Derive the equation for geodesic deviation starting from the geodesic equation

Question

A family of geodesics can be parametrised $x^a = x^a(s, t)$ where $s$ is the distance along a geodesic and the parameter $t$ specifies the geodesic. For each $t$ the geodesic equation is
$$\frac{\partial^2 x^a}{\partial s^2}+\Gamma_{bc}^a\frac{\partial x^b}{\partial s}\frac{\partial x^c}{\partial s}=0\tag{1}$$
Partially differentiate this equation with respect to $t$ to obtain the equation of geodesic deviation
$$\frac{D^2 w^a}{\partial s^2}=-R_{bcd}^a u^bw^cu^d\tag{2}$$
where $u^a = \partial x^a/\partial s\,$and $w^a=\partial x^a/\partial t$.

---

So taking the time derivative of $(1)$ gives $$\frac{\partial}{\partial t}\left(\frac{\partial u^a}{\partial s}\right)+\left(\frac{\partial}{\partial t}\Gamma_{bc}^a\right)u^bu^c+\Gamma_{bc}^a\frac{\partial u^b}{\partial t}u^c+\Gamma_{bc}^au^b\frac{\partial u^c}{\partial t}=0\tag{a}$$
But right away I feel totally stuck as I don't understand what to do next in order to reach the desired expression, $(2)$. Could someone please provide hints or tips on how to proceed?

---

Just for reference, I have typeset the full solution as given by the author below:

Differentiating the geodesic equation with respect to $t$
$$\frac{\partial^2 w^a}{\partial s^2}+\partial_d\Gamma_{bc}^aw^du^bu^c+2\Gamma_{bc}^a\frac{\partial w^b}{\partial s}u^c=0\tag{3}$$
Now
$$\frac{D^2w^a}{\partial s^2}=\frac{\partial}{\partial s}\frac{Dw^a}{\partial s}+\Gamma_{bc}^a\frac{D w^b}{\partial s}u^c$$
$$=\frac{\partial}{\partial s}\left(\frac{\partial w^a}{\partial s}
+\Gamma_{bc}^aw^bu^c\right)+\Gamma_{bc}^a\left(\frac{\partial w^b}{\partial s}+\Gamma_{de}^bw^du^e\right)u^c$$
$$=\frac{\partial^2 w^a}{\partial s^2}+\left(\partial_d\Gamma_{bc}^a\right)u^dw^bu^c+\Gamma_{bc}^aw^b\frac{\partial u^c}{\partial s}+2\Gamma_{bc}^a\frac{\partial w^b}{\partial s}u^c+\Gamma_{bc}^a\Gamma_{de}^bw^du^eu^c$$
$$=\left(\partial_d\Gamma_{bc}^a\right)u^dw^bu^c-\Gamma_{bc}^a\Gamma_{de}^cu^du^ew^b+\Gamma_{bc}^a\Gamma_{de}^bw^du^eu^c+\frac{\partial^2w^a}{\partial s^2}+2\Gamma_{bc}^a\frac{\partial w^b}{\partial s}u^c,$$
using $\partial u^c/\partial s=-\Gamma_{de}^cu^du^e$. Using $(3)$ this can be written as
$$\frac{D^2 w^a}{\partial s^2}=\left(\partial_d\Gamma_{bc}^a\right)u^dw^bu^c-\Gamma_{bc}^a\Gamma_{de}^cu^du^ew^b+\Gamma_{bc}^a\Gamma_{de}^bw^du^eu^c-\partial_d\Gamma_{bc}^aw^du^bu^c.$$
All indices apart from $a$ are dummy indices. In the first two terms on the right-hand side swap $b$ and $c$ indices. In the second two terms swap the $c$ and $d$ indices. This yields
$$\frac{D^2 w^a}{\partial s^2}=\left(\partial_d\Gamma_{bc}^a\right)u^dw^cu^b-\Gamma_{bc}^a\Gamma_{de}^bu^du^ew^c+\Gamma_{bd}^a\Gamma_{ce}^bw^cu^eu^d-\partial_c\Gamma_{bd}^aw^cu^bu^d$$
$$=-\left(\partial_c \Gamma_{bd}^a-\partial_d\Gamma_{bc}^a+\Gamma_{ec}^a\Gamma_{db}^e-\Gamma_{ed}^a\Gamma_{cb}^e\right)u^bw^cu^d=-R_{bcd}^au^bw^cu^d$$

Looking at the solution above, I really don't know how eqn. $(3)$ in the solution was found by differentiating eqn. $(1)$ and moreover, why can I not arrive at the same eqn. via $(\mathrm{a})$?

Answer 1

There are two things you need to use to get from (a) to (3):

• Partial derivatives commute: $$ \frac{\partial}{\partial t} \frac{\partial x^a}{\partial s} = \frac{\partial}{\partial s} \frac{\partial x^a}{\partial t} $$
• When a quantity $A$ is a function of $x, y, z$, etc., which in turn all depend on $t$, then $$ \frac{\partial}{\partial t} A = \frac{\partial A}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial A}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial A}{\partial z} \frac{\partial z}{\partial t} + \mathrm{etc.} $$ Or, the same idea in slightly different notation: when a quantity $A$ is a function of $x^a$, and $x^a$ depends on $t$, then $$ \frac{\partial}{\partial t} A = \partial_a A \frac{\partial x^a}{\partial t} $$