Back

Jacobi's formula:
$det(A) = det [{A^a}_b]$
$=\epsilon_{abcd} {A^a}_0 {A^b}_1 {A^c}_2 {A^d}_3$ ... for 4D spacetime indexed $0-3$
$=-\epsilon^{0123} \epsilon_{abcd} {A^a}_0 {A^b}_1 {A^c}_2 {A^d}_3$
   Summing over all even permutations of $uvrs$ (such that $-\epsilon^{uvrs}=1$) means every time the $uvrs$ indexes are permuted an even number of times, so must the associated $a$'s be rearranged to match the original ${A^a}_0 {A^b}_1 {A^c}_2 {A^d}_3$. For this to happen the $abcd$ indexes must be permuted an even number of times to match the original expression. Because $abcd$ incides have been permuted an even number of times, the $\epsilon_{abcd}$ doesn't change its sign.
   Likewise when summing the odd permutations of $uvrs$, $-\epsilon^{uvrs}=-1$, the $A$'s need to be rearranged to match the original expression, the $abcd$ indexes have to be permuted an odd number of times, and $\epsilon_{abcd}$ changes its sign as well to $-1$, so the term is each equal to the original expression.
   The total number of expressions summed is $\frac{1}{3!}$.
$=-\frac{1}{3!}\epsilon^{uvrs} \epsilon_{abcd} {A^a}_u {A^b}_v {A^c}_r {A^d}_s$
$=-\frac{1}{4!}\epsilon^{uvrse} \epsilon_{abcde} {A^a}_u {A^b}_v {A^c}_r {A^d}_s$
$det(A)=\frac{1}{4!} \delta^{uvrs}_{abcd} {A^a}_u {A^b}_v {A^c}_r {A^d}_s$
$4=\delta^a_a={A^a}_u{(A^{-1})^u}_a=({A^a}_u )(\frac{1}{3!} \delta^{uvrs}_{abcd} {A^b}_v {A^c}_r {A^d}_s / det(A))$
Therefore ${(A^{-1})^u}_a=\frac{1}{3!} \delta^{uvrs}_{abcd} {A^b}_v {A^c}_r {A^d}_s / det(A)$
${(A^{-1})^u}_a det(A)=\frac{1}{3!} \delta^{uvrs}_{abcd} {A^b}_v {A^c}_r {A^d}_s$

$\frac{\delta}{\delta{A^e}_f} det(A)=\frac{1}{4!} \delta^{uvrs}_{abcd} \frac{\delta}{\delta{A^e}_f} ({A^a}_u {A^b}_v {A^c}_r {A^d}_s)$
$=\frac{1}{4!} \delta^{uvrs}_{abcd} \frac{\delta}{\delta{A^e}_f} ({A^a}_u {A^b}_v {A^c}_r {A^d}_s)$
$=\frac{1}{4!} \delta^{uvrs}_{abcd}(\frac{\delta}{\delta{A^e}_f} {A^a}_u {A^b}_v {A^c}_r {A^d}_s + {A^a}_u \frac{\delta}{\delta{A^e}_f} {A^b}_v {A^c}_r {A^d}_s + {A^a}_u {A^b}_v \frac{\delta}{\delta{A^e}_f} {A^c}_r {A^d}_s + {A^a}_u {A^b}_v {A^c}_r \frac{\delta}{\delta{A^e}_f} {A^d}_s)$
$=\frac{1}{4!} \delta^{uvrs}_{abcd}(\delta^a_e \delta^e_u {A^b}_v {A^c}_r {A^d}_s + {A^a}_u \delta^b_e \delta^e_v {A^c}_r {A^d}_s + {A^a}_u {A^b}_v \delta^c_e \delta^e_r {A^d}_s + {A^a}_u {A^b}_v {A^c}_r \delta^d_e \delta^e_s)$
$=\frac{1}{4!} \delta^{uvrs}_{abcd}\delta^a_e \delta^e_u ({A^b}_v {A^c}_r {A^d}_s + {A^b}_v {A^c}_r {A^d}_s + {A^b}_v {A^c}_r {A^d}_s + {A^b}_v {A^c}_r {A^d}_s)$
$=\frac{1}{3!} \delta^{fvrs}_{ebcd} {A^b}_v {A^c}_r {A^d}_s$
$\frac{\delta}{\delta{A^e}_f} det(A)={(A^{-1})^f}_e det(A)$ ...aka... $\frac{\delta}{\delta A} det(A)=A^{-T}det(A) =adj(A^T)$
$(\frac{\delta}{\delta{A^e}_f} det(A))/det(A)={(A^{-1})^f}_e$
$\frac{\delta}{\delta{A^e}_f} ln|det(A)|={(A^{-1})^f}_e$
$\delta ln|det(A)|={(A^{-1})^f}_e \delta{A^e}_f$ ...aka... $\delta ln|det(A)| = tr(A^{-1} dA)$

