Algorithm for finding the Riemann tensor in terms of the stress-energy tensor
Start with any symmetric stress-energy tensor $T_{uv}$
The Einstein Field Equations says:
$G_{uv} = \kappa T_{uv}$
...where $\kappa = 8 \pi \frac{G}{c^4}$, for G the gravitational constant and c the speed of light.
...its trace:
$G = \kappa T$
$T = \frac{1}{\kappa} G$
The definition of the Einstein tensor:
$G_{uv} = R_{uv} - \frac{1}{2} R g_{uv}$
...its trace:
$G = -\frac{n - 2}{2} R$
$R = -\frac{2}{n - 2} G$
Combine:
$G_{uv} = R_{uv} + \frac{1}{2} \frac{2}{n - 2} G g_{uv}$
$R_{uv} = G_{uv} - \frac{1}{n - 2} G g_{uv}$
Combine:
$\kappa T_{uv} = R_{uv} - \frac{1}{2} R g_{uv}$
Combine Traces:
$\kappa T = -\frac{n-2}{2} R$
If $n = 2$ then we get $T = 0$ and cannot solve for R.
For $n \ge 3$:
$R = -\kappa \frac{2}{n-2} T$
Substitute:
$R_{uv} = \kappa (T_{uv} - \frac{1}{n - 2} T g_{uv})$
The Kulkarni-Nomizu product
The Kulkarni-Nomizu product of two (0, 2) symmetric tensors $A_{ab}$ and $B_{ab}$ is defined as:
$\require{enclose}$
$(A \ {\scriptstyle \enclose{circle}{\wedge}} \ B)_{abcd} = A_{ac} B_{bd} + A_{bd} B_{ac} - A_{ad} B_{bc} - A_{bc} B_{ad}$
Notice that the Kulkarni-Nomizu product applied to the same tensor is:
$(A \ {\scriptstyle \enclose{circle}{\wedge}} \ A)_{abcd}$
$= A_{ac} A_{bd} + A_{bd} A_{ac} - A_{ad} A_{bc} - A_{bc} A_{ad}$
$= 2 (A_{ac} A_{bd} - A_{ad} A_{bc})$
For $n = 2$ then there is only one possible Riemann tensor value for the Ricci tensor:
$R_{abcd} = \frac{1}{2} R (g_{ac} g_{bd} - g_{ad} g_{bc})$
$R_{abcd} = \frac{1}{4} R (g \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{abcd}$
For $n \ge 3$:
Find the Schouten tensor in terms of the Ricci tensor
The Schouten tensor is defined as:
$P_{uv} = \frac{1}{n - 2} (R_{uv} - \frac{1}{2 (n - 1)} R g_{uv})$
Substitute our definition of stress-energy:
$P_{uv} = \frac{1}{n - 2} (
\kappa (T_{uv} - \frac{1}{n - 2} T g_{uv})
+ \frac{1}{2 (n - 1)} g_{uv} \kappa \frac{2}{n-2} T
)$
$P_{uv} = \kappa \frac{1}{n - 2} (
T_{uv}
+ (
\frac{1}{2 (n - 1)} \frac{2}{n-2}
- \frac{1}{n - 2}
) T g_{uv}
)$
$P_{uv} = \kappa \frac{1}{n - 2} (
T_{uv}
+ (
\frac{1}{n - 1} \frac{1}{n-2}
- \frac{n - 1}{n - 1} \frac{1}{n - 2}
) T g_{uv}
)$
$P_{uv} = \kappa \frac{1}{n - 2} (
T_{uv}
- \frac{1}{n - 1} T g_{uv}
)$
Choose a Weyl tensor subject to its constraints
$C_{(ab)cd} = C_{ab(cd)} = 0$
$C_{a[bcd]} = 0$
${C^u}_{aub} = 0$
A simple choice, and nearest point projected onto the solution space, is $C_{abcd} = 0$
Find the Riemann tensor in terms of the Schouten tensor and the Weyl tensor
Ricci decomposition:
$R_{abcd} = C_{abcd} + (P \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{abcd}$
$R_{abcd} = C_{abcd} + P_{ac} g_{bd} + P_{bd} g_{ac} - P_{ad} g_{bc} - P_{bc} g_{ad}$
In terms of the Ricci tensor:
