Why, when deriving the Einstein equations, do we want the energy-momentum tensor to be divergence free?

Question

So when deriving the Einstein equation we assume $\nabla_\mu T^{\mu\nu}=0$. Now I get this is not true energy conservation but why do we assume this, it seems vital for the einstein tensor to have the form it does? Or is there another way to get the Einstein tensor which doesn't require it to be divergence free (and the divergence free-ness is sort of a coincidence)?

Answer 1

So when deriving the Einstein equation we assume $\nabla_\mu T^{\mu\nu}=0$. Now I get this is not true energy conservation, but why do we assume this?

Actually it is energy and momentum conservation together.

To see this consider the meaning of the $T^{\mu\nu}$ components, with $m$ and $n$ being spatial indexes ($1,2,3$).
(See also Stress-energy tensor - Identifying the components of the tensor)

• $T^{00}$ is energy density.
• $T^{m0}$ is energy flux density.
• $T^{0n}$ is momentum density
• $T^{mn}$ is momentum flux density.

Therefore the divergence free-ness $\nabla_\mu T^{\mu\nu}=0$ has the following parts:

• For $\nu=0$:
  $$\nabla_0 T^{00}+\nabla_m T^{m0}=0$$ This is energy conservation, written as a continuity equation.
• For $\nu=1,2,3$:
  $$\nabla_0 T^{0n}+\nabla_m T^{mn}=0$$ This is momentum conservation, written as a continuity equation.

Answer 2

It is not necessary to assume that $\nabla_\mu T^{\mu\nu}=0$, for deriving the Einstein equation. The best argument for the form of the Einstein equation is usually considered to be Lovelock's theorem. This theorem states the form of the Einstein equation follows from the assumption that it follows from an action principle, provided that the gravitational action satisfies:

• It depends only on the metric (and its derivatives.)
• It is at most second order in derivatives of the metric.
• It is local.
• Spacetime is four-dimensional.

That the Einstein tensor is divergence free, then follows as a consequence and is not an a priori assumption.

Answer 3

Well, this is energy conservation essentially and a bit more. It also states that momentum is conserved along axis.

I believe, that the answer to this comes from the classical field theory. If you consider a Lagrangian that is translationally invariant $\mathcal{L}(x^\mu)$. By performing a transformation $x^{\mu} \rightarrow x^\mu + \epsilon^\mu$ and by applying Noether's theorem you get that the conserved current is of the following form:

$j^\mu = \epsilon^{\lambda} T^{\mu}_{\lambda}$

But this really gives you four conserved currents, because you have 4 independent choices for $\epsilon^{\lambda}$ (for example $(1,0,0,0)$). Since the current is conserved, we have:

$\partial_{\mu} j^{\mu}=0$

Which is equivalent to your divergence-free condition.

Now, in General Relativity this is true only locally where the space is Minkowski. But, you can always get away with replacing partial with a covariant derivative.

I am not sure if there are theories in which it is sensible not to have divergence free stress tensor. However, a useful thing to remember is that you can always modify your stress-energy tensor by adding $\partial_\lambda K^{\lambda\mu}_{\rho}$ where $K^{\lambda\mu}_{\rho}$ is anti-symmetric in $\lambda$ and $\mu$ which guarantees that the new stress energy tensor is divergence free.

Answer 4

Within a normal neighbourhood, we have that the metric tensor can locally be expressed by the flat metric, and the Christoffel symbols vanish. More specifically, within a distance $x$ from the point $p$, we have

\begin{eqnarray} g &=& \eta + \mathcal{O}(x^2)\\ \Gamma &=& \mathcal{O}(x) \end{eqnarray}

So in that neighbourhood, we have

\begin{eqnarray} \nabla_\mu T ^{\mu\nu} = \partial_\mu T^{\mu\nu} + \mathcal{O}(x) \end{eqnarray}

If the divergence is zero, then imagine the measured four-momentum as measured by an observer $u$, $p^\mu = T^{\mu\nu} u_\nu$ :

\begin{eqnarray} \nabla_\mu (T ^{\mu\nu} u_\nu) &=& T ^{\mu\nu} \nabla_\mu u_\nu + u_\nu \nabla_\mu T ^{\mu\nu} \\ &=& T ^{\mu\nu} u_\mu a_\nu + u_\nu \nabla_\mu T ^{\mu\nu} \end{eqnarray}

Therefore, if we assume an acceleration of zero (so that we're dealing with an inertial observer), we have

\begin{eqnarray} \partial_\mu (p^\mu) &=& \mathcal{O}(x) \end{eqnarray}

This means that locally, the four-momentum is a conserved current, with a deviation from conservation roughly equivalent to the curvature of spacetime times the distance (for any local experiment this is extremely small). So far, any experiment we have conducted has shown local energy conservation.

A non-zero divergence on a scale larger than the curvature of spacetime would imply differences with experiments. A very small divergence is still possible, and indeed part of some theories, but so far none of these have been measured.

Answer 5

OP is right that the equation $\nabla_\mu T^{\mu\nu}=0$ doesn't by itself signify a conservation law (a Killing symmetry of the metric is also needed). Rather the equation $\nabla_\mu T^{\mu\nu}=0$ is a consequence of diffeomorphism symmetry of the matter theory. For details, see e.g. my Phys.SE answer here.

