How does the strength of dark energy compare to the strength of the other forces?

Question

I have read this question:

http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html

So , in a nutshell, it is the fitting of data with a specific standard model that organizes the particle interactions in line with four forces.

How do physicists compare the relative strengths of the four forces?

We do have data, because we do know how fast the galaxies are flying apart, and this is in mainstream physics accounted for to the existence of dark energy.

Is there a way to put dark energy into this table and somehow match its strength compared to the other forces?

Answer 1

Árpád Szendrei asked: Is there a way to put dark energy into this table and somehow match its strength compared to the other forces?

In general relativity dark energy acts gravitationally (as you can see on the $\rm G $ in the equation below), so you have to put it in that category.

Its effect is due to its density, since the square of the Hubble parameter $$ \rm H^2 = \frac{8 \ \pi \ G \ \rho}{3} $$ is proportional to the product of the gravitational constant $\rm G$ times the density $\rm \rho=\rho_r+\rho_m+\rho_{\Lambda}$ for radiation, matter and dark energy.

Since the dark energy density $\rho_{\Lambda}$ is constant the Hubble parameter is also constant when dark energy dominates, while it shrinks when radiation or matter whose densities dilute while space expands dominate.

For better understanding: if you could magically multiply the matter so that its density would stay constant while the space expands, that would have the same gravitational effect on the Hubble parameter as dark matter's constant density.

Answer 2

In the table of "fundamental forces" emphasis should be given to the word "fundamental", not "forces". The table is relevant to models that use quantum field theory and more specifically the Feynman diagram method for computing the crossections and decays of elementary particles, to start with for the three first forces, and the last is an extrapolation to an effective field theory for gravitation.

In the Feynman diagram model, the expression of the word force is in the $dp/dt$ carried by the virtual particles at the vertex of interaction. So at the particle level there are many forces. At the particle level the term "fundamental" is used for the forces resulting from the gauge boson exchanged in the theory of $SU(3)xSU(1)xU(1)$ of the standard model of particle physics.

The fourth column in the table has the gauge bosons . Before symmetry breaking there are three "fundamental forces". The fourth fundamental force in the link you give, gravity , is a hypothetical unified model, a theory of everything. At the moment there are only effective quantum theories for gravity, but it is in these models that gravity occupies the position it does, as a fundamental force.

Is there a way to put dark energy into this table and somehow match its strength compared to the other forces?

As dark energy is a complicated phenomenon depending on cosmological observations of gravitational interactions, at the moment, it is a subset of the possible $dp/dt$ in the existing models. A model which would use dark energy to define a new fundamental force is needed in order for it to have its own "gauge bosons" and thus occupy a place in the particle table in the future. At present the answer is "no".