Vacuum with cosmological constant

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pervect
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I was thinking about constraints on the stress-energy tensor of the vacuum, and came to an interesting conclusion that the vacuum itself should only be isotropic in one rest frame if there is a cosmological constant.

If we start with a vaccuum that is homogeneous and isotropic in some cartesian frame, we demand that

T = diag(rho,P,P,P)

We can demand that T always be diagonal when transformed by a boost. By setting T_01 equal to zero, and doing a boost [tex]T_{cd} = \Lambda^a{}_c \Lambda^b{}_d T_{ab}[/itex], we can find that rho = -P, as [tex]\Lambda^0{}_0 = \Lambda^1{}_1 = \beta[/tex], [tex]\Lambda^1{}_0 = \Lambda^0{}_1 = -\beta \gamma[/tex], so

[tex]T'_{01} = \gamma^2 \beta (T_{00} + T_{01}) = 0[/tex]

Thus T = diag(-P,P,P,P)

However, while T will always be diagonal, it won't be isotropic in the boosted frame. If we boost T, we get something like

diag(-(1-[itex]\beta^2[/itex])P,(1-[itex]\beta^2[/itex])P,P,P)

Thus if there is a cosmological constant, the vacuum itself should have a unique frame in which it is isotropic.
 
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George Jones
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Isn't it also true that boosting from comoving coordiantes in an FRW universe without cosmological constant also destroy isotropy?

Regards,
George
 
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pervect
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George Jones said:
Isn't it also true that boosting from comoving coordiantes in an FRW universe without cosmological constant also destroy isotropy?
Regards,
George
Yep. I thought it was interesting that the vacuum itself apparently becomes anisotropic, though.
 
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Garth
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pervect said:
Yep. I thought it was interesting that the vacuum itself apparently becomes anisotropic, though.
Now apply Mach's Principle - does not a reductio ad absurdum suggest the cosmological constant has to be zero?

Garth
 
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George Jones
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Garth said:
Now apply Mach's Principle - does not a reductio ad absurdum suggest the cosmological constant has to be zero?
Garth
Are you saying that we should also give up on FRW models (without cosmological constant) of the universe?

Why does Mach's principle have be true?

Regards,
George
 
  • #6
Garth
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The FRW models select out a preferred frame of reference, a space-like foliation in which the universe is isotropic and homogeneous. This is not in conflict with Mach's Principle as that is also the frame of reference of the universe's co-moving centroid and can be physically identified with that frame which is stationary w.r.t the surface of last scattering in which the CMB is globally isotropic.

However a Machian view on the anisotropy of vacuum-with-cosmological-constant is: "What on Earth identifies the isotropic frame of reference from all the others?"

What do you 'hang' this coordinate system on?

Garth
 
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George Jones
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What is the universe's comoving centroid?

Also, I just flipped through a few books, and it appears that Penrose and Hartle believe that Mach's principle is not valid in general relativity.

It would be much appriciated if you would give a formulation of Mach's principle as it applies to general relativity in general, and to cosmology in particular.

Assume that I know something about general relativity and differential geometry, but that I am totally unfamiliar with Mach's Principle.

Regards,
George
 
  • #8
Garth
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Hi George.

In Newtonian physics velocities are relative but accelerations are not. There are inertial frames of reference which can be physically determined by an accelerometer, you are in one when the accelerometer reads zero.

Newton was quite happy to think of these frames as 'absolute' and if you asked how is such a frame selected out of all possible frames, accelerating relative to each other, he was quite happy to think that 'God' determined the absolute non-accelerating and non-rotating frame.

Ernst Mach (1838-1916) believed that inertial frames of reference could not be absolute but had to be tied to the matter in the rest of the universe.

"The inertia of any system is the result of the interaction of that system and the rest of the universe. In other words, every particle in the universe ultimately has an effect on every other particle."

Two thought experiments are Newton's bucket and Focault's pendulum.

In the latter case imagine a pendulum closed off from the outside world. Measure how much the pendulum precesses throughout the day and knowing the latitude you can work out the Earth's rate of rotation. Now go outside and observe the stars, you again calculate the Earth's rate of rotation and find the two rates are equal. But why? What is it that connects the pendulum with the distant stars?

Einstein was impressed with the principle but realised that his GR did not fully incorporate it. There are valid solutions of the field equation that accelerate relative to the mass within it. The equivalence principle subsumes the principle of relativity and guarantees that in any general freely falling frame the laws of physics are locally that of SR, even though that frame is accelerating towards a large mass and that freely falling mass is accelerating relative to the rest of the universe.

Dicke extended Mach Principle: "The gravitational constant should be a function of the mass distribution in the universe"

The Brans Dicke theory modified GR to include a scalar field that determines the inertial masses of elementary particles. This field is coupled to matter in motion in the rest of the universe. The presence of this scalar field perturbs the GR field equation and solar system experiments are now inconsistent with BD unless the matter coupling of the scalar field is very weak.

In my work I have modified BD to include the local conservation of energy and find this produces a scalar field force that acts on particles but not photons, which exactly compensates for this perturbation. This theory is called A New Self Creation Cosmology (You can recover the theory free here). The solar system predictions are exactly equal to those of GR in vacuo but a definitive test is being carried out at present in the analysis of the Gravity Probe B experiment.

Garth
 
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  • #9
pervect
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Garth said:
Now apply Mach's Principle - does not a reductio ad absurdum suggest the cosmological constant has to be zero?
Garth
I'll admit that I find the cosmological constant ugly. However, I'm not quite ready to buy into the "freely coasting" model quite yet, though I do find its simplicity much more attractive than the complexity of current cosmological models.

Basically I think we need some more observations to sort the whole mess out.

As far as Mach's principle itself goes, I've never been much of a believer.
 
  • #10
Garth
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George - the comoving centroid is the frame co-moving with the Centre of mass or the Centre of (3) Momentum.

An observer with a 4-velocity [itex]u^{\alpha}[/itex] defines the centroid of the system, 4-momentum [itex]P^{\alpha}[/itex] at his Lorentz time [itex]x^0 = t[/itex] and in his own Lorentz frame, by

[tex]X^j_u(t) = (\frac{1}{P^0}) \int x^j T^{00} d^3x[/tex]


pervect - Consider an empty universe - the Milne model but with a cosmological constant. Through any event [itex]x^{\mu}[/itex] there are many Lorentz frames moving relative to each other; yet only in one frame is space-time isotropic. This frame can be experimentally identified by the isotropic tidal forces produced by the cosmological constant.

But w.r.t. the principle of relativity (no preferred frames) why this frame and not another?

And as far as observations are concerned we await the GP-B results!

Garth
 
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