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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
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Joshua Pasa
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    This is a duplicate or near-duplicate of this I think: Can general relativity be completely described as a field in a flat space? –  Apr 05 '20 at 13:42
  • @tfb, that seems to be a different question, asking about quantum gravity. This one is only asking about the approximation to Newtonian gravity. – Charles Francis Apr 05 '20 at 14:00
  • The force felt by a stationary observer is m·d²x/dτ²·√g^tt, while in free fall there is no force at all, but you can still plot the potential energy or escape velocity or something like that to represent a field. Your equation shows the proper coordinate acceleration of a free falling body, which is coordinate dependent, since x is not a physical distance – Yukterez Apr 05 '20 at 17:40
  • @CharlesFrancis: it's not asking about quantum gravity I think: the person asking the question just wonders whether QG is needed to do this, which I don't think it is. It's obviously a topic of interest to people who are interested in QG however. The answer I think is also 'no, you can't, not in any useful sense'. –  Apr 05 '20 at 17:42

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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
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  • I'm not asking for an approximate solution. I'm asking for an exact vector field the metric would create – Joshua Pasa Apr 05 '20 at 14:30
  • Then the question no longer makes sense, because you can only take spacetime as flat in a local inertial frame which does not include the region surrounding a gravitating body. Otherwise, you use an approximation in which space is taken to be flat, and re-interpret the metric as a potential. – Charles Francis Apr 05 '20 at 14:50
  • Well imagine another vector field cancelling the effect of gravity completely. To that perspective space-time looks flat. However they see the effect gravity has on other objects that are not affected by this new vector field. They would perceive a vector field that gravity would create from their perspective. Using the geodesic equation is it possible to create a vector field using the geodesic equation. – Joshua Pasa Apr 05 '20 at 14:59
  • Those are just mathematical fields with no physical meaning. –  Apr 05 '20 at 15:36
  • It's abstract, but it can lead to physical results. It leaves us with a vector field with real, measurable results. – Joshua Pasa Apr 05 '20 at 15:40
  • makes no sense to me. I think you should refer to the answer linked by tfb. – Charles Francis Apr 05 '20 at 15:45
  • There is a way of doing something like it for Schwarzschild, which I have added. – Charles Francis Apr 05 '20 at 16:06
  • Is there a general 4d vector field including the time component. – Joshua Pasa Apr 05 '20 at 16:31
  • And the derivatives of the field are with respect to proper time – Joshua Pasa Apr 05 '20 at 16:32
  • I don't think so, but it is very difficult to think of and quite alien to the geometric approach one ought to take. People want to do that kind of thing in quantum gravity, (see link given by tfb) but they haven't succeeded, and my own view is that it is not the right way to go. – Charles Francis Apr 05 '20 at 16:35
  • So is there no way to solve the geodesic equation in the form that I asked? – Joshua Pasa Apr 05 '20 at 16:40