Could dark energy simply be gravitational field energy?(adsbygoogle = window.adsbygoogle || []).push({});

If we assume that the Universe is always spatially flat then we have the equation:

H(t)^2 = 8 Pi G rho(t) / 3

where H(t) and rho(t) are the Hubble constant and total matter density at time t.

Now we have the equation

c = H(t) R(t)

where R(t) is the Hubble radius which defines the radius of the observable Universe at time t.

If we assume that the observable Universe is a sphere of radius R we can substitute the second equation into the first to find the relation:

G M(t) / R(t) = c^2 / 2

where M(t) is the mass of the Universe at time t.

Now the gravitational potential energy of a sphere of mass M is:

PE = - 3 G M^2 / 5R

Therefore using the relation GM/R=c^2/2 we find

PE = - 3 * M * c^2 / 10

Therefore the mass due to the Universe's potential energy is:

- 3 M / 10

This is of the same order of magnitude as the mass of the Universe itself, M.

This agrees with the current consensus that dark energy density is of the same order of magnitude as the matter density.

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# Could dark energy simply be gravitational field energy?

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