# How do I calculate the (apparent) gravitational pull with General Relativity?

Assume a static metric with (known) components $g_{\mu\nu}$. I'd like to know what is the gravitational pull $g$ of a test particle placed on an arbitrary point $X$.

The gravitational pull being defined as the acceleration the particle suffers as measured by an observer sitting in a reference frame fixed at the origin. What are (theoretically speaking) the steps one needs to take to find this acceleration? (no need to actually calculate for the general case, just list the general steps)

Furthermore, what is the result for the simple case of the Schwarzschild metric?
In other words, what is the gravitational pull of a star or black hole? Will it matter if the test particle is moving or is static?

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Possible duplicates: physics.stackexchange.com/q/47783/2451 and links therein. – Qmechanic Feb 15 at 18:15

This problem is somewhat ill-formed. It's difficult to even define what 'static at the origin' means--static relative to what? What I will do is show the steps one needs to take to derive the motion of a particle, however. The following will be more constructive than explanatory, as explaining all of this thoroughly would be close to a whole chapter in a relativity/differential geometry textbook.

Given the metric tensor $g_{ab}$, one can define the Christoffel symbols $\Gamma_{ab}{}^{c}$ by the equation$^{1}$ $\Gamma_{ab}{}^{c} = \frac{1}{2}g^{cd}\left(\partial_{a}g_{bd}+\partial_{b}g_{ad}-\partial_{d}g_{ab}\right)$. While the Christoffel symbols serve an array of purposes in relativity, the simplest notion that one can tie to them is that they define a notion of parallel transport in the spacetime--a vector $v^{a}$ is parallel transported along a curve $x^{a}$ if $x^{a} \nabla_{a}v^{b} \equiv x^{a}\partial_{a}v^{b} + x^{a}v^{c}\Gamma_{ac}{}^{b} =0$.

Now, we can talk about paths whose tangent vectors are parallel translated relative to themselves. These paths are known as geodesics. If we parameterize these curves with respect to the arc length subtended by the curve$^{2}$(which we interpret as the clock of the observer following the geodesic), then the curves satisfy the equation $\ddot x^{a} + \Gamma_{bc}{}^{a}\dot x^{b} \dot x^{c} = 0$. If we choose the $x^{a}$ as a locally Cartesian orthonormal set of coordinates, then we can interpret $\ddot x^{a}$ as the acceleration of the $x^{a}$ position along the curve, and then interpret the other term as the acceleration of the curve. Practically, this is a difficult game to interpret in general relativity, since time is one of the coordinates (making acceleration a subtle thing to define), and because our coordinates are completely arbitrary and need not even be (and in fact, usually are not) orthonormal. Nonetheless, this interpretation can be useful for a few special cases--the Schwarzschild black hole metric being one of them.

Here, it's actually easiest to look at the conserved energy of a particle on the geodesic first, and then to define a force from this. Here, in units where $G=c=1$, we have $E^{2} -1= \dot r ^{2} - \frac{2M}{r} + \left(\frac{L}{r}\right)^{2} - \frac{2ML^{2}}{r^{3}}$, where E is the energy per unit mass of the test particle, $M$ is the mass of the central gravitating object, and $L$ is the angular momentum of the test particle. Taking a time derivative of this, and dividing by the common factor $2\dot r$, we get:

$$0=\ddot r + \frac{M}{r^{2}}-\frac{L^{2}}{r^{3}} + \frac{3ML^{2}}{r^{4}}$$

The first three terms should be familiar from Newtonian theory--they are precisely the Newtonian gravitational force term plus the centripetal inertia of the orbit. Relative to the clock of an observer in orbit, so long as $\frac{3ML^{2}}{r^{4}}$ is small, Newtonian motion should be indistinguishable from Relativistic motion. So, the only difference in acceleration comes from the rotation of the orbit. It is precisely this term that created the anamolous acceleration of Mercury that was such a puzzle to the astronomers at the turn of the century. When Einstein observed that this term was of the right sign and magnitude to explain this anamoly was when he truly believed that he had found the correct theory of gravity.

$^{1}$ where $\partial_{a}f$ is defined to be the derivative of $f$ with respect to the a$^{th}$ coordinate: $\partial_{a}f \equiv \frac{\partial f}{\partial x^{a}}$

$^{2}$ this sounds complicated, but is in fact what we do when measuring angles on a circle in radians--how many radians I trace out tells me how much length along the circle I've traced out since I've left the origin. Same basic idea here.

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Thanks for the v very complete answer.I'll start dissecting it now, but first: what I meant as static is that the metric is not time dependant in the observer's reference frame. Not all metrics be able to satisfy that, but I was willing to make that concession. Does that make more sense? – Malabarba Jan 10 '11 at 22:43
Static is actually stronger than that. It also means that there are no $g_{ti}$ terms in the metric, where $i=x,y,z$. So, for example, Schwarzschild is static while Kerr isn't. – Vagelford Jan 11 '11 at 10:37
@Vagelford: A coordinate independend definition would be more appropriate. – NikolajK Mar 5 '13 at 14:28
@NickKidman: strictly, you can't be coordinate invariant when talking about these things. If you want to make things look coordinate invariant-y, you're going to have to define a vector $\partial_{t}$, and then define all of the normals and tangents to this vector, and then state that $\pounds_{\partial_{t}}g_{ab}=0=\partial_{t}^{a}g_{ab}P^{b}_{c}$, where $P_{b}^{c}$ is the normal projection operator. But that's as unsatisfying as just giving metric coordinates, really. – Jerry Schirmer Mar 5 '13 at 15:19
@JerrySchirmer: I think instead of saying "The condition is that the $g_{ti}$-coordinates are zero" you say "the condition is that there is a coordinate system where the $g_{ti}$ are zero" and then it's a coordinate independent statement. – NikolajK Mar 5 '13 at 15:35