# The divergence and conserved quantities

Recall that the divergence of a vector field $${u}$$ can be generalized to a pseudo-Riemannian manifold of signature $${\left(r,s\right)}$$ by defining $${\mathrm{div}(u)\equiv(-1)^{s}*\mathrm{d}(*(u^{\flat}))}$$. Also recalling that $${i_{u}\Omega=(-1)^{s}*(u^{\flat})}$$ and $${A=(*A)\Omega}$$ for $${A\in\Lambda^{n}M^{n}}$$, we have $${\mathrm{d}(i_{u}\Omega)=(-1)^{s}\mathrm{d}(*(u^{\flat}))=(-1)^{s}*\mathrm{d}(*(u^{\flat}))\Omega=\mathrm{div}(u)\Omega}$$. Using $${i_{u}\mathrm{d}+\mathrm{d}i_{u}=L_{u}}$$ we then arrive at $${\mathrm{div}(u)\Omega=L_{u}\Omega}$$, or as it is more commonly written

$$\displaystyle \mathrm{div}(u)\mathrm{d}V=L_{u}\mathrm{d}V.$$

Thus we can say that $${\mathrm{div}(u)}$$ is “the fraction by which a unit volume changes when transported by the flow of $${u}$$,” and if $${\mathrm{div}(u)=0}$$ then we can say that “the flow of $${u}$$ leaves volumes unchanged.” Expanding the volume element in coordinates $${x^{\lambda}}$$ we can obtain an expression for the divergence in terms of these coordinates,

$$\displaystyle \mathrm{div}(u)=\frac{1}{\sqrt{\left|\mathrm{det}(g)\right|}}\partial_{\lambda}\left(u^{\lambda}\sqrt{\left|\mathrm{det}(g)\right|}\right).$$

Note that both this metric-dependent expression and the expression $${\nabla_{a}u^{a}}$$ (sometimes called the covariant divergence) in terms of the Levi-Civita connection are coordinate-independent and equal to $${\partial_{a}u^{a}}$$ in Riemann normal coordinates, confirming our expectation that for zero torsion we have

$$\require{cancel}\displaystyle \mathrm{div}(u)\overset{\cancel{T}}{=}\nabla_{a}u^{a}.$$

Recall however that the connection coefficients do not vanish in Riemann normal coordinates for non-zero torsion; in this case we can use the contorsion tensor contraction $${K^{a}{}_{ba}=T^{a}{}_{ab}}$$ to relate the pseudo-Riemannian divergence to the covariant divergence by

$$\displaystyle \mathrm{div}(u) =\nabla_{a}u^{a}-T^{a}{}_{ab}u^{b}.$$

 Δ Note that the different symbols and names given here for the pseudo-Riemannian divergence versus the covariant divergence are oftentimes not distinguished, since they are the same for zero torsion. The distinction also vanishes if the torsion is completely anti-symmetric, i.e. if it leaves geodesics unchanged.

The divergence measures the change in volume due to the flow. Here we assume the vector field $${u}$$ has unit length at point $${p}$$, and choose an orthonormal frame which aligns $${e_{2}}$$ with $${u}$$. Each covariant derivative extends a face of the volume, with their sum being proportional to the total change in volume. Note that the upper right corner is of order $${\varepsilon^{4}}$$ and so can be neglected, and e.g. any component of $${\nabla_{1}u}$$ orthogonal to $${e_{1}}$$ leaves the volume unchanged, since a more accurate depiction would include the volume with edge $${-\varepsilon e_{1}}$$, where by linearity this component would be in the opposite direction and thus cancel the volume change. Also note that each component of non-zero torsion would subtract from each edge, matching the algebraic result.

