Rotation of a vector field

This article describes the mathematical differential operator "rotation". For the rotational movement of a body see rotation (physics) , for the geometrical representation see rotation .

In vector analysis , a branch of mathematics, a certain differential operator is called rotation or rotor , which assigns a new vector field to a vector field in three-dimensional Euclidean space with the help of differentiation .

The rotation of a flow field indicates twice the angular speed at which a body that is swimming with it rotates ("rotates") for each location . This connection is eponymous.

The speed field of a rotating disk has a constant rotation parallel to the axis of rotation
${\displaystyle \left(\omega <0\right)}$

However, it does not always have to be a speed field and a rotary movement; for example, the law of induction concerns the rotation of the electric field.

A vector field whose rotation is zero everywhere in an area is called eddy-free or, especially in the case of force fields, conservative . If the area is simply connected, then the vector field is the gradient of a function if and only if the rotation of the vector field in the considered area is zero.

The divergence of the rotation of a vector field is zero. Conversely, in areas that are simply connected, a field whose divergence is zero is the rotation of another vector field.

Examples:

• The vector field, which indicates the wind direction and speed of a cyclone at every location, has a rotation other than zero in the vicinity of the eye.
• The vector field that indicates the speed at each point of a rotating disk has the same non-zero rotation at each point. The rotation is twice the angular velocity, ${\displaystyle {\vec {v}}(x,y,z)=\omega \cdot (x\,{\hat {e}}_{y}-y\,{\hat {e}}_{x})\,,}$ ${\displaystyle \operatorname {rot} \,{\vec {v}}(x,y,z)=2\,\omega \,{\hat {e}}_{z}\,.}$
• The force field, which indicates the gravitational force of the sun on a test particle at each point, is eddy-free. The force field is the negative gradient of the potential energy of the particle.

Definition of the rotation in Cartesian coordinates

Are the Cartesian coordinates of the three-dimensional Euclidean space and , and the normalized to unit length, orthogonal basis vectors pointing at any point in the direction of increasing coordinates. ${\displaystyle (x,y,z)}$ ${\displaystyle {\hat {e}}_{x}}$ ${\displaystyle {\hat {e}}_{y}}$ ${\displaystyle {\hat {e}}_{z}}$

The rotation of a three-dimensional, differentiable vector field

${\displaystyle {\vec {F}}(x,y,z)=F_{x}(x,y,z)\,{\hat {e}}_{x}+F_{y}(x,y,z)\,{\hat {e}}_{y}+F_{z}(x,y,z)\,{\hat {e}}_{z}}$

is the three-dimensional vector field

${\displaystyle \operatorname {rot} \,{\vec {F}}(x,y,z)=\left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right){\hat {e}}_{x}+\left({\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\right){\hat {e}}_{y}+\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right){\hat {e}}_{z}\,.}$

One can understand a matrix as a formal determinant , the first column of which contains the Cartesian basis vectors, the second the partial derivatives according to the Cartesian coordinates and the third the component functions to be differentiated ${\displaystyle \operatorname {rot} \,{\vec {F}}}$

${\displaystyle \operatorname {rot} \,{\vec {F}}=\operatorname {det} \,{\begin{pmatrix}{\hat {e}}_{x}&{\frac {\partial }{\partial x}}&F_{x}\\{\hat {e}}_{y}&{\frac {\partial }{\partial y}}&F_{y}\\{\hat {e}}_{z}&{\frac {\partial }{\partial z}}&F_{z}\end{pmatrix}}\,=\operatorname {det} \,{\begin{pmatrix}{\hat {e}}_{x}&{\hat {e}}_{y}&{\hat {e}}_{z}\\{\frac {\partial }{\partial x}}&{\frac {\partial }{\partial y}}&{\frac {\partial }{\partial z}}\\F_{x}&F_{y}&F_{z}\end{pmatrix}}\,.}$

However, the different columns here are not vectors of the same vector space.

