# Incandescence

## Deriving Part of the Kerr Geometry

In Chapter 20 of Incandescence, the Splinterites derive a spacetime geometry that is symmetrical under rotations around a single axis. This is a much harder feat than the case of a geometry that has spherical symmetry. In our own history, the spherically symmetrical case was solved by Schwarzschild in 1915, but the Kerr spacetime, with axial symmetry, was not found until 1963! The Kerr solution describes a black hole formed by the collapse of a rotating star, and the imprint of the star’s angular momentum can be seen in the geometry surrounding the hole.

The spherically symmetrical problem is essentially one-dimensional; everything depends only upon r, the radial coordinate, and so it boils down to solving equations for functions of one variable, known as ordinary differential equations. In the axially symmetrical version, the geometry depends not only upon r, but also upon the “latitude” of each point, which measures its relationship to the axis of symmetry. So we’re forced to solve equations for functions of two variables: partial differential equations. These are not so simple to deal with.

In this web page, we’ll circumvent that difficulty by confining our attention to the equatorial plane of the geometry, a plane of symmetry that lies perpendicular to the axis of symmetry. That way, we can deal with functions of a single variable, r. The results we derive will still be very useful, because the Splinter’s orbit around the Hub lies in (or very close to) the equatorial plane.

This page will refer frequently to our derivation of the Schwarzschild geometry, and will continue to use some slightly non-standard notation that was introduced there, to make the treatment friendlier to readers with no background in general relativity.

Readers who would like to see a derivation of the complete Kerr solution from first principles can find one in The Mathematical Theory of Black Holes, by S. Chandrasekhar. Discussions of the solution’s properties can be found in Gravitation by Misner, Thorne and Wheeler (Chapter 33) and General Relativity by Robert M. Wald (Chapter 12).

### The Coordinate System

To set up our problem, we will start by describing the coordinates on spacetime that we’ll be using. We will name these t, r, θ and φ — the time coordinate, the radial coordinate, the latitudinal coordinate and the longitudinal coordinate — but there will be some subtle differences from the same coordinates as used in the Schwarzschild geometry.

We are assuming that our geometry has three basic kinds of symmetry:

• we can slide everything forwards or backwards in time
• we can rotate everything around an axis in space, and
• we can reflect everything in a plane perpendicular to that axis (the equatorial plane).

We’re assuming that all these symmetries commute: for example, the order in which we combine a time-translation and a rotation around the axis has no effect on the net result. (The fact that two operations are symmetries doesn’t force them to commute: all rotations about the central point in the Schwarzschild geometry are symmetries, but rotations around different axes don’t commute.) Because the first two symmetries commute, it’s possible to find coordinates on spacetime, t and φ, such that increasing the t coordinate of every spacetime event by the same amount gives us a time-translation symmetry, and increasing the φ coordinate of every spacetime event by the same amount gives us a rotational symmetry.

The diagram on the left is a spacetime diagram for the equatorial plane, showing two dimensions of space and one of time, with the dimension of space away from the plane suppressed. Circles of varying φ — which we slide along to get a rotational symmetry — appear in grey, while lines of varying t — which we slide along to get a time-translation symmetry — appear in red.

In the Schwarzschild case, we defined the radial coordinate r such that the surfaces of constant r were just like spheres in Euclidean space with radius r. Without spherical symmetry we can’t do that, and though we could demand that circles of constant r in the equatorial plane have the same circumference as Euclidean circles of radius r, there turns out to be a choice for r that makes life simpler than that: we will require the tidal squeezing measured in the east direction on the equatorial plane to be equal to precisely M/r3, as it is in the Schwarzschild case. Note that we aren’t defining r away from the equatorial plane. Nor will we define the latitude, θ, apart from saying that it is zero on the equatorial plane, and that reversing the sign of θ leaves the whole geometry unchanged.

With our coordinates described, we can now specify four perpendicular directions in spacetime. The first, time, is the direction in which only the t coordinate is changing, and describes time according to a stationary observer. As in the Schwarzschild case, the t coordinate itself doesn’t record the passage of time for such an observer; compared to local clocks, t runs at a different rate in different places, dictated by the need for a uniform increase in t to be a symmetry of the whole geometry.