We can derive many useful relations using the above results. Abbreviating $${\sqrt{g}\equiv\sqrt{\left|\det\left(g\right)\right|}}$$ and including torsion for completeness, we expand both sides of the coordinate divergence expression to get

$$\displaystyle \partial_{\lambda}\sqrt{g}=\sqrt{g}\left(\Gamma^{\mu}{}_{\lambda\mu}-T^{\mu}{}_{\mu\lambda}\right).$$

From $${\det\left(\exp\left(g\right)\right)=\exp\left(\mathrm{tr}\left(g\right)\right)\Rightarrow\ln\left(\det\left(g\right)\right)=\mathrm{tr}\left(\ln\left(g\right)\right)}$$, we can take the derivative of the components to get

\begin{aligned}\frac{1}{\det\left(g\right)}\partial_{\lambda}\left(\det\left(g\right)\right) & =\mathrm{tr}\left(g^{-1}\partial_{\lambda}g\right)\\
& =g^{\mu\nu}\partial_{\lambda}g_{\mu\nu}\\
\Rightarrow\partial_{\lambda}\sqrt{g} & =\frac{1}{2}\sqrt{g}g^{\mu\nu}\partial_{\lambda}g_{\mu\nu}\\
\Rightarrow\Gamma^{\mu}{}_{\lambda\mu}-T^{\mu}{}_{\mu\lambda} & =\frac{1}{2}g^{\mu\nu}\partial_{\lambda}g_{\mu\nu}.
\end{aligned}

Applying the above to the Lie derivative and using its coordinate expression then gives us

\begin{aligned}L_{u}\sqrt{g} & =\sqrt{g}\mathrm{div}\left(u\right)\\
\Rightarrow\mathrm{div}\left(u\right) & =\frac{1}{2}g^{\mu\nu}L_{u}g_{\mu\nu}.
\end{aligned}

Finally, for an anti-symmetric tensor $${F^{\mu\nu}}$$ and symmetric tensor $${G^{\mu\nu}}$$, it is not hard to see that

\begin{aligned}\nabla_{\nu}F^{\mu\nu}-T^{\lambda}{}_{\lambda\nu}F^{\mu\nu} & =\frac{1}{\sqrt{g}}\partial_{\nu}\left(\sqrt{g}F^{\mu\nu}\right)-\frac{1}{2}T^{\mu}{}_{\lambda\nu}F^{\lambda\nu},\\
\nabla_{\nu}G_{\mu}{}^{\nu}-T^{\lambda}{}_{\lambda\nu}G_{\mu}{}^{\nu} & =\frac{1}{\sqrt{g}}\partial_{\nu}\left(G_{\mu}{}^{\nu}\sqrt{g}\right)-\Gamma^{\lambda}{}_{\mu\nu}G_{\lambda}{}^{\nu}\\ & =\frac{1}{\sqrt{g}}\partial_{\nu}\left(G_{\mu}{}^{\nu}\sqrt{g}\right)-\frac{1}{2}\partial_{\mu}g_{\lambda\nu}G^{\lambda\nu}+T^{\lambda}{}_{\mu\nu}G_{\lambda}{}^{\nu}.
\end{aligned}

The left side of these equations can be used to extend the divergence to general tensors by operating on the last index, but other conventions are also in use. All of these expressions are more commonly presented with zero torsion.

Using the relation $${\mathrm{div}(u)\Omega=\mathrm{d}(i_{u}\Omega)}$$ above, along with Stokes’ theorem, we again recover the classical divergence theorem

\displaystyle \begin{aligned}\int_{V}\mathrm{div}(u)\mathrm{d}V & =\int_{\partial V}i_{u}\mathrm{d}V\\ & =\int_{\partial V}\left\langle u,\hat{n}\right\rangle \mathrm{d}S, \end{aligned}

where $${V}$$ is an $${n}$$-dimensional compact submanifold of $${M^{n}}$$, $${\hat{n}}$$ is the unit normal vector to $${\partial V}$$, and $${\mathrm{d}S\equiv i_{\hat{n}}\mathrm{d}V}$$ is the induced volume element (“surface element”) for $${\partial V}$$. In the case of a Riemannian metric, this can be thought of as reflecting the intuitive fact that “the change in a volume due to the flow of $${u}$$ is equal to the net flow across that volume’s boundary.” If $${\mathrm{div}(u)=0}$$ then we can say that “the net flow of $${u}$$ across the boundary of a volume is zero.” We can also consider an infinitesimal $${V}$$, so that the divergence at a point measures “the net flow of $${u}$$ across the boundary of an infinitesimal volume.”

