I am working with the Laplacian on a Riemannian manifold $(M,g)$ (compact, without boundary). In spherical geodesic coordinates $(r, \sigma)$ around some arbitrary $x \in M$ (where $\sigma$ denotes the angular coorinates taken together), the Laplacian looks like $\Delta = \frac {\partial ^2} {\partial r^2} + H(x,r) \frac \partial {\partial r} + \Delta\_{S(x,r)}$ where $S(x,r)$ is the geodesic sphere of radius $r$ around $x$ and $H(x,r)$ is the total mean curvature of $S(x,r)$.


I need to do some calculations at the end of which I must make $r \to 0$ (i.e. evaluate things in $x$). To this end, it would be very helpful to know the first few terms of the development of $H(x,r)$ with respect to $r$. So far, I only know that $H(x,r) = \frac {n-1} r + O(r)$, which is good but not enough. Does anybody know the coefficient of $r$ (and possibly of $r^2$) in the above expansion?



Limited expansion of mean curvature of geodesic spheres
See <http://link.springer.com/article/10.1007%2FBF02395060> Lemma 12.2 (the $\gamma\_i$ terms are defined on page 167, also see $\S2$ for the curvature notations). 


An alternative "do-it yourself" approach to what you want might be to consider the expansion of the metric in normal coordinates, e.g. [Riemann's formula for the metric in a normal neighborhood](https://mathoverflow.net/questions/185527/riemanns-formula-for-the-metric-in-a-normal-neighborhood), which is not that hard to prove. Then plug this into the coordinate expression for the Laplacian and find a power series for the Laplacian. Changing into polar coordinates should then presumably give you what you want.




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EDIT: Here are is the formula written out. 


For $m\in M$, the mean curvature at the point $\exp\_m(ru)$ (for $r >0$ and $u \in T\_mM$ a unit vector) is given by
$$
\frac{n-1}{r} + \alpha\_1r+\alpha\_2r^2+O(r^3)
$$
where
\begin{align\*}
\alpha\_1 & = -\frac13 Ric|\_m(u,u)\\
\alpha\_2 & = -\frac14(\nabla Ric)|\_m (u,u,u).
\end{align\*}


