Divergence of a vector field

The divergence of a vector field is a scalar field that indicates at each point how much the vectors diverge in a small area around the point ( Latin divergere ). If the vector field is interpreted as a flow field of a size for which the continuity equation applies, then the divergence is the source density . Sinks have negative divergence. If the divergence is zero everywhere, the field is called source-free .

The divergence results from the vector field by using a differential operator . Related differential operators provide the rotation of a vector field and the gradient of a scalar field. The mathematical field is vector analysis .

In physics, for example, divergence is used in the formulation of Maxwell's equations or the various continuity equations . In the Ricci calculus , the quantity formed with the help of the covariant derivative is sometimes somewhat imprecisely referred to as the divergence of a tensor (for example, the Gaussian integral theorem does not apply to this quantity on curved manifolds ). ${\displaystyle \nabla _{k}T^{ik}}$ ${\displaystyle T^{ik}}$

Example from physics

For example, consider a calm surface of water that is hit by a thin stream of oil. The movement of the oil on the surface can be described by a two-dimensional (time-dependent) vector field: At any point and at any point in time, the flow rate of the oil is given in the form of a vector. The point at which the jet hits the surface of the water is an “oil well”, since oil flows away from there without there being any inflow on the surface. The divergence at this point is positive. In contrast, a place where the oil flows out of the water basin, for example at the edge, is called a sink. The divergence at this point is negative.

definition

Let be a differentiable vector field. Then the divergence of is defined as ${\displaystyle {\vec {F}}\colon \mathbb {R} ^{n}\to \mathbb {R} ^{n},(x_{1},\dots ,x_{n})\mapsto (F^{1},\dots ,F^{n})}$ ${\displaystyle {\vec {F}}}$

${\displaystyle \operatorname {div} {\vec {F}}:=\nabla \cdot {\vec {F}}=({\tfrac {\partial }{\partial x_{1}}},\dotsc ,{\tfrac {\partial }{\partial x_{n}}})\cdot \left(F^{1},\dotsc ,F^{n}\right)={\tfrac {\partial }{\partial x_{1}}}F^{1}+\dotsb +{\tfrac {\partial }{\partial x_{n}}}F^{n}}$

Explanation

The divergence is the scalar product of the Nabla operator with the vector field . ${\displaystyle \nabla }$ ${\displaystyle {\vec {F}}}$

The divergence is an operator on a vector field that results in a scalar field:

${\displaystyle \operatorname {div} [\cdot ]\ \colon C^{1}\left(\mathbb {R} ^{n},\mathbb {R} ^{n}\right)\to C^{0}\left(\mathbb {R} ^{n},\mathbb {R} \right)}$

For the case of a three-dimensional vector field , the divergence in Cartesian coordinates is defined as ${\displaystyle {\vec {F}}(x_{1},x_{2},x_{3})}$

${\displaystyle {\begin{matrix}\operatorname {div} \colon &C^{1}\left(\mathbb {R} ^{3},\mathbb {R} ^{3}\right)&\to &C^{0}\left(\mathbb {R} ^{3},\mathbb {R} \right)\\&{\vec {F}}=\left(F^{1},F^{2},F^{3}\right)&\mapsto &{\frac {\partial }{\partial x_{1}}}F^{1}+{\frac {\partial }{\partial x_{2}}}F^{2}+{\frac {\partial }{\partial x_{3}}}F^{3}\end{matrix}}}$.

With the notation , it is important to write the multiplication point between and the vector field , otherwise the operator would be understood as a gradient of the vector components (written ). ${\displaystyle \nabla \cdot {\vec {F}}}$ ${\displaystyle \nabla }$ ${\displaystyle {\vec {F}}}$ ${\displaystyle \nabla }$ ${\displaystyle \nabla {\vec {F}}}$

The divergence as "source density"

If a vector field is interpreted as a flow field, its total differential describes an acceleration field. If the acceleration matrix can be diagonalized at a point , each eigenvalue describes the acceleration in the direction of the associated eigenvector . Every positive eigenvalue describes the intensity of a directed source and every negative eigenvalue the directed intensity of a sink. If you add these eigenvalues, you get the resulting intensity of a source or sink. Since the sum of the eigenvalues ​​is just the trace of the acceleration matrix , the source intensity is through ${\displaystyle {\vec {F}}\colon \mathbb {R} ^{n}\to \mathbb {R} ^{n}}$ ${\displaystyle DF\colon \mathbb {R} ^{n}\to \mathbb {R} ^{n\times n}}$ ${\displaystyle x_{0}\in \mathbb {R} ^{n}}$ ${\displaystyle DF(x_{0})}$ ${\displaystyle \lambda _{i}(x_{0})}$ ${\displaystyle u_{i}(x_{0})}$ ${\displaystyle \lambda _{i}(x_{0})}$ ${\displaystyle DF(x_{0})}$

${\displaystyle \operatorname {Spur} (DF)=\sum _{i=1}^{n}{\frac {\partial }{\partial x_{i}}}F^{i}=\operatorname {div} \,{\vec {F}}}$

measured.

In this sense, the divergence can be interpreted as “source density”.