If the vectors are given as column vectors of their Cartesian components, then the formal cross product of the column vector of the partial derivatives according to the Cartesian coordinates, the Nabla operator , with the column vector of the Cartesian component functions ${\displaystyle \operatorname {rot} \,{\vec {F}}}$ ${\displaystyle \nabla }$

${\displaystyle \operatorname {rot} \,{\vec {F}}(x,y,z)=\nabla \times {\vec {F}}={\begin{pmatrix}{\frac {\partial }{\partial x}}\\{\frac {\partial }{\partial y}}\\{\frac {\partial }{\partial z}}\end{pmatrix}}\times {\begin{pmatrix}F_{x}\\F_{y}\\F_{z}\end{pmatrix}}={\begin{pmatrix}{\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\\{\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\\{\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\end{pmatrix}}\,.}$

Other coordinate representations of the rotation

Spherical coordinates

If you write the vector field in spherical coordinates as a linear combination ${\displaystyle (r,\theta ,\varphi )}$

${\displaystyle {\vec {F}}(r,\theta ,\varphi )=F_{r}(r,\theta ,\varphi )\,{\hat {e}}_{r}+F_{\theta }(r,\theta ,\varphi )\,{\hat {e}}_{\theta }+F_{\varphi }(r,\theta ,\varphi )\,{\hat {e}}_{\varphi }}$

of the vectors normalized to unit length

${\displaystyle {\begin{aligned}{\hat {e}}_{r}&={\frac {1}{\sqrt {x^{2}+y^{2}+z^{2}}}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}\,,\\{\hat {e}}_{\theta }&={\frac {1}{\sqrt {(x^{2}+y^{2}+z^{2})(x^{2}+y^{2})}}}{\begin{pmatrix}z\,x\\z\,y\\-x^{2}-y^{2}\end{pmatrix}}\,,\\{\hat {e}}_{\varphi }&={\frac {1}{\sqrt {x^{2}+y^{2}}}}{\begin{pmatrix}-y\\x\\0\end{pmatrix}}\,,\end{aligned}}}$

which point in the direction of increasing coordinates at every point , then is the rotation ${\displaystyle r,\theta ,\varphi }$

${\displaystyle {\begin{aligned}\operatorname {rot} \,{\vec {F}}=\,&{\frac {1}{r\sin \theta }}\left[{\frac {\partial }{\partial \theta }}\left(F_{\varphi }\sin \theta \right)-{\frac {\partial F_{\theta }}{\partial \varphi }}\right]{\hat {e}}_{r}+\left[{\frac {1}{r\sin \theta }}{\frac {\partial F_{r}}{\partial \varphi }}-{\frac {1}{r}}{\frac {\partial }{\partial r}}\left(rF_{\varphi }\right)\right]{\hat {e}}_{\theta }\,+\,{\frac {1}{r}}\left[{\frac {\partial }{\partial r}}\left(rF_{\theta }\right)-{\frac {\partial F_{r}}{\partial \theta }}\right]{\hat {e}}_{\varphi }\,.\end{aligned}}}$

Cylindrical coordinates

If you give the vector field in cylindrical coordinates as a linear combination ${\displaystyle (r,\varphi ,z)}$

${\displaystyle {\vec {F}}(r,\varphi ,z)=F_{r}(r,\varphi ,z)\,{\hat {e}}_{r}+F_{\varphi }(r,\varphi ,z)\,{\hat {e}}_{\varphi }+F_{z}(r,\varphi ,z)\,{\hat {e}}_{z}}$

of the vectors

${\displaystyle {\begin{aligned}{\hat {e}}_{r}&={\frac {1}{\sqrt {x^{2}+y^{2}}}}{\begin{pmatrix}x\\y\\0\end{pmatrix}}\,,\\{\hat {e}}_{\varphi }&={\frac {1}{\sqrt {x^{2}+y^{2}}}}{\begin{pmatrix}-y\\x\\0\end{pmatrix}}\,,\\{\hat {e}}_{z}&={\begin{pmatrix}0\\0\\1\end{pmatrix}}\,,\end{aligned}}}$

which, normalized to unit length, point at each point in the direction of increasing coordinates, then the rotation is ${\displaystyle r,\varphi ,z}$

