- #1
Kevin_spencer2
- 29
- 0
If we have Einstein-Hilbert Lagrangian so:
[tex] \mathcal L = (-g)^{1/2}R [/tex] for R Ricci scalar (¿?) then my question perhaps mentioned before is if we can split the metric into:
[tex] g_{ab}= Ndt^{2}+g_{ij} dx^{i} dx^{j} [/tex] N=N(t) 'lapse function'
then i would like to know if somehow you can split the Lagrangian into:
[tex] \mathcal L = (-g)^{1/2}R=N^{1/2}(g^{3})^{1/2}g^{00}R_{00}+N^{1/2}(g^{3})^{1/2}g^{ij}R_{ij} [/tex]
Following Einstein equations then [tex] R_{00}=T_{00}= \rho [/tex] 'energy density equation' and
[tex] \iiint (g^{3})^{1/2}R^{(3)}d^{3}x = 2T [/tex]
[tex] \iint (g^{(3)})^{1/2}R_{00}=\mathcal H = E [/tex]
Is some kind of Kinetic energy so the Einstein-Hilbert Lagrangian takes the form:
[tex] \mathcal L =\int_{a}^{b} dt(T-V) [/tex] a Kinetic plus potential term.
[tex] \mathcal L = (-g)^{1/2}R [/tex] for R Ricci scalar (¿?) then my question perhaps mentioned before is if we can split the metric into:
[tex] g_{ab}= Ndt^{2}+g_{ij} dx^{i} dx^{j} [/tex] N=N(t) 'lapse function'
then i would like to know if somehow you can split the Lagrangian into:
[tex] \mathcal L = (-g)^{1/2}R=N^{1/2}(g^{3})^{1/2}g^{00}R_{00}+N^{1/2}(g^{3})^{1/2}g^{ij}R_{ij} [/tex]
Following Einstein equations then [tex] R_{00}=T_{00}= \rho [/tex] 'energy density equation' and
[tex] \iiint (g^{3})^{1/2}R^{(3)}d^{3}x = 2T [/tex]
[tex] \iint (g^{(3)})^{1/2}R_{00}=\mathcal H = E [/tex]
Is some kind of Kinetic energy so the Einstein-Hilbert Lagrangian takes the form:
[tex] \mathcal L =\int_{a}^{b} dt(T-V) [/tex] a Kinetic plus potential term.
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