The Kerr metric is an exact solution of the Einstein equations, generalizing the Schwarzschild metric to represent a spinning black hole. Chandrasekhar famously contradicted Landau’s assessment of the determination of this metric to claim that a derivation is “really very simple.” Surely only Chandrasekhar would consider his own presentation in the referenced volume “very simple”: I for one do not.

Knowing something of the features of the final metric, it is instructive to present a derivation that is simpler than Chandrasekhar’s and focuses on the mechanics of teasing out a solution from the Einstein equations. While the initial metric is every bit as general as that used by Chandrasekhar, the process will be guided by foreknowledge of the answer, with only evaluations made that lead to concrete progress. This is useful as an illustration of how to find the right thread to pull to unravel a rather complicated nonlinear problem.

Begin with a stationary axisymmetic metric of the form

ds2 =(Adt +εBdφ )2 -(Cdφ +εDdt )2 -E2 dr2 -F2 dθ2

where coefficients other than ε are functions of only the two spatial variables r and θ. Derivatives with respect to these variables will be denoted with subscripts 1 and 2, respectively.

The constant parameter ε measures deviation from a diagonal metric. In the limit ε0 the spherical symmetry of the Schwarzschild metric must be recovered:

Crsinθ Fr , ε0

In the same limit, A and E are functions of only the radial variable.

Choose basis forms in the Cartan formalism as

θ0 =Adt +εBdφ θ1 =Edr θ2 =Fdθ θ3 =Cdφ +εDdt

which are orthonormal by construction. The first and last basis forms are inverted with

dt =CG θ0 -εBG θ3 dφ =AG θ3 -εDG θ0 G=AC -ε2BD

The function combination G is the square root of the absolute value of the determinant of the submetric in the two variables t and φ.

Indices on forms are raised and lowered with the diagonal Minkowski metric [1,−1,−1,−1] .

The exterior derivatives of the basis forms are

dθ0 =A1 drdt +A2 dθdt +εB1 drdφ +εB2 dθdφ dθ0 =A1C -ε2 B1D EG θ0 θ1 -A2C -ε2 B2D FG θ0 θ2 -εA1B -B1A EG θ1 θ3 -εA2B -B2A FG θ2 θ3 dθ0 P θ0 θ1 -Rθ0 θ2 -Hθ1 θ3 -Kθ2 θ3   dθ1 =E2dθ dr =E2 EF θ1 θ2   dθ2 =F1dr dθ =F1 EF θ1 θ2   dθ3 =C1 drdφ +C2 dθdφ +εD1 drdt +εD2 dθdt dθ3 =εC1D -D1C EG θ0 θ1 +εC2D -D2C FG θ0 θ2 +C1A -ε2 D1B EG θ1 θ3 +C2A -ε2 D2B FG θ2 θ3 dθ3 J θ0 θ1 +Lθ0 θ2 +Qθ1 θ3 +Sθ2 θ3

For convenience of evaluation, a comparative tabulation of the function combinations is useful:

H=ε A1B -B1A EG J=ε C1D -D1C EG K=ε A2B -B2A FG L=ε C2D -D2C FG P=A1C -ε2 B1D EG Q=C1A -ε2 D1B EG R=A2C -ε2 B2D FG S=C2A -ε2 D2B FG

These additional combinations look promising and will turn out to be important:

P+Q =1 lnGE R+S =2 lnGF

Consistent connection forms are identified with Cartan’s first equation of structure:

dθi =ω ik θk

Attempting to do this by inspection tends to miss the crucial nonzero value for ω 03  . The straightforward procedure is to assume nonzero coefficients for all basis forms, expand the equation of structure and solve all nontrivial equations. The resulting connection forms are

ω 01 =ω 10 =Pθ0 -H-J2 θ3 ω 02 =ω 20 =Rθ0 -K-L2 θ3 ω 03 =ω 30 =H+J2 θ1 +K+L2 θ2 ω 12 =ω 21 =E2 EF θ1 -F1 EF θ2 ω 13 =ω 31 =Qθ3 -H-J2 θ0 ω 23 =ω 32 =Sθ3 -K-L2 θ0

It is simple to substitute these expressions back into the first equation of structure to verify the consistency of the set of forms.