Similarly, the direction out is the direction in which only r is changing, but r doesn’t measure distance in that direction; r is defined so that tidal squeezing in the east direction is M/r3.

Now, east itself carries us around the axis of symmetry, and is perpendicular to both out and time. (In the diagram it might not be obvious that east is perpendicular to time, but perpendicular time and space directions on a spacetime diagram are those that slant at the same angle from the vertical and the horizontal respectively.) Unlike the other two cases, we can’t say that only φ changes as you head east; on the diagram we have labelled the direction in which only φ changes with the symbol ∂φ. We demand that east is perpendicular to time, but the spacetime symmetry that defines φ will, in general, describe a motion that is not perpendicular to the symmetry that defines t! So we’re forced to accept that, in general, heading east will change your t coordinate as well as your φ coordinate.

Finally, north is defined to be perpendicular to the other three, pointing out of the equatorial plane.

### The Connection on the Equatorial Plane

We can write a template for the connection on the equatorial plane as follows:

 changetimetime = G(r) out (1a) changetimeout = G(r) time + L(r) east (1b) changetimeeast = – L(r) out (1c) changeouttime = L(r) east (1d) changeouteast = L(r) time (1e) changenorthout = K(r) north (1f) changenorthnorth = – K(r) out (1g) changeeasttime = – L(r) out (1h) changeeastout = H(r) east – L(r) time (1i) changeeasteast = – H(r) out (1j)

If we compare this with the spherically symmetrical version, there are two clear differences. The first is that where we once had a single function, H(r), describing the way our frame turned as we went east or north, we can no longer assume these directions behave in the same way, so there are now two functions, H(r) and K(r). The second is the appearance of the function L(r), which measures how much someone who fixes their frame of reference to east and out will feel themself to be rotating, even though they aren’t moving, and they would see the distant stars as fixed! This shows up initially in (1b) and (1c), where with the passage of time out and east are judged to be rotating, but the fact that the frame is rotating then wends its way into two other pairs of equations: in (1d) and (1e), we see that east picks up a time component as you head out, just as in any rotating frame moving away from the centre gives you a velocity sideways; the same thing happens with the roles of east and out swapped in (1h) and (1i).

To be clear, we should note that though G(r) and H(r) play similar roles here to those they played in the spherically symmetrical spacetime, we certainly can’t take it for granted that the actual functions are identical to the ones we found in that case.

In addition to G(r), H(r), K(r) and L(r), we need to introduce four other functions, T(r), R(r), F1(r) and F2(r), that spell out the relationship between our directions in spacetime and the coordinate system:

 time = T(r) ∂t (2a) out = R(r) ∂r (2b) east = F1(r) ∂t + F2(r) ∂φ (2c) ∂t = (1 / T(r)) time (2d) ∂r = (1 / R(r)) out (2e) ∂φ = – (F1(r) / (F2(r) T(r)) time + (1 / F2(r)) east (2f)

Here the notation “∂t” means the vector we get by holding all coordinates except t fixed, and moving one unit in the t direction; similarly for ∂r and ∂φ. So the relationship between time and the passage of the coordinate t is recorded in the function T(r); the vector time involves 1 second (or whatever our time units are) passing, which equates to t changing by T(r). Similarly, out involves moving one light-second in the radial direction, while the coordinate r changes by R(r). And in the case of east, as we’ve already mentioned, we have to accept that both t and φ will change, with the amounts given by F1(r) and F2(r) respectively. The last three equations here follow from the first three, and just give the inverse relationships: expressing the coordinate vectors in terms of our spacetime directions.

