal force" or "tidal acceleration", that's OK with me as long as I know what you're talking about.

I agree that this is sensible. For the purposes of this discussion, then, let's call it "coffee grounds motion", say CGM for short, rather than pressure of any kind.

But what you must not do is to confuse the concept we have been discussing with the standard defintion of the pressure of a fluid, the one that is used in the GR defintion, the P in rho + 3P.

And if you call the concept that we have been discussing just "pressure" without any other modifier, I don't see any way to avoid confusion, having two different ideas with the same name.

Yes, I may well have been rather stupidly sowing confusion here. My apologies for this.

A solution to the FRW equations where rho > 0 and P = 0 is called a 'dust solution' (per the wikipedia article).

Thanks for this URL about various solutions. With the Wikipedia available, one hardly needs to attend grad school these days!

It's a red herring to talk about whether any partricular grain of dust has mass. The important thing is that in the specific solution I'm talking about, the average density of matter per unit volume is non-zero...Because P is defined to be equal to zero by the standard defintion, the rate of contraction of the volume of the sphere of comoving coffee grounds cannot be considered to be a measure of pressure of the fluid, since the former is non-zero and the later is zero in the example case of a FRW dust solution.

What you say is mostly correct. But, as I discuss below, is the specific dust solution the one that applies to CGM? Or is this so only by definition?

Because rho > 0, the volume of a sphere of comoving coffee grounds will contract due to gravity in the above solution

I agree; as you said earlier, this is why the expansion of the universe slows down, and why Baez and Bunn at

http://math.ucr.edu/home/baez/einstein/node7.html say:

...this is the same as the equation of motion for a particle in an attractive force field. In other words, the equation governing this simplified cosmology is the same as the Newtonian equation for what happens when you throw a ball vertically upwards from the earth!

But in real situations coffee grounds do have mass, even if this can often be neglected in applying an idealised solution. In the universe, for example, the coffee grounds are galaxies. CGM then does have momentum and energy, and hence mass. Surely this must contribute to the gravitating mass?

I believe (firmly at the moment) that what you describe as "the pressure term in the standard definition of the pressure of a fluid, the one that is used in the GR definition, the P in rho + 3P" represents (quite appropriately) nothing more than the mass-equivalent of the kinetic energy in this "standard" situation. If I'm wrong here then I've misunderstood the fundamentals of general relativity quite badly.

The heart of my difficulty seems to be this: you are looking at the situation using a specific established solution in the literature, which is appropriate on the largest scales because the universe is mainly empty, whereas I'm trying to understand things on a scale where one talks of bars being compressed and strains being measured with stick-on strain gauges.

As a result I still can't decide whether the effects we've been discussing actually exist, and if they would be so huge during inflation.

 

[pervect]

I believe (firmly at the moment) that what you describe as "the pressure term in the standard definition of the pressure of a fluid, the one that is used in the GR definition, the P in rho + 3P" represents (quite appropriately) nothing more than the mass-equivalent of the kinetic energy in this "standard" situation. If I'm wrong here then I've misunderstood the fundamentals of general relativity quite badly.

The heart of my difficulty seems to be this: you are looking at the situation using a specific established solution in the literature, which is appropriate on the largest scales because the universe is mainly empty, whereas I'm trying to understand things on a scale where one talks of bars being compressed and strains being measured with stick-on strain gauges.

As a result I still can't decide whether the effects we've been discussing actually exist, and if they would be so huge during inflation.

The pressure term is not the mass-equivalent of the kinetic energy, unfortunately. I think at one time I had this same (incorrect) idea. However, in any given frame, the intergal of T⁰⁰ gives the total energy content, including kinetic energy. More specifically, adopt a locally Minkowskian frame. Then in a unit volume dV in this locally Minkowskain (i.e. flat) frame, the local energy contained in a volume element dV is just.

\int_V T^{00} dV

where T^{00} = \rho does not include any pressure terms.

If you consider any complete closed system, however, the pressure terms will integrate to zero, so they don't contribute to the system mass. Systems with finite volume that aren't closed are very tricky to deal with, and I'd suggest avoiding them as much as possible.

You can "close" an otherwise unclosed system just by adding in neglected components of the system. When you do so, you have a complete, closed system and the pressure terms then will integrate out to zero.

I talk a little bit about some of the fine points regarding mass in

http://en.wikipedia.org/wiki/Mass_i...simple_examples_of_mass_in_general_relativity

the question and answer section is the most relevant.
(note that I'm the author of this article), and that it also discusses some unrealted stuff as well as related stuff. This has a reference to the Carlip paper, for instance, which discusses the relativistic virial theorem which is what makes the pressure terms integrate out to zero. Unfortunately I haven't seen this (the relativistic virial theorem) disucssed much in any of the textbooks I happen to own, so the Carlip reference is the best I have on this somewhat obscure but interesting topic.

As far as the rest goes, I'm personally quite convinced that there is no special difference between calculating tidal forces due to the Schwarzschild metric and tidal forces due to the FRW or De-Sitter metrics.

The only thing to beware of in my opinion is that when one calculates the tidal forces using the geodesic equation, one calculates the tidal forces for a non-rotating observer following a geodesic, that observers following non-geodesics may experince different forces (as may the "centrifugal forces" for rotating observers).

 

[oldman]

The pressure term is not the mass-equivalent of the kinetic energy, unfortunately.

You have seriously undermined my firm belief that this was so! It is now clear to me from your Wikipedia article and the Carlip article that gravitating mass/energy is an altogether more difficult concept in general relativity than I'd ever realized. Indeed my confusion about pressure gravitating (I'd thought that pressure exists only because elements of mass move, and that because the kinetic energy of their motion gravitated, pressure itself gravitated) is not entirely dispelled by statements like those below, which I of course accept:

We can thus tell our students with confidence that kinetic energy has weight, not just as a theoretical expectation, but as an experimental fact.

and

If two objects have the same mass, and we heat one of them up from an external source, does the heated object gain mass? If we put both objects on a sensitive enough balance, would the heated object weigh more than the unheated object? Would the heated object have a stronger gravitational field than the unheated object?

The answer to all of the above questions is yes...

I guess I'll have to accept that the answer to the question in my original post:

Consider an observer... in an inflating flat FRW universe that begins to expand exponentially rapidly after he has set up his local inertial frame. Suppose he ... releases two test particles (joined by string) from rest in this frame... What happens as the scale factor, and its derivatives with respect to time, change exponentially with time? ...

is that the string breaks because expansion in this case speeds up -- very rapidly indeed, since H^2 is so huge!

Thanks for your patience in bringing me to this conclusion.