Intuitive

The continuity equation states that the $\color{red}{\text{total amount of a quantity (like water) that is produced (or destroyed) inside some volume}}$ is proportional to the $\color{blue}{\text{change of the quantity}}$ plus the $\color{magenta}{\text{total amount that flows in minus the amount that flows out of the volume}}$.

Or formulated differently, the total $\color{blue}{\text{change of some quantity}}$ is equal to the $\color{red}{\text{amount that gets produced}}$ plus the amount that $\color{magenta}{\text{flows in minus the amount that flows out of the volume}}$.

Image by Thierry Dugnolle

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.

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.

Abstract

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.