This proliferation of unknown functions is a bit daunting, but we can start to rein things in by calculating some commutators. In the previous page, we discussed the formula for commutators in terms of change relative to parallel transport:

 [a, b] = changeab – changeba

From this, we can use our connection, (1), to find:

 [out, time] = – G(r) time (3a) [east, time] = 0 (3b) [east, out] = – 2L(r) time + H(r) east (3c)

But we can also use the descriptions of our direction vectors in terms of coordinate vectors to calculate their commutators. Two pure coordinate vectors always commute: if you travel one degree of latitude east and then one degree of longitude north, the effect is exactly the same as doing those two things in the reverse order. However, if we have two vectors that are not pure coordinate vectors, but are expressed in terms of coordinate vectors, we can find their commutator by seeing how an arbitrary function on spacetime changes as we move in ways corresponding to each of the vectors, and then subtracting the result that’s obtained when we swap the order in which the vectors are applied. By capturing the way an arbitrary function changes under these operations, what we get back is another vector, which is the commutator. (This might seem a little baffling at first, but the idea that vectors are really derivatives is very powerful and useful. This is discussed a bit more in my Foundations article on general relativity, as well as in any modern textbook on the subject.)

To calculate this way, we simply treat the “∂” symbol in our coordinate vectors as a derivative; for example, ∂r means “take the derivative with respect to the r coordinate”. The [a, b] notation of the commutator means “compute the derivative due to vector a following that of vector b, minus the derivative due to vector b following that of vector a”. So we can find the commutator of out and time as follows:

 [out, time] = [R(r) ∂r, T(r) ∂t] [out, time] f(r,t) = R(r) ∂r( T(r) ∂tf(r,t) ) – T(r) ∂t( R(r) ∂rf(r,t) ) = R(r) T'(r) ∂tf(r,t) + R(r) T(r) ∂r,tf(r,t) – T(r) R(r) ∂t,rf(r,t) = R(r) T'(r) ∂tf(r,t) [out, time] = R(r) T'(r) ∂t = ( R(r) T'(r) / T(r) ) time

Basically we’re just using the product rule for derivatives from elementary calculus, and the fact that partial derivatives commute so all the second-order derivatives cancel out in the end.

Equating this result to our previous one from (3a), and doing the same thing for [east, out], we arrive at these equations:

 G(r) = – R(r) T'(r) / T(r) (4a) L(r) = ( H(r) F1(r) + R(r) F1'(r) ) / ( 2 T(r) ) (4b) H(r) = – R(r) F2'(r) / F2(r) (4c)

Rather than substituting these formulas into the connection immediately, which would complicate all the algebra to follow enormously, we’ll just hold them in reserve for when we really need them.

### Imposing Zak’s Principle

In the “Geodesic Squeeze” section of our derivation of the spherically symmetrical geometry, we came up with a formula for the relative — or “tidal” — acceleration between geodesics. We did this on the surface of the Earth, for lines of longitude, but abstracting away from the specific directions we used there, the formula was:

 tidal(a,b) = changea(changeba) – changeb(changeaa) + change[b, a]a (5)

Here the geodesics point in direction a and are separated by direction b, and the tidal acceleration is itself a vector. Note that if a and b are the same, this formula will always give zero.

We can describe Zak’s principle, which is equivalent to the vacuum Einstein’s equation, as requiring that for every direction a in our set of four perpendicular directions:

 SUM over b of (the b component of tidal(a,b)) = 0 (6)

By “SUM over b” we mean add up the four expressions we get by letting b be each one of our four perpendicular directions; we’re used to talking about adding tidal accelerations for three directions, and of course every sum will have a term where b is the same as a, so it will really only have three terms. So, we have four equations (for the different choices of a) that are formed by stating that the sum of three tidal accelerations will be zero. [In fact, Einstein’s equation generalises all of this a little more, to give ten rather than four equations, but we will have nothing to gain from including those other equations.]

Now, because we have chosen to make life simpler by confining everything to the equatorial plane, we also pay a penalty: we haven’t defined the connection off that plane, so we can’t use (1) to get all the terms in the four equations in (6). We could supplement our connection on the plane with some more unknown functions describing the derivatives of everything as we head north, off the plane, but it will be easier if we just treat the tidal accelerations we can’t obtain from (1) as the unknowns, and then make use of the equations in (6) to solve for them.

