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6.2. A worked example

The easiest way to see what is going on is to work through a specific example, the m2phi2 / 2 potential which we already saw in Section 5.5. We'll see that we don't even have to solve the evolution equations to get our predictions.

  1. Inflation ends when epsilon = 1, so phiend appeq mPl / sqrt(4pi).
  2. We're interested in 60 e-foldings before this, which from Eq. (52) gives phi60 appeq 3mPl.
  3. Substitute this in:

    Equation 58a

  4. Reproducing the COBE result requires [25] deltaH appeq 2×10-5 (provided AG << deltaH), so we need m appeq 10-6mPl.

Because the required value of m is so small, that means it is easy to get sufficient inflation to solve the cosmological problems, without violating the classicality condition V < mPl4. That implies only that phi < mPl2 / m appeq 106mPl, and as Ntot appeq 2piphi2 / mPl2, we can get up to about 1013 e-foldings in principle. This compares extremely favourably with the 70 or so actually required.