Coordinate-free representation

For the interpretation of the divergence as "source density" the following coordinate-free definition in the form of a volume derivative is important (here for the case n = 3)

${\displaystyle \operatorname {div} \,{\vec {F}}=\lim _{|\Delta V|\to 0}\left({\frac {1}{|\Delta V|}}\oint _{\partial (\Delta V)}{\vec {F}}\;\cdot {\vec {n}}\,\mathrm {d} S\right)\,.}$

It is an arbitrary volume, for example a ball or a parallelepiped; is its content. It is integrated over the edge of this volume element , is the outward directed normal and the associated surface element. You can also find the notation with . ${\displaystyle \Delta V}$ ${\displaystyle |\Delta V|}$ ${\displaystyle \partial (\Delta V)}$ ${\displaystyle {\vec {n}}}$ ${\displaystyle \mathrm {d} S}$ ${\displaystyle \mathrm {d} {\vec {S}}:={\vec {n}}\cdot \mathrm {d} S}$

For n > 3 this statement can easily be generalized by considering n -dimensional volumes and their ( n- 1) -dimensional edge areas. If you specialize in infinitesimal cubes or cuboids, you get the familiar representation in Cartesian coordinates

${\displaystyle \operatorname {div} \,{\vec {F}}=\sum _{i=1}^{n}\,{\frac {\partial F_{i}}{\partial x_{i}}}\,.}$

In orthogonal curvilinear coordinates , for example spherical coordinates or elliptical coordinates , (i.e. for , with ), where is, whereby not the , but those have the physical dimension of a "length", is somewhat more general ${\displaystyle \textstyle \mathrm {d} {\vec {r}}=\sum _{i=1}^{n}a_{i}(u_{1},\dots ,u_{n})\cdot \mathrm {d} u_{i}\,{\hat {e}}_{i}(u_{1},\dots ,u_{n})\,}$ ${\displaystyle {\hat {e}}_{i}\cdot {\hat {e}}_{k}=\delta _{i,k}\,}$ ${\displaystyle \textstyle {\vec {F}}=\sum _{i=1}^{n}\,F_{i}(u_{1},\dots ,u_{n})\,{\hat {e}}_{i}}$ ${\displaystyle \mathrm {d} u_{i}}$ ${\displaystyle a_{i}\cdot \mathrm {d} u_{i}}$

${\displaystyle \operatorname {div} \,{\vec {F}}=(a_{1}\cdot a_{2}\cdot \ldots \cdot a_{n})^{-1}\,\left\{{\frac {\partial }{\partial u_{1}}}[(a_{2}\cdot a_{3}\cdot \ldots \cdot a_{n-1}\cdot a_{n})\,F_{1}]+\dots \right\},}$

where the points at the end contain further terms that follow from what has been written by continued cyclic permutations , generated according to the scheme etc. ${\displaystyle 1\to 2,2\to 3,\dots ,(n-1)\to n,n\to 1}$

Derivation of the Cartesian representation

To derive the Cartesian representation of the divergence from the coordinate-free representation, consider an infinitesimal cube . ${\displaystyle [x_{1},x_{1}+\Delta x_{1}],\ldots ,[x_{n},x_{n}+\Delta x_{n}]}$

${\displaystyle {\begin{aligned}{\frac {1}{|\Delta V|}}\int \limits _{\partial (\Delta V)}{\vec {F}}\cdot {\vec {n}}\,\mathrm {d} S&={\frac {1}{\Delta x_{1}\Delta x_{2}\ldots \Delta x_{n}}}\left[\int \limits _{x_{2}}^{x_{2}+\Delta x_{2}}\ldots \int \limits _{x_{n}}^{x_{n}+\Delta x_{n}}\left({\vec {F}}(x_{1}+\Delta x_{1},x_{2},\ldots ,x_{n})-{\vec {F}}(x_{1},x_{2},\ldots ,x_{n})\right)\cdot {\hat {e}}_{1}\,\mathrm {d} x_{2}\ldots \mathrm {d} x_{n}\right]+\ldots \\&+{\frac {1}{\Delta x_{1}\Delta x_{2}\ldots \Delta x_{n}}}\left[\int \limits _{x_{1}}^{x_{1}+\Delta x_{1}}\ldots \int \limits _{x_{n-1}}^{x_{n-1}+\Delta x_{n-1}}\left({\vec {F}}(x_{1},\ldots ,x_{n-1},x_{n}+\Delta x_{n})-{\vec {F}}(x_{1},\ldots ,x_{n-1},x_{n})\right)\cdot {\hat {e}}_{n}\,\mathrm {d} x_{1}\ldots \mathrm {d} x_{n-1}\right]\end{aligned}}}$

Now we apply the mean value theorem of the integral calculus , whereby the deleted quantities are from the interval . ${\displaystyle x'_{i}}$ ${\displaystyle [x_{i},x_{i}+\Delta x_{i}]}$

