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Ich

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Quite accurate on small scales. It's a pseudoforce, like gravity.I have read that "the cosmological constant introduces a force of repulsion between bodies. The force increases in (simple) proportion to the distance between them." Is this description accurate?

Yes. In a Newtonian approximation, you may describe it as the gravitational effect of some uniform "dark energy" with a negative mass density. So, essentially:Is there a classical formula of the force of the cosmological constant like that of gravitational force?

[tex]F=4/3 \pi G \rho r m[/tex]

[tex]\rho[/tex] is the double density of dark energy (2*6.8*10^-26 kg/m³).

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Why do you say on small scales? The force of the cosmological constant is what is suppose to be driving the present acceleration of the entire universe.Quite accurate on small scales. It's a pseudoforce, like gravity.

Why is the cosmological constant force described as an anti-gravitational effect when the force increases inYes. In a Newtonian approximation, you may describe it as the gravitational effect of some uniform "dark energy" with a negative mass density. So, essentially:

[tex]F=4/3 \pi G \rho r m[/tex]

[tex]\rho[/tex] is the double density of dark energy (2*6.8*10^-26 kg/m³).

What does the

Why is [tex]\rho[/tex] taken as the

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Ich

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Yes, but gravity is not a force, it's spacetime curvature. You can treat it as a force if you can reasonably define a flat background. If not, the "force-picture" is not really helpful.Why do you say on small scales? The force of the cosmological constant is what is suppose to be driving the present acceleration of the entire universe.

Gravity behaves exactly the same way. I'm not considering a point mass here, but uniformly distributed matter. The amount of matter within distance r scales as r³, gravity as r^-2, the net effect scales as r.Why is the cosmological constant force described as an anti-gravitational effect when the force increases in simple proportion to distance, while gravity decreases as the inverse square of the distance?

You asked for gravitational force, not acceleration. m is the arbitrary mass of the body that experiences the force. The force is proportional to m, the acceleration is independent of it.What does the m in the cosmological constant force formula stand for? If it is mass, what is its value?

Dark Energy has a negative pressure in each space dimension that equals its energy density. As there are three space dimensions, that's three times the energy density. Negative pressure repels the same amount that positive energy density attracts, so there's a net effect of -2 times energy density.Why is rho taken as thedoubledensity of dark energy?

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Chronos

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Ich

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Have a look at http://math.ucr.edu/home/baez/einstein/" [Broken] (I'm using the simplified form with Lambda absorbed in the stress-energy-tensor).Could you show the above equationally?

Not for a cosmological constant. One might tweak some form of Quintessence to give the desired result (p=-rho/5), but that doesn't seem compatible with observations.Can the opposite be also true: net effect is -2 times pressure?

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Can you give a similar Newtonian approximation of the formula of cosmological constant dark energy, which would be analogous to gravitational potential energy? I am interested to see how dark energy increases in proportion to distance.In a Newtonian approximation, you may describe it (cosmological constant force) as the gravitational effect of some uniform "dark energy" with a negative mass density. So, essentially:

[tex]F=4/3 \pi G \rho r m[/tex]

[tex]\rho[/tex] is the double density of dark energy (2*6.8*10^-26 kg/m³).

Cosmological constant force increases in proportion to distance, while gravitational force decreases as the inverse square of distance. Gravitational potential energy decreases as the inverse of distance, so how does cosmological constant dark energy increase in relation to distance?

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Ich

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Simply integrate and take the negative, that givesCan you give a similar Newtonian approximation of the formula of cosmological constant dark energy, which would be analogous to gravitational potential energy?

[tex]

\Phi=-2/3 \pi G \rho r^2

[/tex]

Gravitational force increases exactly like "Cosmological constant force". We're not talking about a point mass here, rather a homogeneous sphere. Look at http://www.strw.leidenuniv.nl/~franx/college/sterrenstelsels07/handout2.pdf" [Broken], p. 4/7.I am interested to see how dark energy increases in proportion to distance.

Cosmological constant force increases in proportion to distance, while gravitational force decreases as the inverse square of distance.

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Thank you for your response.

The formula of 'cosmological constant dark energy' shows that it increases as r^2. But gravitational potential energy decreases as 1/r. How to make the two same, so that it is consistent with the fact that the 'cosmological constant force' increases exactly as gravitational force decreases. Will gravitational potential energy decrease as 1/r^2 if we look at mass in terms of a homogenous sphere?Simply integrate and take the negative, that gives

[tex]

\Phi=-2/3 \pi G \rho r^2

[/tex]

Gravitational force increases exactly like "Cosmological constant force". We're not talking about a point mass here, rather a homogeneous sphere. Look at http://www.strw.leidenuniv.nl/~franx/college/sterrenstelsels07/handout2.pdf" [Broken], p. 4/7.

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Chronos

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Aye, there's the rub. The universe is not a homogenous sphere.

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Ich, could you answer this question?The formula of 'cosmological constant dark energy' shows that it increases as r^2. But gravitational potential energy decreases as 1/r. How to make the two same, so that it is consistent with the fact that the 'cosmological constant force' increases exactly as gravitational force decreases. Will gravitational potential energy decrease as 1/r^2 if we look at mass in terms of a homogenous sphere?

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Ich

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Sorry.Will gravitational potential energy decrease as 1/r^2 if we look at mass in terms of a homogenous sphere?

I gave you a link with the correct answer: Potential energy increases as r² in a homogenoeus sphere.

You need some handwaving to argue that one may neglegt contributions from outside a sphere in newtonian mechanics. But you can derive the result easily from GR in a Newtonian approximation: from any given center, the attraction towards / repulsion from that center at distance r is caused by the amount of energy and pressure in a sphere with radius r.

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