# Einstein's Basis for Equivalence in his Field Equations

1. Nov 2, 2013

### McGuire

The following is a question regarding the derivation of Einstein's field equations.

Background
In deriving his equations, it is my understanding that Einstein equated the Einstein Tensor Gμv and the Cosmological Constant*Metric Tensor with the Stress Energy Momentum Tensor Tμv term simply because the covariant derivative of all three terms equals zero.

Rμv - (1/2)*gμv*R + $\Lambda$*gμv = (8*pi*G)/(c4)*Tμv

Question
Is this basis for equivalence (that terms are equivalent if their covariant derivatives are equal) standard practice in mathematics, or did Einstein take a leap of faith?

Thank you very much for your time! Please let me know if I can clarify my question.

Last edited: Nov 2, 2013
2. Nov 2, 2013

### WannabeNewton

Well we know that $\nabla^{\mu}T_{\mu\nu} = 0$ has to hold for the total matter field so the expression on the left hand side should obey the same thing as well and this definitely constrains the possible choice of tensor expressions for the left hand side but this is not the end all be all of Einstein's route to the field equation for obvious reasons. You can find tons of historical information about his thought process online just by googling.

This is not how it is done nowadays however. For classical GR you would start with an action for the gravitational field, say the Hilbert action, and derive the Einstein field equation from that using the usual variational principle. The condition $\nabla^{\mu}T_{\mu\nu} = 0$ will also follow suit from diffeomorphism invariance.

3. Nov 2, 2013

### McGuire

Excellent. Thank you!

4. Nov 2, 2013

### WannabeNewton

No problem! For example here is one heuristic route. We know in Newtonian gravity that the relative tidal acceleration of two nearby particles with separation vector $\vec{\xi}$ is given by $\vec{a} = -(\vec{\xi}\cdot \vec{\nabla})\vec{\nabla} \varphi$ where $\varphi$ is the Newtonian gravitational potential. In GR, the relative tidal acceleration of two nearby worldlines is given by $a^{\mu} = u^{\gamma}\nabla_{\gamma}(u^{\nu}\nabla_{\nu}\xi^{\mu}) = -R_{\gamma\beta\nu}{}{}^{\mu}\xi^{\beta}u^{\gamma}u^{\nu}$ where $\xi^{\mu}$ is the separation vector again (now a 4-vector) and $u^{\mu}$ is the 4-velocity of the reference worldline.

So we have a natural correspondence between $R_{\gamma\beta\nu}{}{}^{\mu}u^{\gamma}u^{\nu}$ and $\partial_{\beta}\partial^{\mu}\varphi$ because $(\vec{\xi}\cdot \vec{\nabla})\vec{\nabla} \varphi$ in index notation is just $\xi^{\beta}\partial_{\beta}\partial^{\mu}\varphi$. We also know that the mass density is given by $\rho = T_{\mu\nu}u^{\mu}u^{\nu}$ so Poisson's equation $\nabla^2 \varphi = 4\pi \rho$ suggests that we try the field equation $R_{\mu\nu} = 4\pi T_{\mu\nu}$. This won't really be satisfactory because $\nabla^{\mu}T_{\mu\nu} = 0$ constrains us to consider divergence free Ricci tensors alone: $\nabla^{\mu}R_{\mu\nu} = 0$. But from $\nabla^{\mu}G_{\mu\nu} = 0$ we know this can be true if and only if $\nabla^{\mu}R = 0$ which is a completely unphysical constraint on our field equation. From here we can easily remedy the problem by using $G_{\mu\nu}$ instead and arrive at the Einstein field equation.