${\displaystyle {\begin{aligned}{\frac {1}{|\Delta V|}}\int \limits _{\partial (\Delta V)}{\vec {F}}\cdot {\vec {n}}\,\mathrm {d} S&={\frac {\left({\vec {F}}(x_{1}+\Delta x_{1},x'_{2},\ldots ,x'_{n})-{\vec {F}}(x_{1},x'_{2},\ldots ,x'_{n})\right)\cdot {\hat {e}}_{1}}{\Delta x_{1}}}\underbrace {\left[{\frac {1}{\Delta x_{2}}}\int \limits _{x_{2}}^{x_{2}+\Delta x_{2}}\mathrm {d} x_{2}\right.} _{=1}\ldots \underbrace {\left.{\frac {1}{\Delta x_{n}}}\int \limits _{x_{n}}^{x_{n}+\Delta x_{n}}\mathrm {d} x_{n}\right]} _{=1}+\ldots \\&+{\frac {\left({\vec {F}}(x'_{1},\ldots ,x'_{n-1},x_{n}+\Delta x_{n})-{\vec {F}}(x'_{1},\ldots ,x'_{n-1},x_{n})\right)\cdot {\hat {e}}_{n}}{\Delta x_{n}}}\left[{\frac {1}{\Delta x_{1}}}\int \limits _{x_{1}}^{x_{1}+\Delta x_{1}}\mathrm {d} x_{1}\ldots {\frac {1}{\Delta x_{n-1}}}\int \limits _{x_{n-1}}^{x_{n-1}+\Delta x_{n-1}}\mathrm {d} x_{n-1}\right]\end{aligned}}}$

Thus, only the sum of the remains difference quotient left

${\displaystyle {\frac {1}{|\Delta V|}}\int \limits _{\partial (\Delta V)}{\vec {F}}\cdot {\vec {n}}\,\mathrm {d} S={\frac {F_{1}(x_{1}+\Delta x_{1},x'_{2},\ldots ,x'_{n})-F_{1}(x_{1},x'_{2},\ldots ,x'_{n})}{\Delta x_{1}}}+\ldots +{\frac {F_{n}(x'_{1},\ldots ,x'_{n-1},x_{n}+\Delta x_{n})-F_{n}(x'_{1},\ldots ,x'_{n-1},x_{n})}{\Delta x_{n}}}}$,

which become partial derivatives at the border crossing : ${\displaystyle \Delta x_{i}\to 0}$

${\displaystyle \operatorname {div} \,{\vec {F}}(x_{1},x_{2},\ldots ,x_{n})={\frac {\partial F_{1}}{\partial x_{1}}}(x_{1},x_{2},\ldots ,x_{n})+\ldots +{\frac {\partial F_{n}}{\partial x_{n}}}(x_{1},x_{2},\ldots ,x{}_{n})}$

Covariant behavior with rotations and displacements

The divergence operator interchanges with spatial rotations and displacements of a vector field, i.e. H. the order of these operations makes no difference.

Reason: If the vector field is rotated or (parallel) shifted in space, one only needs to rotate the surface and volume elements in the same way in the coordinate-independent representation given above in order to get back to the same scalar expression. The scalar field rotates and shifts in the same way as the vector field . ${\displaystyle {\vec {F}}}$ ${\displaystyle \operatorname {div} \,{\vec {F}}}$ ${\displaystyle {\vec {F}}}$

A "decomposition theorem"

For n = 3 dimensional vector fields that are continuously differentiable throughout the room at least twice and go at infinity sufficiently rapidly to zero is considered that they are in a vortex-free portion and a source-free part disintegrate . For the eddy-free part it applies that it can be represented by its source density as follows: ${\displaystyle {\vec {F}}({\vec {r}})}$ ${\displaystyle {\vec {E}}}$ ${\displaystyle {\vec {B}}}$ ${\displaystyle {\vec {F}}={\vec {E}}+{\vec {B}}}$

${\displaystyle {\vec {E}}({\vec {r}})=-\nabla \Phi ({\vec {r}})}$, With

${\displaystyle \Phi ({\vec {r}})=\int _{\mathbb {R} ^{3}}\,\mathrm {d} ^{(3)}r'\,\,{\frac {\mathrm {div} \,{\vec {E}}({\vec {r}}')}{4\pi |{\vec {r}}-{\vec {r}}'|}}}$.

For the source-free part,, the same applies if the scalar potential is replaced by a so-called vector potential and at the same time the expressions or (= source density of ) are substituted by the operations or (= vortex density of ). ${\displaystyle {\vec {B}}({\vec {r}})}$ ${\displaystyle \Phi }$ ${\displaystyle {\vec {A}}}$ ${\displaystyle -\nabla \,\Phi }$ ${\displaystyle \operatorname {div} \,{\vec {E}}}$ ${\displaystyle {\vec {E}}}$ ${\displaystyle \nabla \times \,{\vec {A}}}$ ${\displaystyle \nabla \times {\vec {B}}}$ ${\displaystyle {\vec {B}}}$

This procedure is part of the Helmholtz theorem .

properties

In n-dimensional space

Let be a constant, an open subset, a scalar field and two vector fields. Then the following rules apply: ${\displaystyle c\in \mathbb {R} }$ ${\displaystyle \Omega \subset \mathbb {R} ^{n}}$ ${\displaystyle u\colon \Omega \to \mathbb {R} }$ ${\displaystyle {\vec {F}},{\vec {G}}\colon \Omega \to \mathbb {R} ^{n}}$

• The divergence is linear , that is, it applies

${\displaystyle \operatorname {div} \,(c\cdot {\vec {F}})=c\cdot \operatorname {div} \,{\vec {F}}}$ and

${\displaystyle \operatorname {div} \,({\vec {F}}+{\vec {G}})=\operatorname {div} \,{\vec {F}}+\operatorname {div} \,{\vec {G}}.}$

