What is the force of the cosmological constant?

[Ranku]

Main Question or Discussion Point

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? Is there a classical formula of the force of the cosmological constant like that of gravitational force?

 

[Ich]

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?

Quite accurate on small scales. It's a pseudoforce, like gravity.

Is there a classical formula of the force of the cosmological constant like that of gravitational force?

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:
$$F=4/3 \pi G \rho r m$$
$$\rho$$ is the double density of dark energy (2*6.8*10^-26 kg/m³).

 

[Ranku]

Thank you for the response.

Quite accurate on small scales. It's a pseudoforce, like gravity.

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.

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:
$$F=4/3 \pi G \rho r m$$
$$\rho$$ is the double density of dark energy (2*6.8*10^-26 kg/m³).

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?
What does the m in the cosmological constant force formula stand for? If it is mass, what is its value?
Why is $$\rho$$ taken as the double density of dark energy?

 

[Ich]

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.

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 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?

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.

What does the m in the cosmological constant force formula stand for? If it is mass, what is its value?

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.

Why is rho taken as the double density of dark energy?

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.

 

[Ranku]

Thank you for your response.

Negative pressure repels the same amount that positive energy density attracts, so there's a net effect of -2 times energy density.

Could you show the above equationally? Can the opposite be also true: net effect is -2 times pressure?

 

[Chronos]

Force may not be the best of terms. 'Force' is not a very good way to characterize the effects of gravity, as einstein surmised. I suspect the cosmological constant, as is gravity, may be an artifact of spacetime topology - perhaps different manifestations of the same basic effect. Another way of looking at it - does gravity suck, or space push?

 

[Ich]

Could you show the above equationally?

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).

Can the opposite be also true: net effect is -2 times pressure?

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.

 

[Ranku]

Thank you for all your responses.

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:
$$F=4/3 \pi G \rho r m$$
$$\rho$$ is the double density of dark energy (2*6.8*10^-26 kg/m³).

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.
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?

 

[Ich]

Can you give a similar Newtonian approximation of the formula of cosmological constant dark energy, which would be analogous to gravitational potential energy?

Simply integrate and take the negative, that gives
$$\Phi=-2/3 \pi G \rho r^2$$

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.

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.

 

[Ranku]

Thank you for your response.

Simply integrate and take the negative, that gives
$$\Phi=-2/3 \pi G \rho r^2$$

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.

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?

 

[Chronos]

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

 

[Ranku]

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?

Ich, could you answer this question?

 

[Ich]

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

Sorry.
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.