2

The description Einstein gave us about gravity, describes gravity from the perspective of the particle, where it "feels" movement. Instead from that perspective the particle can say space-time it's self is curved and is just following a geodesic trajectory. However, if you want to take space time to be flat, because there is no gravitational force affecting you, is it possible to represent Einstein's field equations as a vector field? Possibly using the geodesic equation?

Even fictitious forces can be described as forces and both the stationary frame and the accelerated frame give the same result of where the ball ends up relative to them? This fictitious force field would need to be approximately equal to Newton's gravitational law at slow speeds, but needs to be relativistic in general. This would need to be described as a four force.

If this field can be constructed from the geodesic equation, what would the acceleration field look like as a four-vector?

If we assume a Schwarzschild metric: $$ds^2 = - (1-\frac{r_s}{r})c^2 dt^2 + \frac{1}{1-\frac{r_s}{r}} + r^2 d\Omega^2, $$

And substitute the metric into the geodesic equation: $$ {d^2 x^\mu \over d\tau^2} =- \Gamma^\mu {}_{\alpha \beta}{d x^\alpha \over d\tau}{d x^\beta \over d\tau}, $$

Is it possible to get into the form:

$$ {d^2 x^\mu \over d\tau^2} = F(x^\mu)~? $$

Qmechanic
  • 220,844
Joshua Pasa
  • 825

1 Answers1

3

Yes. This is the meaning of the equivalence principle, and it is recognised in the weak field limit. We take space as flat, and the result of small variations in time is that gravity appears as an inertial force causing acceleration to appear in the geodesic equation. The Newtonian approximation then shows a Newtonian potential as an approximate solution to Einstein's equation for gravity.

Consider a Schwarzschild geometry. Then you can write the equations of motion $$ r^2 \dot\phi = h = \mathrm {const}$$ $$ {\dot r}^2 = {2\mu\over r} - \bigg(1-{2m\over r}\bigg){h^2\over r^2} $$ and differentiating wrt $r$ would give you a vector field for a fictitious force, but this is strictly coordinate dependent and treats Schwarzschild as though it were flat when it isn't. I doubt whether something like this would work in a general solution.

Charles Francis
  • 11,871