Acceleration equation for the geodesic deviation equation

Question

Defining a family of geodesics by $\gamma_s$, parametrized by the affine parameter $t$, the coordinates are defined by $x^{\alpha}(s,t)$. The vector tangent to the geodesics is defined as $u^{\alpha} = \frac{d x^{\alpha}}{dt}$ and the family of tangent vectors to the family of geodesics is $\xi^{\alpha} = \frac{d x^{\alpha}}{s}$.

Following Eric Poisson's book (The mathematics of black hole mechanics, equation 1.43), he writes: we wish to derive an expression for its acceleration

$$ \frac{D^2 \xi^{\alpha}}{dt^2} \equiv (\xi^{\alpha}_{; \beta} u^{\beta})_{;\gamma}u^{\gamma} $$

and the rest of the section is showing how this may be written in terms of the Riemann tensor. However, he does not show where this equation comes from. How do we deduce this acceleration equation?

Answer 1

This combines the definition of the relative acceleration between the observers along the congruence, as well as an equivalent way of writing a derivative along the vector field.

In essence, the derivative along the vector field $ \mathbf{u} = \partial_{t} $ can be applied to a scalar field as a simple derivative operator. Because the field is affinely parameterized by $t$, $\mathbf{u} $ can be indeed expressed as $\partial_{t} $.

However, in order to differentiate tensor fields (including vector fields, such as $\xi$) in the same sense - along a vector field, we must employ the covariant derivative. $$ \frac{D T}{dt} = \nabla_{u}T$$

The covariant derivative will of course reduce to a regular derivative when applied to a scalar.

$$ \frac{D^{2}\xi^{\alpha}}{dt^{2}} = (\nabla_{\mathbf{u} } (\nabla_{\mathbf{u} } \xi )) ^{\alpha} $$

The latter is the same as your right hand side, except yours is in semicolon notation. I have added extra parentheses to show clearly the order of operations. You take the covariant derivative twice and you get the relative acceleration.