${\displaystyle {\begin{aligned}\operatorname {rot} \,{\vec {F}}=\left[{\frac {1}{r}}{\frac {\partial F_{z}}{\partial \varphi }}-{\frac {\partial F_{\varphi }}{\partial z}}\right]{\hat {e}}_{r}+\left[{\frac {\partial F_{r}}{\partial z}}-{\frac {\partial F_{z}}{\partial r}}\right]{\hat {e}}_{\varphi }\,+\,{\frac {1}{r}}\left[{\frac {\partial }{\partial r}}\left(r\cdot F_{\varphi }\right)-{\frac {\partial F_{r}}{\partial \varphi }}\right]{\hat {e}}_{z}\,.\end{aligned}}}$

Coordinate-free representation of the rotation as a volume derivative

With the help of Stokes' classical integral theorem , the rotation, similar to the divergence (source density), can be represented as a volume derivative . This representation has the advantage that it is independent of coordinates. For this reason, rotation is often defined directly in the field of engineering .

If there is a spatial area with a piecewise smooth edge and the volume , then the rotation of the vector field in the point can be done by means of the volume derivative ${\displaystyle {\mathcal {V}}}$ ${\displaystyle \partial {\mathcal {V}}}$ ${\displaystyle V}$ ${\displaystyle {\vec {F}}\colon {\mathcal {V}}\to \mathbb {R} ^{3}}$ ${\displaystyle p\in {\mathcal {V}}}$

${\displaystyle \mathrm {rot} \,{\vec {F}}(p)=\lim _{V\rightarrow 0}{\frac {\oint _{\partial {\mathcal {V}}}\mathrm {d} {\vec {A}}\times {\vec {F}}}{V}}}$

be calculated. In this case, referred to the outer surface vector element of which the outward facing normal vector and the scalar element surface. To form the limit, the spatial area is contracted to point p, so that its content approaches zero. ${\displaystyle \mathrm {d} {\vec {A}}={\frac {\vec {n}}{\mid {\vec {n}}\mid }}\mathrm {d} A}$ ${\displaystyle \partial {\mathcal {V}},}$ ${\displaystyle {\vec {n}}}$ ${\displaystyle \mathrm {d} A}$ ${\displaystyle {\mathcal {V}}}$ ${\displaystyle V}$

If you replace it with a flow velocity , the rotation appears as vortex density . Similar synonyms also exist for divergence (source density) and gradient (force density). The coordinate representations of the previous section result from the volume derivation if the respective volume element is selected as the spatial area . ${\displaystyle {\vec {F}}}$ ${\displaystyle {\mathcal {V}}}$

Axial vector field

The rotation of a vector field is a pseudo vector field . When mirrored at the origin, a vector field merges into its negative at the mirrored location, the rotation of the vector field does not change its sign during this mirroring ,

${\displaystyle {\begin{aligned}{\vec {F}}^{\prime }({\vec {x}})&=-{\vec {F}}(-{\vec {x}})\,,\\{\bigl (}\operatorname {rot} \,{\vec {F}}^{\prime }{\bigr )}({\vec {x}})&={\bigl (}\operatorname {rot} \,{\vec {F}}{\bigr )}(-{\vec {x}})\,.\end{aligned}}}$

Vector field in two dimensions

A vector field in two-dimensional, Euclidean space can be used as a vector field

${\displaystyle {\vec {F}}(x,y,z)=F_{x}(x,y)\,{\hat {e}}_{x}+F_{y}(x,y)\,{\hat {e}}_{y}}$

can be understood in three dimensions, which does not depend on the third coordinate and whose third component disappears. Its rotation is not a vector field of this kind, but has a third component,

${\displaystyle \operatorname {rot} \,{\vec {F}}(x,y,z)=\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right){\hat {e}}_{z}\,.}$

Define the rotation as the differential operator in two dimensions

${\displaystyle \operatorname {rot} :\ {\vec {F}}\mapsto {\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\,,}$

then the result is a scalar function, not a vector field.