$R_{abcd} = C_{abcd} + \frac{1}{n - 2} (
(R_{ac} - \frac{1}{2 (n-1)} R g_{ac}) g_{bd}
+ (R_{bd} - \frac{1}{2 (n-1)} R g_{bd}) g_{ac}
- (R_{ad} - \frac{1}{2 (n-1)} R g_{ad}) g_{bc}
- (R_{bc} - \frac{1}{2 (n-1)} R g_{bc}) g_{ad}
)$
$R_{abcd} = C_{abcd} + \frac{1}{n - 2} (
R_{ac} g_{bd}
+ R_{bd} g_{ac}
- R_{ad} g_{bc}
- R_{bc} g_{ad}
- \frac{1}{2 (n-1)} R g_{ac} g_{bd}
- \frac{1}{2 (n-1)} R g_{bd} g_{ac}
+ \frac{1}{2 (n-1)} R g_{ad} g_{bc}
+ \frac{1}{2 (n-1)} R g_{bc} g_{ad}
)$
$R_{abcd} = C_{abcd} + \frac{1}{n - 2} (
R_{ac} g_{bd}
+ R_{bd} g_{ac}
- R_{ad} g_{bc}
- R_{bc} g_{ad}
+ \frac{1}{n - 1} R (g_{ad} g_{bc} - g_{ac} g_{bd})
)$
$R_{abcd} = C_{abcd} + \frac{1}{n - 2} (
(R \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{abcd}
- \frac{1}{2 (n - 1)} R \cdot (g \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{abcd}
)$
In terms of the stress-energy tensor:
$R_{abcd} = C_{abcd} + \frac{1}{n - 2} (
\kappa (T_{ac} - \frac{1}{n-2} T g_{ac}) g_{bd}
+ \kappa (T_{bd} - \frac{1}{n-2} T g_{bd}) g_{ac}
- \kappa (T_{ad} - \frac{1}{n-2} T g_{ad}) g_{bc}
- \kappa (T_{bc} - \frac{1}{n-2} T g_{bc}) g_{ad}
+ \frac{1}{n - 1} (-\kappa \frac{2}{n - 2} T) (g_{ad} g_{bc} - g_{ac} g_{bd})
)$
$R_{abcd} = C_{abcd} + \frac{1}{n - 2} \kappa (
T_{ac} g_{bd}
+ T_{bd} g_{ac}
- T_{ad} g_{bc}
- T_{bc} g_{ad}
- \frac{1}{n-2} T g_{ac} g_{bd}
- \frac{1}{n-2} T g_{bd} g_{ac}
+ \frac{1}{n-2} T g_{ad} g_{bc}
+ \frac{1}{n-2} T g_{bc} g_{ad}
- \frac{2}{(n - 1)(n - 2)} T (g_{ad} g_{bc} - g_{ac} g_{bd})
)$
$R_{abcd} = C_{abcd} + \kappa \frac{1}{n - 2} (
T_{ac} g_{bd}
+ T_{bd} g_{ac}
- T_{ad} g_{bc}
- T_{bc} g_{ad}
+ (\frac{2}{n - 2} - \frac{2}{(n - 1)(n - 2)}) T (g_{ad} g_{bc} - g_{ac} g_{bd})
)$
$R_{abcd} = C_{abcd} + \kappa \frac{1}{n - 2} (
T_{ac} g_{bd}
+ T_{bd} g_{ac}
- T_{ad} g_{bc}
- T_{bc} g_{ad}
+ \frac{2}{n - 1} T (g_{ad} g_{bc} - g_{ac} g_{bd})
)$
$R_{abcd} = C_{abcd} + \kappa \frac{1}{n - 2} (
(T \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{abcd}
- \frac{1}{n - 1} T \cdot (g \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{abcd}
)$
Using $R^{\sharp\sharp\flat\flat}$ to remove the appearance of the metric:
${R^{ab}}_{cd} = {C^{ab}}_{cd} + \kappa \frac{1}{n - 2} (
{T^a}_c \delta^b_d
+ {T^b}_d \delta^a_c
- {T^a}_d \delta^b_c
- {T^b}_c \delta^a_d
+ \frac{2}{n - 1} T (\delta^a_d \delta^b_c - \delta^a_c \delta^b_d)
)$
${R^{ab}}_{cd} = {C^{ab}}_{cd} + \kappa \frac{1}{n - 2} (
{T^a}_c \delta^b_d
- {T^a}_d \delta^b_c
+ {T^b}_d \delta^a_c
- {T^b}_c \delta^a_d
- \frac{1}{n - 1} T (
\delta^a_c \delta^b_d
- \delta^a_d \delta^b_c
+ \delta^b_d \delta^a_c
- \delta^b_c \delta^a_d
)
)$
${R^{ab}}_{cd} = {C^{ab}}_{cd} + \kappa \frac{4}{n - 2} (
{T^{[a}}_{[c} \delta^{b]}_{d]}
- \frac{1}{n - 1} T (
\delta^{[a}_{[c} \delta^{b]}_{d]}
)
)$