• The product rule applies to the divergence

${\displaystyle \operatorname {div} (u{\vec {F}})=\operatorname {grad} (u)\cdot {\vec {F}}+u\operatorname {div} {\vec {F}}}$

• The divergence of the vector field corresponds to the trace of the covariant derivative of in any coordinates , that is, it applies . This representation is coordinate-invariant, since the trace of a linear mapping is invariant to a base change . ${\displaystyle {\vec {F}}}$ ${\displaystyle \nabla {\vec {F}}}$ ${\displaystyle {\vec {F}}}$
  ${\displaystyle \operatorname {div} \,{\vec {F}}=\operatorname {Spur} (\nabla {\vec {F}})}$
  

In three-dimensional space

If there is also a product rule for the cross product , this is ${\displaystyle n=3}$ ${\displaystyle \times }$

${\displaystyle \operatorname {div} ({\vec {F}}\times {\vec {G}})={\vec {G}}\cdot \operatorname {rot} \,{\vec {F}}-{\vec {F}}\cdot \operatorname {rot} \,{\vec {G}}\,,}$

whereby the rotation is meant. Because of all differentiable it follows ${\displaystyle \operatorname {rot} }$ ${\displaystyle \operatorname {rot(grad} (f))=0}$ ${\displaystyle f}$

${\displaystyle \operatorname {div(grad} f_{1}\times \operatorname {grad} f_{2})=0}$

for any differentiable . ${\displaystyle f_{1},\,f_{2}}$

Examples

One finds immediately in Cartesian coordinates

${\displaystyle \operatorname {div} \,{\vec {r}}=\operatorname {div} \left({\begin{array}{c}x\\y\\z\end{array}}\right)=3}$

For the Coulomb field can be found when the first product rule ,   and is set ${\displaystyle u\!\,=1/r^{3}}$ ${\displaystyle \operatorname {grad} \,u=-3{\hat {e}}_{r}/r^{4}}$ ${\displaystyle {\vec {F}}={\vec {r}}}$

${\displaystyle \operatorname {div} \,{\frac {{\hat {e}}_{r}}{r^{2}}}=\operatorname {div} \,{\frac {\vec {r}}{r^{3}}}=-{\frac {3{\hat {e}}_{r}}{r^{4}}}\cdot {\vec {r}}+{\frac {3}{r^{3}}}=0\qquad r=|{\vec {r}}|\neq 0,\quad {\vec {r}}=r{\hat {e}}_{r}\,.}$

This result can also be obtained with the formula for the divergence in spherical coordinates.

According to the corollary, fields of the following type are source-free: ${\displaystyle {\vec {f}}}$

${\displaystyle {\vec {f}}={\frac {{\vec {c}}\times {\vec {r}}}{r^{3}}}\quad {\vec {c}}={\text{const}}\,,\quad {\vec {f}}={\vec {r}}\times \operatorname {grad} \,Y_{lm}(\vartheta ,\varphi )}$

Gaussian integral theorem

→ Main article : Gaussian integral theorem

statement

The divergence in the statement of the Gaussian integral theorem plays an important role. It says that the flow through a closed surface is equal to the integral over the divergence of the vector field inside this volume, and thus allows the conversion of a volume integral into a surface integral :

${\displaystyle \iiint \limits _{V}\operatorname {div} {\vec {F}}\;\mathrm {d} V=\iint \limits _{\partial V}{\vec {F}}\;\cdot {\vec {n}}\,\mathrm {d} S\,,}$

where is the normal vector of the surface . In the case of a flow, it clearly describes the relationship between the flow through this area and the flow sources and sinks within the associated volume. ${\displaystyle {\vec {n}}}$ ${\displaystyle \partial V}$

Point source

If one uses the Coulomb-like field in the Gaussian integral theorem and one chooses a spherical surface with a radius around the origin as the integration surface , then and the integrand becomes constant . Because the surface of the sphere is followed ${\displaystyle {\vec {F}}_{C}={\hat {e}}_{r}/r^{2}}$ ${\displaystyle \partial V}$ ${\displaystyle r\!\,}$ ${\displaystyle {\vec {n}}={\hat {e}}_{r}}$ ${\displaystyle 1/\!\,r^{2}}$ ${\displaystyle 4\pi \!\,r^{2}}$

${\displaystyle \iint \limits _{\partial K}{\vec {F}}_{C}\cdot {\vec {n}}\,\mathrm {d} S={\frac {1}{r^{2}}}\iint \limits _{\partial K}\mathrm {d} S=4\pi }$

Thus, the integral theorem provides information that, in contrast to the derivation expressions (product rule or spherical coordinates), also includes the point : The volume integral of is . This can be summarized with the result of the derivative calculation to form a distribution equation: ${\displaystyle \operatorname {div} \,{\vec {F}}_{C}}$ ${\displaystyle {\vec {r}}=0}$ ${\displaystyle \operatorname {div} \,{\vec {F}}_{C}}$ ${\displaystyle 4\pi }$

${\displaystyle \operatorname {div} \,{\frac {{\hat {e}}_{r}}{r^{2}}}=4\pi \delta ({\vec {r}})}$

Cylinder and spherical coordinates

In cylindrical coordinates the following applies to the divergence of a vector field : ${\displaystyle {\vec {F}}(\rho ,\varphi ,z)}$