Relation to the angular velocity

For the sake of simplicity, we consider the rotation of a rigid body around the axis with constant angular velocity . The angle of rotation increases evenly with time, and each point follows a path ${\displaystyle z}$ ${\displaystyle \omega \,.}$ ${\displaystyle \varphi }$ ${\displaystyle \varphi =\omega \,t\,,}$

${\displaystyle {\begin{pmatrix}x(t)\\y(t)\\z(t)\end{pmatrix}}={\begin{pmatrix}\cos(\omega \,t)\,x(0)-\sin(\omega \,t)\,y(0)\\\sin(\omega \,t)\,x(0)+\cos(\omega \,t)\,y(0)\\z(0)\end{pmatrix}}\,.}$

The speed is

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \,t}}{\begin{pmatrix}x(t)\\y(t)\\z(t)\end{pmatrix}}=\omega \,{\begin{pmatrix}-\sin(\omega \,t)\,x(0)-\cos(\omega \,t)\,y(0)\\\ \cos(\omega \,t)\,x(0)-\sin(\omega \,t)\,y(0)\\0\end{pmatrix}}=\omega \,{\begin{pmatrix}-y(t)\\x(t)\\0\end{pmatrix}}\,.}$

The velocity field of a rigid rotation around the -axis is, as given in the example above, ${\displaystyle z}$

${\displaystyle {\vec {v}}(x,y,z)=\omega \,(-y\,{\hat {e}}_{x}+x\,{\hat {e}}_{y})\,.}$

Its rotation is twice the angular velocity

${\displaystyle \operatorname {rot} \,{\vec {v}}=2\,\omega \,{\hat {e}}_{z}\,.}$

Illustration through torque

A sphere with the radius (and the associated volume ) experiences the torque in a surface force density field , which impresses the force on every body surface element with its content, regardless of its orientation ${\displaystyle {\vec {f}}}$ ${\displaystyle \mathrm {d} A}$ ${\displaystyle {\vec {f}}\mathrm {d} A}$ ${\displaystyle R}$ ${\displaystyle V}$

${\displaystyle {\vec {M}}=RV\mathrm {rot} \,{\vec {f}}.}$

It is assumed that is constant in the area of ​​the sphere. The equation follows with the integral theorem with and, which follows directly from the coordinate-free representation of the rotation ${\displaystyle \mathrm {rot} \,{\vec {f}}}$ ${\displaystyle \int _{\mathcal {V}}\mathrm {rot} \,{\vec {f}}\mathrm {d} V=\oint _{\mathcal {A}}\mathrm {d} {\vec {A}}\times {\vec {f}}}$ ${\displaystyle \int _{\mathcal {V}}\mathrm {rot} \,{\vec {f}}\mathrm {d} V=V\mathrm {rot} \,{\vec {f}}}$ ${\displaystyle \oint _{\mathcal {A}}R\mathrm {d} {\vec {A}}\times {\vec {f}}={\vec {M}}.}$

Decomposition into source and eddy-free part

Twice, continuously differentiable vector fields , which with their derivatives approach zero sufficiently quickly for large distances, can be clearly divided into an eddy-free part and a source-free part , ${\displaystyle {\vec {v}}({\vec {r}})}$ ${\displaystyle {\vec {E}}\,,\ {\textsf {mit}}\ \operatorname {rot} \,{\vec {E}}={\vec {0}}\,,}$ ${\displaystyle {\vec {B}}\,,\ {\textsf {mit}}\ \operatorname {div} \,{\vec {B}}=0\,,}$

${\displaystyle {\begin{array}{lll}{\vec {v}}&={\vec {E}}+{\vec {B}}\,,&{\vec {E}}=-\operatorname {grad} \,\phi \,,\ {\vec {B}}=\operatorname {rot} \,{\vec {A}}\,,\\\phi ({\vec {x}})&={\frac {1}{4\,\pi }}\int \!\mathrm {d} ^{3}y\,\,{\frac {\operatorname {div} \,{\vec {v}}({\vec {y}})}{|{\vec {x}}-{\vec {y}}|}}\,,\ &{\vec {A}}({\vec {x}})={\frac {1}{4\,\pi }}\int \!{\mathrm {d} }^{3}y\,\,{\frac {\operatorname {rot} \,{\vec {v}}({\vec {y}})}{|{\vec {x}}-{\vec {y}}|}}\,.\end{array}}}$

And denote the divergence or gradient operator , the definition being the conventional convention in physics. Mathematically is: ${\displaystyle \operatorname {div} }$ ${\displaystyle \operatorname {grad} }$ ${\displaystyle E=-\operatorname {grad} \,\phi \,}$ ${\displaystyle E=\operatorname {grad} \,\phi \,}$

This decomposition is part of the Helmholtz theorem .

