Continuity Equation

Prerequisites: Vector Calculus. Full path (3 pages)

Best explanations

Intuitive The Perfection of the Continuity Equation, Key to the Foundations of Reality

Intuitive
Concrete
Abstract
Why is it interesting?

$\color{blue}{\frac{\partial \rho}{\partial t}} = \color{red}{\sigma} - \color{magenta}{\rho \vec{\nabla} \cdot \vec{v}}$

Intuitive

The continuity equation states that the total $\color{blue}{\text{change of some quantity}}$ is equal to the $\color{red}{\text{amount that gets produced}}$ minus the amount that $\color{magenta}{\text{flows out of the volume}}$ .

If we are dealing with a conserved quantity, like energy or electric charge, the total amount that is produced or destroyed is exactly zero.

Whenever a system possesses some symmetry we know from Noether's theorem that some corresponding quantity is conserved. Using Noether's theorem, we can then also derive the corresponding continuity equation that describes how the conserved quantity flows through the system.

Recommended further reading

The Perfection of the Continuity Equation, Key to the Foundations of Reality by WIRED magazine

Concrete

The continuity equation in hydrodynamics describes the flow of mass. Here $ρ$ is fluid density and $\vec v$ the fluid flow velocity. In this context it is also known as one of the Euler equations of fluid dynamics.
The continuity equation in electrodynamics describes the flow of electric charge. Here $ρ$ is the charge density and $\vec v$ the electric flow velocity, such that $ρ \vec v = j$ is the electric current.
The continuity equation in quantum mechanics describes the flow of probability. Here $ρ = \Psi^\dagger \Psi$ is the probability density and $\vec v$ the probability flow velocity, such that $ρ \vec v = j = \frac { \hbar } { 2m i } [ \Psi ^ { * } ( \nabla \Psi ) - \Psi ( \nabla \Psi ^ { * } )]$ is the probability current .

In general, continuity equations can be derived by using Noether's theorem.

Derivation of the continuity equation in electrodynamics

We start with Ampere's law, which is one of the Maxwell equations $\nabla \times H = J + \frac { \partial D } { \partial t }.$

Next we take the divergence of this equation, which yields $\nabla \cdot ( \nabla \times H ) = \nabla \cdot J + \frac { \partial ( \nabla \cdot D ) } { \partial t }$ The divergence of a curl is zero, and therefore we get $\nabla \cdot J + \frac { \partial ( \nabla \cdot D ) } { \partial t } = 0.$ Finally, we use another Maxwell equation, namely Gauss law, $\nabla \cdot D = \rho$ and substitute it into the previous equation $\nabla \cdot J + \frac { \partial \rho } { \partial t } = 0 .$ This is exactly the continuity equation.

Derivation of the continuity equation in quantum mechanics

The probability density is quantum mechanics is $\rho = \Psi^* \Psi$ . The partial derivative of $\rho$ with respect to time is therefore $\frac { \partial \rho } { \partial t } = \frac { \partial | \Psi | ^ { 2} } { \partial t } = \frac { \partial } { \partial t } ( \Psi ^ { * } \Psi ) = \Psi ^ { * } \frac { \partial \Psi } { \partial t } + \Psi \frac { \partial \Psi ^ { * } } { \partial t } .$

Next, we consider the Schrödinger equation

$- \frac { \hbar ^ { 2} } { 2m } \nabla ^ { 2} \Psi ^ { * } + U \Psi ^ { * } = - i \hbar \frac { \partial \Psi ^ { * } } { \partial t }.$

Taking the complex conjugate of it yields

$- \frac { \hbar ^ { 2} } { 2m } \nabla ^ { 2} \Psi ^ { * } + U \Psi ^ { * } = - i \hbar \frac { \partial \Psi ^ { * } } { \partial t }.$

Multiplying the Schrödinger equation with $\Psi^\star$ and the complex conjugated Schrödinger equation with $\Psi$ yields the two equations

