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Consider a point mass $A$ with mass $m$ in empty space. The point mass $A$ does not have a velocity and does not rotate. Since gravity is symmetric for nonmoving objects, the spacetime curvature around $A$ is also symmetric.

So at a distance $d$ from the point mass $A$ how strong is the curvature $C$ ?

$$ C = f(d,m) $$ $$f = ???$$

mick
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  • (1) how exactly are you defining curvature? e.g. are you looking for the gravitational field strength? the Riemann tensor? the Ricci scalar? (2) Could you clarify what aspect of this question you were unable to figure out by reading the Wikipedia page on the Schwarzschild metric? – David Z Jan 07 '14 at 22:32
  • I do not understand tensors or metrics very well. I did not get the Scharzschild metric. I hear relativity dictates that mass creates curvature in spacetime. Im unaware of different forms of curvature ? I think I mean gravitational field strength ? Im no expert, but I want to know the local acceleration due too gravity at a distance $d$ from $A$. – mick Jan 07 '14 at 22:36
  • @David Z I think he means Ricci scalar. – Anixx Jan 08 '14 at 11:28

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It sounds as if you just want the acceleration given by the non-relativistic equation from Newton's law:

$$ a = \frac{GM}{r^2} $$

where $M$ is the mass of the object generating the gravitational field (strictly speaking this equation only applies when the mass of the accelerating object is much less than $M$).

For the GR version of this have a look at twistor59's answer to What is the weight equation through general relativity?:

$$ a = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}} $$

This is the simplest treatment of the problem I've seen, but even so I suspect you'll have problems with this unless your maths is reasonably advanced.

Community
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John Rennie
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  • ALthough Im only 14 and have yet still to learn much math, I have 2000+ on MSE. Is that advanced enough ? +1 from me. – mick Jan 08 '14 at 12:09
  • In special relativity there is a velocity addition formula. Is adding acellerations in GR equivalent to taking the integral over the velocity addition formula ? The reason I ask is because accelerations in normal newtonian use leads to faster than light speeds after long enough times. – mick Jan 08 '14 at 12:15
  • @mick: I could not have done this stuff at 14. In fact I didn't learn enough maths to understand it until I was 21. If you want to have a go I would start with the book by Schutz. If you manage to work all the way through the book you'll have a better understand of GR than me :-) – John Rennie Jan 08 '14 at 12:19
  • I think I will ask a followup question ... – mick Jan 08 '14 at 12:22