${\displaystyle \operatorname {div} \,{\vec {F}}={\frac {1}{\rho }}{\frac {\partial }{\partial \rho }}(\rho F_{\rho })+{\frac {1}{\rho }}{\frac {\partial F_{\varphi }}{\partial \varphi }}+{\frac {\partial F_{z}}{\partial z}}}$

In spherical coordinates the following applies to the divergence of a vector field : ${\displaystyle {\vec {F}}(r,\theta ,\varphi )}$

${\displaystyle \operatorname {div} \,{\vec {F}}={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}(r^{2}F_{r})+{\frac {1}{r\sin \theta }}{\frac {\partial }{\partial \theta }}(F_{\theta }\sin \theta )+{\frac {1}{r\sin \theta }}{\frac {\partial F_{\varphi }}{\partial \varphi }}}$

The latter formula can be derived without differentiating basis vectors: A test function is introduced and a volume integral is written once in Cartesian and once in spherical coordinates. With known expressions for gradient and volume element, this results after multiplying out the basis vectors ${\displaystyle g}$

${\displaystyle {\begin{aligned}\int {\vec {F}}\cdot \operatorname {grad} \,g\,\mathrm {d} V&=\int \left(F_{x}{\frac {\partial g}{\partial x}}+F_{y}{\frac {\partial g}{\partial y}}+F_{z}{\frac {\partial g}{\partial z}}\right)\mathrm {d} x\,\mathrm {d} y\,\mathrm {d} z\\&=\int \left(F_{r}{\frac {\partial g}{\partial r}}+{\frac {F_{\theta }}{r}}\,{\frac {\partial g}{\partial \theta }}+{\frac {F_{\varphi }}{r\sin \theta }}\,{\frac {\partial g}{\partial \varphi }}\right)r^{2}\mathrm {d} r\,\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi \end{aligned}}}$

The derivatives of are partially integrated, whereby boundary terms vanish. On the right-hand side, the volume element must also be differentiated and then restored in two terms (expand). That makes ${\displaystyle g}$

${\displaystyle \int g\,\left(\operatorname {div} \,{\vec {F}}\right)_{\mathrm {kartesisch} }\mathrm {d} V=\int g\,\left(\operatorname {div} \,{\vec {F}}\right)_{\mathrm {Kugelkoordinaten} }\mathrm {d} V}$

From the equality of the integrals for all test functions it follows that the expressions for the divergence are the same.

Inverse

According to the Poincaré lemma , for every scalar field there is a vector field whose divergence it is. This vector field is not clearly defined, because a locally constant vector can be added without changing the divergence and thus the scalar field.

Under certain conditions there is a right or left inverse of the divergence. So there is an operator for an open and restricted area with Lipschitz's continuous edge , so that for each with ${\displaystyle \Omega \subset \mathbb {R} ^{n}}$ ${\displaystyle B\colon W^{m,p}(\Omega )\rightarrow W^{m,p}(\Omega )}$ ${\displaystyle f\in W^{m,p}(\Omega )}$ ${\displaystyle \textstyle \int _{\Omega }f\,\mathrm {d} \lambda =0}$

${\displaystyle \operatorname {div} (Bf)=f}$

applies, where the corresponding Sobolew space denotes for and . is called the Bogowskii operator. ${\displaystyle W^{m,p}(\Omega )}$ ${\displaystyle m\in \mathbb {N} }$ ${\displaystyle 1<p<\infty }$ ${\displaystyle B}$

Divergence on Riemannian manifolds

In the section Properties it was already said that the divergence can be expressed with the help of the trace of the Jacobian matrix and that this representation is coordinate-invariant. For this reason one uses this property to define the divergence on Riemannian manifolds . With the help of this definition one can, for example, define the Laplace operator on Riemannian manifolds without coordinates. This is then called the Laplace-Beltrami operator .

definition

Let be a Riemannian manifold and a - vector field with . Then the divergence is over ${\displaystyle M}$ ${\displaystyle F\in \Gamma (TM)}$ ${\displaystyle C^{k}}$ ${\displaystyle k\geq 1}$

${\displaystyle \operatorname {div} (F):=\operatorname {Spur} (\xi \mapsto \nabla _{\xi }F)}$

Are defined. There is a vector field and the operator is the Levi-Civita relationship , which generalizes the Nabla operator. If one evaluates an , then is and one can form the trace known from linear algebra for all of them . ${\displaystyle \xi \in \Gamma (TM)}$ ${\displaystyle \nabla }$ ${\displaystyle \nabla _{\xi }F}$ ${\displaystyle p\in M}$ ${\displaystyle \xi |_{p}\mapsto \nabla _{\xi |_{p}}F|_{p}\in \operatorname {End} (T_{p}M)}$ ${\displaystyle p}$

Transport theorem and geometric interpretation

The transport theorem applies to the flow of a vector field ${\displaystyle \varphi \colon U\subseteq \mathbb {R} \times M\to M,\;(t,x)\mapsto \varphi _{t}(x)}$ ${\displaystyle F}$

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\int _{\varphi _{t}(\Omega )}\varrho (t,x)\mathrm {d} \lambda (x)=\int _{\varphi _{t}(\Omega )}(\partial _{t}\varrho (t,x)+\operatorname {div} (\varrho (t,x)F(x)))\mathrm {d} \lambda (x).}$

It is the Riemann-Lebesgue volume measure on the manifold, a relatively compact measurable subset and a smooth function . If one interprets the density of a conserved quantity , then the continuity equation follows . For one receives ${\displaystyle \lambda }$ ${\displaystyle \Omega }$ ${\displaystyle \varrho }$ ${\displaystyle \varrho }$ ${\displaystyle \varrho =1}$

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\lambda (\varphi _{t}(\Omega ))=\int _{\varphi _{t}(\Omega )}\operatorname {div} F(x)d\lambda (x).}$

The divergence is therefore the density of the rate of volume change with respect to the flow. The divergence at a point indicates how quickly the content of an infinitesimal volume element changes at that point when it moves with the flow. The result is that a vector field is divergence-free precisely when the generated flow is volume-preserving.

