Theory of relativity/Schwarzschild metric

In Schwarzschild coordinates, the Schwarzschild solution is
 * $$ds^2 = \left(1-\frac{2GM}{rc^2}\right)dct^2 - \frac{dr^2 }{\left(1-\frac{2GM}{rc^2}\right)} - r^2 d\theta ^2 - r^2 sin^2 \theta d\phi ^2 $$

It is an exact vacuum solution to General relativity/Einstein equations, and according to the Birkoff theorem, all spherically symmetric exact vacuum solutions are equivalent to this solution, related through mere frame transformation.

Working out the solution
We are looking for a static spherically symmetric vacuum solution so it is reasonable to write down as a trial solution
 * $$ds^2 = \left(f\left(r\right)\right)dct^2 - \left(g\left(r\right)\right)dr^2 - \left(h\left(r\right)\right) r^2 \left(d\theta ^2 + sin^2 \theta d\phi ^2 \right)$$

But note that a mere r coordinate transformation can be done to drop the $$h\left(r\right)$$ absorbing it into the radial coordinate. So we will start with a trial solution of
 * $$ds^2 = \left(f\left(r\right)\right)dct^2 - \left(g\left(r\right)\right)dr^2 - r^2 d\theta ^2 - r^2 sin^2 \theta d\phi ^2 $$

The Einstein tensor $$G^{\mu \nu} $$ for this line element has a time time element of
 * $$G^{00} = \frac{r\frac{dg}{dr}+g^2 -g }{fg^2 r^2 } $$

For this to be an exact vacuum solution we must have
 * $$r\frac{dg}{dr}+g^2 -g =0 $$

The $$g$$ that is the solution to that differential equation is
 * $$g=\frac{1}{\left(1-\frac{r_{0}}{r}\right)} $$

where $$r_{0}$$ comes in as an integration constant. So thus far our trial solution is
 * $$ds^2 = \left(f\left(r\right)\right)dct^2 - \frac{dr^2}{\left(1-\frac{r_{0}}{r}\right)} - r^2 d\theta ^2 - r^2 sin^2 \theta d\phi ^2 $$

For this line element the metric tensor yields an r-r element of
 * $$G^{rr} = \frac{\left(1-\frac{r_{0}}{r}\right)\left(-r^2 \frac{df}{dr} +r_{0}r\frac{df}{dr}+r_{0}f\right)}{f r^3 } $$

For this to be an exact vacuum solution we must have
 * $$ -r^2 \frac{df}{dr} +r_{0}r\frac{df}{dr}+r_{0}f=0 $$

The $$f$$ that is the solution to that differential equation is
 * $$f=k\left(1-\frac{r_{0}}{r}\right) $$

where $$k$$ comes in an an integration constant. But $$k$$ can be absorbed by the time in a time coordinate transformation, a mere scaling of time, so doing so thus far our trial solution is
 * $$ds^2 = \left(1-\frac{r_{0}}{r}\right)dct^2 - \frac{dr^2}{\left(1-\frac{r_{0}}{r}\right)} - r^2 d\theta ^2 - r^2 sin^2 \theta d\phi ^2 $$

Now it turns out that for this line element all of the Einstein tensor elements are zero, so we have an exact vacuum solution. Finding the Riemann tensor though several elements are not zero and so what we have has spacetime curvature and is thus not a mere frame transformation of the metric of special relativity.

For the moment lets define a constant $$M$$ according to
 * $$r_{0}=\frac{2GM}{c^2} $$

The exact vacuum Schwarzschild solution is then written
 * $$ds^2 = \left(1-\frac{2GM}{rc^2}\right)dct^2 - \frac{dr^2 }{\left(1-\frac{2GM}{rc^2}\right)} - r^2 d\theta ^2 - r^2 sin^2 \theta d\phi ^2 $$

An exact calculation of the radial case of Geodesic motion of a test mass yields
 * $$\frac{\gamma^2 - 1}{2}c^2 =\frac{1}{2}\left(\frac{dr}{d\tau}\right)^2 -\frac{GM}{r}$$

where $$\gamma $$ is a constant of the motion called the energy parameter, the conserved energy per $$mc^2$$ of the test mass, and $$\tau $$ is the proper time for the test mass. Differentiating this equation with respect to proper time and simplifying results in
 * $$\frac{d^2 r}{d\tau ^2} = -\frac{GM}{r^2 }$$

So we see that this mass $$M$$ which was essentially an integration constant for this spacetime geometry is what we think of as the active gravitational mass.

Geodesic Motion
The exact equations of geodesic motion for a test mass in Schwarzschild coordinates with the test mass's proper time being $$\tau $$ are
 * $$\frac{dt}{d\tau} =\frac{\gamma}{\left(1-\frac{2GM}{rc^2 }\right)}$$

where $$\gamma$$ is a constant of the motion called the energy parameter, the energy per $$mc^2 $$ for the test mass.
 * $$\frac{d\phi}{d\tau} =\frac{l_z}{r^2 sin^2 \theta}$$

where $$l_z $$ is the conserved angular momentum per mass $$m$$ for the test mass.
 * $$\frac{d}{d\tau}\left(r^2 \frac{d\theta}{d\tau}\right) = r^2 sin\theta cos\theta \left(\frac{d\phi}{d\tau}\right)^2$$

and finally
 * $$\frac{\gamma^2 - 1}{2}c^2 =\frac{1}{2}\left(\frac{dr}{d\tau}\right)^2 -\frac{GM}{r}+\frac{1}{2}\left(\frac{l_z ^2}{r^2 sin^2 \theta} +r^2 \left(\frac{d\theta}{d\tau}\right)^2 \right)\left(1-\frac{2GM}{rc^2 }\right)$$

