Energy scale of Hubble constant for dark energy scalar field

[victorvmotti]

Hello All,

In Carroll's there is a brief introduction to a dynamical dark energy in which the equation of motion for slowly rolling scalar field is discussed.

Then to give an idea about the mass scale of this field it is compared to the Hubble constant, saying that it has an energy of almost $H_0= 10^-33 eV$.

If we assume that the Hubble constant is almost 70 km/s/Mpc can someone please show how we infer and compute that energy scale in particle physics?

 

[Mordred]

I'm not familiar with how Carrol derived his value. This article by Peebles although older may provide some insight.


http://arxiv.org/pdf/astro-ph/0207347v2.pdf

 

[George Jones]

If we assume that the Hubble constant is almost 70s km/s/Mpc can someone please show how we infer and compute that energy scale in particle physics?

I haven't done it, but let's try the following.

First, in H_0, convert either km to Mpc, or Mpc to km, thus converting H_0 to units of s^{-1}. Next, to get H_0 in units of energy, multiply by Planck's constant. If uints of Joule-seconds are used for Plank's constant, convert J to eV. If units of eV-seconds are used for Planck's constant, there is no need for this last step.

 

[Chalnoth]

I haven't done it, but let's try the following.

First, in H_0, convert either km to Mpc, or Mpc to km, thus converting H_0 to units of s^{-1}. Next, to get H_0 in units of energy, multiply by Planck's constant. If uints of Joule-seconds are used for Plank's constant, convert J to eV. If units of eV-seconds are used for Planck's constant, there is no need for this last step.

I somewhat doubt it. My bet is he was using the Friedmann equation:

$$H^2 = {8\pi G \over 3} \rho$$

The $\rho$ on the right is a mass density, so multiply by $c^2$ to get an energy density.

The reason I think this is the way the calculation was done is because if we have a universe with only dark energy in it, then it is the Friedmann equation above that relates the energy density of the dark energy to the expansion rate.

 

[George Jones]

The reason I think this is the way the calculation was done is because if we have a universe with only dark energy in it, then it is the Friedmann equation above that relates the energy density of the dark energy to the expansion rate.

But using units such that $\hbar = 1$, which is standard in particle physics, makes energy and inverse time equivalent in all of physics, not just in the Friedmann equation. I have now done the unit conversion calculation, and it works.

 

[Mordred]

Not sure on either method as the OP used the term dynamically dark energy..

sounds like he is suggesting an evolving dark energy in relation the H_o.

This is one of his earlier papers on dark energy but it doesn't appear to fit the OPs descriptive. However I could be reading to much into it

http://arxiv.org/abs/astro-ph/0301273

the paper may or may not have the correct metrics Carrol is using. Perhaps the OP can provide a reference as to which article of Carrol's he saw that line in

Edit just saw George Jones post

 

[victorvmotti]

Thanks a lot, all clear now, for the record the reference is his book Spacetime and Geometry, page 360.

 

[strangerep]

I have now done the unit conversion calculation, and it works.

Me too.

(In the course of that calculation, I also noticed Wikipedia's value for the Hubble time looks wrong by 1 order of magnitude.)

 

[Fran1886]

Effectively as Chalnoth said, Carroll is using the Friedmann equation $$ H^2 = \frac{8\pi G}{3}\rho = \frac{1}{3M_P^2}\rho = H_0^2 \frac{\rho}{\rho_c}. $$ The last equality states a relation between $H_0, M_P$ and $\rho_c$ as follows $$ H_0 = \frac{1}{M_P}\ \sqrt{\frac{\rho_c}{3}}. $$ Using the critical density and Planck mass values in eV given by $\rho_c = 3.649\times 10^{-11}$eV$^{4}$ and $M_P = 2.435\times 10^{27}$eV, you obtain $$ H_0 = 1.43228\times 10^{-33} eV\, . $$