Taking that approach, the four equations we obtain are:

 aout + anorth + aeast = 0 (7a) – aout + bnorth + beast = 0 (7b) – anorth + bnorth + cnorth = 0 (7c) – aeast + beast + cnorth = 0 (7d)

where we have defined the abbreviations:

 aout = out component of tidal(time, out) = – L(r)2 – G(r)2 – R(r) G'(r) (8a) anorth = north component of tidal(time, north) (8b) aeast = east component of tidal(time, east) = – L(r)2 – G(r) H(r) (8c) bnorth = north component of tidal(out, north) (8d) beast = east component of tidal(out, east) = – 3L(r)2 + H(r)2 + R(r) H'(r) (8e) cnorth = north component of tidal(east, north) (8f)

In comparing (6) and (7), note that we’ve exploited the symmetries implicit in the definition of the tidal acceleration (including the symmetries of the connection discussed here) to reduce the number of distinct terms from twelve down to six. We have already computed three of these terms directly from the connection — aout, aeast and beast — but we will solve for beast along with the unknowns anorth, bnorth, and cnorth, using the four equations of (7):

 anorth = – aeast – aout = 2L(r)2 + G(r)2 + G(r) H(r) + R(r) G'(r) (9a) bnorth = – aeast = L(r)2 + G(r) H(r) (9b) cnorth = – aout = L(r)2 + G(r)2 + R(r) G'(r) (9c) beast = aeast + aout = – 2L(r)2 – G(r)2 – G(r) H(r) – R(r) G'(r) (9d)

Subtracting (8e) from (9d) gives us:

 L(r)2 – G(r)2 – H(r)2 – G(r) H(r) – R(r) (G'(r) + H'(r)) = 0 (10)

When we described our coordinate system, we declared that the definition of the r coordinate would be that the tidal squeezing in the east direction would equal M/r3; in other words, aeast = –M/r3. Along with (8c), this gives us:

 L(r)2 = M / r3 – G(r) H(r) (11)

which lets us simplify (10) to:

 M / r3 – (G(r) + H(r))2 – R(r) (G'(r) + H'(r)) = 0 (12)

Unfortunately, we can’t proceed any further without making some more assumptions, since even if we treat G(r) + H(r) as a single function, we’re left with a single equation for two functions: G(r) + H(r) and R(r).

The two assumptions we will make are that two quantities in the Kerr geometry are, on the equatorial plane, exactly the same as functions of r as they are in the Schwarzschild geometry. Now, since we can actually look up the full Kerr solution in a textbook, we can be sure that we’re not making false assumptions, but whether they’re really justified a priori is another question.

The first assumption will be that T(r) takes its Schwarzschild value:

 T(r) = 1 / √(1 – 2M / r) (13)

Physically, what this means is that the gravitational time dilation for a stationary observer on the equatorial plane is completely unaffected by the spin of the central body that collapsed to form the black hole. This is not saying that we’re neglecting the energy content of rotational motion; that energy must be included in M. What we are saying is that for equal values of M and r (and where equal values of r means equal values of eastwards tidal stretching, M/r3), there is equal time dilation, regardless of how much or how little of the mass-energy M happens to come from the central body’s spin.

This might seem a little more intuitively reasonable if we recall that the time dimension is always associated with energy. The energy of a particle is proportional to the component of the tangent to its world line in the time direction, and if a particle falls in towards the black hole, the ratio between its energy as measured by two different stationary observers at r1 and r2 will simply be the ratio T(r1) / T(r2), regardless of any sideways component to its motion. In that sense, measurements confined to the time dimension are sensitive only to the particle’s energy, and are blind to its angular momentum. So although we have no rigorous proof (short of consulting the full Kerr solution) that T(r) itself depends only on the mass-energy of the hole, it does seem like a reasonable guess that one might make, and then proceed to test for consistency.