Answer 6

When expressed in terms of the stress tensor density $$𝔗^μ_ν = \sqrt{|g|} g^{μρ} T_{ρν},$$ rather than the stress tensor $T_{ρν}$, ifself, the continuity equation is just a transport law, such as you would see in fluid dynamics - but with the added twist of the connection coefficients: $$0 = ∇_μ 𝔗^μ_ν = ∂_μ 𝔗^μ_ν - Γ^ρ_{μν} 𝔗^μ_ρ.$$

This is the continuum form of the geodesic law, which can be recovered by setting: $$𝔗^μ_ν(x) = p_ν(s) \frac{dx^μ}{ds} δ³(x - x(s)),$$ where $p_ν(s)$ is the momentum of a body on a worldline given by $x = x(s)$, where $s$ is the proper time. Take the divergence, apply the chain rule and integrate by parts: $$\begin{align} ∂_μ 𝔗^μ_ν &= p_ν(s) \frac{dx^μ}{ds} ∂_μ\left(δ³(x - x(s))\right) \\ &= -p_ν(s) \frac{d}{ds}\left(δ³(x - x(s))\right) \\ &= \frac{dp_ν}{ds} δ³(x - x(s)). \end{align}$$ Subtract the extra contribution containing the connection coefficients: $$\begin{align} Γ^ρ_{μν} 𝔗^μ_ρ &= Γ^ρ_{μν}(x) p_ρ(s) \frac{dx^μ}{ds} δ³(x - x(s)) \\ &= Γ^ρ_{μν}(x(s)) p_ρ(s) \frac{dx^μ}{ds} δ³(x - x(s)). \end{align}$$ Combine the results: $$ 0 = ∂_μ 𝔗^μ_ν - Γ^ρ_{μν} 𝔗^μ_ρ = \left(\frac{dp_ν}{ds} - Γ^ρ_{μν}(x(s)) p_ρ(s) \frac{dx^μ}{ds}\right) δ³(x - x(s)), $$ and integrate over the the 3-surface layered at $s$. This assumes the space-time is foliated into space-like layers in the vinicity of the worldline: $$\frac{dp_ν}{ds} - Γ^ρ_{μν}(x(s)) p_ρ(s) \frac{dx^μ}{ds} = 0.$$

That's the momentum-space version of the geodesic law. Substituting in the relation: $$p_ν = m g_{μν} \frac{dx^μ}{ds},$$ then after a little reworking, using the chain rule and the metrical relation for the connection: $$\frac{dg_{μν}}{ds} = ∂_σg_{μν} \frac{dx^σ}{ds} = \left(g_{ρν} Γ^ρ_{σμ} + g_{μρ} Γ^ρ_{σν}\right) \frac{dx^σ}{ds},$$ setting $m = 1$, which you can do because of the equivalence principle, and inverting the metric, you can convert this to the coordinate-space version of the geodesic law: $$\begin{align} 0 &= \frac{d}{ds}\left(g_{μν} \frac{dx^μ}{ds}\right) - Γ^ρ_{μν} g_{ρσ} \frac{dx^σ}{ds} \frac{dx^μ}{ds} \\ &= \left(g_{ρν} Γ^ρ_{σμ} + g_{μρ} Γ^ρ_{σν} - g_{ρσ} Γ^ρ_{μν}\right) \frac{dx^σ}{ds} \frac{dx^μ}{ds} + g_{μν} \frac{d^2x^μ}{ds^2} \\ &= g_{ρν} \left(\frac{d^2x^ρ}{ds^2} + Γ^ρ_{σμ} \frac{dx^σ}{ds} \frac{dx^μ}{ds}\right) \\ 0 &= \frac{d^2x^ρ}{ds^2} + Γ^ρ_{σμ} \frac{dx^σ}{ds} \frac{dx^μ}{ds}. \end{align}$$

It bears pointing out that the attributes "density" belong to the components of the tensor density $𝔗^μ_ν$, not the tensor $T_{ρν}$! Unfortunately, this error and misunderstanding is very widespread in the literature; and leads to further errors down the line, such as the wrong power of $c$ being attached to the coupling coefficient in Einstein's equations (it's $c^3$, not $c^4$).

The stress tensor density and canonical stress tensor density, in its diagonal components, $𝔗^μ_μ$ and space-like components $𝔗^i_j$ ($i, j = 1, 2, 3$), all have the dimension of action divided by space-time volume. That is

(Mass×Length²/Time)/(Length³×Time).

This is equivalent

to energy density: (Mass×Length²/Time²)/Length³,
to momentum flux density: (Mass×Length/Time)/Length³×Length/Time
and to stress and pressure: (Mass×Length/Time²)/Length².

Call this dimension P.

The components $𝔗^0_j$ ($j = 1, 2, 3$) all have the dimension of

P×Time/Length = (Mass×Length/Time)/Length³: momentum density.

Finally, the components $𝔗^i_0$ ($i = 1, 2, 3$) all have the dimension of

P×Length/Time = (Mass×Length²/Time²)/Length³×Length/Time: energy flux.