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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2024

Paper 357

With $R=r+M$ and $dR=dr$,
$$
1-\frac{2M}{R}=\frac{r-M}{r+M},
\qquad
\frac{R^2}{(R-M)^2}=\left(\frac{r+M}{r}\right)^2.
$$
Defining the <conformal factor> by
$$
\psi^4=\left(\frac{r+M}{r}\right)^2,
$$
also gives $R^2=\psi^4r^2$. Hence
$$
\boxed{
ds^2=-\frac{r-M}{r+M}\,dt^2
+\frac{2M}{r}\,dt\,dr
+\psi^4\left(dr^2+r^2d\Omega^2\right).
}
$$
The constant-time spatial metric is therefore <conformally flat>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 357 / 2 / b / Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 357 / 2 / b / Solution

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Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 357 / 2 / b / Solution

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