${R^{ab}}_{cd} = {C^{ab}}_{cd} + \kappa \frac{4}{n - 2} (
{\delta^{[a|}}_u \delta^v_{[c} \delta^{|b]}_{d]}
- \frac{1}{n - 1} \delta_u^v (
\delta^{[a}_{[c} \delta^{b]}_{d]}
)
) {T^u}_v$
Find the double-dual Riemann tensor in terms of the stress-energy tensor, in $n=4$ dimensions
Kulkarni-Nomizu double-dual in $n=4$ dimensions:
$\frac{1}{4} {\epsilon_{ab}}^{pq} (A \ {\scriptstyle \enclose{circle}{\wedge}} \ B)_{pqrs} {\epsilon^{rs}}_{cd}$
$= \frac{1}{4} {\epsilon_{ab}}^{pq} (A_{pr} B_{qs} + A_{qs} B_{pr} - A_{ps} B_{qr} - A_{qr} B_{ps}) {\epsilon^{rs}}_{cd}$
$=
\frac{1}{4} {\epsilon_{ab}}^{pq} A_{pr} B_{qs} {\epsilon^{rs}}_{cd}
+ \frac{1}{4} {\epsilon_{ab}}^{pq} A_{qs} B_{pr} {\epsilon^{rs}}_{cd}
- \frac{1}{4} {\epsilon_{ab}}^{pq} A_{ps} B_{qr} {\epsilon^{rs}}_{cd}
- \frac{1}{4} {\epsilon_{ab}}^{pq} A_{qr} B_{ps} {\epsilon^{rs}}_{cd}
$
$=
\frac{1}{2} {\epsilon_{ab}}^{pq} A_{pr} B_{qs} {\epsilon^{rs}}_{cd}
- \frac{1}{2} {\epsilon_{ab}}^{pq} A_{ps} B_{qr} {\epsilon^{rs}}_{cd}
$
$= {\epsilon_{ab}}^{pq} A_{pr} B_{qs} {\epsilon^{rs}}_{cd} $
$= \epsilon_{abpq} {A^p}_r {B^q}_s {\epsilon^{rs}}_{cd} $
$= 4 (\star A^{\sharp\flat} B^{\sharp\flat} \star)_{abcd} $
TODO Kulkarni-Nomizu of two dual-tensors:
TODO duals of different indexes of Kulkarni-Nomizu...
Kulkarni-Nomizu double-dual of a symmetric tensor and the metric, in $n=4$ dimensions:
$(\star (T \ {\scriptstyle \enclose{circle}{\wedge}} \ g) \star)_{abcd}$
$= \frac{1}{4} {\epsilon_{ab}}^{pq} (T \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{pqrs} {\epsilon^{rs}}_{cd}$
...using the Kulkarni-Nomizu double-dual definition above:
$= \epsilon_{abpq} {T^p}_r \delta^q_s {\epsilon^{rs}}_{cd}$
$= \epsilon_{abpq} {T^p}_r \delta^q_s \epsilon^{rsef} g_{ce} g_{df}$
$= \epsilon_{pabq} {T^p}_r \epsilon^{refq} g_{ce} g_{df}$
$= -\delta^{ref}_{pab} {T^p}_r g_{ce} g_{df}$
$= -{T^p}_r (
\delta^r_p g_{ca} g_{db}
+ \delta^r_a g_{cb} g_{dp}
+ \delta^r_b g_{cp} g_{da}
- g_{dp} g_{ca} \delta^r_b
- g_{da} g_{cb} \delta^r_p
- g_{db} g_{cp} \delta^r_a
)$
$= T_{ac} g_{bd}
- T_{ad} g_{bc}
+ T_{bd} g_{ac}
- T_{bc} g_{ad}
- T (g_{ac} g_{bd} - g_{ad} g_{bc})
$
$= (T \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{abcd}
- \frac{1}{2} T (g \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{abcd}
$
Kulkarni-Nomizu double-dual of the metric and itself, in $n=4$ dimensions:
$(\star (g \ {\scriptstyle \enclose{circle}{\wedge}} \ g) \star)_{abcd}$
$= \frac{1}{4} {\epsilon_{ab}}^{pq} (g \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{pqrs} {\epsilon^{rs}}_{cd}$
$= \epsilon_{abpq} \delta^p_r \delta^q_s {\epsilon^{rs}}_{cd}$
$= 2 (g_{ac} g_{bd} - g_{ad} g_{bc})$
$= (g \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{abcd}$
...so the Kulkarni-Nomizu of the metric and itself is its own self-double-dual.
