Rotor (differential operator)

Rotor , rotation or vortex - a vector differential operator over a vector field .

It is designated in different ways:

$\operatorname {rot}$ ${\ displaystyle \ operatorname {rot}}$ $\ operatorname {rot}$ (in Russian-language ^[1] literature),
$\operatorname {curl}$ ${\ displaystyle \ operatorname {curl}}$ $\ operatorname {curl}$ (in English literature, proposed by Maxwell ^[2] ),
$\mathbf {\nabla } \times$ ${\ displaystyle \ mathbf {\ nabla} \ times}$ ${\ displaystyle \ mathbf {\ nabla} \ times}$ - as a differential operator nabla , vectorly multiplied by a vector field, i.e. for a vector field F, the result of the action of the rotor operator, written in this form, will be the vector product of the nabla operator and this field: $\mathbf {\nabla } \times \mathbf {F} .$ ${\ displaystyle \ mathbf {\ nabla} \ times \ mathbf {F}.}$ ${\ displaystyle \ mathbf {\ nabla} \ times \ mathbf {F}.}$

The result of the action of the rotor operator on a specific vector field F is called the rotor of the field F or simply the rotor F and is a new vector ^[3] field:

\operatorname {rot} \mathbf {F} \equiv \mathbf {\nabla } \times \mathbf {F}

{\ displaystyle \ operatorname {rot} \ mathbf {F} \ equiv \ mathbf {\ nabla} \ times \ mathbf {F}}

{\ displaystyle \ operatorname {rot} \ mathbf {F} \ equiv \ mathbf {\ nabla} \ times \ mathbf {F}}

The field rot F (the length and direction of the vector rot F at each point in space) characterizes in a sense ( see below ) the rotational component of the field F at the corresponding points.

Content

1 Definition
2 Intuitive look
3 Expression in specific coordinates
- 3.1 The formula of the rotor in Cartesian coordinates
- 3.2 the formula of the rotor in curvilinear coordinates
- 3.3 Rotor formula in orthogonal curved coordinates
- 3.4 Generalizations
4 Basic properties
5 Physical interpretation
6 Stokes Theorem
7 Examples
- 7.1 Simple example
- 7.2 A more complex example
- 7.3 Explanatory examples
- 7.4 Important counterintuitive example
8 Notes
9 See also

Definition

Rotor $\operatorname {rot} \,\mathbf {a}$ ${\ displaystyle \ operatorname {rot} \, \ mathbf {a}}$ $\operatorname{rot}\, \mathbf a$ vector field $\mathbf {a}$ ${\ displaystyle \ mathbf {a}}$ $\mathbf {a}$ Is a vector whose projection $\operatorname {rot} _{\mathbf {n} }\mathbf {a}$ ${\ displaystyle \ operatorname {rot} _ {\ mathbf {n}} \ mathbf {a}}$ $\operatorname {rot} _{\mathbf {n} }\mathbf {a}$ for each direction n there is a limit to the ratio of the circulation of the vector field along the contour L , which is the edge of a flat area Δ S perpendicular to this direction, to the value of this area (area), when the dimensions of the area tend to zero, and the area itself contracts to a point ^[4] :

\operatorname {rot} _{\mathbf {n} }\mathbf {a} =\lim _{\Delta S\to 0}{\frac {\oint \limits _{L}\mathbf {a\cdot \,dr} }{\Delta S}}

{\ displaystyle \ operatorname {rot} _ {\ mathbf {n}} \ mathbf {a} = \ lim _ {\ Delta S \ to 0} {\ frac {\ oint \ limits _ {L} \ mathbf {a \ cdot \, dr}} {\ Delta S}}}

\operatorname {rot} _{\mathbf {n} }\mathbf {a} =\lim _{\Delta S\to 0}{\frac {\oint \limits _{L}\mathbf {a\cdot \,dr} }{\Delta S}}

.

The bypass path is selected so that when viewed in the direction $\mathbf {n}$ ${\ displaystyle \ mathbf {n}}$ $\mathbf n$ , the circuit L is circumvented clockwise ^[5] .

An operation defined in this way exists strictly speaking only for vector fields over three-dimensional space. For generalizations to other dimensions, see below .

An alternative definition may be a direct computational definition of a differential operator, which reduces to

\operatorname {rot} \mathbf {a} =\nabla \times \mathbf {a} ,

{\ displaystyle \ operatorname {rot} \ mathbf {a} = \ nabla \ times \ mathbf {a},}

what can be written in specific coordinates as shown below .

Sometimes you can come across such an alternative ^[6] definition ^[7]

\mathrm {rot} \ \mathbf {a} {\Big |}_{O}=\lim _{S\rightarrow O}{\frac {\oint \limits _{S}[\mathbf {a} \times \mathbf {dS} ]}{V}},

{\ displaystyle \ mathrm {rot} \ \ mathbf {a} {\ Big |} _ {O} = \ lim _ {S \ rightarrow O} {\ frac {\ oint \ limits _ {S} [\ mathbf {a } \ times \ mathbf {dS}]} {V}},}

where O is the point at which the rotor of the field a is determined,

S - some closed surface containing the point O inside and in the limit contracting to it,

dS is the vector of an element of this surface whose length is equal to the area of the surface element orthogonal to the surface at a given point,

familiar

\times

{\ displaystyle \ times}

vector product

V is the volume inside the surface S.

