I Exact expression for the tensor-to-scalar ratio

Whitehole
Messages
128
Reaction score
4
Based on the vast cosmology texts, I have seen that the expression for the tensor-to-scalar ratio ##r## in cold inflation is,

##r=16\epsilon_H = -16\frac{\dot H}{H^2}\qquad## where, ##~~~\epsilon_H = -\frac{\dot H}{H^2}##

##H## is the Hubble parameter, and ##\epsilon_H## is the Hubble slow roll parameter.

I want to know if this expression is exact or approximation only? I want to numerically compute ##r## but in the case of warm inflation.
 
Last edited:
Space news on Phys.org
Whitehole said:
Based on the vast cosmology texts, I have seen that the expression for the tensor-to-scalar ratio ##r## in cold inflation is,

##r=16\epsilon_H = -16\frac{\dot H}{H^2}\qquad## where, ##~~~\epsilon_H = -\frac{\dot H}{H^2}##

##H## is the Hubble parameter, and ##\epsilon_H## is the Hubble slow roll parameter.

I want to know if this expression is exact or approximation only? I want to numerically compute ##r## but in the case of warm inflation.
It's a lowest-order approximation. To get an exact value, you must numerically compute the amplitudes of the tensor and scalar perturbation spectra at the scale of interest.
 
##\tilde \omega##
bapowell said:
It's a lowest-order approximation. To get an exact value, you must numerically compute the amplitudes of the tensor and scalar perturbation spectra at the scale of interest.
Oh, since ##~r=\frac{P_T}{P_S}~## where ##P_T## is the tensor amplitude and ##P_S## is the scalar amplitude. But where can I find a reference for ##P_S## in warm inflation? I only know of that in cold inflation, and I think ##P_T## is the same as in the cold inflation since ##P_T## doesn't couple strongly to the thermal background so gravitational waves are only generated by quantum fluctuations.
 
Last edited:
Back
Top