Divergence of second order tensors

In engineering, the divergence in Cartesian coordinates is also introduced for second order tensors and then supplies vector fields. For example, the divergence of the stress tensor is included in the local momentum balances of continuum mechanics .

definition

Second level tensors are formed with the dyadic product “ ” of vectors to which the divergence can be applied. In this way the divergence can also be generalized to tensors. Be ${\displaystyle \otimes }$

${\displaystyle \mathbf {T} =\sum _{j=1}^{n}{\vec {t}}_{j}\otimes {\hat {e}}_{j}=\sum _{i,j=1}^{n}T_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}}$

a tensor with column vectors with components T _ij with respect to the standard basis of n-dimensional space. Then the divergence of the tensor can be defined as: ${\displaystyle {\vec {t}}_{j}}$ ${\displaystyle {\hat {e}}_{1,...,n}}$

${\displaystyle \operatorname {div} (\mathbf {T} ):=\sum _{k=1}^{n}{\hat {e}}_{k}\cdot {\frac {\partial }{\partial x_{k}}}\mathbf {T} =\sum _{j,k=1}^{n}{\hat {e}}_{k}\cdot {\frac {\partial }{\partial x_{k}}}{\vec {t}}_{j}\otimes {\hat {e}}_{j}=\sum _{j=1}^{n}\operatorname {div} ({\vec {t}}_{j}){\hat {e}}_{j}=\sum _{i,j=1}^{n}{\frac {\partial T_{ij}}{\partial x_{i}}}{\hat {e}}_{j}\,.}$

In three dimensions, a Cartesian coordinate system with x, y and z coordinates results:

${\displaystyle \operatorname {div} \mathbf {T} ={\begin{pmatrix}{\frac {\partial T_{xx}}{\partial x}}+{\frac {\partial T_{yx}}{\partial y}}+{\frac {\partial T_{zx}}{\partial z}}\\{\frac {\partial T_{xy}}{\partial x}}+{\frac {\partial T_{yy}}{\partial y}}+{\frac {\partial T_{zy}}{\partial z}}\\{\frac {\partial T_{xz}}{\partial x}}+{\frac {\partial T_{yz}}{\partial y}}+{\frac {\partial T_{zz}}{\partial z}}\end{pmatrix}}\,.}$

With the Nabla operator , the divergence of a tensor is written:

${\displaystyle \operatorname {div} \mathbf {T} :=\nabla \cdot \mathbf {T} }$

In the literature, however, the transposed version with the line vectors also occurs ${\displaystyle {\vec {z}}_{i}}$

${\displaystyle \mathbf {T} =\sum _{i=1}^{n}{\hat {e}}_{i}\otimes {\vec {z}}_{i}\quad \rightarrow \quad {\tilde {\operatorname {div} }}\mathbf {T} =\nabla \cdot (\mathbf {T} ^{\top })=\operatorname {div} (\mathbf {T} ^{\top })=\sum _{i=1}^{n}\operatorname {div} ({\vec {z}}_{i}){\hat {e}}_{i}=\sum _{i,j=1}^{n}{\frac {\partial T_{ij}}{\partial x_{j}}}{\hat {e}}_{i}\,,}$

which differs from the definition here by the transposition of the argument.

Expansion rate

Primordial image space V, which is transformed into image space v by the motion function χ

The divergence of a vector field in this formalism reads: ${\displaystyle {\vec {v}}}$

${\displaystyle \operatorname {div} \,{\vec {v}}=\operatorname {Sp} (\operatorname {grad} \,{\vec {v}})\quad \leftrightarrow \quad \sum _{i=1}^{n}{\frac {\mathrm {d} v_{i}}{\mathrm {d} x_{i}}}=\operatorname {Sp} \left(\sum _{i,j=1}^{n}{\frac {\mathrm {d} v_{i}}{\mathrm {d} x_{j}}}{\hat {e}}_{i}\otimes {\hat {e}}_{j}\right)\,.}$

If specifically the speed field of a movement (image space) of points from a time-independent volume V (original image space), see figure, then the gradient of the vector field is the speed gradient l ${\displaystyle {\vec {v}}({\vec {x}},t)={\dot {\vec {\chi }}}({\vec {X}},t)}$ ${\displaystyle {\vec {\chi }}({\vec {X}},t)={\vec {x}}\in v}$ ${\displaystyle {\vec {X}}\in V}$

${\displaystyle \operatorname {div} \,{\vec {v}}=\operatorname {Sp} (\operatorname {grad} \,{\vec {v}})=\operatorname {Sp} (\mathbf {l} )=\operatorname {Sp} ({\dot {\mathbf {F} }}\cdot \mathbf {F} ^{-1})\,,}$

which is related to the time derivative of the deformation gradient F and its inverse . The determinant of the deformation gradient transforms the volume shapes (red in the picture) into one another:

${\displaystyle \mathrm {d} v=\operatorname {det} (\mathbf {F} )\,\mathrm {d} V\,.}$

The time derivative of this equation with the Frobenius scalar product results in ":" (see derivatives of the main invariants )

${\displaystyle (\mathrm {d} v){\dot {}}=\operatorname {det} (\mathbf {F} )\mathbf {F} ^{\top -1}:{\dot {\mathbf {F} }}\,\mathrm {d} V=\operatorname {Sp} (\mathbf {F} ^{-1}\cdot {\dot {\mathbf {F} }})\,\mathrm {d} v=\operatorname {Sp} ({\dot {\mathbf {F} }}\cdot \mathbf {F} ^{-1})\,\mathrm {d} v=\operatorname {div} ({\vec {v}})\,\mathrm {d} v\,,}$

because the form of volume in the archetypal space does not depend on time. When the divergence disappears, the movement is locally volume-preserving. A positive divergence means expansion, which in reality is associated with a decrease in density .

properties

In n-dimensional space

Let be a constant, an open subset, a scalar field, two vector fields and T a tensor field. Then the following rules apply: ${\displaystyle c\in \mathbb {R} }$ ${\displaystyle \Omega \subset \mathbb {R} ^{n}}$ ${\displaystyle u\colon \Omega \to \mathbb {R} }$ ${\displaystyle {\vec {F}},{\vec {G}}\colon \Omega \to \mathbb {R} ^{n}}$

${\displaystyle {\begin{array}{rclcl}\operatorname {div} (u\mathbf {T} )&=&{\hat {e}}_{i}\cdot (u_{,i}\mathbf {T} +u\mathbf {T} _{,i})&=&\operatorname {grad} (u)\cdot \mathbf {T} +u\operatorname {div} \mathbf {T} \\\operatorname {div} (\mathbf {T} \cdot {\vec {F}})&=&{\hat {e}}_{i}\cdot \left(\mathbf {T} _{,i}\cdot {\vec {F}}+\mathbf {T} \cdot {\vec {F}}_{,i}\right)&=&\operatorname {div} (\mathbf {T} )\cdot {\vec {F}}+\mathbf {T} ^{\top }:\operatorname {grad} {\vec {F}}=\operatorname {div} ({\vec {F}}\cdot \mathbf {T} ^{\top })\\\operatorname {div} ({\vec {F}}\otimes {\vec {G}})&=&{\hat {e}}_{i}\cdot \left({\vec {F}}_{,i}\otimes {\vec {G}}+{\vec {F}}\otimes {\vec {G}}_{,i}\right)&=&\operatorname {div} ({\vec {F}}){\vec {G}}+\operatorname {grad} ({\vec {G}})\cdot {\vec {F}}\end{array}}}$

This is the Frobenius scalar product for vectors or tensors and a derivation according to the coordinate x _i in a Cartesian coordinate system with base vectors is abbreviated with an index _{, i} , over which one to three can also be added ( Einstein's sum convention ). ${\displaystyle \cdot ,:}$ ${\displaystyle {\hat {e}}_{i}}$

In three-dimensional space

For the derivation of the second Cauchy-Euler law of motion , which ensures the conservation of the angular momentum in a continuum, the product rule is

${\displaystyle \operatorname {div} (\mathbf {T} \times {\vec {F}})=\operatorname {div} (\mathbf {T} )\times {\vec {F}}-\operatorname {grad} ({\vec {F}})\cdot \times \mathbf {T} }$

second hand. There is a vectorial and T a tensorial, differentiable field and the product “· ×” is defined by dyads ${\displaystyle {\vec {F}}}$ ${\displaystyle ({\vec {a}}\otimes {\vec {g}})\cdot \times ({\vec {h}}\otimes {\vec {b}}):=({\vec {g}}\cdot {\vec {h}}){\vec {a}}\times {\vec {b}}\,.}$

Gaussian integral theorem

This integral theorem is also used in continuum mechanics for tensor fields, e.g. B. of stress tensors , required: ${\displaystyle {\boldsymbol {\sigma }}}$

${\displaystyle \iiint \limits _{V}\operatorname {div} {\boldsymbol {\sigma }}\;\mathrm {d} V=\iint \limits _{\partial V}{\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}\,\mathrm {d} S=\iint \limits _{\partial V}{\vec {t}}\,\mathrm {d} S\,.}$

According to Cauchy's fundamental theorem, the normal vector to the surface transformed by the transposed stress tensor is the stress vector acting on the surface (a vector with the dimension force per surface). In the case of its disappearance, this equation is already the momentum balance of deformable bodies in the static case in the absence of a volume force. ${\displaystyle {\vec {t}}}$

Cylinder and spherical coordinates

For tensors of the second order a transformation in cylindrical coordinates results