Using Jacobi's formula: $ \delta det(A) = tr(adj(A) \delta A) $
$ \delta g = \delta det(g_{ab}) = g g^{ab} \delta g_{ab} $
$ 0 = \delta tr(n) = \delta (g_{ab} g^{ab}) = g^{ab} \delta g_{ab} + g_{ab} \delta g^{ab} $
$ g^{ab} \delta g_{ab} = -g_{ab} \delta g^{ab} $
$ \delta g = -g g_{ab} \delta g^{ab} $

Differential of covariant derivative of connection:
$ \nabla_d \delta {\Gamma^a}_{bc} = e_d (\delta {\Gamma^a}_{bc}) + \delta {\Gamma^e}_{bc} {\Gamma^a}_{de} - \delta {\Gamma^a}_{ec} {\Gamma^e}_{db} - \delta {\Gamma^a}_{be} {\Gamma^e}_{dc} $
$ \nabla_u \delta {\Gamma^u}_{ba} - \nabla_b \delta {\Gamma^u}_{ua} = e_u (\delta {\Gamma^u}_{ba}) + \delta {\Gamma^v}_{ba} {\Gamma^u}_{uv} - \delta {\Gamma^u}_{va} {\Gamma^v}_{ub} - \delta {\Gamma^u}_{bv} {\Gamma^v}_{ua} - e_b (\delta {\Gamma^u}_{ua}) - \delta {\Gamma^v}_{ua} {\Gamma^u}_{bv} + \delta {\Gamma^u}_{va} {\Gamma^v}_{bu} + \delta {\Gamma^u}_{uv} {\Gamma^v}_{ba} $
$ = e_u (\delta {\Gamma^u}_{ba}) - e_b (\delta {\Gamma^u}_{ua}) + {\Gamma^u}_{uv} \delta {\Gamma^v}_{ba} + \delta {\Gamma^u}_{uv} {\Gamma^v}_{ba} - \delta {\Gamma^v}_{ua} {\Gamma^u}_{bv} - {\Gamma^v}_{ua} \delta {\Gamma^u}_{bv} - \delta {\Gamma^u}_{va} ({\Gamma^v}_{ub} - {\Gamma^v}_{bu}) $
$ = e_u (\delta {\Gamma^u}_{ba}) - e_b (\delta {\Gamma^u}_{ua}) + {\Gamma^u}_{uv} \delta {\Gamma^v}_{ba} + \delta {\Gamma^u}_{uv} {\Gamma^v}_{ba} - \delta {\Gamma^v}_{ua} {\Gamma^u}_{bv} - {\Gamma^v}_{ua} \delta {\Gamma^u}_{bv} - \delta {\Gamma^u}_{va} ({T^v}_{ub} + {c_{ub}}^v) $

Differential of Ricci curvature tensor:
$ \delta R_{ab} = \delta {R^u}_{aub}$
$ = \delta ( e_u ({\Gamma^u}_{ba}) - e_b ({\Gamma^u}_{ua}) + {\Gamma^u}_{uv} {\Gamma^v}_{ba} - {\Gamma^u}_{bv} {\Gamma^v}_{ua} - {\Gamma^u}_{va} {c_{ub}}^v )$
$ = \delta e_u ({\Gamma^u}_{ba}) - \delta e_b ({\Gamma^u}_{ua}) + \delta {\Gamma^u}_{uv} {\Gamma^v}_{ba} + {\Gamma^u}_{uv} \delta {\Gamma^v}_{ba} - \delta {\Gamma^u}_{bv} {\Gamma^v}_{ua} - {\Gamma^u}_{bv} \delta {\Gamma^v}_{ua} - \delta {\Gamma^u}_{va} {c_{ub}}^v - {\Gamma^u}_{va} \delta {c_{ub}}^v $
...using $\delta e_a (...) = e_a (\delta ...)$ (TODO prove this) ...
$ = e_u (\delta {\Gamma^u}_{ba}) - e_b (\delta {\Gamma^u}_{ua}) + {\Gamma^u}_{uv} \delta {\Gamma^v}_{ba} + \delta {\Gamma^u}_{uv} {\Gamma^v}_{ba} - \delta {\Gamma^u}_{bv} {\Gamma^v}_{ua} - {\Gamma^u}_{bv} \delta {\Gamma^v}_{ua} - \delta {\Gamma^u}_{va} {c_{ub}}^v - {\Gamma^u}_{va} \delta {c_{ub}}^v $
So $\delta R_{ab} + {\Gamma^u}_{va} \delta {c_{ub}}^v = \nabla_u \delta {\Gamma^u}_{ba} - \nabla_b \delta {\Gamma^u}_{ua} + \delta {\Gamma^u}_{va} {T^v}_{ub}$

