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Dear all,
I'm taking a second look at the Newtonian limit of GR and the covariance group. My main interest is to see how the Newton potential transforms under the covariance group of Newtonian gravity. I know that the gradient of the potential is given by the Christoffel symbol [tex]\Gamma^i_{00} = \partial^i \Phi [/tex]
from which you can derive how this gradient transforms under linear accelerations [itex] x^i \rightarrow x^{'i} = x^i + a^i(t) [/itex] (x^0 is called t):
[tex] \Gamma^{'i}_{00} = \Gamma^{i}_{00} + \frac{\partial^2 a^i}{\partial t^2} \ \ \ \ (1) [/tex]
However, from the Poisson equation [tex] \partial_i \partial ^i \Phi = 4 \pi G \rho[/tex] I know that the gradient is not only allowed to transform under only time-dependent accelerations in order to keep the Poisson eqn. covariant, but under accelerations with vanishing divergence:
[tex] \partial_i a^i (r,t) = 0 [/tex]
Apart from the Galilei transformations these restricted accelerations also keep the Poisson eqn. covariant. This makes good sense, since a freely falling observer in a Newtonian gravitational field caused by a mass M undergoes an acceleration
[tex]
a(t,r) = \frac{1}{2} g(r) t^2, \ \ \ \ g(r) = - \frac{GM}{r} \ \ \ \ \ \ (2)
[/tex]
which indeed is divergenceless. My question is: how does Phi transform under these accelerations? The covariance group of Newtonian gravity would then be the Galilei group plus 'divergence-free' but otherwise arbitrary accelerations, where the coefficient of the acceleration is given by g(r) of (2). So I'd say we cannot just integrate eqn.(1) to find the transformation of \Phi. I also tried to find the transformation from [itex]g_{00} = [/itex] but I'm not getting any sensible answer. Apparently, all the literature I know about of the Newtonian limit doesn't mention how the general coordinate transformations are broken down to this "acceleration-extended Galilei symmetries". Does anybody have an idea?
I'm taking a second look at the Newtonian limit of GR and the covariance group. My main interest is to see how the Newton potential transforms under the covariance group of Newtonian gravity. I know that the gradient of the potential is given by the Christoffel symbol [tex]\Gamma^i_{00} = \partial^i \Phi [/tex]
from which you can derive how this gradient transforms under linear accelerations [itex] x^i \rightarrow x^{'i} = x^i + a^i(t) [/itex] (x^0 is called t):
[tex] \Gamma^{'i}_{00} = \Gamma^{i}_{00} + \frac{\partial^2 a^i}{\partial t^2} \ \ \ \ (1) [/tex]
However, from the Poisson equation [tex] \partial_i \partial ^i \Phi = 4 \pi G \rho[/tex] I know that the gradient is not only allowed to transform under only time-dependent accelerations in order to keep the Poisson eqn. covariant, but under accelerations with vanishing divergence:
[tex] \partial_i a^i (r,t) = 0 [/tex]
Apart from the Galilei transformations these restricted accelerations also keep the Poisson eqn. covariant. This makes good sense, since a freely falling observer in a Newtonian gravitational field caused by a mass M undergoes an acceleration
[tex]
a(t,r) = \frac{1}{2} g(r) t^2, \ \ \ \ g(r) = - \frac{GM}{r} \ \ \ \ \ \ (2)
[/tex]
which indeed is divergenceless. My question is: how does Phi transform under these accelerations? The covariance group of Newtonian gravity would then be the Galilei group plus 'divergence-free' but otherwise arbitrary accelerations, where the coefficient of the acceleration is given by g(r) of (2). So I'd say we cannot just integrate eqn.(1) to find the transformation of \Phi. I also tried to find the transformation from [itex]g_{00} = [/itex] but I'm not getting any sensible answer. Apparently, all the literature I know about of the Newtonian limit doesn't mention how the general coordinate transformations are broken down to this "acceleration-extended Galilei symmetries". Does anybody have an idea?
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