Schwarzschild's solution to Einstein's field equations
During the first world war, in 1916, an educated German fighting on the
Eastern front derived a free-field solution to Einstein's 1915 field equation of
gravitation, with no electromagnetic field. This was the first solution to the
field equations, and it's the simplest. It's named after its
discoverer, Karl
Schwarzschild (1873–1916), but is more commonly referred to as a black
hole. Because we're dealing with the free-field equation with zero
electromagnetic field, the field equation reduces to: the Ricci tensor is a
scalar field multiplied by the metric, and its trace is zero.
Schwarzschild chose to look for spherically symmetric solutions for the sake
of simplicity, so we naturally deal with one time co-ordinate, c.t, one spatial
radial co-ordinate, r, and two spatial angular co-ordinates, n and m (measured
in radians). Note that it might be better to describe this as a 4-cylindrical
symmetry, since t is really an axial co-ordinate; but I'll stick with the usual
language for now, if only because we aren't assuming time-independence
(i.e. 4-axial translation invariance). We specify that the rotations forming
the spherical symmetry don't change r or t co-ordinates, that m serves as a
longitude co-ordinate – i.e. there is an axis about which rotations vary
only m – and that n serves as a latitude co-ordinate relative to the same
axis, taking value ±π/2 on the axis and zero on a plane perpendicular
to it through the centre from which r is measured.
We now so parameterize r and t as to ensure that each 2-surface of constant
r and t has area 4.π.r.r. Our metric will then only involve n and m
as
−r.r.(dn×dn +cos(n).dm×cos(n).dm)
which will be added to a simple quadratic form in dt and dr, whose
coefficients will depend only on t and r. By adjusting our parameterization of
t we can arrange for this quadratic form to be diagonal, yielding
g = A(t,r).c.dt×c.dt −C(t,r).dr×dr
−r.r.(dn×dn +cos(n).dm×cos(n).dm)
for some scalar fields A and C – which need not be positive, but
their signs in the above are chosen so that they will be so anywhere r and t
behave the way intuition expects. Reparameterizing t (in a way that doesn't
depend on r) still leaves us scope for applying a purely t-dependent positive
scaling to A, but that is our only remaining freedom. We must now compute the
covariant derivative (under which this metric is constant), its Riemann tensor
and the resulting Ricci tensor. Then we can plug the last into the Einstein
field equation and reveal the constraints on A and C that result.
Covariant Derivative
A smooth manifold (such as space-time) has
a natural derivative on scalar fields, yielding gradient (a.k.a. covector)
fields and a natural differentiation on trajectories, yielding a tangent vector
at each point on any smooth trajectory. However, it can support many different
differential operators. These can be characterized
as Leibniz operators that agree with
the natural differentiation on scalar fields. Any two of them necessarily
differ, on each rank of tensor field, by a linear map acting on that rank. We
are only interested in differential operators which regard the space-time
metric, g, as constant; and it so happens
that this suffices to completely specify a
unique differential operator. This differential operator is known as the
covariant derivative.
Thanks to Leibniz's product rule and its specified action on scalar fields,
a differential operator is entirely determined by its action on the gradient
fields of the scalar fields which are the co-ordinates of a chart; our chart's
co-ordinates are c.t, r, n and m; so let b = [c.dt,dr,dn,dm] and take p =
[s,q,u,w] be the vector fields obtained as, at each point, the dual of b. From
the constancy (and symmetry) of the metric,
we can
infer
and, since dm usually appears with a cos(n) factor, it seems worth
noting:
D(cos(n).dm)
= cos(n).D(dm) −sin(n).dn×dm
= sin(n).dm×dn −(dr×cos(n).dm +cos(n).dm×dr)/r
and it's worth noting that s·dX = ∂X/∂t/c and
q·dX = ∂X/∂r, for X as either A or C (and, later, either of
these derivatives of either).