$\Psi \cdot \frac { \partial \Psi } { \partial t } = \frac { 1} { i \hbar } [ - \frac { \hbar ^ { 2} \Psi ^ { * } } { 2m } \nabla ^ { 2} \Psi + U \Psi ^ { * } \Psi ]$ $\Psi \frac { \partial \Psi ^ { * } } { \partial t } = - \frac { 1} { i \hbar } [ - \frac { \hbar ^ { 2} \Psi } { 2m } \nabla ^ { 2} \Psi ^ { * } + U \Psi \Psi ^ { * } ] .$

Putting these two equations into our equation for $\frac { \partial \rho } { \partial t }$ from above yields $\frac { \partial \rho } { \partial t } = \frac { 1} { i \hbar } [ - \frac { \hbar ^ { 2} \Psi ^ { * } } { 2m } \nabla ^ { 2} \Psi + U \Psi ^ { * } \Psi ] - \frac { 1} { i \hbar } [ - \frac { \hbar ^ { 2} \Psi } { 2m } \nabla ^ { 2} \Psi ^ { * } + U \Psi \Psi ^ { * } ]$ $= \frac { \hbar } { 2i m } [ \Psi \nabla ^ { 2} \Psi ^ { * } - \Psi ^ { * } \nabla ^ { 2} \Psi ] .$

The second puzzle piece that appears in the continuity equation is the current $j$ which in quantum mechanics is given by $j = \frac { \hbar } { 2m i } [ \Psi ^ { * } ( \nabla \Psi ) - \Psi ( \nabla \Psi ^ { * } )].$ Taking the divergence of it (since $\nabla j$ is what appears in the continuity equation) yields $\nabla \cdot j = \nabla \cdot [ \frac { \hbar } { 2m i } ( \Phi ^ { * } ( \nabla \Phi ) - \Psi ( \nabla \Psi ^ { * } ) ) ]$ $= - \frac { \hbar } { 2m i } [ \Psi ( \nabla ^ { 2} \Psi ^ { * } ) - \Psi ^ { * } ( \nabla ^ { 2} \Psi ) ].$

This is exactly, except for the minus sign what we derived above for $\frac { \partial \rho } { \partial t }$ and therefore we can conclude

$\frac { \partial \rho } { \partial t } = - \nabla \cdot j$ $\therefore \frac { \partial \rho } { \partial t } + \nabla \cdot j =0.$

This is exactly the continuity equation that we wanted to derive.

Integral Form of the continuity equation

By integrating the continuity equation over some volume $V$ , we get

$\int_V \frac { \partial \rho } { \partial t } + \int_V \nabla \cdot j =0$

For the second term, we can then use Gauss divergence theorem that tells us that we can replace the volume integral over some divergence by a surface integral

$\int_V \frac { \partial \rho } { \partial t } + \int_S j =0 ,$

where $S$ denotes the surface of our volume $V$ . This is the integral form of the continuity equation.

The first term simply describes the total amount of the quantity, e.g. electric charge or mass, inside our volume.
The second term describes the amount that flows into the surface minus the amount that flows out of the surface.
No-Nonsense Electrodynamics by Jakob Schwichtenberg, section "The continuity equation"

Abstract

In relativistic theories, the charge density and the current live in one object called the current four-vector (or four-current) $j_\mu = (c \rho, \vec j),$ where $c$ denotes the speed of light and is inserted such that all components have the same dimensions. Using this definition, the continuity equation reads

$\partial_\mu j^\mu = 0.$

The continuity equation can also be written using the 3-form of charge density

$d \gamma = 0$

where $\gamma = - \frac { 1} { c } ( 1,d x ^ { 2} \wedge d x ^ { 3} + i _ { 2} d x ^ { 3} \wedge d x ^ { 1} + j _ { 3} d x ^ { 1} \wedge d x ^ { 2} ) \wedge d x ^ { 0} + \rho d x ^ { \prime } \wedge d x ^ { 2} \wedge d x ^ { 3}.$

Why is it interesting?

A continuity equation is important whenever we are dealing with a system where some quantity is conserved. The continuity helps us to keep track how the conserved quantity moves through the system.

Important examples are

the continuity equation in electrodynamics that encodes the conservation of electric charge,
the continuity equation in hydrodynamics that encodes the conservation of mass.
the continuity equation in quantum mechanics that encodes the conservation of probability.

Continuity Equation

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