Calculation rules

The rotation is linear. For all constants and differentiable vector fields and applies ${\displaystyle c\in \mathbb {R} }$ ${\displaystyle {\vec {F}}}$ ${\displaystyle {\vec {G}}}$

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

The rotation of a vector field disappears exactly when it is locally a gradient field . The divergence of a vector field vanishes exactly when it is locally the rotation of another field,

${\displaystyle \operatorname {rot} ~\operatorname {grad} \,f=0\,,\ \operatorname {div} ~\operatorname {rot} \,{\vec {F}}=0\,,}$

and the other implications are special cases of the Poincaré lemma .

For differentiable functions and vector fields and the product rules ${\displaystyle f\,}$ ${\displaystyle {\vec {F}}}$ ${\displaystyle {\vec {G}}}$

${\displaystyle {\begin{aligned}\operatorname {rot} \,(f\,{\vec {F}})=&f\,\operatorname {rot} \,{\vec {F}}+(\operatorname {grad} \,f)\,\times {\vec {F}}\\\operatorname {rot} \,({\vec {F}}\times {\vec {G}})=&\left({\vec {G}}\cdot \nabla \right){\vec {F}}-\left({\vec {F}}\cdot \nabla \right){\vec {G}}+{\vec {F}}\,(\nabla \cdot {\vec {G}})-{\vec {G}}\,(\nabla \cdot {\vec {F}})\\=&\left(\operatorname {grad} {\vec {F}}\right)\cdot {\vec {G}}-\left(\operatorname {grad} {\vec {G}}\right)\cdot {\vec {F}}+{\vec {F}}\,(\operatorname {div} \,{\vec {G}})-{\vec {G}}\,(\operatorname {div} \,{\vec {F}})\,.\end{aligned}}}$

There is the Nabla operator and in the last formula grad forms the vector gradient . The following applies to the double application of the rotation ${\displaystyle \nabla }$

${\displaystyle \operatorname {rot} \,\operatorname {rot} \,{\vec {F}}=\operatorname {grad} \,\operatorname {div} \,{\vec {F}}-\operatorname {div} \,\operatorname {grad} \,{\vec {F}}}$

The chain rule applies to a vector that depends on a scalar and this in 3D on the location ${\displaystyle {\vec {v}}}$ ${\displaystyle s\!\,}$

${\displaystyle \operatorname {rot} \,{\vec {v}}(s)=\operatorname {grad} \,s\times {\frac {\mathrm {d} {\vec {v}}}{\mathrm {d} s}}.}$

Stokes' integral theorem

Area with border ${\displaystyle {\mathcal {F}}}$ ${\displaystyle \partial {\mathcal {F}}}$

→ Main article : Stokes theorem

The integral over an area over the rotation of a vector field is, according to Stokes' theorem, equal to the curve integral over the boundary curve over ${\displaystyle {\mathcal {F}}}$ ${\displaystyle {\vec {A}}}$ ${\displaystyle \partial {\mathcal {F}}}$ ${\displaystyle {\vec {A}}\,,}$

${\displaystyle \iint _{\mathcal {F}}\!\!(\operatorname {rot} \,{\vec {A}})\cdot \mathrm {d} {\vec {f}}=\oint _{\partial {\mathcal {F}}}\!\!{\vec {A}}\cdot \mathrm {d} {\vec {x}}\,.}$