For radial motion this reduces to
 * $$\frac{dt}{d\tau} =\frac{\gamma}{\left(1-\frac{2GM}{rc^2 }\right)}$$


 * $$\frac{\gamma^2 - 1}{2}c^2 =\frac{1}{2}\left(\frac{dr}{d\tau}\right)^2 -\frac{GM}{r}$$

The final equation of motion prior to the radial motion case, looks much like a Newtonian gravitation conservation of energy equation with the exception of the $$\left(1-\frac{2GM}{rc^2 }\right)$$ term multiplying the angular part and the time derivatives being with respect to the test mass time. This factor multiplying the angular part perturbs the motion of nearly elliptical orbits so that they process. Orienting the coordinates so that the motion of an orbit is equatorial and defining $$u=\frac{1}{r}$$ the equations of motion yield
 * $$\frac{d^2 u}{d\phi ^2}-3\frac{GM}{c^2}u^2 +u-\frac{GM}{l_z ^2 }=0$$

which in a weak field the solution can be approximated by
 * $$u=\frac{GM}{l_z ^2 }\left(1+ecos\left(\left(1-3\frac{\left(GM\right)^2}{l_z ^2 c^2 }\right)\phi \right) \right)$$

where $$e$$ is the eccentricity So perihelion occurs at
 * $$\left(1-3\frac{\left(GM\right)^2}{l_z ^2 c^2 }\right)\phi _p = 2n\pi$$

which for the weak field can be approximated by
 * $$\phi _p = 2n\pi +\frac{6n\pi \left(GM\right)^2}{l_z ^2 c^2 }$$

and given an orbital period of T this implies that after a time t the orbit will have processed by an amount given by
 * $$\Delta \phi _p =\frac{6\pi \left(GM\right)^2}{l_z ^2 c^2 }\frac{t}{T}=\frac{6\pi GM}{a\left(1-e^2 \right)c^2}\frac{t}{T}$$

where $$a$$ is the semi-major axis. This effect was first observed in nature for Mercury which processes 575" per 100 earth years, 534" of which are accounted for by the gravitational effects from other planets.

Geodesics For Light
Writing the geodesic motion in terms of Schwarzschild time t instead of time for the test mass $$\tau$$ and taking the limit as $$\gamma$$ and $$l_z $$ go to $$\infty$$ yields the motion for a massless test particle such as a photon. Orienting the coordinates so that the orbital plane of the photon is equatorial and defining $$u=\frac{1}{r}$$ results in
 * $$\frac{d^2 u}{d\phi ^2}-3\frac{GM}{c^2}u^2 +u=0$$

and
 * $$\frac{dr}{dt}=\pm \left(1-\frac{2GM}{rc^2 }\right)\sqrt{1-\left(\frac{r_m}{r}\right)^2 \frac{\left(1-\frac{2GM}{rc^2 }\right)}{\left(1-\frac{2GM}{r_m c^2}\right)}}$$

where $$r_m$$ is the distance of closest approach for a deflected photon. For small deflection of light the first of these two yields a deflection angle of
 * $$\Delta \phi = \frac{4GM}{r_m c^2}$$

And for circular orbit of a photon yields
 * $$r_p = \frac{3GM}{c^2}$$

which is a location referred to as the photon sphere. Integrating the second with appropriate weak field approximation and writting the result in terms of a lab's time instead of remote observer Schwarzschild time for a photon following geodesics between earth and another planet at superior conjunction yields the round trip Shapiro delay equation of
 * $$\Delta t_l =4\frac{GM}{c^3}\left(-\frac{1+\delta _1}{2}+\frac{1+\delta _2}{2}ln\left(\frac{4ab}{r_m ^2}\right) \right)$$

where $$a$$ and $$b$$ are the orbital distance from sun of the planets and the $$\delta$$s are curve fit parameters allowing for perturbances such as the gravitational time dilation from the earth's mass itself.

Untransversable Wormhole Structure
Consider a spacelike hypersurface described by the Schwarzschild solution as a constant Schwarzschild time slice given cutting through for example at $$\theta=\frac{\pi}{2}$$ by
 * $$ds^2 = - \frac{dr^2 }{\left(1-\frac{r_{0}}{r}\right)} - r^2 d\phi ^2 $$

In order to get a conceptual image of the way this spacelike hypersurface is curved lets write it as an imbedding in a higher dimensional hyperspace with an extradimensional spatial axis of w.
 * $$ds^2 = - \frac{dr^2 }{\left(1-\frac{r_{0}}{r}\right)} - r^2 d\phi ^2 = - dw^2 - dr^2 - r^2 d\phi ^2 $$

This gives us the differential equation for the hypersurface as
 * $$dw=\pm \frac{dr}{\sqrt{\frac{r}{r_{0}}-1}}$$

The integration of which gives us the equation for the hypersurface as
 * $$w^2 = 4r_{0}\left(r-r_{0} \right)$$

Drawing this swept around for all $$\phi $$ the surface looks like

Here we see the two external regions and a wormhole cross section example is drawn on, but one can see now that there is no way to transverse from one side to the other. In Kruskal-Szekeres coordinates the coordinate speed of radial moving light is everywhere c, the solution expressed as
 * $$ds^2 = 4\left(\frac{r_{0}}{r}\right)\left(\frac{r_{0}}{R}\right)^2 e^{-\frac{r}{r_{0}}}\left(dct'^2 -dr'^2 \right)-r^2 \left(d\theta ^2 +sin^2 \theta d\phi ^2 \right)$$

and light like paths described by ds=0, so on the diagram one can not travel a path more than 45 degrees from the vertical. Starting from inside one external region, one can't get to the other external region without dipping past 45 degrees and thus traveling faster than light with respect to nearby observers.