2
$\begingroup$

I know from "Gravitation" of Misner, Thorne and Wheeler(MTW) that Newtonian gravity can be formulated in a geometric language. Can we do the reverse for General Relativity? That is, could General Relativity be formulated or translated to a language of forces?

$\endgroup$
  • $\begingroup$ insofar as I'm aware, it really can't be done without dropping the equivalence principle to some extent. check out this paper, he does a good job discussing just this (I'd just come across it the other day): arxiv.org/abs/0809.2323 $\endgroup$ – R. Rankin Nov 28 '16 at 11:25
  • $\begingroup$ I don't know enough to make an answer, but proponents of teleparallel gravity claim that it can be made fully equivalent to GR: link. In teleparallel gravity, a flat, but "torsionful" connection is used instead of a curved but torsionless. The information regarding gravity is contained in the torsion, and particle-field interactions are described by a Lorentz-force like term involving torsion instead of $F_{\mu\nu}$. This might be of interest to you. $\endgroup$ – Bence Racskó Nov 29 '16 at 10:43
  • $\begingroup$ @Uldreth How well-accepted is this theory among scientists? $\endgroup$ – TheQuantumMan Nov 29 '16 at 23:45
  • $\begingroup$ @TheQuantumMan I don't know! I have no doubt the proof that it is equivalent is sound, (though feel free to check for yourself) whether there is any benefit to using this formalism over the standard one, I don't know. $\endgroup$ – Bence Racskó Nov 30 '16 at 14:58
2
$\begingroup$

In some special cases like the Schwarzschild-case it can, the equation of motion for a massless test particle then becomes:

$$\ddot{r}(t) = -\frac{G\cdot M}{r(t)^2} + r(t)\cdot \dot{{\theta}}(t)^2 {\color{red} {- \frac{3\cdot G\cdot M\cdot \dot{{\theta}}(t)^2}{c^2}}}$$

for the radial acceleration,

$$\ddot{{\theta}}(t) = -2\cdot \dot{r}(t)/r(t)\cdot \dot{{\theta}}(t)$$

for the angular acceleration and

$${\color{red} {\dot{{\tau}}(t) = \frac{\sqrt{c^2\cdot r(t)-c^2\cdot r_s +r(t)\cdot r \dot{(t)}^2- r_s\cdot r(t)^2\cdot \dot{\theta}(t)^2+r(t)^3\cdot \dot{\theta}(t)^2}}{c\cdot \sqrt{r(t)- r_s } \cdot\sqrt{1-\frac{ r_s }{r(t)}}}}}$$

for the time dilation.

The newtonian equation of motion is the same, but without the red terms.

The transformation from the local velocity to the local coordinate speed is

$$\dot{r} = \frac{v_{\parallel}\color{green}{\cdot \sqrt{1-r_s/r}}}{\color{blue}{ \sqrt{ 1-v^2/c^2}}}$$

with an extra term in the numerator for the gravitational length contraction for the radial, and

$$\dot{\theta} = \frac{ v_{\perp}}{\color{blue}{ \sqrt{ 1-v^2/c^2}}\cdot r}$$

for the angular component (with newton the blue and green terms vanish) where $\perp$ stands for the transversal and $\parallel$ for the radial component, so by Pythagoras $v^2=v_{\perp}^2+v_{\parallel}^2$ .

To get the shapiro-delayed velocity as observed by a field-free observer you multiply $v_{\parallel}$ with $1-r_s/r$ and $v_{\perp}$ with $\sqrt{1-r_s/r}$.

In other cases where you have more than 2 bodies of which one's mass is neglible it might be a different story, but if you have only one significant mass and almost massless test particles you can describe it in an almost newtonian way, like in the Schwarzschild oder Kerr case.

But I have no idea how you would deal with a full relativistic n-body simulation, that might be a little more complicated than with Newton.

$\endgroup$
1
$\begingroup$

This is like asking whether Statistical mechanics can be reformulated in terms of Thermodynamics. Statistical mechanics is the microscopic description that puts Thermodynamics on a firmer ground. In the same fashion what Newton called a force is what you get when curvatures in space-time are low and change very slowly and velocities are well below the speed of light. One of the major lessons of GR is that even gravity is a fictitious force. It is on the same grounds as Coriolis force and centrifugal force. What you are asking for is a reformulation of a fundamental theory of space-time in terms of fictitious forces.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.