The double integral on the left emphasizes that a two-dimensional integration is assumed. On the right-hand side, the circle symbol in the integral symbol is intended to underline that it is an integral over a closed path. The orientation corresponds to the three-finger rule , see figure on the right: the following three vectors, namely first the vector in the direction of the surface normal, second the vector in the tangential direction of the curve and third the vector pointing from the edge into the surface, correspond to thumb, The index finger and middle finger of the right hand, that is, they form a legal system. Often one writes by emphasizing the direction of the size with the normal vector . ${\displaystyle \mathrm {d} {\vec {f}}}$ ${\displaystyle \mathrm {d} {\vec {x}}}$ ${\displaystyle \mathrm {d} {\vec {f}}={\vec {n}}\,\mathrm {d} f\,}$ ${\displaystyle {\vec {n}}}$

Rotation of second order tensors

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

${\displaystyle \mathbf {T} ={\vec {t}}_{j}\otimes {\hat {e}}_{j}=T_{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}}$

a tensor with column vectors with components T _ij . Einstein's summation convention was used in the equation, according to which indices occurring twice in a product, here i and j, have to be summed from one to three. Then the rotation of the tensor can be defined as: ${\displaystyle {\vec {t}}_{j}}$

${\displaystyle \operatorname {rot} (\mathbf {T} ):={\hat {e}}_{k}\times {\frac {\partial }{\partial x_{k}}}\mathbf {T} ={\hat {e}}_{k}\times {\frac {\partial }{\partial x_{k}}}{\vec {t}}_{j}\otimes {\hat {e}}_{j}=\operatorname {rot} ({\vec {t}}_{j})\otimes {\hat {e}}_{j}=T_{ij,k}({\hat {e}}_{k}\times {\hat {e}}_{i})\otimes {\hat {e}}_{j}\,.}$

The index after a comma is the shorthand notation for the derivation according to this coordinate:

${\displaystyle T_{ij,k}:={\frac {\partial T_{ij}}{\partial x_{k}}}\,.}$

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

${\displaystyle \operatorname {rot} (\mathbf {T} ):=\nabla \times \mathbf {T} }$

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

${\displaystyle \mathbf {T} ={\hat {e}}_{i}\otimes {\vec {z}}_{i}\quad \rightarrow \quad {\tilde {\operatorname {rot} }}(\mathbf {T} )=\nabla \times (\mathbf {T} ^{\top })=\operatorname {rot} (\mathbf {T} ^{\top })=\operatorname {rot} ({\vec {z}}_{i})\otimes {\hat {e}}_{i}\,,}$

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

In the context of tensors, brackets are an important aid to clarify the order of application and the arguments of the various operators, which has a decisive influence on the result. It is for example

${\displaystyle \operatorname {rot} (\mathbf {T} ^{\top })=T_{ij,k}({\hat {e}}_{k}\times {\hat {e}}_{j})\otimes {\hat {e}}_{i}\neq T_{ij,k}{\hat {e}}_{j}\otimes ({\hat {e}}_{k}\times {\hat {e}}_{i})=\operatorname {rot} (\mathbf {T} )^{\top }\,.}$

properties

If the tensor is symmetric , then its rotation is trackless :

${\displaystyle \mathbf {T} =\mathbf {T} ^{\top }\quad \rightarrow \quad \operatorname {Sp(rot} (\mathbf {T} ))=T_{ij,k}({\hat {e}}_{k}\times {\hat {e}}_{i})\cdot {\hat {e}}_{j}=T_{ij,k}({\hat {e}}_{i}\times {\hat {e}}_{j})\cdot {\hat {e}}_{k}=0}$

because terms with interchanged indices and are of the same size, but have the opposite sign and therefore cancel each other out in the sum. ${\displaystyle i}$ ${\displaystyle j}$

The product rule states in the product with a scalar , vectors and the tensor : ${\displaystyle f}$ ${\displaystyle {\vec {f}},{\vec {g}}}$ ${\displaystyle \mathbf {T} }$