The second assumption is a bit more subtle: we will assume that G(r)2L(r)2 also takes its Schwarzschild value. Since L(r) is zero for the Schwarzschild solution, what this gives us is:

 G(r)2 = G(r)Schwarz2 + L(r)2 = M2 / [ r3 (r – 2M) ] + L(r)2 (14)

What is G(r)2L(r)2, physically? We can see from (1b) that it is minus the squared length of the spacetime vector that is the rate of change with time of out for a stationary observer. In other words, it measures the combined effect of the fact that a stationary observer is accelerating, which is quantified by G(r), and the fact that their choice of frame is rotating, which is quantified by L(r). Our assumption then amounts to saying that, although these individual effects will be altered by the spin of the black hole, the size of the combined effect will depend only on its mass.

[A note for hardcore relativists: if we wanted to, we could replace either (but not both) of our assumptions with the assumption that the Kerr spacetime possesses a certain abstract symmetry that is classified as Petrov type D.]

Why do we need these supplementary assumptions at all? The reason we have too few equations without them is that we’re only making use of Einstein’s equation on the equatorial plane itself, whereas in reality it holds everywhere — and the fact that it holds in a small region close to the equatorial plane is needed to pin down the geometry there. If we wished to get rid of our assumptions, we could take derivatives of Einstein’s equation in the north direction, which would give us the additional information we need, at the cost of substantially more difficult calculations.

Equation (13) is of most use to us if we combine it with equation (4a), to give:

 G(r) = M R(r) / [r (r – 2M)] (15)

Making use of (11) and (14) we then have:

 H(r) = (r–M) / [r2 R(r)] – M R(r) / [r (r – 2M)] (16) G(r) + H(r) = (r–M) / [r2 R(r)]

Putting G(r) + H(r) in terms of R(r) into (12) yields the differential equation in R(r):

 r R(r)2 + r2 R(r) R'(r) + r – M = 0 (17)

which is solved by:

 R(r)2 = 1 – 2M / r + C / r2 (18)

for some constant C. Then via (14) and (15) we obtain:

 L(r)2 = M2 C / [r4 (r – 2M)2] (19)

Clearly C is related to the angular momentum of the central body, and since L(r) will always be real-valued but might be positive or negative, it makes sense to introduce a new parameter, a, with C=a2. We then have:

 R(r) = √(1 – 2M / r + a2 / r2) (20a) L(r) = M a / [r2 (r – 2M)] (20b) G(r) = M √[r (r – 2M) + a2] / [r2 (r – 2M)] (20c) H(r) = ( r (r – 2M)2 – M a2 ) / ( r2 (r – 2M) √[r (r – 2M) + a2] ) (20d)

We can substitute these functions into (8a) and (9a) to obtain, along with the value for aeast that comes to us by definition of r, the full set of tidal accelerations for a stationary observer:

 aout = 2M / r3 + 3 M a2 / [r4 (r – 2M)] (21a) anorth = – M / r3 – 3 M a2 / [r4 (r – 2M)] (21b) aeast = – M / r3 (21c)

Of course, a frame locked to the out and east directions experiences a spin of L(r), so in addition to these purely tidal accelerations there is a centrifugal acceleration of L(r)2 in the out and east directions. We also get, from (9bd):

 bnorth = – aeast = M / r3 (21d) cnorth = – aout = – 2M / r3 – 3 M a2 / [r4 (r – 2M)] (21e) beast = aeast + aout = – anorth = M / r3 + 3 M a2 / [r4 (r – 2M)] (21f)

### Circular Orbits

We now consider the motion of a point on the Splinter as it orbits the Hub. If the velocity of the point is measured by a stationary observer to be v(r), then the 4-velocity of the point — the unit-length tangent to its world line — will be:

 u(r) = ( v(r) east + time ) / √(1 – v(r)2) (22)

Using the connection to compute the acceleration for this motion, we find:

 acc(r) = changeu(r)u(r) = 1 / (1 – v(r)2) [-H(r) v(r)2 – 2 L(r) v(r) + G(r)] out (23)

For the centre of the Splinter at r=r0 the acceleration will be zero, and we can solve for v(r0) by solving the quadratic acc(r0)=0. There will be two distinct roots of the quadratic, with orbits in different directions having different speeds so long as L(r0) is not equal to zero:

 v(r0) = [ – L(r0) ± √(L(r0)2 + G(r0) H(r0)) ] / H(r0) = [ – L(r0) ± √(M / r03) ] / H(r0) (24)

Here we’ve used (8c) and (21c) to substitute for the discriminant of the quadratic.

