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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2018

Paper 1

Differentiating the ansatz gives
$$
\dot\phi
=\frac{2\lambda\phi_0}{\sinh(2\lambda t)}.
$$
The <Raychaudhuri equation> therefore becomes
$$
\dot H
=-2\lambda^2\frac{\phi_0^2}{M_{\rm Pl}^2}
\operatorname{csch}^2(2\lambda t).
$$
Since $d[\coth(2\lambda t)]/dt=-2\lambda\operatorname{csch}^2(2\lambda t)$, the requested specific solution takes the integration constant to be zero:
$$
\boxed{\
H(t)=\lambda\frac{\phi_0^2}{M_{\rm Pl}^2}\coth(2\lambda t)\
}.
$$
Now $\dot a/a=H$, so
$$
\log a
=\frac{\phi_0^2}{2M_{\rm Pl}^2}\log\sinh(2\lambda t)+\log a_0,
$$
and
$$
\boxed{\
a(t)=a_0[\sinh(2\lambda t)]^{\phi_0^2/(2M_{\rm Pl}^2)}\
}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 15B / b / Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 15B / b / Solution

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Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 15B / b / Solution

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