# I Exact expression for the tensor-to-scalar ratio

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1. Feb 28, 2017

### Whitehole

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: Feb 28, 2017
2. Mar 1, 2017

### bapowell

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.

3. Mar 2, 2017

### Whitehole

$\tilde \omega$
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: Mar 2, 2017
4. Mar 2, 2017