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Proving one field equation leads to the other

+ 5 like - 0 dislike
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Assume that the universe is homogenous and isotropic, and the following equation holds:

\begin{equation}R_{00}-\frac{1}{2}g_{00}R=8\pi GT_{00}; \space \space \nabla_{\mu}T^{\mu 0}=0.\end{equation}

How do I prove that the following equations are identically satisfied provided that the above two are satisfied?

\begin{equation}R_{0i}-\frac{1}{2}g_{0i}R=8\pi GT_{0i}; \space \space R_{ij}-\frac{1}{2}g_{ij}R=8\pi GT_{ij}; \space \space \nabla_{\mu}T^{\mu i}=0.\end{equation}

My approach was to write $g_{00}=1$ and $g_{ij}=-a^2\gamma_{ij}$ and evaluate the Ricci tensors and so on, but I know this is not the way to do it. Can anyone suggest me the way?

This post imported from StackExchange Physics at 2014-06-21 08:52 (UCT), posted by SE-user titanium
asked Jun 20, 2014 in Theoretical Physics by titanium (55 points) [ no revision ]

2 Answers

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Here is a simple approach that might work. Start by defining

$$F_{\mu\nu} \equiv G_{\mu\nu} - T_{\mu\nu}$$

where $G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R$ is the Einstein tensor. Now from what you know we have $$F_{00} = 0$$

$$\nabla^{\mu}F_{\mu0} = 0$$

You must show that $F_{\mu\nu} = 0$. Writing out the last equation gives

$$0 = \partial_{i} F^{i 0} + \Gamma^{\mu}_{\mu \alpha}F^{\alpha 0} + \Gamma^{0}_{\mu \alpha}F^{\mu \alpha}$$

Homogenous and isotropic implies that gradients vanish and that $F^{11}=F^{22}=F^{33}$ so

$$0 = a^2 H \delta_{ij} F^{i j}$$

This shows that $F_{00} = F_{11} = F_{22} = F_{33} = 0$. Now to show $F_{ij} = 0$ for $i\not =j$ you might need some additional assumptions on the energy momentum tensor $T^{\mu\nu}$, for example $T^{ij} = 0$.

This post imported from StackExchange Physics at 2014-06-21 08:52 (UCT), posted by SE-user Winther
answered Jun 20, 2014 by Winther (30 points) [ no revision ]
+ 1 like - 0 dislike

First what we know: $G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2}g_{\mu \nu} R$ and $T_{\mu \nu}$ are tensors in that they transform properly under coordinate transformations ($G_{\mu \nu}$ by construction and $T_{\mu \nu}$ because of the EFEs), so it doesn't matter which frame we do our measurements in, this tensor equation will always hold.

Suppose that a comoving observer takes careful measurements in his frame and finds the first equation to be true. This is a special case of how we determine what an observer with an arbitrary four-velocity will measure, which is the contraction with that four-velocity $$G_{\mu \nu} u^{\mu} u^{\nu} = 8\pi T_{\mu \nu} u^{\mu} u^{\nu}$$ Then imagine other observers with four velocities of the form $u_i^{\alpha} = Ae_0^{\alpha} + B e_i^{\alpha}$, where $e_0$ denotes the unit vector in the time direction, $e_1$ denotes the unit vector in the $1/x/r/$whathaveyou direction, etc., and $A$ and $B$ are normalization factors. The above equation is an invariant scalar equation and from this fact and a plethora of observers we can build up the rest of the relations. The same procedure can be applied to the energy conservation equation, only now we are contracting $u_{\nu}\nabla_{\mu} T^{\mu \nu}$

This post imported from StackExchange Physics at 2014-06-21 08:52 (UCT), posted by SE-user Jordan
answered Jun 20, 2014 by Jordan (15 points) [ no revision ]

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