Quintessence in CAMB - Units of the potential?
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- Posts: 2
- Joined: December 17 2013
- Affiliation: Institute of Theoretical Astrophysics, University of Oslo
Quintessence in CAMB - Units of the potential?
Hi all
I have been using the Quintessence equations for CAMB and have been having some doubts on the way the potential is defined.
From what I can see in the code, the potential should be given in units of 1/Mpc^2 as it is multiplied by 8*Pi*G.
Those units are considered in the norm parameter, but looking carefully at it, I don't understand how it changes from a dimensionless quantity to a quantity with 1/Mpc^2 by multiplying it with a factor (Mpc/c)^2/Tpl^2.
Shouldn't it remain dimensionless?
Thanks in advance,
Best
Paulo
I have been using the Quintessence equations for CAMB and have been having some doubts on the way the potential is defined.
From what I can see in the code, the potential should be given in units of 1/Mpc^2 as it is multiplied by 8*Pi*G.
Those units are considered in the norm parameter, but looking carefully at it, I don't understand how it changes from a dimensionless quantity to a quantity with 1/Mpc^2 by multiplying it with a factor (Mpc/c)^2/Tpl^2.
Shouldn't it remain dimensionless?
Thanks in advance,
Best
Paulo
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- Posts: 2
- Joined: December 17 2013
- Affiliation: Institute of Theoretical Astrophysics, University of Oslo
Quintessence in CAMB - Units of the potential?
Hi
Could anyone provide some insight on this subject?
I still couldn't figure out exactly how to make the code work for an m^2\phi^2/2 potential as I need to.
I guess that the problems likely lie in the normalisations considered in the potential.
I was also checking that in the code and there is one more thing that confuses me a bit.
Why does the normalisation factor presented features a power of the Planck Time?
Since the potential depends usually on a mass, wouldn’t it make more sense to input the Planck Mass in the normalisation?
Thanks in advance,
All the best,
Paulo
Could anyone provide some insight on this subject?
I still couldn't figure out exactly how to make the code work for an m^2\phi^2/2 potential as I need to.
I guess that the problems likely lie in the normalisations considered in the potential.
I was also checking that in the code and there is one more thing that confuses me a bit.
Why does the normalisation factor presented features a power of the Planck Time?
Since the potential depends usually on a mass, wouldn’t it make more sense to input the Planck Mass in the normalisation?
Thanks in advance,
All the best,
Paulo
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- Posts: 6
- Joined: April 26 2013
- Affiliation: inr
Quintessence in CAMB - Units of the potential?
I suppose the potential is given in units of (1/Mpc)^2 really because there isn't kappa=8*pi*G (ussually [V]=(1/Mpc)^4). I made this decision as in the code I saw
grhoex_t=phidot**2/2 + a2*Vofphi(phi,0)
grho=grho+grhoex_t
But we know that
grho = a^2 kappa rho
therefore Vofphi(phi,0) is in fact kappa*V.
Hitherto I don't understand the sence of constant "norm" which equals 1d-122 (it's very strange value). In my mind there is to be no "norm" constant.
grhoex_t=phidot**2/2 + a2*Vofphi(phi,0)
grho=grho+grhoex_t
But we know that
grho = a^2 kappa rho
therefore Vofphi(phi,0) is in fact kappa*V.
Hitherto I don't understand the sence of constant "norm" which equals 1d-122 (it's very strange value). In my mind there is to be no "norm" constant.
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- Posts: 29
- Joined: June 10 2012
- Affiliation: University of Mexico
Quintessence in CAMB - Units of the potential?
Dear all. I think that the motivation of the definition
phi_code= kappa*phi
is for leaving the phi_code variable with no units.
So, if anyone wants to work with a potential of the form
m*phi**2
and if phi is originally measured in GeV, one has to change to the variable phi_code described above. The value of the normalisation factor is determined by the conditions of the equation of state today (close to -1) and by the current acceleration of the expansion, but otherwise I think there are no more restriction for the choice of a particular value.
Regards
phi_code= kappa*phi
is for leaving the phi_code variable with no units.
So, if anyone wants to work with a potential of the form
m*phi**2
and if phi is originally measured in GeV, one has to change to the variable phi_code described above. The value of the normalisation factor is determined by the conditions of the equation of state today (close to -1) and by the current acceleration of the expansion, but otherwise I think there are no more restriction for the choice of a particular value.
Regards