Why positive pressure causes deceleration
Selasa, 19 November 2013
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komentar
...explaining a puzzling sign to Mr Joe Public...
Sean Carroll shouted a completely wrong claim about basic general relativity, namely that the "accelerated expansion has nothing to do with the negative pressure", and I clarified his misguided remarks yesterday.
But it may be useful to mention that Carroll was provoked to write his text by Peter Coles' challenge in the article:
That may look strange because we think that it's the positive pressure, like the positive pressure of some gas in a bottle, that is trying to make the bottle explode and expand (at an accelerating rate). So why is the sign in front of pressure in the second Friedmann equation opposite to what we expect based on the experience with gases in the PET bottles?
Coles asked his readers to produce an explanation for Mr Joe Public. I don't know this man – except for knowing that he has 2.3 children and didn't manage to sign up for Obamacare. ;-) He has also recorded the song arguing that "you've got to live and learn before your bridges burn" along with his clones and namesakes (see above). But because all the answers at Coles' blog are completely wrong (for example, Phil Gibbs is building on a non-existent conserved total energy in the FRW cosmology), let me try to clarify some of the reasons.
Well, recall that the second Friedmann equation is\[
\frac{\ddot a}{a} =-\frac{4\pi G}{3} \left(\rho + \frac{3p}{c^2}\right)
\] Here, \(\rho\) is the mass density and I added the factor of \(1/c^2\) to convert the pressure (energy density) to the units of mass density. Also, \(a\) is a scale factor that may literally be understood as the distance between two particular galaxies (which is a function of time \(t\)). It's helpful to return to the \(c\neq 1\) units because we may immediately recognize that the pressure is a (special) relativistic correction of a sort.
Before we begin, I want to say that in locally Minkowskian coordinates, \(T_{00}=T^{00}=\rho\) is non-negative everywhere in Nature while the doubly spatial components \(T_{ii}=T^{ii}=p\) are the pressure if we talk about isotropic environments. Note that we got no minus signs from raising or lowering the indices because two indices of the same kind were raised or lowered simultaneously. Only \(T^{0i}\) and \(T_{0i}\) would have opposite signs. Of course, the components \(T^0_0\) or \(T^i_i\) with one upper or one lower index would have the opposite signs than the doubly upper or doubly lower components.
Fine. Let me return to the mission.
First, let us answer the following question: What's wrong with the argument that a positive pressure, like the pressure of gas in the PET bottle, wants to repel the walls of the bottle and cause an accelerated expansion of the bottle (and the Universe)?
Well, there's nothing wrong with the claim about the bottle. But this simple effect of pressure is a completely different effect than the effect of the pressure in Friedmann's equations. There are two main differences. In the case of the bottle,
\frac{d\vec F}{dV} = -\nabla p
\] If you realize that \(-\nabla p\) is simply \(\rho_{\rm water} g\) pointing in the positive \(z\) direction, you may easily integrate this force density and get the total upward force \(V\rho_{\rm water} g\) acting on the cube. In other words, we have just derived Archimedes' principle. Note that the upward force only exists because the pressure depends on the depth. If it didn't depend on the depth (or location), all the forces would cancel each other.
To understand the second point, let me mention that the water has a nonzero mass and this mass therefore induces a gravitational field as well. But we actually never measure the gravitational force caused by this water. This force is proportional to the tiny constant \(G\) and decreases as \(1/r^2\) with the distance from the molecules of the water. This has clearly nothing to do with the reason why the cube is being pushed upward. The upward force is proportional to \(\nabla p\), just the gradient of the pressure, rather than \(Gp\), the pressure multiplied by Newton's constant.
Well, the hydrostatic pressure depends on the depth of the water because of gravity but it's the gravity caused by the Earth. If you remove the Earth, the water will lose the ability to push the cube upward (there won't be any preferred upward direction, after all).
Fine, so this was a negative explanation, an explanation why the term that depends on the pressure in the Friedmann equation is a completely different term than the usual pressure that we know from gases and liquids.