If we choose an orthonormal frame with $${e_{1}=\hat{n}}$$, the divergence theorem can be written

\begin{aligned}\int_{V}\mathrm{div}(u)\mathrm{d}V & =\int_{\partial V}u^{1}\mathrm{d}S,\end{aligned}

and if we can choose coordinates with $${x^{1}}$$ constant on $${\partial V}$$ and normal to it, the divergence theorem can be written

\begin{aligned}\int_{V}\partial_{\lambda}\left(u^{\lambda}\sqrt{g}\right)\mathrm{d}^{n}x & =\int_{\partial V}\sqrt{g}dx^{1}\left(u\right)\mathrm{d}^{n-1}x\\
& =\int_{\partial V}u^{1}\sqrt{g}\mathrm{d}^{n-1}x,
\end{aligned}

where $${\mathrm{d}^{n}x\equiv\mathrm{d}x^{1}\wedge\cdots\wedge\mathrm{d}x^{n}}$$ and $${\mathrm{d}^{n-1}x\equiv\mathrm{d}x^{2}\wedge\cdots\wedge\mathrm{d}x^{n}}$$. Alternatively, we get the same result if we choose a cylindrical volume $${V}$$ whose boundary is the flow of $${u}$$, and then (if possible) choose coordinates with $${x^{1}}$$ parallel to this flow, with the second integral therefore over the “caps” of $${\partial V}$$.

Since the divergence of a tensor $${T}$$ with order greater than $${2}$$ is tensor-valued, and the parallel transport of tensors is path-dependent, we cannot in general integrate to get a divergence theorem for tensors. In the case of a flat metric and zero torsion however, we are able to integrate to get a divergence theorem for each component, e.g.

\begin{aligned}\int_{V}\nabla_{b}T^{ab}\mathrm{d}V & =\int_{\partial V}T_{b}{}^{a}\hat{n}^{b}\mathrm{d}S.\end{aligned}

For a metric which is not necessarily flat, but where we can choose coordinates as above, we can write coordinate-dependent component-wise divergence theorems

\begin{aligned}\int_{V}\partial_{\nu}\left(\sqrt{g}T^{\mu\nu}\right)\mathrm{d}^{n}x & =\int_{\partial V}T^{\mu1}\sqrt{g}\mathrm{d}^{n-1}x.\end{aligned}

In particular, if the left hand side vanishes, then the right hand side does as well, for any such coordinates. In the special case of an anti-symmetric tensor with zero torsion, we can write

\begin{aligned}\int_{V}\nabla_{\nu}F^{\mu\nu}\mathrm{d}V & =\int_{\partial V}F^{\mu1}\sqrt{g}\mathrm{d}^{n-1}x.\end{aligned}

In physics, the vector field $${u}$$ often represents the current vector (AKA current density, flux, flux density) $${j\equiv\rho u}$$ of an actual physical flow, where $${\rho}$$ is the density of the physical quantity $${Q}$$ and $${u}$$ is thus a velocity field; e.g. in $${\mathbb{R}^{3}}$$, $${j}$$ has units $${Q/(\mathrm{length})^{2}(\mathrm{time})}$$. There are several quantities that can be defined around this concept:

QuantityDefinitionMeaning
Current vector$${j\equiv\rho u}$$The vector whose length is the amount of $${Q}$$ per unit time crossing a unit area perpendicular to $${j}$$
Current form
\begin{aligned}\zeta & \equiv i_{j}\mathrm{d}V\\
& =\left\langle j,\hat{n}\right\rangle \mathrm{d}S
\end{aligned}
The $${(n-1)}$$-form which gives the amount of $${Q}$$ per unit time crossing the area defined by the argument vectors
Current density
\begin{aligned}\mathfrak{j} & \equiv\sqrt{g}\,j\\
\Rightarrow\zeta & =\mathfrak{j}^{1}\mathrm{d}^{2}x
\end{aligned}
The vector whose coordinate length is the amount of $${Q}$$ per unit time crossing a unit coordinate area perpendicular to $${j}$$
Current
\begin{aligned}I & \equiv\int_{S}\zeta\\
& =\int_{S}\left\langle j,\hat{n}\right\rangle \mathrm{d}S\\
& =\int_{S}\mathfrak{j}^{1}\mathrm{d}^{2}x
\end{aligned}
The amount of $${Q}$$ per unit time crossing $${S}$$
Four-current$${J\equiv(\rho,j^{\mu})}$$Current vector on the spacetime manifold

Notes: $${\rho}$$ is the density of the physical quantity $${Q}$$, $${u}$$ is a velocity field, $${\hat{n}}$$ is the unit normal to a surface $${S}$$, and $${\mathrm{d}^{3}x}$$ are coordinates with $${x^{1}}$$ constant on $${S}$$ and normal to it. The four-current can be generalized to other Lorentzian manifolds, and can also be turned into a form $${\xi\equiv i_{J}\mathrm{d}V}$$ or a density $${\mathfrak{J}\equiv\sqrt{g}\,J}$$.

 Δ Note that the terms flux and current (as well as flux density and current density) are not used consistently in the literature.

The current density $${\mathfrak{j}}$$ is an example of a tensor density, which in general takes the form $${\mathfrak{T}\equiv\left(\sqrt{\left|\mathrm{det}(g)\right|}\right)^{W}T}$$, where $${T}$$ is a tensor and $${W}$$ is called the weight. Note that tensor densities are not coordinate-independent quantities, and $${\sqrt{g}}$$ itself can thus be called a scalar density. From the previous expressions we get

\begin{aligned}\partial_{\lambda}\left(\mathfrak{T}\right) & =\sqrt{g}^{W}\partial_{\lambda}T+W\left(\Gamma^{\mu}{}_{\lambda\mu}-T^{\mu}{}_{\mu\lambda}\right)\mathfrak{T}\\
& =\sqrt{g}^{W}\partial_{\lambda}T+\frac{W}{2}g^{\mu\nu}\partial_{\lambda}g_{\mu\nu}\mathfrak{T},\\
L_{u}\left(\mathfrak{T}\right) & =\sqrt{g}^{W}L_{u}T+W\mathrm{div}\left(u\right)\mathfrak{T}\\
& =\sqrt{g}^{W}L_{u}T+\frac{W}{2}g^{\mu\nu}L_{u}g_{\mu\nu}\mathfrak{T},\\
\nabla_{\lambda}\left(\mathfrak{T}\right) & =\sqrt{g}^{W}\nabla_{\lambda}T,\end{aligned}

where the last is due to the covariant derivative of the metric vanishing. In particular, this means that for zero torsion the divergence of a vector density is

\begin{aligned}\nabla_{\lambda}\mathfrak{J}^{\lambda} & =\sqrt{g}\nabla_{\lambda}J^{\lambda},\\
& =\partial_{\lambda}\mathfrak{J}^{\lambda}.\end{aligned}

For a Riemannian metric on space, we now define the continuity equation (AKA equation of continuity)

$$\displaystyle \frac{\mathrm{d}q}{\mathrm{d}t}=\Sigma-\int_{\partial V}\left\langle j,\hat{n}\right\rangle \mathrm{d}S,$$

where $${q}$$ is the amount of $${Q}$$ contained in $${V}$$, $${t}$$ is time, and $${\Sigma}$$ is the rate of $${Q}$$ being created within $${V}$$. The continuity equation thus states the intuitive fact that the change of $${Q}$$ within $${V}$$ equals the amount generated less the amount which passes through $${\partial V}$$. Using the divergence theorem, we can then obtain the differential form of the continuity equation

