The Gravity of Dark Energy

1. Sep 10, 2008

McHeathen

According to an article on the CERN website 'dark energy' is detected by its 'gravitational effect':
Most of the Universe is made up of invisible substances known as 'dark matter' (26%) and 'dark energy' (70%). These do not emit electromagnetic radiation, and we detect them only through their gravitational effects.
http://public.web.cern.ch/public/en/...Recipe-en.html

If E=mc2, then mass would appear to be formed from a concentration of energy. Yet a quantity of energy does have mass - so how can it have gravity?

2. Sep 11, 2008

neutralseer

Our best description of gravity is General Relativity. Within this framework, mass/energy tells spacetime how to warp, and spacetime tells mass/energy how to move. Mass isn't "formed from a concentration of energy," it is a form of energy. Energy can have plenty of different forms, and all of them (including dark energy) warp spacetime.

3. Sep 11, 2008

McHeathen

If this be the case, then why do energy particles such as photons (are there any other type of energy particle?) have zero mass?

4. Sep 11, 2008

neutralseer

Yes they have zero rest mass, but they don't have zero energy. Anything with energy is affected by gravity and causes gravity (mass or energy warp spacetime, and spacetime tell mass/energy how to move).

5. Sep 11, 2008

McHeathen

Light waves being pulled/defected towards/by a star would be a good example of energy being effected by gravity. However there are no examples (to the best of my knowledge) of bodies of mass being pulled towards energy.

6. Sep 12, 2008

mysearch

Hi McHeathen,
I am also looking for some answers about dark energy in terms of its energy density, negative pressure and any gravitational effects, so I hope you don’t mind if I append some additional issues to yours.

From general reading I get the impression that dark energy acts as an effective anti-gravity in that it is said to expand space and push things apart by virtue of its negative pressure. However, I am not sure whether this implies that dark energy is also negative, i.e. like potential energy (?), therefore has negative effective mass by virtue of E=mc^2 and negative pressure is really the net result of anti-gravity?

I have appended some equations and issues linked to the Wikipedia page on the Friedmann equations for further reference and clarification. Thanks

http://en.wikipedia.org/wiki/Friedmann_equations
On the assumption that dark energy equates to the cosmological constant in the Friedmann equation, it seems possible to make the following assumptions about its energy density based on the equivalence of units in the Friedmann equation, i.e.

[1] $$\rho_\Lambda \equiv \frac {\Lambda c^2}{8 \pi G}$$

As an aside, the units of Lambda are 1/metres^2, does this suggest some correlation to the radius of the visible universe (?)

Given that energy density is energy per unit volume and any energy can be equated to mass via Einstein’s equation, does this suggests that dark energy must have some sort of effective mass? However, I am not sure whether this scalar quantity is considered positive or negative, especially in light of the following pressure and energy density relationship?

[2] $$P = \omega \rho c^2$$ where $$\omega_\Lambda= -1$$

Equation [1] allows the cosmological constant in Friedmann’s equation to be replaced by an equivalent energy density. E=mc^2 suggests that dark energy must have an effective mass by virtue of its energy density and therefore some sort of gravitational effect, but I am not sure of the sign of this energy. In contrast, equation [2] seems to suggest that the energy density and pressure must be of different sign due to the sign of omega [w].

Consequently, I am having problems resolving the implied direction of [H] in the Friedman equation plus similar problems with the Fluid equation, which only references energy density with no direct inference to pressure. The Acceleration equation is offset by a pressure factor $$[\rho + 3P]$$ that seems to allow [-P] to overcome energy density giving a net positive acceleration, however I am not sure that I have a clear picture as to what is really implied here.