- #1
Frank Castle
- 580
- 23
As I understand it, a curve ##x^{\mu}(\lambda)## (parametrised by some parameter ##\lambda##) connecting two spacetime events is a geodesic if it is locally the shortest path between the two events. It can be found by minimising the spacetime distance between these two events: $$s_{AB}=\int_{A}^{B}\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}=\int_{\lambda_{A}}^{\lambda_{B}}\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}d\lambda$$ Minimising ##s_{AB}## (i.e. requiring that ##\delta s_{AB}=0##) results in the following equation of motion: $$\frac{d^{2}x^{\mu}}{d\lambda^{2}}+\Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}=a(\lambda)\frac{dx^{\mu}}{d\lambda}\qquad (1)$$ where ##a(\lambda)=\left(-\frac{1}{L}\frac{dL}{d\lambda}\right)## and ##L=\frac{ds}{d\lambda}=\sqrt{g_{\mu\nu}(x(\lambda))\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}## is the appropriate Lagrangian.
I've read that if ##\lambda## is an affine parameter then the right hand side of ##(1)## is zero and we have that $$\frac{d^{2}x^{\mu}}{d\lambda^{2}}+\Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}=0\qquad (2)$$ Why is this the case though?!
How does one show that if the parameter is affine then the right-hand side of ##(1)## vanishes?
I can see that if ##\lambda\propto s##, where ##s## is the arc-length of the curve, then
$$\frac{d^{2}x^{\mu}}{d\lambda^{2}}+\Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}=\left(-\frac{1}{L}\frac{dL}{d\lambda}\right)\frac{dx^{\mu}}{d\lambda}=0$$ since ##L=\frac{ds}{d\lambda}=\text{constant}## and so ##\frac{dL}{d\lambda}=0##, so I guess that my confusion is over the definition of an affine parameter. Is it any parameter, ##\lambda## that is proportional to the arc-length, ##s## of the curve, such that ##\lambda=as+b##, where ##a## and ##b## are constants?!
I've read that if ##\lambda## is an affine parameter then the right hand side of ##(1)## is zero and we have that $$\frac{d^{2}x^{\mu}}{d\lambda^{2}}+\Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}=0\qquad (2)$$ Why is this the case though?!
How does one show that if the parameter is affine then the right-hand side of ##(1)## vanishes?
I can see that if ##\lambda\propto s##, where ##s## is the arc-length of the curve, then
$$\frac{d^{2}x^{\mu}}{d\lambda^{2}}+\Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}=\left(-\frac{1}{L}\frac{dL}{d\lambda}\right)\frac{dx^{\mu}}{d\lambda}=0$$ since ##L=\frac{ds}{d\lambda}=\text{constant}## and so ##\frac{dL}{d\lambda}=0##, so I guess that my confusion is over the definition of an affine parameter. Is it any parameter, ##\lambda## that is proportional to the arc-length, ##s## of the curve, such that ##\lambda=as+b##, where ##a## and ##b## are constants?!
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