It is now straightforward to evaluate curvature 2-forms using Cartan’s second equation of structure:

R ik =dω ki +ω im ω mk

The curvature 2-forms are related to the Riemann tensor in so-called flat indices by

R ik =R iklm θl θm

The Ricci tensor in the same indices is the contraction

Rik =R mimk

Since vacuum solutions to the Einstein equation can be determined with either the Ricci or Einstein tensor, the former will be the goal of this presentation, bypassing listing all components of the Riemann tensor as unnecessary.

The six explicit equations for the curvature 2-forms are

R 01 =R 10 =dω 01 +ω 02 ω 21 +ω 03 ω 31 R 02 =R 20 =dω 02 +ω 01 ω 12 +ω 03 ω 32 R 03 =R 30 =dω 03 +ω 01 ω 13 +ω 02 ω 23 R 12 =R 21 =dω 12 +ω 10 ω 02 +ω 13 ω 32 R 13 =R 31 =dω 13 +ω 10 ω 03 +ω 12 ω 23 R 23 =R 32 =dω 23 +ω 20 ω 03 +ω 21 ω 13

Running through the combinations of basis forms that can appear on the right-hand sides of these equations indicates that the third and fourth are simpler than the others. The possible resulting indices for the Riemann tensor can be represented as

R 01[ 01,02, 13,23 ] =R 10[ 01,02, 13,23 ] R 02[ 01,02, 13,23 ] =R 20[ 01,02, 13,23 ] R 03[ 03,12 ] =R 30[ 03,12 ] R 12[ 03,12 ] =R 21[ 03,12 ] R 13[ 01,02, 13,23 ] =R 31[ 01,02, 13,23 ] R 23[ 01,02, 13,23 ] =R 32[ 01,02, 13,23 ]

where alternatives are listed in the square subscripted brackets. This is already enough information to determine that four of the components of the Ricci tensor are identically zero,

R01 =R02 =R13 =R23 =0

since the components of the Riemann tensor that would compose them do not appear.

Now one must dig in and evaluate the curvature 2-forms one by one. Immediate simplifications using function combinations above are included: further simplifications are possible.

R 01 = i=1,2 i(AP -εD H-J 2) dxi (CG θ0 -εBG θ3) + i=1,2 i( εBP -CH-J 2) dxi (AG θ3 -εDG θ0) -(R θ0 -K-L2 θ3) (E2 EF θ1 -F1 EF θ2) +( H+J2 θ1 +K+L2 θ2) (Q θ3 +H-J2 θ0) R 01 =[ P1E +P2 +(H+3J) (H-J)4 +E2R EF] θ0 θ1 -[ P2F +PR +(K+3L) (H-J)4 -F1R EF] θ0 θ2 -[ H1 -J1 2E +PH -QJ +E2 (K-L) 2EF] θ1 θ3 -[ H2 -J2 2F +PK +S(H-J) -Q(K+L) 2 -F1 (K-L) 2EF] θ2 θ3

R 02 = i=1,2 i(AR -εD K-L 2) dxi (CG θ0 -εBG θ3) + i=1,2 i( εBR -CK-L 2) dxi (AG θ3 -εDG θ0) +(P θ0 -H-J2 θ3) (E2 EF θ1 -F1 EF θ2) +( H+J2 θ1 +K+L2 θ2) (S θ3 +K-L2 θ0) R 02 =[ R1E +PR +(H+3J) (K-L)4 -E2P EF] θ0 θ1 -[ R2F +R2 +(K+3L) (K-L)4 +F1P EF] θ0 θ2 -[ K1 -L1 2E +RH +P(K-L) -S(H+J) 2 -E2 (H-J) 2EF] θ1 θ3 -[ K2 -L2 2F +RK -SL +F1 (H-J) 2EF] θ2 θ3

