# Understading the integral form of a conservation law

When I think of a conservation law I think of a continuity equation like the following

$$\partial_t \rho = -\nabla \cdot \vec j$$

But now I'm reading a book on electrodynamics (that's honestly a bit over my head) and its version of a conservation law is as an integral equation that vanishes over any closed submanifold. For instance conservation of charge is given as follows: Given the twisted charge current $3$-form $J$, then $$\oint_{C_3} J=0$$ where $C_3$ is an arbitrary closed $3$-dimensional submanifold [of our $4$-dimensional spacetime].

I'm not looking for any physical intuition, I'm just asking how does this equation imply conservation of a quantity? Is this equation equivalent to a continuity equation?

Depending on your background, the following might help:

The most elementary way to introduce the charge conservation in the relativistic context (4-dimensional spacetime) is via the 4-current $$j^\mu = (\rho, \vec {j})$$ where $\mu=0,1,2,3$. The continuity equation then reads $$\partial_\mu j^\mu = \frac{\partial j^\mu}{\partial x^\mu} =0 \tag{1}$$ where we use the Einstein summation convention.

Borrowing from differential geometry, we can also introduce the current 3-form $J$ that can be written as $$J= \frac{1}{3!} J_{\mu\nu\sigma} dx^\mu \wedge dx^\nu \wedge dx^\sigma$$ with the antisymmetric tensor $J_{\mu\nu\sigma}$.

The connection to the 4-current is given by the duality $$j^\mu = \epsilon^{\mu\nu\sigma\tau} J_{\nu\sigma\tau}$$ with $\epsilon^{\mu\nu\sigma\tau}$ the completely antisymmetric pseudotensor (thus $J_{\nu\sigma\tau}$ is a twisted 3-form).

The continuity equation (1) then assumes the simple form $$d J= 0$$ meaning that the 3-form is closed.

Now you can apply Stokes' theorem on an arbitrary region $R$ of the 4-space to obtain $$0 = \int_R dJ = \oint_{C} J,$$ with $C=\partial R$ the boundary of $R$.

The continuity equation is the local form of the integral form of electric charge conservation. First let us recall how to obtain the conservation integral from simple physical arguments without invoking 4th-dimensional spacetime manifold nor differential forms over it. Imagine a control volume $\mathcal V$ in space enclosed by surface boundary $\mathcal S$. Conservation expresses that the time-variation rate of electric charge within volume $V$ is balanced by (opposed to) the outward flux of current density across boundary $\mathcal S$:

$$\frac{d}{dt} \int_{\mathcal V}\rho\, dV = - \int_S \vec j.d\vec S$$

Since is the control volume is fixed in space not varying with time one gets, $$\frac{d}{dt} \int_{\mathcal V}\rho\, dV = \int_{\mathcal V} \frac{\partial\rho}{\partial t} \, dV$$

From the application of Stokes theorem (Ostrogradski formula) applied to volume $\mathcal V$ on also gets, $$\int_{\mathcal S} \vec j.d\vec S = \int_{\mathcal V} \vec\nabla_3.\vec j\, dV$$

Resulting in some integral form, $$\int_{\mathcal V} [\frac{\partial\rho}{\partial t} + \vec \nabla_3.\vec j] \, dV =0$$

or in local differential form. $$\frac{\partial\rho}{\partial t} + \vec \nabla_3.\vec j = 0.$$

where $\vec \nabla_3=(\frac{\partial}{\partial x^1},\frac{\partial}{\partial x^2},\frac{\partial}{\partial x^3})$ is the gradient vector in space.

Now Let us reinterpret the formula in 4th-dimensional Minkowski's (relativistic) space-time with event coordinates $(x^0=ct,x^1=x,x^2=y,x^3=z)$. One can show that the analogue of the 3th-dimensional current density is the Lorentz-covariant four-vector defined by $$\vec J = (c\rho,\vec j) = (J_0, (J_1,J_2,J_3))$$ $c$ indicating the speed of light. Similarly, the analogue of gradient is the four-gradient, $$\vec \nabla_4=(\frac{\partial .}{\partial x^0},\nabla_{3})$$

The local equation for charge conservation simply writes, $$\vec \nabla_4 .\vec J = 0$$

To the electric current four-vector $\vec J$ one can associate a twisted differential 3-form: $$J = J_0 dx^1\wedge dx^2\wedge dx^3 -J_1 dx^0\wedge dx^2\wedge dx^3 + J_2 dx^0\wedge dx^1\wedge dx^3 - J_3 dx^0\wedge dx^1\wedge dx^2.$$

such that its exterior derivative $dJ$ when computed gives

$$dJ = \vec \nabla_4. \vec J$$

In 4th-dimensional spacetime the local form of charge conservation expresses then as $dJ=0$. If one considers a 4th-dimensional Riemanian manifold $\mathcal M$ of closed boundary $\partial M$ and applies Stokes theorem over it one obtains the 4-th dimensional integral form of charge conservation. From the previous equation one gets, $\int_{\mathcal M} dJ =0$ then Stokes theorem gives $\int_{\mathcal M} dJ = \int_{\partial M} J=0$.

My feeling is that continuity equation and 4th-dimensional integral form are equivalent. Conversely starting with J as the differential twisted 3-form and integral equation $\int_{\mathcal \partial M} J =0$, Stokes-theorem renders, $$\frac{\partial J_0}{\partial x^0} + \vec \nabla_3.(J_1,J_2,J_3) = 0.$$ as the associated continuity equation.

Hope it helps.

Note: it is customary to set $c=1$.

• Nicely explained – Simon S Dec 31 '15 at 11:14