But why is there the term \(+3p\) in the Friedmann equation for the acceleration?
As I have already indicated by the restored factor \(1/c^2\), it is a relativistic correction, something that only matters if you appreciate all the (special) relativistic phenomena that occur when the velocities are no longer negligible relatively to the speed of light \(c\).
We may talk about the mass density and pressure and say that the non-relativistic limit is only applicable if \(p/c^2\) is negligible relatively to the mass density \(\rho\). For everyday materials, this is the case because \(p,\rho\) are numbers of order one in the SI units while \(1/c^2\) is a tiny number. But the corrections scaling like \(1/c^2\) are there, anyway. If you want to be accurate enough (for example because you are able to perform very accurate measurements), you can't omit them. And for all forms of matter except for dust, \(p/c^2\) is actually "comparable" to \(\rho\).
Now, I must emphasize that the only "truly controllable" way to derive the Friedmann equation is the full calculation in GR. You start with Einstein's equations and simply insert the Ansatz for the metric tensor – the Ansatz that defines the FRW cosmology (a uniform, isotropic universe whose geometric properties – well, the overall size dictated by the linear scale factor \(a\) – only depends on time). With this Ansatz, Einstein's equations reduce to the simpler Friedmann equations.
The problem is that Mr Joe Public didn't understand the previous paragraph because he had enough trouble with the Obamacare website. Can we please give a physical justification of the \(\rho+3p/c^2\) pattern while avoiding words like Einstein's equations and Ansatz? The answer is the same as the answer that turned out to be wrong in the case of Obamacare: Yes, we can. ;-)
I don't want to perform the full nonlinear, exact relativistic calculation – only to estimate the leading correction to \(\rho\) in a particular environment with some particular (positive) pressure, namely some gas.
What we have is a grid of galaxies that may be arranged to cubic boxes. Look at one edge of a box. The vertices host two galaxies whose relative distance is \(a(t)\). That's the proper length of a "spatially" straight line through the \(t={\rm const}\) slice. (It is not a geodesic in the whole spacetime because a geodesic would prefer to visit other values of \(t\) as well.) Why is the expansion of the Universe accelerating or decelerating in the first place?
We often hear (and say) that the Hubble expansion is something miraculous, intrinsically general relativistic, that cannot be visualized as a simple explosion that shot galaxies in different directions. But the truth is that when we talk to Mr Joe Public, it is actually totally legitimate to visualize the Hubble expansion as a simple explosion. The reason is that a sufficiently thin hypercylindrical "strip" of the curved four-dimensional spacetime (in our case, one of the small boxes in the cosmological grid, taken at all times \(t\)) may always be flattened i.e. embedded into the Minkowski space. And once the galaxies are embedded in this auxiliary Minkowski space, their receding motion simply has to have the usual special relativistic interpretation – after all, it's just some relative motion.
(This special relativistic visualization of the expansion of the Universe has one undesirable feature: it invites us to think that one of the galaxies simply has to "at rest", in the center of the primordial explosion, while others are not at the center. But this conclusion is actually unnecessary even in special relativity – all motion is relative even in special relativity. So it's not such an insurmountable problem. The real virtue of a proper general relativistic description is that it allows us to consistently avoid the subtle questions how to "cancel" the infinite force from the galaxies on the left side with the infinite force from the galaxies on the right side and similar issues.)
Why is the expansion of the Universe decelerating for \(p=0\), \(\rho\gt 0\)? It's simple: it's because of the gravitational field created by the matter inside the box of volume \(V=a^3\). This box has mass \(m=\rho V=\rho a^3\). Because of the inverse square law, we may see that the acceleration at the distance of order \(a\) from the center of the cube, e.g. at one of the vertices, is of order \[
a_{\rm acc} = \frac{Gm}{a^2}=\frac{G\rho a^3}{a^2} =G\rho a.
