The cosmological constant introduces a force of repulsion between bodies, which increases in direct proportion to the distance between them. In a Newtonian approximation, this force can be expressed as F = (4/3) π G ρ r m, where ρ represents the double density of dark energy (2 * 6.8 * 10^-26 kg/m³). This pseudoforce behaves similarly to gravity on small scales, but unlike gravity, which decreases with the inverse square of distance, the force associated with the cosmological constant increases linearly with distance. The discussion clarifies that the cosmological constant's effects can be understood through the lens of spacetime curvature and energy density.
PREREQUISITESAstronomers, physicists, and cosmologists interested in the dynamics of the universe, particularly those studying dark energy and its effects on cosmic acceleration.
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?
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.Ich said: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 in simple proportion to distance, while gravity decreases as the inverse square of the distance?Ich said: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).
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 the double density of dark energy?
Ich said:Negative pressure repels the same amount that positive energy density attracts, so there's a net effect of -2 times energy density.
Have a look at http://math.ucr.edu/home/baez/einstein/" (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?
Ich said: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).
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?
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" , 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.
Ich said:Simply integrate and take the negative, that gives
<br /> \Phi=-2/3 \pi G \rho r^2<br />
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" , p. 4/7.
Ranku said: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?
Sorry.Will gravitational potential energy decrease as 1/r^2 if we look at mass in terms of a homogenous sphere?