${\displaystyle {\begin{aligned}\operatorname {div} (\mathbf {T} )=&\left(T_{\rho \rho ,\rho }+{\frac {1}{\rho }}(T_{\varphi \rho ,\varphi }+T_{\rho \rho }-T_{\varphi \varphi })+T_{z\rho ,z}\right){\hat {e}}_{\rho }\\&+\left(T_{\rho \varphi ,\rho }+{\frac {1}{\rho }}(T_{\varphi \varphi ,\varphi }+T_{\rho \varphi }+T_{\varphi \rho })+T_{z\varphi ,z}\right){\hat {e}}_{\varphi }\\&+\left(T_{\rho z,\rho }+{\frac {1}{\rho }}(T_{\varphi z,\varphi }+T_{\rho z})+T_{zz,z}\right){\hat {e}}_{z}\end{aligned}}}$

and in spherical coordinates:

${\displaystyle {\begin{aligned}\operatorname {div} (\mathbf {T} )=&\left({\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}(r^{2}T_{rr})+{\frac {1}{r\sin \vartheta }}T_{\varphi r,\varphi }+{\frac {1}{r\sin \vartheta }}{\frac {\partial }{\partial \vartheta }}(T_{\vartheta r}\sin \vartheta )-{\frac {1}{r}}(T_{\vartheta \vartheta }+T_{\varphi \varphi })\right){\hat {e}}_{r}\\&+\left({\frac {1}{r^{3}}}{\frac {\partial }{\partial r}}(r^{3}T_{r\vartheta })+{\frac {1}{r\sin \vartheta }}T_{\varphi \vartheta ,\varphi }+{\frac {1}{r\sin \vartheta }}{\frac {\partial }{\partial \vartheta }}(T_{\vartheta \vartheta }\sin \vartheta )+{\frac {1}{r}}(T_{\vartheta r}-T_{r\vartheta }-T_{\varphi \varphi }\cot \vartheta )\right){\hat {e}}_{\vartheta }\\&+\left({\frac {1}{r^{3}}}{\frac {\partial }{\partial r}}(r^{3}T_{r\varphi })+{\frac {1}{r\sin \vartheta }}T_{\varphi \varphi ,\varphi }+{\frac {1}{r\sin \vartheta }}{\frac {\partial }{\partial \vartheta }}(T_{\vartheta \varphi }\sin \vartheta )+{\frac {1}{r}}(T_{\varphi r}-T_{r\varphi }+T_{\varphi \vartheta }\cot \vartheta )\right){\hat {e}}_{\varphi }\,.\end{aligned}}}$

literature

• Adolf J. Schwab : Conceptual World of Field Theory , Springer Verlag, ISBN 3-540-42018-5
• H. Altenbach: Continuum Mechanics . Springer, 2012, ISBN 978-3-642-24118-5 .
• LP Kuptsov: Divergence . In: Michiel Hazewinkel (Ed.): Encyclopaedia of Mathematics . Springer-Verlag , Berlin 2002, ISBN 978-1-55608-010-4 (English, online ).

Web links

Commons : Divergence  - collection of images, videos and audio files

• How do I "bend" Nabla and Delta? - Derivation of the nabla operator for orthonormal curvilinear coordinates , matheplanet.com

Footnotes

1. ↑ Using Einstein's summation convention and the abbreviation " _{, i} " for a derivation according to the coordinate x _i in a Cartesian coordinate system with basis vectors, the following is calculated step by step and thus: ${\displaystyle \mathbf {T} =:{\hat {e}}_{k}\otimes {\vec {z}}_{k}}$ ${\displaystyle {\hat {e}}_{i}}$
   ${\displaystyle {\hat {e}}_{i}\cdot (\mathbf {T} \times {\vec {F}}_{,i})={\hat {e}}_{i}\cdot ({\hat {e}}_{k}\otimes {\vec {z}}_{k}\times {\vec {F}}_{,i})={\vec {z}}_{i}\times {\vec {F}}_{,i}=-{\vec {F}}_{,i}\times {\vec {z}}_{i}=-({\vec {F}}_{,i}\otimes {\hat {e}}_{i})\cdot \times ({\hat {e}}_{k}\otimes {\vec {z}}_{k})=-\operatorname {grad} ({\vec {F}})\cdot \times \mathbf {T} }$
   
   ${\displaystyle \operatorname {div} (\mathbf {T} \times {\vec {F}})={\hat {e}}_{i}\cdot (\mathbf {T} _{,i}\times {\vec {F}}+\mathbf {T} \times {\vec {F}}_{,i})={\hat {e}}_{i}\cdot \mathbf {T} _{,i}\times {\vec {F}}+{\hat {e}}_{i}\cdot (\mathbf {T} \times {\vec {F}}_{,i})=\operatorname {div} (\mathbf {T} )\times {\vec {F}}-\operatorname {grad} ({\vec {F}})\cdot \times \mathbf {T} }$

Individual evidence

1. ^ GP Galdi, An introduction to the mathematical theory of the Navier-Stokes equations. Vol. I, Springer Tracts in Natural Philosophy, vol. 38, Springer-Verlag, New York, 1994, ISBN 0-387-94172-X
2. ↑ Isaac Chavel: Eigenvalues ​​in Riemannian Geometry , Academic Press, 1984, 2nd edition ISBN 978-0-12-170640-1 , page 3.
3. ^ Herbert Amann, Joachim Escher: Analysis III . 2nd Edition. Birkhäuser, Basel 2008, ISBN 978-3-7643-8883-6 , p. 438 (Chapter XII).

1. ↑ http://www.mb.uni-siegen.de/fkm/lehre/konti2012_files/skript1-3.pdf script, University of Siegen