R 03 =2( EH+J 2) dθdr +1( FK+L 2) drdθ -(P θ0 -H-J2 θ3) (Q θ3 +H-J2 θ0) -(R θ0 -K-L2 θ3) (S θ3 +K-L2 θ0) R 03 =[ PQ+RS +14 (H-J )2 +14 (K-L )2] θ0 θ3 +1EF [1( FK+L 2) -2( EH+J 2)] θ1 θ2

R 12 =2( E2F) dθdr -1( F1E) drdθ +(P θ0 -H-J2 θ3) ( Rθ0 -K-L2 θ3) -(Q θ3 +H-J2 θ0) (S θ3 +K-L2 θ0) R 12 =12 [(Q-P) (K-L) +(R-S) (H-J)] θ0 θ3 -1EF [1( F1E) +2( E2F)] θ1 θ2

R 13 = i=1,2 i( εDQ +AH-J 2) dxi (CG θ0 -εBG θ3) - i=1,2 i(CQ +εB H-J 2) dxi (AG θ3 -εDG θ0) +(P θ0 -H-J2 θ3) ( H+J2 θ1 +K+L2 θ2) -(E2 EF θ1 -F1 EF θ2) (S θ3 +K-L2 θ0) R 13 =[ H1 -J1 2E +PH -QJ +E2 (K-L) 2EF] θ0 θ1 +[ H2 -J2 2F -QL +R(H-J) +P(K+L) 2 -F1 (K-L) 2EF] θ0 θ2 -[ Q1E +Q2 -(3H+J) (H-J)4 +E2S EF] θ1 θ3 -[ Q2F +QS -(3K+L) (H-J)4 -F1S EF] θ2 θ3

R 23 = i=1,2 i( εDS +AK-L 2) dxi (CG θ0 -εBG θ3) - i=1,2 i(CS +εB K-L 2) dxi (AG θ3 -εDG θ0) +( Rθ0 -K-L2 θ3) ( H+J2 θ1 +K+L2 θ2) +(E2 EF θ1 -F1 EF θ2) (Q θ3 +H-J2 θ0) R 23 =[ K1 -L1 2E -SJ +P(K-L) +R(H+J) 2 -E2 (H-J) 2EF] θ0 θ1 +[ K2 -L2 2F +RK -SL +F1 (H-J) 2EF] θ0 θ2 -[ S1E +QS -(3H+J) (K-L)4 -E2Q EF] θ1 θ3 -[ S2F +S2 -(3K+L) (K-L)4 +F1Q EF] θ2 θ3

Some of the symmetries of the Riemann tensor pop out in these forms as written, but not all do. This can be used to advantage in determining relationships among metric functions.

Constructing the remaining components of the Ricci tensor is a simple matter of reading off components of the Riemann tensor. Components of the latter as written are antisymmetric in the first two indices when one is nonzero and always antisymmetric in the last two indices.

R00 =R 1010 +R 2020 +R 3030 R00 =1 (FP) EF +2 (ER) EF +P(P+Q) +R(R+S) +(H+J) (H-J)2 +(K+L) (K-L)2

R11 =R 0101 +R 2121 +R 3131 R11 =P1 +Q1E -P2 -Q2 -E2 (R+S) EF -1EF [1( F1E) +2( E2F)] +(H-J )22

R22 =R 0202 +R 1212 +R 3232 R22 =R2 +S2F -R2 -S2 -F1 (P+Q) EF -1EF [1( F1E) +2( E2F)] +(K-L )22

R33 =R 0303 +R 1313 +R 2323 R33 =1 (FQ) EF -2 (ES) EF -Q(P+Q) -S(R+S) +(H+J) (H-J)2 +(K+L) (K-L)2

R03 =R 1013 +R 2023 =R 1310 +R 2320 =R30 R03 =1 [F(H-J)] 2EF -2 [E(K-L)] 2EF -PH +QJ -RK +SL

R12 =R 0102 +R 3132 =R 0201 +R 3231 =R21 R12 =P2 +Q2F -PR -QS +F1 (R+S) EF +(H-J) (K-L)2 R12 =R1 +S1E -PR -QS +E2 (P+Q) EF +(H-J) (K-L)2