Riemann Tensor
Next we must look at D&on;D applied to gradient fields, antisymmetrizing
over the two leading tensor terms thereby generated; that is, Riemann(D). Since
this acts as a linear map at
each point, it suffices to look at its action on our
basis:
Ricci(D) has to be parallel to the metric, g. This immediately implies that
∂C/∂t/c = s·dC (the co-efficient of the one off-diagonal term)
must be zero, i.e. C depends only on r. That, in turn, eliminates half the
terms in the coefficient of (A.c.dt×c.dt −C.dr×dr). The two
stray c.dt×c.dt and dr×dr terms have to have their co-efficients in
the same ratio as A to −C, implying q·dA/A = −q·dC/C,
whence
∂(A.C)/∂r = q·d(A.C) = q·dA.C +q·dC.A = 0
so A.C doesn't depend on r. By re-choosing our t parameterization, we
can alter A by any positive scaling that depends only on t; this doesn't affect
C, so we can make A.C constant and substitute A(r) = E/C(r) for some constant E;
furthermore, we can chose the magnitude of E (though not its sign). Thus our
solution is time-independent; this is Birkhoff's theorem –
spherical symmetry in a vacuum implies time-independence (but this would appear
to depend on the vacuum being devoid of electromagnetic activity as well as
matter). Since A and C are now simply functions of r, we no longer have to
trouble ourselves with partial differentiation; we can refer to q·dA as
A', the derivative of A, and likewise for q·dC = C' =
−A'.C/A. This, together with A.C = E, reduces the above
to
Next, for Ricci(D) to be parallel to g, the co-efficients of
(A.c.dt×c.dt −C.dr×dr) and (dn×dn
+cos(n).dm×cos(n).dm) must be in the ratio 1 to −r.r,
whence
0
= r.r.(A''/2 +2.A'/r) −(A −E +A'.r)
= r.r.A''/2 +r.A' −A +E −A'.r
= r.r.A''/2 −A +E
or r.r.A'' = 2.(A−E). Since powers of r match orders of
differentiation, it makes sense to consider polynomials; so consider A(r) =
sum(: a(i).power(i,r) ←i :) for some ({reals}: a :{integers}). Equating
coefficients of power(n,r), for given integer r, we obtain:
0 = a(0)−E for n = 0;
0 = a(1) for n = 1; otherwise
0 = a(n).(n.(n−1) −2) = a(n).(n−2).(n+1)
from which we infer a(0) = E, a(−1) and a(2) are arbitrary and
all other a(n) are zero. Thus, for some k and h, A = E.(1 +k/r +h.r.r),
yielding A'/E = 2.h.r −k/r/r, A'' = 2.E.(h+k/r/r/r) = 2.(A−E)/r/r,
as required. Our remaining co-efficients reduce to A'.r +A −E = 3.E.h.r.r
and A''/2 +A'/r = 3.E.h. We finally have
with h and k arbitrary. The magnitude of E merely defines our units of
time with respect to our units of length, so only its sign matters; for
space-time to have three space-like and one time-like component, E must be
positive (so that the dt×dt term's sign differs from the
others). Consequently, we can take E = +1. Einstein's field equation insists on
Ricci(D)/g having trace equal to a particular (dimension-specific, but non-zero
– unless dimension is two) multiple of the energy density (specifically,
the trace of the energy-momentum-stress tensor, whose spatial diagonal entries
are pressure terms); since this is zero in the vacuum solution, −3.h/2
must be zero, i.e. h = 0.
Thus Schwarzschild's solution takes E = 1 and h = 0, so that Ricci(D) is
zero; and interprets k as −2.G.m/c/c for a mass m, since the resulting
metric then matches up with the weak field limit
obtained by approximating Newtonian gravitation. Note that A/E =
1−2.G.m/c/c/r is zero at r = 2.G.m/c/c and negative when r is less than
this, causing t to become space-like and r time-like. The boundary is called
the event horizon; and 2.G.m/c/c is known as the Schwarzschild
radius for the given mass, m.
The (scalar curvature of space-time, a.k.a.) trace of Ricci(D)/g is
−3.h/2 times the dimension of space-time, i.e. −6.h. The Riemann
tensor contracted with the metric is
so the measure (which mediates integration) will be, simply,
μ = c.dt^dr^(dn.r)^(dm.cos(n).r).√E
Now I just have to work out what non-zero h would mean !
Geodesics
An object small enough to not disturb our metric (i.e. gravitational field)
significantly shall, unless interfered with by some other force, follow a
geodesic – i.e. a trajectory of extremal length, in our metric's
terms. The length of (more formally: the proper time experienced by a
body traversing) a trajectory (M:x:{scalars}) between two points on it at input
parameter values a and b, with a <b, is integral(: dp.√g(x'(p),x'(p))
←p; a≤p≤b :), with the √ taken positive when its input is
positive and positive imaginary when its input is negative; the integral is
potentially complex.
Two-body problem
I should also examine the general relativistic analogue
of the classical two-body problem. This shall have
two black holes (as above) in mutual orbit – and the solution isn't quite
periodic, as the tidal stresses cause it to decay. This may be somewhat
difficult.
Written by Eddy.