\] So this box filled with some mass of density \(\rho\) will simply attract the galaxies at the vertices with the acceleration comparable to \(g\rho a\). To derive the numerical factor, one would either have to deal with messy questions about the gravitational fields of cubes or do the general relativistic calculation. But the acceleration \(a_{\rm acc}\) is nothing else than the \(-d^2 a / dt^2\). I included a minus sign because a positive mass inside the box will tend to shrink \(a\) in the far future (deceleration i.e. attractive gravity) while \(a_{\rm acc}\) was defined to be positive by the displayed equation above.
(Note that you don't have to worry about the attractive forces exerted by the other boxes on the same galaxy. They will be responsible for the evolving size of the other cubes – which will of course evolve in the same way as our box. But the assumption of a uniform Universe allows us to "localize" the whole problem and study the effect of a single box only.)
Up to the numerical constant \(4\pi/3\) and assuming \(p=0\), I have just derived the equation\[
\frac{d^2 a}{dt^2} = - \frac{4\pi}{3} G\rho a.
\] We usually multiply it by \(1/a\) so that \(a\) disappears from the right hand side while the left hand side becomes \(\ddot a / a\). Great. So a part of the Friedmann equation could now be clear to Mr Joe Public. It is nothing else than the motion of probes – the galaxies – in the gravitational field produced by some distribution of mass – and it's only the mass in the single cube that mattered for our problem. Again, let me stress that this explanation involving the usual "gravitational attraction" you know from the basic school isn't just an analogy: general relativity is still a theory of gravity so even when we discuss the expansion of the Universe, the physics still has to agree with some basic mechanisms of gravity (up to some nonlinear corrections for strong fields and fast motion etc.).
Finally, we would like to see why \(\rho\) should be replaced by the relativistically corrected \(\rho+3p/c^2\). The previous section already boasted a headline that promised such an explanation but we're finally getting there.
Imagine that the cubic box in between the galaxies (the edge has length \(a\)) is filled with gas of mass density \(\rho\) whose molecules have the velocity \(v\) and mass \(m\). What is the pressure? Well, start with a molecule bouncing from the left to the right and back. It takes the time \(t=a/v\) for the molecule to get to the other side and during each collision, the molecule changes its velocity by \(2v\) so it deposits the momentum \(2mv\) to the wall. The momentum deposited to the walls per unit time is \(2mv/(a/v) = 2mv^2/a\) and that's nothing else than the definition of the (average) force. Here, \(v\) was really \(v_x\) but we must add the motion (and momentum transfer) to the remaining \(6-2=4\) faces of the cube which is equivalent to replacing \(v_x^2\) by simply \(v^2\), the total one. Finally, the total force \(2mv^2/a\) must be divided by \(6a^2\), the surface of the cube, to get the pressure.
We see that the pressure contributed by one molecule is \(mv^2/3a^3\). Multiply it by the number of molecules \(N\) to get the total pressure \[
p= \frac{Nmv^2}{3a^3}=\frac{M_{\rm total}v^2}{3a^3}=\frac{\rho v^2}{3}.
\] We explicitly see that \((p/c^2)/\rho\) goes like \(v^2/3c^2\). It approaches a number of order one exactly when the speed of the molecules (or whatever the gas is composed of) approaches a speed comparable to the speed of light. Note that \(3p\) is nothing else than \(\rho v^2\) and\[
\frac{\rho+ \frac{3p}{c^2}}{\rho} = 1+\frac{v^2}{c^2}.
\] We're assuming \(v\ll c\) but we want to derive the first subleading correction. Why is the force going like \(1+v^2/c^2\) and not just \(1\)?
The answer may be found by a careful counting of the Lorentz factors\[
\gamma = \frac{1}{\sqrt{1-v^2/c^2}} \approx 1+\frac{v^2}{2c^2}+\dots
\] How does the counting go? It's relatively simple:
The rest mass of the molecule is actually smaller than the total mass, by a factor of \(\gamma\) – that's the usual relativistic increase of the total mass with the velocity. So if we switch to the rest frame of the molecule where the gravitational field caused by the molecule is calculable by the usual methods, we could think that a higher speed (and nonzero pressure) makes the total force that is decelerating the Universe smaller.