This last definition is such that it immediately gives the rotor vector, without the need to define projections on three axes separately.

Intuitive Image

If v (x, y, z) is the gas velocity (or fluid flow) field, then rot v is a vector proportional to the angular velocity vector of a very small and light dust particle (or ball) in the flow (and carried away by the movement of gas or liquid ; although the center of the ball can be fixed if desired, if only it could freely rotate around it).

Specifically, rot v = 2 ω , where ω is this angular velocity.

A simple illustration of this fact - see below .

This analogy can be drawn quite strictly ( see below ). The main definition through circulation given above can be considered equivalent to that obtained in this way.

Expression in specific coordinates

The rotor formula in Cartesian coordinates

In the three-dimensional Cartesian coordinate system, the rotor (as defined above) is calculated as follows (here F is the vector field with the Cartesian components $(F_{x},F_{y},F_{z})$ ${\ displaystyle (F_ {x}, F_ {y}, F_ {z})}$ , but $\mathbf {e} _{x},\mathbf {e} _{y},\mathbf {e} _{z}$ ${\ displaystyle \ mathbf {e} _ {x}, \ mathbf {e} _ {y}, \ mathbf {e} _ {z}}$ - unit Cartesian coordinates):

\operatorname {rot} \;(F_{x}\mathbf {e} _{x}+F_{y}\,\mathbf {e} _{y}+F_{z}\mathbf {e} _{z})=

{\ displaystyle \ operatorname {rot} \; (F_ {x} \ mathbf {e} _ {x} + F_ {y} \, \ mathbf {e} _ {y} + F_ {z} \ mathbf {e} _ {z}) =}

=\left(\partial _{y}F_{z}-\partial _{z}F_{y}\right)\mathbf {e} _{x}+\left(\partial _{z}F_{x}-\partial _{x}F_{z}\right)\mathbf {e} _{y}+\left(\partial _{x}F_{y}-\partial _{y}F_{x}\right)\mathbf {e} _{z}\equiv

{\ displaystyle = \ left (\ partial _ {y} F_ {z} - \ partial _ {z} F_ {y} \ right) \ mathbf {e} _ {x} + \ left (\ partial _ {z} F_ {x} - \ partial _ {x} F_ {z} \ right) \ mathbf {e} _ {y} + \ left (\ partial _ {x} F_ {y} - \ partial _ {y} F_ { x} \ right) \ mathbf {e} _ {z} \ equiv}

\equiv \left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right)\mathbf {e} _{x}+\left({\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\right)\mathbf {e} _{y}+\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right)\mathbf {e} _{z}.

{\ displaystyle \ equiv \ left ({\ frac {\ partial F_ {z}} {\ partial y}} - {\ frac {\ partial F_ {y}} {\ partial z}} \ right) \ mathbf {e } _ {x} + \ left ({\ frac {\ partial F_ {x}} {\ partial z}} - {\ frac {\ partial F_ {z}} {\ partial x}} \ right) \ mathbf { e} _ {y} + \ left ({\ frac {\ partial F_ {y}} {\ partial x}} - {\ frac {\ partial F_ {x}} {\ partial y}} \ right) \ mathbf {e} _ {z}.}

or

(\operatorname {rot} \mathbf {F} )_{x}=\partial _{y}F_{z}-\partial _{z}F_{y}\equiv {\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}

{\ displaystyle (\ operatorname {rot} \ mathbf {F}) _ {x} = \ partial _ {y} F_ {z} - \ partial _ {z} F_ {y} \ equiv {\ frac {\ partial F_ {z}} {\ partial y}} - {\ frac {\ partial F_ {y}} {\ partial z}}}

(\operatorname {rot} \mathbf {F} )_{y}=\partial _{z}F_{x}-\partial _{x}F_{z}\equiv {\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}

{\ displaystyle (\ operatorname {rot} \ mathbf {F}) _ {y} = \ partial _ {z} F_ {x} - \ partial _ {x} F_ {z} \ equiv {\ frac {\ partial F_ {x}} {\ partial z}} - {\ frac {\ partial F_ {z}} {\ partial x}}}

(\operatorname {rot} \mathbf {F} )_{z}=\partial _{x}F_{y}-\partial _{y}F_{x}\equiv {\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}

{\ displaystyle (\ operatorname {rot} \ mathbf {F}) _ {z} = \ partial _ {x} F_ {y} - \ partial _ {y} F_ {x} \ equiv {\ frac {\ partial F_ {y}} {\ partial x}} - {\ frac {\ partial F_ {x}} {\ partial y}}}

(which can be considered an alternative definition, which essentially coincides with the definition at the beginning of the paragraph, at least under the condition that the field components are differentiable).