These last two expressions for the off-diagonal component are necessarily identical. Equating the apparently different parts leads to

2[ E(P+Q)] =1[ F(R+S)] 2[ 1lnG] =1[ 2lnG]

which in general is simply a statement of the commutativity of partial differentiation. While this necessarily holds and shows the consistency of the two expressions, it also suggests a path toward solution: if each side of this statement is separately zero, then the combination G of metric functions is separable in the two independent variables:

G=AC -ε2BD =g(r) h(θ)

With this in mind, consider the combination of Ricci tensor components

R00 -R33 =1[ F(P+Q)] EF +2[ E(R+S)] EF +(P+Q )2 +(R+S )2 R00 -R33 =1EF 1( FE 1lnG) +1EF 2( EF 2lnG) +1E2 (1lnG )2 +1F2 (2lnG )2 R00 -R33 =1EF 1( FE 1lng) +1EF 2( EF 2lnh) +1E2 (1lng )2 +1F2 (2lnh )2

When equated to zero for a vacuum solution,

FE 1( FE 1lng) +FE 2( EF 2lnh) +(FE )2 (1lng )2 +(2lnh )2 =0

one can achieve a separable equation by assuming

FE =f(r)

Then the angular part of the equation, the second and fourth terms, can be set equal to a constant

22lnh +(2lnh )2 =h22h =constant

and the function h is clearly a circular function in the compact variable θ. With the choice h=sinθ , the remainder of the equation is

f1( f1lng) +f2 (1lng )2 =fg 1( fg1) =1

The simplest way for this equation to hold is for the two functions to be equal, for which the solution is

1( ff1) =12 12 f2 =1 f2 =g2 =r2 +αr+β Δ

and one can already begin to see the Kerr metric taking shape, just from assuming the simplest way toward a nontrivial solution.

Now look again at the statement of separability

G=f(r) sinθ =AC -ε2BD =AD(CD -ε2 BA)

The difference of metric function ratios in brackets is unlikely to be equal to a separable product, but clearly is significant. Since it has to balance the product of two metric functions, denote it as a squared function:

ρ2 (r,θ) =CD -ε2 BA =c(r,θ) -ε2 b(r,θ)

The factor of the sine function can be assigned to both C and D in order to match the spherical symmetry of the Schwarzschild solution. Knowing also that the coefficient A in the Schwarzschild limit is a function of the radial variable, the factor f can be assigned with confidence to the this metric function. With these assumptions, one has

A=fρ B=bfρ C=csinθ ρ D=sinθρ

where denominators have been assigned equally as the simplest choice possible.

With a focus on specific ratios of metric functions, the first set of intermediate metric function combinations becomes

H=ε A1 (B/A) DEρ2 =ε b1 f2 ρ2Fsinθ J=ε D1 (C/D) AEρ2 =ε c1sinθ ρ2F K=ε A2 (B/A) DFρ2 =ε b2f ρ2Fsinθ L=ε D2 (C/D) AFρ2 =εc2 sinθ fρ2F

Differentiating the definition of ρ2 ,

2ρρ1 =c1 -ε2 b1 2ρρ2 =c2 -ε2 b2

the second set of metric function combinations can be written

P=f1F -(c1 +ε2 b1)f 2ρ2F Q=(c1 +ε2 b1)f 2ρ2F R=( c2 +ε2 b2) 2ρ2F S=cotθF +(c2 +ε2 b2) 2ρ2F

All of the metric function combinations have a factor of F in the denominator. This means that when a vacuum equation is formed by for example setting R03 equal to zero, then that particular metric function can be eliminated from the equation. This allows writing an equation that determines the interrelationship of b and c as functions of the two independent variables.

