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I Exact expression for the tensor-to-scalar ratio

  1. Feb 28, 2017 #1
    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. jcsd
  3. Mar 1, 2017 #2

    bapowell

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    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.
     
  4. Mar 2, 2017 #3
    ##\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
  5. Mar 2, 2017 #4

    bapowell

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