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We know that special relativity tends to become equivalent to classical theory of relativity as the speed limit of nature becomes infinite.

If this happens, clock will tick at the same rate everywhere in a variable gravitational field. Also all objects will feel equal acceleration of gravity irrespective of their velocity (light falls faster than stationary things).

Since two of the predictions of GR is already gone, does it mean that GR will actually become newton's theory of gravity?

I ask because I am still few months away from understanding Einstein's field equation.

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    $\begingroup$ Isn't the lack of infinitely extended inertial frames a problem with Newtonian Mechanics which has no apparent direct links with the finiteness of the speed of light? $\endgroup$
    – user87745
    Jun 3, 2016 at 13:51
  • $\begingroup$ It has 1 link with speed of light- if speed of light is made such that it makes special relativity become equivalent to classical relativity, then Einstein's equivalence principle will become unnecessary to explain the universe. It will become nothing more than an imaginative complication... $\endgroup$
    – Prem
    Jun 3, 2016 at 16:09

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That is the case for central gravity fields. You can derive Newtonian gravitation from a spacetime metric $$ ds^2 = c^2d\tau^2 = c^2\left(1 - \frac{2GM}{rc^2}\right)dt^{2} - dr^2 - r^2d\Omega^2 $$ where the only metric element different than unity is $g_{tt} = c^2 - GM/r$. The relevant Christoffel symbol is $$ \Gamma^r_{tt} = \frac{1}{2}g^{rr}\frac{\partial g_{tt}}{\partial r} = \frac{GM}{r^2} $$ and the geodesic equation is $$ \frac{d^2r}{dt^2} + \Gamma^r_{tt}U^tU^t = 0 $$ for $U^t\simeq 1$ recovers Newton's second law of motion with gravity. If a $g_{rr}$ term exists there will then be a term $O(1/c^2)$ or larger. The standard Schwarzschild metric term $$ g_{rr} = \frac{1}{\left(1 - \frac{2GM}{rc^2}\right)} = 1 + \frac{2GM}{rc^2} +\left(\frac{2GM}{rc^2}\right)^2 + \dots $$ introduces deviations of $O(1/c^2)$, $O(1/c^4)$ etc. If $c\rightarrow\infty$ these terms are zero.

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  • $\begingroup$ To be super pedantic, you also have $g_{\theta \theta}$ and $g_{\phi \phi}$ different from unity. $\endgroup$ Jun 3, 2016 at 14:07
  • $\begingroup$ The $r^2d\Omega^2$ in the plane $\theta = \pi/2$ is $r^2d\phi^2$ or $r^2(d\phi/ds)^2ds^2$ that in the weak limit is $r^2\omega^2ds^2$ and is for $U^t \simeq 1$ an $O(c^0)$ contribution. $\endgroup$ Jun 3, 2016 at 14:16
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Yes. Any general relativity textbook will have a section where they derive that the proportionality constant between the Einstein tensor and the stress-energy tensor is $8 \pi G$. They do so by showing that in the limit $c \rightarrow \infty$, all the components of Einstein's field equation become zero except for the 00 component. In this limit, $G^{00}$ approaches $-2 \nabla^2 \phi$ (where $\phi$ is the gravitational potential) and $T^{00}$ equals the mass density $\rho$. Thus Einstein's equation reduces to $\nabla^2 \phi = -4 \pi G \rho$, which is equivalent to Newton's law of universal gravitation. The $c \rightarrow \infty$ limit of the geodesic equation is $d^2x/dt^2 = -\nabla \phi$, which is equivalent to Newton's second law in the case of a gravitational force.

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