Such an equation cannot, unfortunately, be written in a completely integrable form. The difficulty essentially arises because the quantities

c1 +ε2 b1 ρ2 =c1 +ε2 b1 c-ε2 b c2 +ε2 b2 ρ2 =c2 +ε2 b2 c-ε2 b

are not in general integrable due to the sign difference between the numerator and denominator. One way to make these expressions integrable is to assume that the two function b and c each depend on only one of the independent variables, and there are two ways to do so:

ρ2 =c(θ) -ε2 b(r) or ρ2 =c(r) -ε2 b(θ)

For the first choice, J=K=0.  The vacuum equation formed by setting R03 equal to zero is in that case

1 (FH) 2EF +2 (EL) 2EF -PH+SL =0 f2 1( b1f2 ρ2 sinθ) +f2 2( c2sinθ f2 ρ2) +b1 f2 ρ2sinθ (f1- ε2 b1f 2ρ2) +c2sinθ fρ2 (cotθ +c2 2ρ2) =0 1( b1f4) +1sinθ 2( c2sin3 θ) =0

which is a separable equation. Setting the second term equal to a constant implies

2( c2sin3 θ) sinθ c(θ) 1 sin2θ

This solution would not match the Schwarzschild solution in the limit ε0 and so cannot be correct. Turn to the second choice, for which H=L=0.  The vacuum equation formed by setting R03 equal to zero is in this case

1 (FJ) 2EF -2 (EK) 2EF +QJ-RK =0 f2 1( c1 sinθ ρ2) +f2 2( b2 ρ2 sinθ) +c1 sinθ ρ2 c1f 2ρ2 -b2f ρ2sinθ ε2 b2 2ρ2 =0 12c +1sinθ 2( b2 sinθ) =0

This equation is also separable and simpler than that of the first choice. The function b can written either in terms of sine or cosine. Selecting the former one has

b(θ) =sin2θ c(r) =r2 +γr+δ

To determine the constants of integration, an independent vacuum equation with F eliminated is needed. Consider the combination of Ricci tensor components

R00 +R33 =1[ F(P-Q)] EF +2[ E(R-S)] EF +(P-Q) (P+Q) +(R-S) (R+S) +(H-J) (H+J) +(K-L) (K+L) R00 +R33 =f1[ F(P-Q)] F2 +2[ F(R-S)] F2 +(P-Q) f1 lnfF +(R-S) 2 lnsinθF -J2 +K2 R00 +R33 =1[f F(P-Q)] F2 +2[ F(R-S) sinθ] F2sinθ -J2 +K2

Setting this equal to zero and substituting the angular function b gives

1[ ff1 -c1 f2 ρ2] -1sinθ 2[cosθ +ε2 b2 ρ2 sinθ] -ε2 c12 sin2θ ρ4 +ε2 b22 f2 ρ4 sin2θ =0 2ρ4 -ρ4 1( c1 f2 ρ2) -2ε2 ρ4 sinθ 2( sin2θ cosθ ρ2) -ε2 c12 sin2θ +4ε2 f2cos2θ =0

This looks messy, but by grouping on powers of sine it simplifies somewhat to

[1( c1f2) -c12 -4f2 +2c] ε2 sin2θ +4ε2 (f2 -c) -c1( c1f2) +c12 f2 +2c2 =0

While further simplification is possible here, it is straightforward to substitute the radial functions f and c and do a bit of algebra. There are significant cancellations in the process:

[γ(α-γ) -2(β-δ) ] ε2 sin2θ +[2(β-δ) -γ(α-γ) ]r2 +[4α (ε2-δ) +2γ(β+δ -2ε2) ]r +[2(β -δ) (2ε2 -δ) -γ(αδ -βγ)] =0

Each of the quantities in square brackets must be separately zero for this to hold. Two of the resulting equations are identical, so only three of the four constants can be determined. From the given grouping it is clear that a good choice is β=δ=ε2  . For a third constant one can choose either γ=0 or γ=α  : the Kerr metric uses the former, but clearly a slightly different solution is possible.