However, there are actually 3 factors of \(\gamma\) that go in the opposite direction, so that the total deceleration is \(\gamma^3/\gamma\approx 1+v^2/c^2\) times the non-relativistic expectation. Why are there three "positive" factors of \(\gamma\)? All of them are related to the Lorentz contraction.
One of them appears because we want to switch to the rest frame of the moving molecules (those are the source of the gravitational field and we only know how to describe this gravitational field in the sources' rest frame) but in this frame, the boxes are Lorentz contracted and we may therefore squeeze \(\gamma\) times more molecules into the same cube in the coordinate space.
Two additional powers of \(\gamma\) emerge because the gravitational acceleration goes like \(1/r^2\) and this \(r\) is Lorentz-contracted (reduced) which means that \(1/r^2\), the acceleration, is increased by the factor \(\gamma^2\approx 1+v^2/c^2\). Just to be sure, only the component of the force that goes in the same direction as the separation of the two galaxies – separation whose magnitude we call \(a\) – will be nonzero.
At any rate, when you combine the factors together, the enhancement from the nonzero velocities is by \[
\gamma^2\approx 1+\frac{v^2}{c^2}\approx \zav{\rho+\frac{3p}{c^2}}\frac{1}{\rho}
\] as we wanted to prove. Once Mr Joe Public learns something about general relativity, he will notice that the stress-energy tensor enters Einstein's equations linearly so the pressure must enter the second Friedmann equation linearly, too. That's why the form of the acceleration involving \(\rho+3p/c^2\) is actually exact even for \(v/c\sim 1\).
The environment with negative pressure cannot be visualized as a gas of molecules (we would need imaginary velocities to obtain a negative pressure) but due to the linearity in \(p\), it's clear that the environment with a negative pressure will contribute to the accelerated expansion of the Universe.
Sean Carroll shouted a completely wrong claim about basic general relativity, namely that the "accelerated expansion has nothing to do with the negative pressure", and I clarified his misguided remarks yesterday.
But it may be useful to mention that Carroll was provoked to write his text by Peter Coles' challenge in the article:
A Dark Energy Mission.Coles points out that we need a negative pressure (e.g. one coming from the cosmological constant) to achieve a repulsive gravity producing the accelerated expansion in the FRW cosmology.
That may look strange because we think that it's the positive pressure, like the positive pressure of some gas in a bottle, that is trying to make the bottle explode and expand (at an accelerating rate). So why is the sign in front of pressure in the second Friedmann equation opposite to what we expect based on the experience with gases in the PET bottles?
Coles asked his readers to produce an explanation for Mr Joe Public. I don't know this man – except for knowing that he has 2.3 children and didn't manage to sign up for Obamacare. ;-) He has also recorded the song arguing that "you've got to live and learn before your bridges burn" along with his clones and namesakes (see above). But because all the answers at Coles' blog are completely wrong (for example, Phil Gibbs is building on a non-existent conserved total energy in the FRW cosmology), let me try to clarify some of the reasons.
Well, recall that the second Friedmann equation is\[
\frac{\ddot a}{a} =-\frac{4\pi G}{3} \left(\rho + \frac{3p}{c^2}\right)
\] Here, \(\rho\) is the mass density and I added the factor of \(1/c^2\) to convert the pressure (energy density) to the units of mass density. Also, \(a\) is a scale factor that may literally be understood as the distance between two particular galaxies (which is a function of time \(t\)). It's helpful to return to the \(c\neq 1\) units because we may immediately recognize that the pressure is a (special) relativistic correction of a sort.
Before we begin, I want to say that in locally Minkowskian coordinates, \(T_{00}=T^{00}=\rho\) is non-negative everywhere in Nature while the doubly spatial components \(T_{ii}=T^{ii}=p\) are the pressure if we talk about isotropic environments. Note that we got no minus signs from raising or lowering the indices because two indices of the same kind were raised or lowered simultaneously. Only \(T^{0i}\) and \(T_{0i}\) would have opposite signs. Of course, the components \(T^0_0\) or \(T^i_i\) with one upper or one lower index would have the opposite signs than the doubly upper or doubly lower components.
