The Kerr Metric Tensor and Its Functions
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We have described Kerr spacetime, which is normal spacetime deformed by the presence of a gravitating and spinning spherical body of uniform composition. That description consists of four transformation equations and a metric equation that enable observers occupying different locations around the body to translate measurements made on pairs of events back and forth. Now we want to translate our transformation equations into their equivalent metric tensor and devise the associated Christoffel symbols.
The Metric Tensor
The metric tensor is a 4x4 matrix that displays the spatio-temporal deformations of the four-dimensional analogue of the Pythagorean theorem, which is encoded in the metric equation. This takes us well beyond "the square on the hypotenuse is equal to the sum of the squares on the sides", but it still expresses the same basic idea, albeit with a weirdly deformed version of a right triangle.
Let痴 start with a restatement of our transformation equations;
in which we have defined for convenience,
Those equations apply to space and time in the vicinity of a uniform spherical body pondering mass M and spinning as a rigid body at an angular speed ofΩ. The gravitational mass gives the body a Schwarzschild radius (RS=2MG/c2) and the angular motion gives it a Kerr radius (RK=R02Ω/c, in which R0 denumerates the physical radius of the sphere). The upper-case variables represent measurements made between two events by an observer located far enough from the body M that the body痴 gravitational field has negligible effect on that observer (that is, the observer occupies a region of spacetime effectively indistinguishable from flat Minkowski spacetime). The lower-case variables in Equations 1 represent measurements made between the same two events by an observer located close to the surface of the sphere.
Note also that R and RsinΘ appear in the coordinates and not in the coefficients in the transformation equations. That happens because the transformation coefficients must be unit-free: they must be pure numbers.
Equations 1 become a single equation if we use matrix multiplication on the measured variables represented as a column four-vector;
In that equation I added the imaginary coefficient in order to make the appropriate minus sign appear in the metric equation. We obtain the metric equation, which is a Lorentz invariant, by multiplying Equation 3 by itself;
In reversing the order of the first matrix and four-vector we had to transpose them, converting the column vector into a row vector. Note also that I transferred the imaginary coefficient from the time measurements to the temporal transformation coefficient. Multiplying the two matrices together then gives us the metric tensor, a matrix that describes the shape of spacetime around the body M;
I have depicted the metric tensor as covariant (subscripted) because the elements of the four-vectors represent measurements and are, therefore, contravariant (superscripted). Using the generalized coordinates dQi for the elements of the distance-duration four vectors enables us to write Equation 4 as
in which we employ the Einstein convention of multiplying over the values of the indices that are repeated in the superscripts and subscripts. Those values run from one to four.
We also want to have the contravariant version of the metric tensor. To obtain that we simply divide the cofactor matrix of gik by
The ik-th element of the cofactor matrix comes from calculating the determinant of gik from which we have removed the i-th row and the k-th column. Thus we have
We know, from the essay The Metric Tensor, that multiplying the covariant metric tensor by its contravariant counterpart must yield the identity matrix. So our first test of the validity of our Kerr metric tensor carries out that calculation:
That result passes the test, so we can continue to the next step.
I should point out that the first time I derived the Kerr transformation and the associated metric tensor, the result failed that test (see Appendix). My first derivation of the transformation equations put an additional term, representing a temporal offset, into the time-transformation equation. That put a non-zero off-diagonal element into the metric tensor and that caused the test to fail. When I rederived the transformation equations, I found that the temporal offset term was illegitimate and dropped it. The equations listed above are the result of that rederivation and they yield a metric tensor that passes this first test.
We can also extract a metric tensor from the Boyer-Lindquist metric equation (Equation A-18 in The Kerr Spacetime) and work out its contravariant equivalent. Astonishingly, that metric tensor fails the above test.
The Christoffel Symbols
Begin by defining the Christoffel symbols through the parallel transport of a contravariant vector Ai along a coordinate line xk. Parallel transport does not change the magnitude or the direction of the vector, so relative to a non-Cartesian coordinate frame the vector changes in accordance with
The upper-case gamma represents the Christoffel symbol and the delta indicates a four-vector derivative.