For the standard Kerr metric, the two radial functions are now

f(r) =r2 +αr +ε2 c(r) =r2 +ε2

All that remains is to determine the form of the metric function F relative to the others. For that one could consider the combination of Ricci tensor components

R11 -R22 -R00 +R33 =1EF [1( F1E) +2( E2F)] -2PQ-2RS -(H-J )22 -(K-L )22

but the expression in square brackets is less simple than it appears. Instead use a sum of the two forms of the off-diagonal component:

2R12 =FE 1( R+SF) +EF 2( P+QE) +2PR +2QS -(H-J) (K-L) 2 R12 =fcotθ 1( 1F2) +f1 2( 1F2) -ε2 b2f1 ρ2 F2 +c1 fcotθ ρ2 F2

The somewhat surprising reduction in the number of terms here is because the sine factors in the nonzero functions J and K cancel against each other. This simplification would not occur if H and L were also nonzero.

Equating to zero for a vacuum solution and regrouping,

fcotθ[ 1( 1F2) +c1 ρ2 F2] +f1[ 2( 1F2) -ε2 b2 ρ2 F2] =0

it is clear by inspection that F and ρ are identical functions, which completes the determination of all metric functions. A final check of consistency would be to evaluate all components of the Ricci tensor individually to confirm that they are all zero. This can be done with a computer algebra system for efficiency and will not be included here.

To summarize, the coefficient functions determined are

A=fρ B=f sin2θρ C=(r2 +ε2) sinθρ D=sinθρ   E=ρf F=ρ   ρ2=r2 +ε2 cos2θ f2=r2 +αr +ε2 =Δ

The physical interpretation of the two constants can be achieved by comparing an expansion of the metric in inverse powers of r to the known weak-field limit of a rotating body, given for example as Equation (2) of this excerpt. The coefficient of dt2 as developed here is

A2-ε2D2 =r2 +αr +ε2 cos2θ ρ2 =1+αr ρ2 =1+αr +O(1 r2)

The numerator of the second term must be equal to the negative of twice the gravitational mass of the body: α=2m  . The coefficient of the cross term dtdφ as developed here is

2ε(AB -CD) =2αεr sin2θ ρ2 =2αε sin2θ r +O(1 r2)

The constant part of the numerator here must equal four times the angular momentum of the body:

2αε=4J ε=Jm =a

With these identifications, the metric functions become

A=fρ B=f sin2θρ C=(r2 +a2) sinθρ D=sinθρ   E=ρf F=ρ   ρ2=r2 +a2 cos2θ f2=r2 -2mr +a2 =Δ

The complete Kerr metric with these coefficient functions is

ds2 =Δρ2 [dt-a sin2θdφ ]2 -sin2θ ρ2 [(r2 +a2)dφ -adt ]2 -ρ2Δ dr2 -ρ2 dθ2

Seeing the metric expressed this way in this discussion is what inspired the entire derivation. It struck me that such evenly matched powers of ρ2 could only balance one another through ratios of metric functions. From there it was simple to assume similar behavior for the functions leading to cross terms, so that paying attention to ratios was the key to the entire process.

As stated at the outset, the initial metric form is every bit as general as that employed by Chandrasekhar. The inclusion of two cross terms instead of one makes the whole process as well as the final statement of the metric much simpler, in my opinion. Not that it is particularly simple...

Uploaded 2018.02.10 — Updated 2018.04.06