- #1
Pentaquark5
- 17
- 2
Let us assume a "toy-metric" of the form
$$ g=-c^2 \mathrm{d}t^2+\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2-\frac{4GJ}{c^3 r^3} (c \mathrm{d}t) \left( \frac{x\mathrm{d}y-y\mathrm{d}x}{r} \right)$$
where ##J## is the angular-momentum vector of the source.
Consider the curve
$$ \gamma(\tau)=(x^\mu (\tau))=\bigg(t(\tau),0,0,z(\tau)\bigg) $$
To see wether this is a geodesic, we need to check if the geodesic equation in local coordinates
$$ \frac{\mathrm{d}^2x^\mu}{\mathrm{d}\tau^2}=-\Gamma^\mu_{\:\:\alpha\beta} \frac{\mathrm{d}x^\alpha}{\mathrm{d}\tau}\frac{\mathrm{d}x^\beta}{\mathrm{d}\tau}$$
is fulfilled.
In order to do that, we will need to calculate the Christoffel symbols where ##\alpha, \beta## equals ##r## or ##z##:
Along ##\gamma## we have ##g_{\mu \nu}=\eta_{\mu \nu}## and
$$ \Gamma^\mu_{\:\: z\alpha}=\frac{1}{2}\eta^{\mu \nu}\big( \partial_z g_{\mu \alpha}+\partial_\alpha g_{\mu z}-\partial_\nu g_{z \alpha} \big)=0 $$
Indeed,
$$ \partial_z g_{t y}|_\gamma = \partial_z \left( -\frac{4GJx}{c^2r^3} \right)|_\gamma = x\partial_x\left(-\frac{4GJ}{c^2r^3}\right)|_\gamma=0$$There are a few questions I have about this computation, and I would be really thankful for input:
Firstly, can we assume ##g_{\mu \nu}=\eta_{\mu \nu}## along the curve, since spacetime is locally Minkowski along ##\gamma##?
Secondly, am I missing something, or should the denominator of ##g_{t y}## be proportional to ##r^{-4}## instead of ##r^{-3}##?
And thirdly, since ##r=r(x,y,z)=\sqrt{x^2+y^2+z^2}##, shouldn't the derivation ##\partial_z g_{t y}|_\gamma## be nonzero?
$$ g=-c^2 \mathrm{d}t^2+\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2-\frac{4GJ}{c^3 r^3} (c \mathrm{d}t) \left( \frac{x\mathrm{d}y-y\mathrm{d}x}{r} \right)$$
where ##J## is the angular-momentum vector of the source.
Consider the curve
$$ \gamma(\tau)=(x^\mu (\tau))=\bigg(t(\tau),0,0,z(\tau)\bigg) $$
To see wether this is a geodesic, we need to check if the geodesic equation in local coordinates
$$ \frac{\mathrm{d}^2x^\mu}{\mathrm{d}\tau^2}=-\Gamma^\mu_{\:\:\alpha\beta} \frac{\mathrm{d}x^\alpha}{\mathrm{d}\tau}\frac{\mathrm{d}x^\beta}{\mathrm{d}\tau}$$
is fulfilled.
In order to do that, we will need to calculate the Christoffel symbols where ##\alpha, \beta## equals ##r## or ##z##:
Along ##\gamma## we have ##g_{\mu \nu}=\eta_{\mu \nu}## and
$$ \Gamma^\mu_{\:\: z\alpha}=\frac{1}{2}\eta^{\mu \nu}\big( \partial_z g_{\mu \alpha}+\partial_\alpha g_{\mu z}-\partial_\nu g_{z \alpha} \big)=0 $$
Indeed,
$$ \partial_z g_{t y}|_\gamma = \partial_z \left( -\frac{4GJx}{c^2r^3} \right)|_\gamma = x\partial_x\left(-\frac{4GJ}{c^2r^3}\right)|_\gamma=0$$There are a few questions I have about this computation, and I would be really thankful for input:
Firstly, can we assume ##g_{\mu \nu}=\eta_{\mu \nu}## along the curve, since spacetime is locally Minkowski along ##\gamma##?
Secondly, am I missing something, or should the denominator of ##g_{t y}## be proportional to ##r^{-4}## instead of ##r^{-3}##?
And thirdly, since ##r=r(x,y,z)=\sqrt{x^2+y^2+z^2}##, shouldn't the derivation ##\partial_z g_{t y}|_\gamma## be nonzero?
Last edited: