Toward the spectral index in slow roll

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SUMMARY

The discussion focuses on deriving the spectral index from the power spectrum of curvature perturbations in slow roll inflation. The key relationship established is that the spectral index \( n_s - 1 \) can be expressed as \( (2\eta_V - 6\epsilon_V) \big{|}_{k=aH} \), linking observable quantities to potential slow roll parameters \( \epsilon_V \) and \( \eta_V \). The transformation of the operator \( \frac{d}{d(\ln{k})} \) into \( -M_{pl}^2 \frac{V'}{V} \frac{d}{d\phi} \) is crucial for this derivation. The discussion concludes with the importance of evaluating these parameters at horizon crossing, \( k=aH \).

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  • Understanding of slow roll inflation dynamics
  • Familiarity with curvature perturbations and their power spectrum
  • Knowledge of potential slow roll parameters \( \epsilon_V \) and \( \eta_V \)
  • Proficiency in manipulating derivatives in the context of cosmological equations
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  • Study the derivation of the spectral index in inflationary cosmology
  • Learn about the implications of the slow roll conditions on inflationary models
  • Investigate the relationship between the potential \( V(\phi) \) and the slow roll parameters
  • Explore the role of horizon crossing in cosmological perturbation theory
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ergospherical
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Getting a bit confused over a manipulation; I have a power spectrum ##P_{R}(k)## in terms of ##V## and ##\epsilon_V## (to be evaluated at ##k=aH##) for the curvature perturbation ##R## and from this need to get a spectral index, i.e. apply ##d/d(\ln{k})##. So we need to transform this operator into "something" ##\times d/d\phi##. From slow roll we get ##3H \dot{\phi} = -V'## and ##H^2 = V/(3M_{pl}^2)##, combining to$$\frac{\dot{\phi}}{H} = -\frac{M_{pl}^2 V'}{V}$$Then from the LHS I need to convert into something related to ##d\phi/dk##. We know ##k=aH## so could try something like$$\frac{\dot{\phi}}{H} = a \frac{d\phi}{da} = a \frac{d\phi}{dk} \frac{dk}{da}$$but from there on it's not clear to me how to manipulate ##\frac{dk}{da}## into something without more time derivatives? The answer should be$$\frac{d}{d\ln k} = - M_{pl}^2 \frac{V'}{V} \frac{d}{d\phi}$$i.e. suggesting that $$\frac{a}{k} = \frac{da}{dk}$$ but why is that the case, as ##H = H(t)## is not fixed?
 
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Closing old (unanswered) threads. It's been a couple months since I've needed this, but I figured out the general procedure eventually. Let's see if I remember. The relation between the spectral index and the power spectrum is$$n_s - 1 = \frac{d \log P_R}{d\log k} \bigg{|}_{k=aH}$$and the play is to break this into:
$$n_s -1 = \frac{d \log P_R}{dN} \frac{dN}{d\log k} \bigg{|}_{k=aH}$$The quantity ##N := \log a## is the number of ##e##-foldings of inflation. For the comoving curvature perturbation ##R##, the power spectrum goes like ##P_R \sim H^2 / (M_{\mathrm{pl}}^2 \epsilon)## evaluated at horizon crossing, so
$$\frac{d \log P_R}{dN} = 2 \frac{d\log H}{dN} - \frac{d\log \epsilon}{dN} = -2\epsilon -2(\epsilon - \eta)$$
where the slow roll parameters ##\epsilon := -d\log H/dN## and ##\eta := -d\log H_{,\phi}/dN## respectively. As for the second factor: at the horizon crossing ##k=aH##, or $$\log k = \log a + \log H$$Then$$\frac{dN}{d\log k} = \left( \frac{d\log k}{dN} \right)^{-1} = \left(1 + \frac{d\log H}{dN}\right)^{-1} \approx 1+\epsilon$$Therefore to linear order in the slow roll parameters ##n_s - 1 = (2\eta - 4\epsilon) \big{|}_{k=aH}##. Further, given that the potential slow roll parameters ##\epsilon_V## and ##\eta_V## (defined in terms of ##V(\phi)##) are related to the actual slow roll parameters via ##\epsilon \approx \epsilon_V## and ##\eta \approx \eta_V - \epsilon_V##, we get $$n_s - 1 = (2\eta_V - 6\epsilon_V) \big{|}_{k=aH}$$which relates the spectral index (observable!) to the potential slow roll parameters that you can calculate directly from the form of the potential ##V(\phi)##.
 
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