# Equality between derivatives of the metric

In one of my lecture, it is said:

Let us use the freedom of the choice of parametrization to demand that the variation of $$\lambda$$ after a small displacement along the curve is proportional to the length of the displacement. Such parameter $$\lambda$$ is called affine. One has thus $$d\lambda^2 = K\,g_{ab}\,dx^a dx^b \quad (\mbox{with K constant})\;\rightarrow\;g_{ab}{x'}^a{x'}^b = K^{-1}$$

The previous equation is then reduced to $$\boxed{ \frac{d}{d\lambda}(g_{ac}{x'}^a) = \tfrac12 {x'}^a{x'}^b \,\partial_cg_{ab} }$$

This is the first form of the equations of the geodesic curves, or curve of extremal length, in the affine parametrization.

Remark $$\rightarrow$$ For such a definition of geodesics to make sense, it is necessary that a metric exists (only case considered here), and it has to be positive definite.

The geodesic equation can take another form by expanding the left hand side: \begin{aligned} g_{ac}{x''}^a + &\underbrace{\frac{dg_{ac}}{d\lambda} {x'}^a} = \tfrac12 {x'}^a {x'}^b \,\partial_c g_{ab}\\ &=\partial_b g_{ac}\frac{dx^b}{d\lambda} {x'}^a = {x'}^a{x'}^b \partial_b g_{ac} = \tfrac12 (\partial_a g_{bc} + \partial_c g_{ac}){x'}^a{x'}^b \end{aligned} $$g_{ac}{x''}^a+\tfrac12(\partial_a g_{bc} + \partial_c g_{ac}-\partial_cg_{ab}){x'}^a{x'}^b = 0$$ multiply by $$g^{cd}$$ $$\boxed{ {x''}^d + \tfrac12 g^{cd}(\partial_a g_{bc} + \partial_c g_{ac}-\partial_cg_{ab}) {x'}^a{x'}^b = 0 }$$

How to prove at the bottom of page that:

$$x'^{a}\,x'^{b}\,\partial_{b}g_{ac} = \dfrac{1}{2}(\partial_{a}g_{bc}+\partial_{b}g_{ac})\,x'^{a}\,x'^{b}$$

that would imply that: $$\partial_{b}g_{ac}=\partial_{a}g_{bc}$$ , is it right?

Perhaps $$x^a x^b \partial_b g_{ac} =\\ =1/2\cdot x^a x^b \partial_b g_{ac}+1/2\cdot x^a x^b \partial_b g_{ac}\\ =1/2\cdot x^a x^b \partial_b g_{ac}+1/2\cdot x^b x^a \partial_b g_{ac}\\ =1/2\cdot x^a x^b \partial_b g_{ac}+1/2\cdot x^a x^b \partial_a g_{bc}$$