${\displaystyle {\begin{array}{rclcl}\operatorname {rot} ({\vec {f}}\otimes {\vec {g}})&=&{\hat {e}}_{i}\times ({\vec {f}}_{,i}\otimes {\vec {g}}+{\vec {f}}\otimes {\vec {g}}_{,i})=({\hat {e}}_{i}\times {\vec {f}}_{,i})\otimes {\vec {g}}-{\vec {f}}\times ({\hat {e}}_{i}\otimes {\vec {g}}_{,i})&=&\operatorname {rot} ({\vec {f}})\otimes {\vec {g}}-{\vec {f}}\times \operatorname {grad} ({\vec {g}})^{\top }\\\operatorname {rot} (f\mathbf {T} )&=&{\hat {e}}_{k}\times (f_{,k}\mathbf {T} +f\mathbf {T} _{,k})=f_{,k}{\hat {e}}_{k}\times \mathbf {T} +f{\hat {e}}_{k}\times \mathbf {T} _{,k}&=&\operatorname {grad} (f)\times \mathbf {T} +f\operatorname {rot} (\mathbf {T} )\\\operatorname {rot} ({\vec {f}}\times \mathbf {I} )&=&{\hat {e}}_{k}\times (({\vec {f}}_{,k}\times {\hat {e}}_{i})\otimes {\hat {e}}_{i})=({\hat {e}}_{k}\times ({\vec {f}}_{,k}\times {\hat {e}}_{i}))\otimes {\hat {e}}_{i}\\&=&({\hat {e}}_{k}\cdot {\hat {e}}_{i}){\vec {f}}_{,k}\otimes {\hat {e}}_{i}-({\hat {e}}_{k}\cdot {\vec {f}}_{,k}){\hat {e}}_{i}\otimes {\hat {e}}_{i}={\vec {f}}_{,k}\otimes {\hat {e}}_{k}-\operatorname {div} ({\vec {f}})\mathbf {I} &=&\operatorname {grad} ({\vec {f}})-\operatorname {div} ({\vec {f}})\mathbf {I} \,.\end{array}}}$

When the rotation is linked with other differential operators with the participation of a tensor, formulas similar to those known from vector analysis arise:

${\displaystyle {\begin{array}{rclcl}\operatorname {div(rot} (\mathbf {T} ))&=&\nabla \cdot ((\nabla \times {\vec {t}}_{i})\otimes {\hat {e}}_{i})=(\nabla \cdot (\nabla \times {\vec {t}}_{i})){\hat {e}}_{i}&=&{\vec {0}}\\\operatorname {rot(grad} ({\vec {f}})^{\top })&=&\displaystyle \nabla \times (\nabla \otimes {\vec {f}})&=&\mathbf {0} \\\operatorname {rot(rot} (\mathbf {T} ))&=&\nabla \times (\nabla \times ({\vec {t}}_{i}\otimes {\hat {e}}_{i}))=(\nabla \times (\nabla \times {\vec {t}}_{i}))\otimes {\hat {e}}_{i}\\&=&\nabla \otimes (\nabla \cdot {\vec {t}}_{i}){\hat {e}}_{i}-(\nabla \cdot \nabla ){\vec {t}}_{i}\otimes {\hat {e}}_{i}&=&\operatorname {grad(div} (\mathbf {T} ))^{\top }-\Delta \mathbf {T} \end{array}}}$

See also

• Formula collection tensor algebra
• Tensoranalysis formula collection

Individual evidence

1. ↑ How can one get an idea of ​​the rotor (vortex) of a vector field and of the vector potentials? , Walter Rogowski, Archive for Electrical Engineering
2. ^ Mathematics for natural scientists and engineers: Tensor calculus , Hans Karl Iben
3. ^ Bronstein, Semendjajew, Musiol, Mühlig: Taschenbuch der Mathematik , Verlag Harri Deutsch, Frankfurt, 8th edition 2012, section 13.2, Spatial differential operators
4. ↑ Mechanics formula collection ( Memento from August 10, 2016 in the Internet Archive )

literature

• Adolf J. Schwab: Conceptual world of field theory. Springer Verlag, ISBN 3-540-42018-5
• Siegfried Großmann: Mathematical introductory course for physics , Teubner-Verlag, ISBN 3-519-43028-2
• H. Altenbach: Continuum Mechanics . Springer, 2012, ISBN 978-3-642-24118-5 .

Web links

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