Fine. Let me return to the mission.
First, let us answer the following question: What's wrong with the argument that a positive pressure, like the pressure of gas in the PET bottle, wants to repel the walls of the bottle and cause an accelerated expansion of the bottle (and the Universe)?
Well, there's nothing wrong with the claim about the bottle. But this simple effect of pressure is a completely different effect than the effect of the pressure in Friedmann's equations. There are two main differences. In the case of the bottle,
- the force actually depends on the gradient of the pressure rather than the pressure itself
- the force isn't proportional to Newton's constant \(G\) at all because it is not a gravitational force at all
- it's actually the force we calculate quickly; in the case of the Universe, it's the acceleration and we have to multiply it by the mass of the "probe" as well to get the force
\frac{d\vec F}{dV} = -\nabla p
\] If you realize that \(-\nabla p\) is simply \(\rho_{\rm water} g\) pointing in the positive \(z\) direction, you may easily integrate this force density and get the total upward force \(V\rho_{\rm water} g\) acting on the cube. In other words, we have just derived Archimedes' principle. Note that the upward force only exists because the pressure depends on the depth. If it didn't depend on the depth (or location), all the forces would cancel each other.
To understand the second point, let me mention that the water has a nonzero mass and this mass therefore induces a gravitational field as well. But we actually never measure the gravitational force caused by this water. This force is proportional to the tiny constant \(G\) and decreases as \(1/r^2\) with the distance from the molecules of the water. This has clearly nothing to do with the reason why the cube is being pushed upward. The upward force is proportional to \(\nabla p\), just the gradient of the pressure, rather than \(Gp\), the pressure multiplied by Newton's constant.
Well, the hydrostatic pressure depends on the depth of the water because of gravity but it's the gravity caused by the Earth. If you remove the Earth, the water will lose the ability to push the cube upward (there won't be any preferred upward direction, after all).
Fine, so this was a negative explanation, an explanation why the term that depends on the pressure in the Friedmann equation is a completely different term than the usual pressure that we know from gases and liquids.
But why is there the term \(+3p\) in the Friedmann equation for the acceleration?
As I have already indicated by the restored factor \(1/c^2\), it is a relativistic correction, something that only matters if you appreciate all the (special) relativistic phenomena that occur when the velocities are no longer negligible relatively to the speed of light \(c\).
We may talk about the mass density and pressure and say that the non-relativistic limit is only applicable if \(p/c^2\) is negligible relatively to the mass density \(\rho\). For everyday materials, this is the case because \(p,\rho\) are numbers of order one in the SI units while \(1/c^2\) is a tiny number. But the corrections scaling like \(1/c^2\) are there, anyway. If you want to be accurate enough (for example because you are able to perform very accurate measurements), you can't omit them. And for all forms of matter except for dust, \(p/c^2\) is actually "comparable" to \(\rho\).
Now, I must emphasize that the only "truly controllable" way to derive the Friedmann equation is the full calculation in GR. You start with Einstein's equations and simply insert the Ansatz for the metric tensor – the Ansatz that defines the FRW cosmology (a uniform, isotropic universe whose geometric properties – well, the overall size dictated by the linear scale factor \(a\) – only depends on time). With this Ansatz, Einstein's equations reduce to the simpler Friedmann equations.
The problem is that Mr Joe Public didn't understand the previous paragraph because he had enough trouble with the Obamacare website. Can we please give a physical justification of the \(\rho+3p/c^2\) pattern while avoiding words like Einstein's equations and Ansatz? The answer is the same as the answer that turned out to be wrong in the case of Obamacare: Yes, we can. ;-)
I don't want to perform the full nonlinear, exact relativistic calculation – only to estimate the leading correction to \(\rho\) in a particular environment with some particular (positive) pressure, namely some gas.