...
$G_{abcd} = \frac{1}{4} {\epsilon_{ab}}^{pq} R_{pqrs} {\epsilon^{rs}}_{cd}$
$G_{abcd} = \frac{1}{4} {\epsilon_{ab}}^{pq} {\epsilon^{rs}}_{cd} (
C_{pqrs} + \kappa \frac{1}{2} (
(T \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{pqrs}
- \frac{1}{n - 1} T \cdot (g \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{pqrs}
)
)$
$G_{abcd} = \frac{1}{4} {\epsilon_{ab}}^{pq} {\epsilon^{rs}}_{cd} C_{pqrs}
+ \kappa \frac{1}{2} (
(T \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{abcd}
- \frac{5}{6}
T \cdot (g \ {\scriptstyle \enclose{circle}{\wedge}} \ g)_{abcd}
)
$
Fluid Stress-Energy:
${T^u}_v = (\rho + P) u^u u_v + P \delta^u_v$
${T^u}_u = (\rho + P) u^u u_u + P \delta^u_u$
${T^u}_u = -(\rho + P) + 4 P$
${T^u}_u = 3 P - \rho$
${R^{ab}}_{cd} = {C^{ab}}_{cd} + \kappa \frac{1}{n - 2} (
\delta^b_d ((\rho + P) u^a u_c + P \delta^a_c)
+ \delta^a_c ((\rho + P) u^b u_d + P \delta^b_d)
- \delta^b_c ((\rho + P) u^a u_d + P \delta^a_d)
- \delta^a_d ((\rho + P) u^b u_c + P \delta^b_c)
+ \frac{2}{n - 1} (\delta^a_d \delta^b_c - \delta^a_c \delta^b_d)
( (\rho + P) u^u u_u + 4 P)
)$
${R^{ab}}_{cd} = {C^{ab}}_{cd} + \kappa \frac{1}{n - 2} (
\rho (
u^a u_c \delta^b_d
+ u^b u_d \delta^a_c
- u^a u_d \delta^b_c
- u^b u_c \delta^a_d
)
- \frac{2}{n - 1} (\delta^a_d \delta^b_c - \delta^a_c \delta^b_d) \rho
+ P (
u^a u_c \delta^b_d
+ u^b u_d \delta^a_c
- u^a u_d \delta^b_c
- u^b u_c \delta^a_d
)
+ (\frac{-2 n + 8}{n - 1}) (\delta^a_d \delta^b_c - \delta^a_c \delta^b_d) P
)$
Electromagnetism Stress-Energy:
${T^u}_v = \frac{1}{\mu_0} (-{F^u}_a {F^a}_v + \frac{1}{4} \delta^u_v {F^a}_b {F^b}_a)$
Let ${(F^2)^u}_v = {F^u}_a {F^a}_v$
${T^u}_v = \frac{1}{\mu_0} (-{(F^2)^u}_v + \frac{1}{4} \delta^u_v {(F^2)^a}_a)$
${T^u}_v = -\frac{1}{\mu_0} {TF(F^2)^u}_v$
... where $TF(x)$ is the trace-free operator.