Analysis of the algebraic expression describing a geodesic path, the path of shortest invariant distance between two points, gives us the means to calculate the Christoffel symbols associated with the coordinate frame to which we have referred the path. Written in terms of the metric tensor associated with the coordinate frame, the calculation is
In that equation we have ∂i= ∂/∂xi.
In our four-dimensional spacetime the Christoffel symbol has 64 components. Fortunately most of them equal zero and we can ignore them. For the multiplication by the contravariant metric tensor only four elements of that tensor do not make the elements of the Christoffel symbol equal zero; they are g11, g22, g33, and g44. As for the partial derivatives of the elements of the covariant metric tensor, only six do not zero out: we have
1. For ∂1= ∂/∂R all of the non-zero terms are ∂1g11, ∂1g22, ∂1g33, and ∂1g44;
2. For ∂2= ∂/∂Θ all of the non-zero terms are ∂2g11 and ∂2g22;
3. For ∂3= ∂/∂Φ all of the derivatives equal zero; and
4. For ∂4= ∂/c∂T all of the derivatives equal zero.
We can substitute those components, in their proper order, into Equation 11 and thereby work out our Christoffel symbols. The non-zero terms of the contravariant metric tensor give us four matrices, which we can fill in with the appropriate derivatives. We have, then, for the first matrix,
In assembling that matrix I followed a fairly straightforward procedure. I wanted to lay out the first contravariant component of the Christoffel symbol, so I knew that the component of the contravariant metric tensor in Equation 11 would be only g11. For each element of the derivatives matrix I represented each of the derivatives by its indices. In this case n=1, so for the element in the second column of the first row I wrote the indices as +-. I then checked the above list of nonzero derivatives and found that only the index set  (∂2g11) had a match, which I added to the matrix. I was thus able to fill out the matrix, element by element. It痴 more bookkeeping than physics, but it ensures that we keep our numbers straight in order to devise a proper and correct description of Reality.
For the other three parts of the Christoffel symbol we have:
If you want to see the Christoffel symbol in all of its algebraic glory, expressed in terms of the coordinates and the basic properties of the body M, make the appropriate substitutions from Equations 5 and 8 and then make the next substitutions from Equations 2.
Appendix: On Correcting an Error
In devising the equations of the Kerr Transformation I made an error and it痴 worth taking a look at how I discovered and corrected it. When we work out an algebraic description of the four-geometry of spacetime, we have certain tests that we can use to ensure that we have devised a correct description. The success of those tests in guiding us to a correct description of spacetime implies an intimate relationship between mathematics and Reality. The operations of mathematics appear to conform closely, if not identically, with the laws of Reality, which implies a perfectly rational Universe.
At first I devised the transformation equations of the Kerr spacetime as
in which I defined, for convenience,
In matrix form the transformation equations become
Squaring the transformation matrix, as the metric equation requires, yields the covariant metric tensor as
Using the standard technique for transforming a covariant tensor into its contravariant equivalent, dividing the cofactor matrix by the determinant of the covariant tensor, I obtained the contravariant metric tensor as
When I multiplied those matrices together, I got
That result fails the test, but not by much. The matrix is diagonal, as required, and the nonzero elements all have the same value, as required. But the value of the nonzero elements does not equal one, so we infer that something is wrong with the transformation equations that led to this result. Notice, though, as I did, that the matrix takes the correct form if E=0. I looked again at my derivation of the transformation equations and saw that the time equation should have no dependence upon latitude or longitude. When I removed those dependencies, I got a time equation identical to the time equation in the Schwarzschild Transformation. I recalculated the metric tensor, both covariant and contravariant, applied the multiplication test, and got the unit matrix that I needed. By thus going wrong and then correcting my error, I gained more confidence in my derivation of the Kerr Transformation.
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