$$\displaystyle \frac{\partial\rho}{\partial t}=\sigma-\mathrm{div}(j),$$

where $${\sigma}$$ is the amount of $${Q}$$ generated per unit volume per unit time. This equation then states the intuitive fact that at a point, the change in density of $${Q}$$ equals the amount generated less the amount that moves away. Positive $${\sigma}$$ is referred to as a source of $${Q}$$, and negative $${\sigma}$$ a sink. If $${\sigma=0}$$ then we say that $${Q}$$ is a conserved quantity and refer to the continuity equation as a (local) conservation law.

Under a flat Lorentzian metric, we can combine $${\rho}$$ and $${j}$$ into the current 4-vector $${J}$$ and express the continuity equation with $${\sigma=0}$$ as

$$\displaystyle \mathrm{div}(J)=0,$$

whereupon $${J}$$ is called a conserved current. Note that if any curvature is present, when we split out the time component we recover a Riemannian divergence but introduce a source due to the non-zero Christoffel symbols

\displaystyle \begin{aligned}\nabla_{\mu}J^{\mu} & =\partial_{\mu}J^{\mu}+\Gamma^{\mu}{}_{\nu\mu}J^{\nu}\\ & =\partial_{t}\rho+\nabla_{i}j^{i}+\left(\Gamma^{\mu}{}_{t\mu}\rho+\Gamma^{t}{}_{it}j^{i}\right), \end{aligned}

where $${t}$$ is the negative signature component and the index $${i}$$ goes over the remaining positive signature components. Thus, since the Christoffel symbols are coordinate-dependent, in the presence of curvature there is in general no coordinate-independent conserved quantity associated with a vanishing Lorentzian divergence.

In the integral form, we may identify a coordinate-dependent conserved quantity for a Lorentzian conserved current by integrating over a space-like volume $${S}$$ with coordinates such that $${x^{0}}$$ is constant on $${S}$$ and normal to it, while $${x^{1}}$$ is constant on $${\partial S}$$ and normal to it:

\begin{aligned}0 & =\int_{S}\sqrt{g}\mathrm{div}(J)\mathrm{d}^{3}x\\
& =\int_{S}\partial_{\mu}\mathfrak{J}^{\mu}\mathrm{d}^{3}x\\
& =\partial_{t}\left(\int_{S}\mathfrak{J}^{0}\mathrm{d}^{3}x\right)+\int_{S}\partial_{i}\mathfrak{J}^{i}\mathrm{d}^{3}x\\
& =\partial_{t}\left(\int_{S}\mathfrak{J}^{0}\mathrm{d}^{3}x\right)+\int_{\partial S}\mathfrak{J}^{1}\mathrm{d}^{2}x
\end{aligned}

 Δ Note that the coordinate-dependent factor $${\sqrt{g}}$$ in $${\mathfrak{J}=\sqrt{g}J}$$ cannot be absorbed into either $${\mathrm{d}^{3}x}$$ or $${\mathrm{d}^{2}x}$$ to yield a coordinate-independent quantity. A Lorentzian conserved current $${\mathrm{div}(J)=0}$$ nevertheless means that the quantity is conserved in finite volumes of spacetime, in the sense that $${\int_{\partial V}\left\langle J,\hat{n}\right\rangle \mathrm{d}S=0}$$ over any spacetime volume $${V}$$; in particular, if the current only enters and exits via two space-like cylinder “caps” $${C_{i}}$$ orthogonal to unit time tangents in a given frame, then the quantity $${\int_{C_{i}}J^{0}\mathrm{d}S}$$ is equal at the corresponding two points in time (even if the density is not). Finally, at any given point we can always choose Riemann normal coordinates, to recover a conserved quantity at that point in those coordinates.
 ◊ Noether’s theorem derives conserved currents from transformations (“symmetries”) on the variables of an expression called the action that leave it unchanged.