What we have is a grid of galaxies that may be arranged to cubic boxes. Look at one edge of a box. The vertices host two galaxies whose relative distance is \(a(t)\). That's the proper length of a "spatially" straight line through the \(t={\rm const}\) slice. (It is not a geodesic in the whole spacetime because a geodesic would prefer to visit other values of \(t\) as well.) Why is the expansion of the Universe accelerating or decelerating in the first place?
We often hear (and say) that the Hubble expansion is something miraculous, intrinsically general relativistic, that cannot be visualized as a simple explosion that shot galaxies in different directions. But the truth is that when we talk to Mr Joe Public, it is actually totally legitimate to visualize the Hubble expansion as a simple explosion. The reason is that a sufficiently thin hypercylindrical "strip" of the curved four-dimensional spacetime (in our case, one of the small boxes in the cosmological grid, taken at all times \(t\)) may always be flattened i.e. embedded into the Minkowski space. And once the galaxies are embedded in this auxiliary Minkowski space, their receding motion simply has to have the usual special relativistic interpretation – after all, it's just some relative motion.
(This special relativistic visualization of the expansion of the Universe has one undesirable feature: it invites us to think that one of the galaxies simply has to "at rest", in the center of the primordial explosion, while others are not at the center. But this conclusion is actually unnecessary even in special relativity – all motion is relative even in special relativity. So it's not such an insurmountable problem. The real virtue of a proper general relativistic description is that it allows us to consistently avoid the subtle questions how to "cancel" the infinite force from the galaxies on the left side with the infinite force from the galaxies on the right side and similar issues.)
Why is the expansion of the Universe decelerating for \(p=0\), \(\rho\gt 0\)? It's simple: it's because of the gravitational field created by the matter inside the box of volume \(V=a^3\). This box has mass \(m=\rho V=\rho a^3\). Because of the inverse square law, we may see that the acceleration at the distance of order \(a\) from the center of the cube, e.g. at one of the vertices, is of order \[
a_{\rm acc} = \frac{Gm}{a^2}=\frac{G\rho a^3}{a^2} =G\rho a.
\] So this box filled with some mass of density \(\rho\) will simply attract the galaxies at the vertices with the acceleration comparable to \(g\rho a\). To derive the numerical factor, one would either have to deal with messy questions about the gravitational fields of cubes or do the general relativistic calculation. But the acceleration \(a_{\rm acc}\) is nothing else than the \(-d^2 a / dt^2\). I included a minus sign because a positive mass inside the box will tend to shrink \(a\) in the far future (deceleration i.e. attractive gravity) while \(a_{\rm acc}\) was defined to be positive by the displayed equation above.
(Note that you don't have to worry about the attractive forces exerted by the other boxes on the same galaxy. They will be responsible for the evolving size of the other cubes – which will of course evolve in the same way as our box. But the assumption of a uniform Universe allows us to "localize" the whole problem and study the effect of a single box only.)
Up to the numerical constant \(4\pi/3\) and assuming \(p=0\), I have just derived the equation\[
\frac{d^2 a}{dt^2} = - \frac{4\pi}{3} G\rho a.
\] We usually multiply it by \(1/a\) so that \(a\) disappears from the right hand side while the left hand side becomes \(\ddot a / a\). Great. So a part of the Friedmann equation could now be clear to Mr Joe Public. It is nothing else than the motion of probes – the galaxies – in the gravitational field produced by some distribution of mass – and it's only the mass in the single cube that mattered for our problem. Again, let me stress that this explanation involving the usual "gravitational attraction" you know from the basic school isn't just an analogy: general relativity is still a theory of gravity so even when we discuss the expansion of the Universe, the physics still has to agree with some basic mechanisms of gravity (up to some nonlinear corrections for strong fields and fast motion etc.).
Finally, we would like to see why \(\rho\) should be replaced by the relativistically corrected \(\rho+3p/c^2\). The previous section already boasted a headline that promised such an explanation but we're finally getting there.