Now, we’re also interested in the rate of spin measured on the Splinter, which is found by considering:

 changeu(r0)out = [ (G(r0) – v(r0) L(r0)) time + (L(r0) + v(r0) H(r0)) east ] / √(1 – v(r0)2) = ± √(L(r0)2 + G(r0) H(r0)) rarb = ± √(M / r03) rarb (25) where rarb = ( east + v(r0) time ) / √(1 – v(r0)2)

For a tidally-locked body in orbit, the tidal squeezing in the rarb direction will always be cancelled out by the centrifugal acceleration due to spin, so that tidal squeezing is given by minus the square of the spin:

 ararb = – M / r3 (26)

This is the same as aeast, for a stationary observer! In other words, whether you’re in a circular orbit or standing still, the tidal squeezing in the spatial direction that points around the axis of symmetry turns out to be the same.

We will find asard, the tidal stretching in the out direction as measured on the Splinter, by using (5) to evaluate tidal(u(r0), out). One wonderful thing about (5) is that, despite a and b being vector fields — vectors defined over regions of spacetime, not just at single points — all the variation in those vector fields cancels out between the terms, and the final result only depends on the values of the vectors at a single point. This means we don’t have to calculate any derivatives of v(r0); we simply treat the coefficients of east and time in (22) as fixed values, and use the linearity of the connection to expand out the result of (5).

What we get is an expression that involves two quantities we’ve already dealt with, and one that we haven’t:

 asard = [aout + 2 v(r0) (east component of tidal(out, time)) + v(r0)2 beast] / [1 – v(r0)2] = [aout + 2 v(r0) (2G(r) L(r) + R(r) L'(r)) + v(r0)2 beast] / [1 – v(r0)2] (27)

Ultimately we obtain, for the two choices of direction for v(r0):

 asard,± = M [r (2r – 3M) ± (- 2 a) √(M r) + 3 a2] / (r3 d±) = 2M / r3 + 3M [√(M r) ± (–a)]2 / (r3 d±) (28) where d± = r (r – 3M) ± 2 a √(M r)

Using Zak’s principle, we then find ashomal by the requirement that asard+ashomal+ararb=0:

 ashomal,± = – M [r2 ± (– 4 a) √(M r) + 3 a2] / (r3 d±) = – M / r3 – 3M [√(M r) ± (–a)]2 / (r3 d±) (29)

Of course the total acceleration experienced in the garm-sard direction also includes the centrifugal acceleration due to the spin of the Splinter:

 Asard,± = asard,± + M / r3 = 3M [r (r – 2M) + a2] / (r3 d±) (30)

Finally, we will determine the orbital angular velocity, as measured by clocks on board the Splinter. We first need to solve (4c) as a differential equation for the function F2(r), which tells us the relationship between east and the angular coordinate φ. There is a constant factor in the solution which we can determine by requiring that the asymptotic result as r goes to infinity, and also the result when the angular momentum parameter a=0, yield the usual ratio of r between the length of an arc and the angle it subtends. The solution is then:

 F2(r) = √[ (1 – 2M / r) / (r (r – 2M) + a2) ] (31)

The orbital angular velocity is then obtained by converting the east component of u(r0) in (22) into a coordinate vector; the ∂φ part, which tells us the rate at which observers on the Splinter consider φ to be changing, will be proportional to F2(r). It’s more convenient to state the squared result:

 ωorbit2 = F2(r)2 v(r0)2 / (1 – v(r0)2) = M / (r d±) (32)

The results we’ve found by this approach do agree with calculations based on the complete Kerr solution. Some implications of these formulas are discussed in the Kerr geometry section of the companion page on orbits and tidal accelerations.

Incandescence / Deriving Part of the Kerr Geometry / created Saturday, 5 January 2008 / revised Tuesday, 25 March 2008