Matterwave said:
So, ...
In GR, is divT=0 sufficient to find the FULL trajectory of particles moving in curved space-time?
Thanks.
When Einstein first presented his GR, he postulated the geodesic as the equation of motion. In 1927, he noted that the equations of motion are in fact contained with in the field equations and therefore do not need to be postulated separately. In 1951, Papapetrou managed to show that the equations of motion of spinning particles (which are not geodesics) can also be derived from Einstein’s field equations.
There have been two different approaches to the problem of obtaining the equations of motion in GR. The first (due to Einstein and Infeld) starts with left hand side of the field equations and is a very complicated geometrical treatment. The other approach (the one we are interested in) starts with right hand side of the field equations and is due to Papapetrou and Fock.
OK, let us begin. The field equations G^{ab} = T^{ab} imply
\nabla_{b}T^{ab} = 0.
This can be written as
<br />
\partial_{b}\left( \sqrt{-g} T^{ab}\right) + \sqrt{-g}\Gamma^{a}{}_{bc}T^{bc} = 0. \ \ (1)<br />
We integrate this over a small 3-volume and write
<br />
\int d^{3}x \partial_{b}\left( \sqrt{-g} \ T^{ab}\right) + \int d^{3}x \sqrt{-g}\Gamma^{a}{}_{bc}T^{bc} = 0, \ \ (2)<br />
or
<br />
\frac{d}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{a0} + \int d^{3}x \sqrt{-g}\Gamma^{a}{}_{bc}T^{bc} = - F^{a}, \ \ (3)<br />
where the object F^{a} is defined by
<br />
F^{a} = \int d^{3}x \partial_{j}(\sqrt{-g} \ T^{aj}).<br />
We will identify F^{a} with external non-gravitational forces. So, for the case of gravitational forces alone, one sets F^{a} = 0; one writes F^{a} as surface integral and take a large enough volume for which T^{ab} vanishes on the surface.
Let us now expand the connection about a point X^{a} which will be taken as the coordinates of the “particle”; a function of time or other single parameter,
<br />
x^{a} = X^{a} + \delta x^{a}, \ \ (4)<br />
<br />
\Gamma^{a}{}_{bc}(x) = \Gamma^{a}{}_{bc}(X) + \delta x^{d}\partial_{d}\Gamma^{a}{}_{bc} + … \ \ (5)<br />
Let us spell out what we mean by particle. We define a single pole (non-spining) particle by the followings;
i) For some a and b
<br />
\int d^{3}x \sqrt{-g} \ T^{ab} \neq 0. \ \ \ (6)<br />
In particular; when evaluated in a locally inertial rest frame, the integral
<br />
\int d^{3}\bar{x} \sqrt{-\bar{g}} \ \bar{T}^{00} = m \ > \ 0, \ \ \ (6a)<br />
represents the rest mass of our particle.
ii) For all a, b and c
<br />
\int d^{3}x \sqrt{-g} \ \delta x^{c} \ T^{ab} = 0. \ \ \ (7)<br />
iii) The particle is sufficiently small so that the forces acting on it are large compared to the first moment of these forces on its surface,
<br />
\int d^{3}x \partial_{j}\left(\delta x^{c}\sqrt{-g} \ T^{aj}\right) = \int dS_{j} \delta x^{c}\sqrt{-g} \ T^{aj} \approx 0. \ \ \ (8)<br />
For later convenience, let us write eq(2) and eq(3) taking into account eq(5) and eq(7)
<br />
\int d^{3}x \partial_{b}\left( \sqrt{-g} \ T^{ab}\right) + \Gamma^{a}{}_{bd}(X)\int d^{3}x \sqrt{-g} \ T^{bd} = 0, \ \ (9)<br />
<br />
\frac{d}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{a0} + \Gamma^{a}{}_{bd}(X) \int d^{3}x \sqrt{-g} \ T^{bd} = - F^{a}. \ \ (10)<br />
Now multiply eq(1) by x^{c}, integrate over 3-volume and write the result as
<br />
\int d^{3}x \{ \partial_{0}(\sqrt{-g} \ x^{c}T^{a0}) + \partial_{j}(\sqrt{-g} \ x^{c}T^{aj}) + \sqrt{-g} \ x^{c}\Gamma^{a}{}_{bd}T^{bd}\} = \int d^{3}x \sqrt{-g} \ T^{ac}.<br />
Substituting eq(4) and eq(5) into the above equation gives, with the help of eq(7) and eq(8),
<br />
\frac{dX^{c}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{a0} + X^{c}\int d^{3}x \partial_{b}\left( \sqrt{-g} \ T^{ab}\right) + X^{c}\Gamma^{a}{}_{bd}\int d^{3}x \sqrt{-g} \ T^{bd}= \int d^{3}x \sqrt{-g} \ T^{ac}.<br />
Making use of eq(9), the above equation reduces to the following nice looking equation
<br />
\frac{dX^{c}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{a0} = \int d^{3}x \sqrt{-g} \ T^{ac}. \ \ (11)<br />
For a = 0, and since T^{ab}=T^{ba}, we find
<br />
\int d^{3}x\sqrt{-g} \ T^{a0} = \frac{dX^{a}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00}. \ \ (12)<br />
Inserting eq(12) back into eq(11), we find
<br />
\int d^{3}x \sqrt{-g} \ T^{ac}= \frac{dX^{a}}{dx^{0}}\frac{dX^{c}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00}. \ \ (13)<br />
Substituting eq(12) and eq(13) back into eq(10), we obtain
<br />
\frac{d}{dx^{0}}\left( \frac{dX^{a}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00}\right) + \Gamma^{a}{}_{bc}\frac{dX^{b}}{dx^{0}}\frac{dX^{c}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00} = - F^{a}. \ \ (14)<br />
Using
\frac{d}{dx^{0}} = \frac{ds}{dx^{0}}\frac{d}{ds},
we rewrite eq(14) as
<br />
\frac{d}{ds}\left( \frac{dX^{a}}{ds}\frac{ds}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00}\right) + \Gamma^{a}{}_{bc}(X)\frac{dX^{a}}{ds}\frac{dX^{c}}{ds}\frac{ds}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00} = - F^{a}\frac{dx^{0}}{ds}. \ \ (15)<br />
In this equation we recognize the rest mass of our particle,
<br />
m = \frac{ds}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00} = \int d^{3}\bar{x} \sqrt{-\bar{g}} \ \bar{T}^{00},<br />
Thus, eq(15) becomes
<br />
\frac{d}{ds}\left( m \frac{dX^{a}}{ds} \right) + m \Gamma^{a}{}_{bc}(X) \frac{dX^{a}}{ds}\frac{dX^{c}}{ds} = - F^{a}\frac{dx^{0}}{ds}. \ \ (16)<br />
Finally, assuming that the rest mass does not change with time, we find the equation of motion a pole (non-spining) particle subjected to gravitational and non-gravitational forces,
<br />
\frac{d^{2}X^{a}}{d s^{2}} + \Gamma^{a}{}_{bc}\frac{dX^{b}}{ds}\frac{dX^{c}}{ds} = \frac{f^{a}}{mc^{2}}<br />
where
f^{a} = - c^{2}\frac{dx^{0}}{ds}F^{a}
This method can be extended by relaxing the assumption (ii). Doing this leads to equations of motion for spinning particles.
Regards
Sam
.