MTW states the Palatini identity is: $ \delta R_{ab} = \nabla_u \delta {\Gamma^u}_{ab} - \nabla_b \delta {\Gamma^u}_{au} $
TODO what is it in the presence of torsion and commutation?
TODO save the torsion and non-coordinate basis stuff for the tetrad aka Einstein-Cartan section

TODO show $ \int \sqrt{-g} g^{ab} (\nabla_u \delta {\Gamma^u}_{ab} - \nabla_b \delta {\Gamma^u}_{au}) dx^4 = 0 $

Einstein-Hilbert Action Principle:
$ S_{vacuum} = \int \sqrt{-g} R dx^4 $
$ \delta S_{vacuum} = \delta \int \sqrt{-g} R dx^4 $
$ = \int \delta (\sqrt{-g} R) dx^4 $
$ = \int ( \delta \sqrt{-g} R + \sqrt{-g} \delta R ) dx^4 $
$ = \int ( \frac{1}{2} \frac{1}{\sqrt{-g}} \delta (-g) R + \sqrt{-g} \delta (R_{ab} g^{ab}) ) dx^4 $
$ = \int ( \frac{1}{2g} \sqrt{-g} \delta g R + \sqrt{-g} (R_{ab} \delta g^{ab} + \delta R_{ab} g^{ab})) dx^4 $
$ = \int \sqrt{-g} ( \frac{1}{2g} \delta g R + R_{ab} \delta g^{ab} + g^{ab} \delta R_{ab} ) dx^4 $
$ = \int \sqrt{-g} ( -\frac{1}{2g} g g^{ab} + R \delta g^{ab} + g^{ab} \delta R_{ab}) dx^4 $
$ = \int \sqrt{-g} ( \delta g^{ab} (R_{ab} - \frac{1}{2} R g_{ab}) + g^{ab} \delta R_{ab} ) dx^4 $
$ = \int \sqrt{-g} ( \delta g^{ab} G_{ab} + g^{ab} \delta R_{ab} ) dx^4 $
$ = \int \sqrt{-g} ( \delta g^{ab} G_{ab} + g^{ab} (\nabla_u \delta {\Gamma^u}_{ab} - \nabla_b \delta {\Gamma^u}_{au}) ) dx^4 $
$ \delta S_{vacuum} = \int \sqrt{-g} ( \delta g^{ab} G_{ab} ) dx^4 $

$ S_{matter} = -16 \pi \int \sqrt{-g} \mathcal{L}_{matter} dx^4 $
$ \delta S_{matter} = -16 \pi \int \delta (\sqrt{-g} \mathcal{L}_{matter}) dx^4 $
$ = -16 \pi \int ( \frac{1}{2} \frac{1}{\sqrt{-g}} \delta(-g) \mathcal{L}_{matter} + \sqrt{-g} \delta \mathcal{L}_{matter} ) dx^4 $
$ = -8 \pi \int \sqrt{-g} ( 2 \delta \mathcal{L}_{matter} - \delta g^{ab} \mathcal{L}_{matter} ) dx^4 $
Let $ T_{ab} = 2 \frac{\delta}{\delta g^{ab}} \mathcal{L}_{matter} - \mathcal{L}_{matter} $
$ \delta S_{matter} = -8 \pi \int \sqrt{-g} T_{ab} \delta g^{ab} dx^4 $

$ S_{EH} = S_{vacuum} + S_{matter} $
$ 0 = \delta S_{EH} = \int \sqrt{-g} \delta g^{ab} ( G_{ab} - 8 \pi T_{ab} ) dx^4 $
$ 0 = G_{ab} - 8 \pi T_{ab} $
Einstein Field Equation:
$ G_{ab} = 8 \pi T_{ab} $
TODO consider boundary terms as well?

As one-forms:
${G^a}_b e^b = 8 \pi {T^a}_b e^b$
$G^a = 8 \pi T^a$

$-\frac{1}{2} {\epsilon^{uac}}_e (\star e^e(d^2 e_c)) (e_u) = 8 \pi {T^a}_b e^b$


Back