Imagine that the cubic box in between the galaxies (the edge has length \(a\)) is filled with gas of mass density \(\rho\) whose molecules have the velocity \(v\) and mass \(m\). What is the pressure? Well, start with a molecule bouncing from the left to the right and back. It takes the time \(t=a/v\) for the molecule to get to the other side and during each collision, the molecule changes its velocity by \(2v\) so it deposits the momentum \(2mv\) to the wall. The momentum deposited to the walls per unit time is \(2mv/(a/v) = 2mv^2/a\) and that's nothing else than the definition of the (average) force. Here, \(v\) was really \(v_x\) but we must add the motion (and momentum transfer) to the remaining \(6-2=4\) faces of the cube which is equivalent to replacing \(v_x^2\) by simply \(v^2\), the total one. Finally, the total force \(2mv^2/a\) must be divided by \(6a^2\), the surface of the cube, to get the pressure.
We see that the pressure contributed by one molecule is \(mv^2/3a^3\). Multiply it by the number of molecules \(N\) to get the total pressure \[
p= \frac{Nmv^2}{3a^3}=\frac{M_{\rm total}v^2}{3a^3}=\frac{\rho v^2}{3}.
\] We explicitly see that \((p/c^2)/\rho\) goes like \(v^2/3c^2\). It approaches a number of order one exactly when the speed of the molecules (or whatever the gas is composed of) approaches a speed comparable to the speed of light. Note that \(3p\) is nothing else than \(\rho v^2\) and\[
\frac{\rho+ \frac{3p}{c^2}}{\rho} = 1+\frac{v^2}{c^2}.
\] We're assuming \(v\ll c\) but we want to derive the first subleading correction. Why is the force going like \(1+v^2/c^2\) and not just \(1\)?
The answer may be found by a careful counting of the Lorentz factors\[
\gamma = \frac{1}{\sqrt{1-v^2/c^2}} \approx 1+\frac{v^2}{2c^2}+\dots
\] How does the counting go? It's relatively simple:
The rest mass of the molecule is actually smaller than the total mass, by a factor of \(\gamma\) – that's the usual relativistic increase of the total mass with the velocity. So if we switch to the rest frame of the molecule where the gravitational field caused by the molecule is calculable by the usual methods, we could think that a higher speed (and nonzero pressure) makes the total force that is decelerating the Universe smaller.
However, there are actually 3 factors of \(\gamma\) that go in the opposite direction, so that the total deceleration is \(\gamma^3/\gamma\approx 1+v^2/c^2\) times the non-relativistic expectation. Why are there three "positive" factors of \(\gamma\)? All of them are related to the Lorentz contraction.
One of them appears because we want to switch to the rest frame of the moving molecules (those are the source of the gravitational field and we only know how to describe this gravitational field in the sources' rest frame) but in this frame, the boxes are Lorentz contracted and we may therefore squeeze \(\gamma\) times more molecules into the same cube in the coordinate space.
Two additional powers of \(\gamma\) emerge because the gravitational acceleration goes like \(1/r^2\) and this \(r\) is Lorentz-contracted (reduced) which means that \(1/r^2\), the acceleration, is increased by the factor \(\gamma^2\approx 1+v^2/c^2\). Just to be sure, only the component of the force that goes in the same direction as the separation of the two galaxies – separation whose magnitude we call \(a\) – will be nonzero.
At any rate, when you combine the factors together, the enhancement from the nonzero velocities is by \[
\gamma^2\approx 1+\frac{v^2}{c^2}\approx \zav{\rho+\frac{3p}{c^2}}\frac{1}{\rho}
\] as we wanted to prove. Once Mr Joe Public learns something about general relativity, he will notice that the stress-energy tensor enters Einstein's equations linearly so the pressure must enter the second Friedmann equation linearly, too. That's why the form of the acceleration involving \(\rho+3p/c^2\) is actually exact even for \(v/c\sim 1\).
The environment with negative pressure cannot be visualized as a gas of molecules (we would need imaginary velocities to obtain a negative pressure) but due to the linearity in \(p\), it's clear that the environment with a negative pressure will contribute to the accelerated expansion of the Universe.
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