I have to present an exercise in class and want to make sure that my approach and understanding of the problem is correct.
Let $\Sigma$ be the hypersurface in Schwarzschild spacetime (metric $g$) defined by $\big( t,r,\theta,\phi \big) = \big( f(r),r,\theta,\phi \big)$. Find $f$ such that the metric $h$ induced on $\Sigma$ is flat.
Now, my approach is to calculate the metric $h$ induced on $\Sigma$ by replacing $dt = df$ within the metric $g = -a(r) \, dt^2 + 1/a(r) \, dr^2 + r^2 \, d\Omega^2$. After that, I calculate the Christoffel symbols in order to finally calculate the Riemann curvature tensor. If the Riemann curvature tensor vanishes, then the metric is considered flat. I therefore have to solve the differential equation for $f(r)$, obtained by calculating the Riemann curvature tensor and assuming it to vanish.
Is this approach/idea correct? The calculations to obtain the Riemann curvature tensor is quite lengthy I suppose, and it will contain some third-degree derivative of $f$. Is there another (easier) way to show flatness of a metric?
Best Answer
As suggested by ANDREW, I will write up the answer to the question here myself.
Given the hypersurface $\Sigma = \{ t = f(r) \}$, the induced metric is obtained by replacing $dt = df$ within the expression for the Schwarzschild metric $g$. The induced metric turns out to be
$$h \enspace = \enspace \Big[ \tfrac{1}{a(r)} - a(r) \, \big( \tfrac{\partial f}{\partial r} \big)^2 \Big] \, dr^2 + r^2 \, d\Omega^2$$
Following ANDREW's and JAVIER's proposal, I now identify the induced metric with some flat metric I already know. Since I am using spherical coordinates, the Euclidean metric in spherical coordinates volunteers for such an approach. In order for the induced metric to coincide with the Euclidean metric in spherical coordinates, the property
$$ \Big[ \tfrac{1}{a(r)} - a(r) \, \big( \tfrac{\partial f}{\partial r} \big)^2 \Big] \enspace = \enspace 1$$
must hold. This leads to
$$ \frac{\partial f}{\partial r} \enspace = \enspace \sqrt{\frac{1-a(r)}{a^2(r)}} \enspace = \enspace \sqrt{\frac{\tfrac{2m}{r}}{\big(1-\tfrac{2m}{r} \big)^2}} \quad ,$$
where $a(r) = 1 - \tfrac{2m}{r}$. The solution for this integral is
$$ f \enspace = \enspace 2m \cdot \bigg( 2\sqrt{\tfrac{r}{2m}} + \ln \Big( \big| 1 - \sqrt{\tfrac{r}{2m}} \big| \Big) - \ln \Big( 1 + \sqrt{\tfrac{r}{2m}} \Big) \bigg)$$
up to some constant. The hypersurface is now the set of points with coordinates $(t,r,\theta,\phi) = (f(r), r, \theta, \phi)$, where $f$ is the above function.