What does Divergence mean?divergence is a vector operator that produces a scalar field, giving the quantity of a vector field's source at each point. More technically, the divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point.

As an example, consider air as it is heated or cooled. The velocity of the air at each point defines a vector field. While air is heated in a region, it expands in all directions, and thus the velocity field points outward from that region. The divergence of the velocity field in that region would thus have a positive value. While the air is cooled and thus contracting, the divergence of the velocity has a negative value.

Physical interpretation of divergence
In physical terms, the divergence of a three-dimensional vector field is the extent to which the vector field flux behaves like a source at a given point. It is a local measure of its "outgoingness" – the extent to which there is more of some quantity exiting an infinitesimal region of space than entering it. If the divergence is nonzero at some point then there is compression or expansion at that point. (Note that we are imagining the vector field to be like the velocity vector field of a fluid (in motion) when we use the terms flux and so on.)

More rigorously, the divergence of a vector field F at a point p can be defined as the limit of the net flux of F across the smooth boundary of a three-dimensional region V divided by the volume of V as V shrinks to p. Formally,

{\displaystyle \left.\operatorname {div} \mathbf {F} \right|_{p}=\lim _{V\rightarrow \{p\}}\iint _{S(V)}{\frac {\mathbf {F} \cdot \mathbf {\hat {n}} }{|V|}}\,dS,} {\displaystyle \left.\operatorname {div} \mathbf {F} \right|_{p}=\lim _{V\rightarrow \{p\}}\iint _{S(V)}{\frac {\mathbf {F} \cdot \mathbf {\hat {n}} }{|V|}}\,dS,}
where |V| is the volume of V, S(V) is the boundary of V, and the integral is a surface integral with n̂ being the outward unit normal to that surface. The result, div F, is a function of p. From this definition it also becomes obvious that div F can be seen as the source density of the flux of F.

In light of the physical interpretation, a vector field with zero divergence everywhere is called incompressible or solenoidal – in which case any closed surface has no net flux across it.

The intuition that the sum of all sources minus the sum of all sinks should give the net flux outwards of a region is made precise by the divergence theorem.

Cartesian coordinates
In three-dimensional Cartesian coordinates, the divergence of a continuously differentiable vector field {\displaystyle \mathbf {F} =F_{x}\mathbf {i} +F_{y}\mathbf {j} +F_{z}\mathbf {k} } {\displaystyle \mathbf {F} =F_{x}\mathbf {i} +F_{y}\mathbf {j} +F_{z}\mathbf {k} } is defined as the scalar-valued function:

{\displaystyle \operatorname {div} \mathbf {F} =\nabla \cdot \mathbf {F} =\left({\frac {\partial }{\partial x}},{\frac {\partial }{\partial y}},{\frac {\partial }{\partial z}}\right)\cdot (F_{x},F_{y},F_{z})={\frac {\partial F_{x}}{\partial x}}+{\frac {\partial F_{y}}{\partial y}}+{\frac {\partial F_{z}}{\partial z}}.} {\displaystyle \operatorname {div} \mathbf {F} =\nabla \cdot \mathbf {F} =\left({\frac {\partial }{\partial x}},{\frac {\partial }{\partial y}},{\frac {\partial }{\partial z}}\right)\cdot (F_{x},F_{y},F_{z})={\frac {\partial F_{x}}{\partial x}}+{\frac {\partial F_{y}}{\partial y}}+{\frac {\partial F_{z}}{\partial z}}.}
Although expressed in terms of coordinates, the result is invariant under rotations, as the physical interpretation suggests. This is because the trace of the Jacobian matrix of an N-dimensional vector field F in N-dimensional space is invariant under any invertible linear transformation.

The common notation for the divergence ∇ · F is a convenient mnemonic, where the dot denotes an operation reminiscent of the dot product: take the components of the ∇ operator (see del), apply them to the corresponding components of F, and sum the results. Because applying an operator is different from multiplying the components, this is considered an abuse of notation.

The divergence of a continuously differentiable second-order tensor field ε is a first-order tensor field:[1]

{\displaystyle {\overrightarrow {\operatorname {div} }}(\mathbf {\varepsilon } )={\begin{bmatrix}{\dfrac {\partial \varepsilon _{xx}}{\partial x}}+{\dfrac {\partial \varepsilon _{yx}}{\partial y}}+{\dfrac {\partial \varepsilon _{zx}}{\partial z}}\\{\dfrac {\partial \varepsilon _{xy}}{\partial x}}+{\dfrac {\partial \varepsilon _{yy}}{\partial y}}+{\dfrac {\partial \varepsilon _{zy}}{\partial z}}\\{\dfrac {\partial \varepsilon _{xz}}{\partial x}}+{\dfrac {\partial \varepsilon _{yz}}{\partial y}}+{\dfrac {\partial \varepsilon _{zz}}{\partial z}}\end{bmatrix}}.} {\displaystyle {\overrightarrow {\operatorname {div} }}(\mathbf {\varepsilon } )={\begin{bmatrix}{\dfrac {\partial \varepsilon _{xx}}{\partial x}}+{\dfrac {\partial \varepsilon _{yx}}{\partial y}}+{\dfrac {\partial \varepsilon _{zx}}{\partial z}}\\{\dfrac {\partial \varepsilon _{xy}}{\partial x}}+{\dfrac {\partial \varepsilon _{yy}}{\partial y}}+{\dfrac {\partial \varepsilon _{zy}}{\partial z}}\\{\dfrac {\partial \varepsilon _{xz}}{\partial x}}+{\dfrac {\partial \varepsilon _{yz}}{\partial y}}+{\dfrac {\partial \varepsilon _{zz}}{\partial z}}\end{bmatrix}}.}
Cylindrical coordinates
For a vector expressed in local unit cylindrical coordinates as

{\displaystyle \mathbf {F} =\mathbf {e} _{r}F_{r}+\mathbf {e} _{\theta }F_{\theta }+\mathbf {e} _{z}F_{z},} {\displaystyle \mathbf {F} =\mathbf {e} _{r}F_{r}+\mathbf {e} _{\theta }F_{\theta }+\mathbf {e} _{z}F_{z},}
where ea is the unit vector in direction a, the divergence is[2]

{\displaystyle \operatorname {div} \mathbf {F} =\nabla \cdot \mathbf {F} ={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(rF_{r}\right)+{\frac {1}{r}}{\frac {\partial F_{\theta }}{\partial \theta }}+{\frac {\partial F_{z}}{\partial z}}.} {\displaystyle \operatorname {div} \mathbf {F} =\nabla \cdot \mathbf {F} ={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(rF_{r}\right)+{\frac {1}{r}}{\frac {\partial F_{\theta }}{\partial \theta }}+{\frac {\partial F_{z}}{\partial z}}.}
The use of local coordinates is vital for the validity of the expression. If we consider x the position vector and the functions {\displaystyle r(\mathbf {x} )} {\displaystyle r(\mathbf {x} )}, {\displaystyle \theta (\mathbf {x} )} {\displaystyle \theta (\mathbf {x} )}, and {\displaystyle z(\mathbf {x} )} {\displaystyle z(\mathbf {x} )}, which assign the corresponding global cylindrical coordinate to a vector, in general {\displaystyle r(\mathbf {F} (\mathbf {x} ))\neq F_{r}(\mathbf {x} )} {\displaystyle r(\mathbf {F} (\mathbf {x} ))\neq F_{r}(\mathbf {x} )}, {\displaystyle \theta (\mathbf {F} (\mathbf {x} ))\neq F_{\theta }(\mathbf {x} )} {\displaystyle \theta (\mathbf {F} (\mathbf {x} ))\neq F_{\theta }(\mathbf {x} )}, and {\displaystyle z(\mathbf {F} (\mathbf {x} ))\neq F_{z}(\mathbf {x} )} {\displaystyle z(\mathbf {F} (\mathbf {x} ))\neq F_{z}(\mathbf {x} )}. In particular, if we consider the identity function {\displaystyle \mathbf {F} (\mathbf {x} )=\mathbf {x} } {\displaystyle \mathbf {F} (\mathbf {x} )=\mathbf {x} }, we find that:

{\displaystyle \theta (\mathbf {F} (\mathbf {x} ))=\theta \neq F_{\theta }(\mathbf {x} )=0} {\displaystyle \theta (\mathbf {F} (\mathbf {x} ))=\theta \neq F_{\theta }(\mathbf {x} )=0}.
Spherical coordinates
In spherical coordinates, with θ the angle with the z axis and φ the rotation around the z axis, and {\displaystyle \mathbf {F} } \mathbf {F} again written in local unit coordinates, the divergence is[3]

{\displaystyle \operatorname {div} \mathbf {F} =\nabla \cdot \mathbf {F} ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}F_{r}\right)+{\frac {1}{r\sin \theta }}{\frac {\partial }{\partial \theta }}(\sin \theta \,F_{\theta })+{\frac {1}{r\sin \theta }}{\frac {\partial F_{\varphi }}{\partial \varphi }}.} {\displaystyle \operatorname {div} \mathbf {F} =\nabla \cdot \mathbf {F} ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}F_{r}\right)+{\frac {1}{r\sin \theta }}{\frac {\partial }{\partial \theta }}(\sin \theta \,F_{\theta })+{\frac {1}{r\sin \theta }}{\frac {\partial F_{\varphi }}{\partial \varphi }}.}
General coordinates
Using Einstein notation we can consider the divergence in general coordinates, which we write as x1, ..., xi, ...,xn, where n is the number of dimensions of the domain. Here, the upper index refers to the number of the coordinate or component, so x2 refers to the second component, and not the quantity x squared. The index variable i is used to refer to an arbitrary element, such as xi. The divergence can then be written via the Voss- Weyl formula[4], as:

{\displaystyle \operatorname {div} (\mathbf {F} )={\frac {1}{\rho }}{\frac {\partial \left(\rho \,F^{i}\right)}{\partial x^{i}}},} {\displaystyle \operatorname {div} (\mathbf {F} )={\frac {1}{\rho }}{\frac {\partial \left(\rho \,F^{i}\right)}{\partial x^{i}}},}
where {\displaystyle \rho } \rho is the local coefficient of the volume element and Fi are the components of F with respect to the local unnormalized covariant basis (sometimes written as {\displaystyle \mathbf {e} _{i}=\partial \mathbf {x} /\partial x^{i}} {\displaystyle \mathbf {e} _{i}=\partial \mathbf {x} /\partial x^{i}}). The Einstein notation implies summation over i, since it appears as both an upper and lower index.

The volume coefficient {\displaystyle \rho } \rho is a function of position which depends on the coordinate system. In Cartesian, cylindrical and spherical coordinates, using the same conventions as before, we have {\displaystyle \rho =1} \rho =1, {\displaystyle \rho =r} {\displaystyle \rho =r} and {\displaystyle \rho =r^{2}\sin {\theta }} {\displaystyle \rho =r^{2}\sin {\theta }}, respectively. It can also be expressed as {\displaystyle \rho ={\sqrt {\operatorname {det} g_{ab}}}} {\displaystyle \rho ={\sqrt {\operatorname {det} g_{ab}}}}, where {\displaystyle g_{ab}} g_{ab} is the metric tensor. Since the determinant is a scalar quantity which doesn't depend on the indices, we can suppress them and simply write {\displaystyle \rho ={\sqrt {\operatorname {det} g}}} {\displaystyle \rho ={\sqrt {\operatorname {det} g}}}. Another expression comes from computing the determinant of the Jacobian for transforming from Cartesian coordinates, which for n = 3 gives {\displaystyle \rho =\left|{\frac {\partial (x,y,z)}{\partial (x^{1},x^{2},x^{3})}}\right|.} {\displaystyle \rho =\left|{\frac {\partial (x,y,z)}{\partial (x^{1},x^{2},x^{3})}}\right|.}

Some conventions expect all local basis elements to be normalized to unit length, as was done in the previous sections. If we write {\displaystyle {\hat {\mathbf {e} }}_{i}} {\hat {\mathbf {e} }}_{i} for the normalized basis, and {\displaystyle {\hat {F}}^{i}} {\displaystyle {\hat {F}}^{i}} for the components of F with respect to it, we have that

{\displaystyle \mathbf {F} =F^{i}\mathbf {e} _{i}=F^{i}{\lVert {\mathbf {e} _{i}}\rVert }{\frac {\mathbf {e} _{i}}{\lVert {\mathbf {e} _{i}}\rVert }}=F^{i}{\sqrt {g_{ii}}}\,{\hat {\mathbf {e} }}_{i}={\hat {F}}^{i}{\hat {\mathbf {e} }}_{i},} {\displaystyle \mathbf {F} =F^{i}\mathbf {e} _{i}=F^{i}{\lVert {\mathbf {e} _{i}}\rVert }{\frac {\mathbf {e} _{i}}{\lVert {\mathbf {e} _{i}}\rVert }}=F^{i}{\sqrt {g_{ii}}}\,{\hat {\mathbf {e} }}_{i}={\hat {F}}^{i}{\hat {\mathbf {e} }}_{i},}
using one of the properties of the metric tensor. By dotting both sides of the last equality with the contravariant element {\displaystyle {\hat {\mathbf {e} }}^{i}} {\displaystyle {\hat {\mathbf {e} }}^{i}}, we can conclude that {\displaystyle F^{i}={\hat {F}}^{i}/{\sqrt {g_{ii}}}} {\displaystyle F^{i}={\hat {F}}^{i}/{\sqrt {g_{ii}}}}. After substituting, the formula becomes:

{\displaystyle \operatorname {div} (\mathbf {F} )={\frac {1}{\rho }}{\frac {\partial \left({\frac {\rho }{\sqrt {g_{ii}}}}{\hat {F}}^{i}\right)}{\partial x^{i}}}={\frac {1}{\sqrt {\operatorname {det} g}}}{\frac {\partial \left({\sqrt {\frac {\operatorname {det} g}{g_{ii}}}}\,{\hat {F}}^{i}\right)}{\partial x^{i}}}} {\displaystyle \operatorname {div} (\mathbf {F} )={\frac {1}{\rho }}{\frac {\partial \left({\frac {\rho }{\sqrt {g_{ii}}}}{\hat {F}}^{i}\right)}{\partial x^{i}}}={\frac {1}{\sqrt {\operatorname {det} g}}}{\frac {\partial \left({\sqrt {\frac {\operatorname {det} g}{g_{ii}}}}\,{\hat {F}}^{i}\right)}{\partial x^{i}}}}