In vector calculus, a **Laplacian vector field** is a vector field which is both irrotational and incompressible. If the field is denoted as **v**, then it is described by the following differential equations:

- ${\begin{aligned}\nabla \times \mathbf {v} &=\mathbf {0} ,\\\nabla \cdot \mathbf {v} &=0.\end{aligned}}$

From the vector calculus identity $\nabla ^{2}\mathbf {v} \equiv \nabla (\nabla \cdot \mathbf {v} )-\nabla \times (\nabla \times \mathbf {v} )$ it follows that

- $\nabla ^{2}\mathbf {v} =0$

that is, that the field **v** satisfies Laplace's equation.

However, the converse is not true; not every vector field that satisfies Laplace's equation is a Laplacian vector field, which can be a point of confusion. For example, the vector field ${\bf {v}}=(xy,yz,zx)$ satisfies Laplace's equation, but it has both nonzero divergence and nonzero curl and is not a Laplacian vector field.

A Laplacian vector field in the plane satisfies the Cauchy–Riemann equations: it is holomorphic.

Since the curl of **v** is zero, it follows that (when the domain of definition is simply connected) **v** can be expressed as the gradient of a scalar potential (see irrotational field) *φ* :

- $\mathbf {v} =\nabla \phi .\qquad \qquad (1)$

Then, since the divergence of **v** is also zero, it follows from equation (1) that

- $\nabla \cdot \nabla \phi =0$

which is equivalent to

- $\nabla ^{2}\phi =0.$

Therefore, the potential of a Laplacian field satisfies Laplace's equation.

## See also