${T^u}_u = 0$
${R^{ab}}_{cd} = {C^{ab}}_{cd} + \kappa \frac{1}{n - 2} (
\delta^b_d \frac{1}{\mu_0} (-{F^a}_u {F^u}_c + \frac{1}{4} \delta^a_c {F^u}_v {F^v}_u)
+ \delta^a_c \frac{1}{\mu_0} (-{F^b}_u {F^u}_d + \frac{1}{4} \delta^b_d {F^u}_v {F^v}_u)
- \delta^b_c \frac{1}{\mu_0} (-{F^a}_u {F^u}_d + \frac{1}{4} \delta^a_d {F^u}_v {F^v}_u)
- \delta^a_d \frac{1}{\mu_0} (-{F^b}_u {F^u}_c + \frac{1}{4} \delta^b_c {F^u}_v {F^v}_u)
)$
${R^{ab}}_{cd} = {C^{ab}}_{cd} - \kappa \frac{1}{n - 2} \frac{1}{\mu_0} (
\delta^b_d {F^a}_u {F^u}_c
+ \delta^a_c {F^b}_u {F^u}_d
- \delta^b_c {F^a}_u {F^u}_d
- \delta^a_d {F^b}_u {F^u}_c
+ \frac{1}{2} {F^u}_v {F^v}_u (
\delta^a_d \delta^b_c
- \delta^a_c \delta^b_d
)
)$
Substituting:
${F^u}_v = n^u E_v - n_v E^u + {\epsilon^u}_{vab} n^a B^b$
... where $E_u n^u = B_u n^u = 0$
... and $n_u n^u = -1$
${F^u}_a {F^a}_v = (n^u E_a - n_a E^u + {\epsilon^u}_{abc} n^b B^c) (n^a E_v - n_v E^a + {\epsilon^a}_{vde} n^d B^e)$
$=
n^u E_a n^a E_v
- n^u E_a n_v E^a
+ n^u E_a {\epsilon^a}_{vde} n^d B^e
- n_a E^u n^a E_v
+ n_a E^u n_v E^a
- n_a E^u {\epsilon^a}_{vde} n^d B^e
+ {\epsilon^u}_{abc} n^b B^c n^a E_v
- {\epsilon^u}_{abc} n^b B^c n_v E^a
+ {\epsilon^u}_{abc} n^b B^c {\epsilon^a}_{vde} n^d B^e
$
$=
E^u E_v
- n^u n_v E_a E^a
- n^u \epsilon_{daev} n^d E^a B^e
- \epsilon^{bacu} n_b E_a B_c n_v
- \epsilon^{ubca} \epsilon_{vdea} n_b B_c n^d B^e
$
Let $S_u = n^a \epsilon_{abcu} E^b B^c$
so $S_u n^u = 0$
$=
E^u E_v
- n^u n_v E_a E^a
- n^u S_v
- S^u n_v
+ \delta^{ubc}_{vde} n_b B_c n^d B^e
$
$=
E^u E_v
- n^u n_v E_a E^a
- n^u S_v
- S^u n_v
+ (
\delta^u_v \delta^b_d \delta^c_e
+ \delta^u_d \delta^b_e \delta^c_v
+ \delta^u_e \delta^b_v \delta^c_d
- \delta^u_e \delta^b_d \delta^c_v
- \delta^u_d \delta^b_v \delta^c_e
- \delta^u_v \delta^b_e \delta^c_d
) n_b B_c n^d B^e
$
$=
E^u E_v
- n^u n_v E_a E^a
+ B^u B_v
- n_v n^u B_a B^a
- \delta^u_v B_a B^a
- n^u S_v
- S^u n_v
$
${F^u}_a {F^a}_u =
E^u E_u
- n^u n_u E_b E^b
+ B^u B_u
- n_u n^u B_b B^b
- \delta^u_u B_b B^b
- n^u S_u
- S^u n_u
$
${F^u}_a {F^a}_u =
2 E^u E_u
- (n - 2) B^u B_u
$
${R^{ab}}_{cd} = {C^{ab}}_{cd} - \kappa \frac{1}{n - 2} \frac{1}{\mu_0} (
\delta^b_d {F^a}_u {F^u}_c
+ \delta^a_c {F^b}_u {F^u}_d
- \delta^b_c {F^a}_u {F^u}_d
- \delta^a_d {F^b}_u {F^u}_c
- (E^u E_u - (\frac{n}{2} - 1) B^u B_u) (
\delta^a_d \delta^b_c
- \delta^a_c \delta^b_d
)
)$