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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?

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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} $

I am not entirely confident that substitutions are allowed on subformulas like that but if they are this seems to do the trick.

So the answer is for most instances no; the equality is for the fully contracted term, not the partial derivatives of the tensor components in isolation.

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  • $\begingroup$ thanks a lot ! that was the trick $\endgroup$ – youpilat13 May 4 '19 at 16:48

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