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ozone
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General Relativity -- Motion in Newtonian Limit
Assume that the metric of space-time in a weak gravitational field, [itex] \frac{\phi}{c^2} << 1 [/itex] is [itex] ds^2 = (1 + \frac{2\phi}{c^2})c^2 dt^2 - (dx^1)^2 - (dx^2)^2 - (dx^3)^2 [/itex] for some arbitrary scalar function phi.
Use the variational principle to derive the equation of motion of particles, in the approximation that the velocity is small compared to c. Compare to the equations of Newtonian gravity.
Note that my teacher has been using the squared version of the metric in our actionals to derive things, so I wanted to follow his example. This however means that our parameterization [itex] \tau [/itex] is not arbitrary and that we need to solve for it from the metric.
I have no trouble solving for my spatial components as soon as I make the assumption that [itex] \frac{dt}{d\lambda} \approx 1 [/itex]. However I'm having trouble solving for this quantity from my euler-lagrange equations. I could really use some help in figuring out what approximations I am allowed to make here. Specifically I derive the equation
[itex] \frac{d}{d\tau} [ c^2 \frac{dt}{d\tau} + 2\phi(x) \frac{dt}{d\tau}] - \frac{\partial \phi}{\partial t} (\frac{dt}{d\tau})^2 = 0[/itex]
which we can divide through and by c squared and do a tiny bit of algebra to find something which looks ready to approximate--
[itex] \frac{d}{d\tau} [ \frac{dt}{d\tau} +\frac{2\phi}{c^2} \frac{dt}{d\tau}] = \frac{1}{c^2} \frac{\partial \phi}{\partial \tau} (\frac{dt}{d\tau}) [/itex]
clearly this must solve such that [itex] \frac{dt}{d\tau} = 1 + O(c^{-1}) [/itex] but getting there is a bit of a headache for me.. I was wondering what approximations I might make to reach this final form.
Edit: My best thought right now is to throw away the right hand side, and for whatever arbitrary reason integrate so that [itex] \frac{dt}{d\tau} = constant - \frac{2\phi}{c^2} [/itex] where our second term is negligible and can be neglected.. however I have no idea what to set the constant or what the justification for throwing away the RHS might be.
Homework Statement
Assume that the metric of space-time in a weak gravitational field, [itex] \frac{\phi}{c^2} << 1 [/itex] is [itex] ds^2 = (1 + \frac{2\phi}{c^2})c^2 dt^2 - (dx^1)^2 - (dx^2)^2 - (dx^3)^2 [/itex] for some arbitrary scalar function phi.
Use the variational principle to derive the equation of motion of particles, in the approximation that the velocity is small compared to c. Compare to the equations of Newtonian gravity.
Note that my teacher has been using the squared version of the metric in our actionals to derive things, so I wanted to follow his example. This however means that our parameterization [itex] \tau [/itex] is not arbitrary and that we need to solve for it from the metric.
I have no trouble solving for my spatial components as soon as I make the assumption that [itex] \frac{dt}{d\lambda} \approx 1 [/itex]. However I'm having trouble solving for this quantity from my euler-lagrange equations. I could really use some help in figuring out what approximations I am allowed to make here. Specifically I derive the equation
[itex] \frac{d}{d\tau} [ c^2 \frac{dt}{d\tau} + 2\phi(x) \frac{dt}{d\tau}] - \frac{\partial \phi}{\partial t} (\frac{dt}{d\tau})^2 = 0[/itex]
which we can divide through and by c squared and do a tiny bit of algebra to find something which looks ready to approximate--
[itex] \frac{d}{d\tau} [ \frac{dt}{d\tau} +\frac{2\phi}{c^2} \frac{dt}{d\tau}] = \frac{1}{c^2} \frac{\partial \phi}{\partial \tau} (\frac{dt}{d\tau}) [/itex]
clearly this must solve such that [itex] \frac{dt}{d\tau} = 1 + O(c^{-1}) [/itex] but getting there is a bit of a headache for me.. I was wondering what approximations I might make to reach this final form.
Edit: My best thought right now is to throw away the right hand side, and for whatever arbitrary reason integrate so that [itex] \frac{dt}{d\tau} = constant - \frac{2\phi}{c^2} [/itex] where our second term is negligible and can be neglected.. however I have no idea what to set the constant or what the justification for throwing away the RHS might be.
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