For convenience, we can formally represent the rotor as a vector product of the nabla operator (left) and the vector field:

\operatorname {rot} \mathbf {F} =\mathbf {\nabla } \times \mathbf {F} ={\begin{pmatrix}\partial _{x}\\\partial _{y}\\\partial _{z}\end{pmatrix}}\times \mathbf {F} ={\begin{vmatrix}\mathbf {e} _{x}&\mathbf {e} _{y}&\mathbf {e} _{z}\\\partial _{x}&\partial _{y}&\partial _{z}\\F_{x}&F_{y}&F_{z}\end{vmatrix}}

{\ displaystyle \ operatorname {rot} \ mathbf {F} = \ mathbf {\ nabla} \ times \ mathbf {F} = {\ begin {pmatrix} \ partial _ {x} \\\ partial _ {y} \\ \ partial _ {z} \ end {pmatrix}} \ times \ mathbf {F} = {\ begin {vmatrix} \ mathbf {e} _ {x} & \ mathbf {e} _ {y} & \ mathbf {e } _ {z} \\\ partial _ {x} & \ partial _ {y} & \ partial _ {z} \\ F_ {x} & F_ {y} & F_ {z} \ end {vmatrix}}}

(The last equality formally represents a vector product as a determinant ).

Rotor formula in curved coordinates

A convenient general rotor expression suitable for arbitrary curvilinear coordinates in three-dimensional space is an expression using the Levi-Civita tensor (using upper and lower indices and the Einstein summation rule ):

(\mathrm {rot} \mathbf {v} )_{i}=\varepsilon _{ijk}g^{jm}{\frac {\partial }{\partial x^{m}}}v^{k},

{\ displaystyle (\ mathrm {rot} \ mathbf {v}) _ {i} = \ varepsilon _ {ijk} g ^ {jm} {\ frac {\ partial} {\ partial x ^ {m}}} v ^ {k},}

Where $\varepsilon _{ijk}$ ${\ displaystyle \ varepsilon _ {ijk}}$ - coordinate record of the Levi-Civita tensor, including the factor ${\sqrt {g}},$ ${\ displaystyle {\ sqrt {g}},}$ $g^{jm}$ ${\ displaystyle g ^ {jm}}$ - metric tensor in a representation with superscripts, $g\equiv \mathrm {det} (g_{rs})$ ${\ displaystyle g \ equiv \ mathrm {det} (g_ {rs})}$

This expression can also be rewritten as:

(\mathrm {rot} \ \mathbf {v} )^{n}=g^{ni}\varepsilon _{ijk}g^{jm}{\frac {\partial }{\partial x^{m}}}v^{k}

{\ displaystyle (\ mathrm {rot} \ \ mathbf {v}) ^ {n} = g ^ {ni} \ varepsilon _ {ijk} g ^ {jm} {\ frac {\ partial} {\ partial x ^ { m}}} v ^ {k}}

Rotor Formula in Orthogonal Curved Coordinates

\operatorname {rot} \mathbf {A} =\operatorname {rot} (\mathbf {q_{1}} A_{1}+\mathbf {q_{2}} A_{2}+\mathbf {q_{3}} A_{3})=

{\ displaystyle \ operatorname {rot} \ mathbf {A} = \ operatorname {rot} (\ mathbf {q_ {1}} A_ {1} + \ mathbf {q_ {2}} A_ {2} + \ mathbf {q_ {3}} A_ {3}) =}

={\frac {1}{H_{2}H_{3}}}\left[{\frac {\partial }{\partial q_{2}}}(A_{3}H_{3})-{\frac {\partial }{\partial q_{3}}}(A_{2}H_{2})\right]\mathbf {q_{1}} \ +

{\ displaystyle = {\ frac {1} {H_ {2} H_ {3}}} \ left [{\ frac {\ partial} {\ partial q_ {2}}} (A_ {3} H_ {3}) - {\ frac {\ partial} {\ partial q_ {3}}} (A_ {2} H_ {2}) \ right] \ mathbf {q_ {1}} \ +}

+\ {\frac {1}{H_{3}H_{1}}}\left[{\frac {\partial }{\partial q_{3}}}(A_{1}H_{1})-{\frac {\partial }{\partial q_{1}}}(A_{3}H_{3})\right]\mathbf {q_{2}} \ +

{\ displaystyle + \ {\ frac {1} {H_ {3} H_ {1}}} \ left [{\ frac {\ partial} {\ partial q_ {3}}} (A_ {1} H_ {1} ) - {\ frac {\ partial} {\ partial q_ {1}}} (A_ {3} H_ {3}) \ right] \ mathbf {q_ {2}} \ +}

+\ {\frac {1}{H_{1}H_{2}}}\left[{\frac {\partial }{\partial q_{1}}}(A_{2}H_{2})-{\frac {\partial }{\partial q_{2}}}(A_{1}H_{1})\right]\mathbf {q_{3}}

{\ displaystyle + \ {\ frac {1} {H_ {1} H_ {2}}} \ left [{\ frac {\ partial} {\ partial q_ {1}}} (A_ {2} H_ {2} ) - {\ frac {\ partial} {\ partial q_ {2}}} (A_ {1} H_ {1}) \ right] \ mathbf {q_ {3}}}

$={\frac {1}{H_{1}H_{2}H_{3}}}{\begin{vmatrix}\mathbf {(} H_{1}{e}_{1})&\mathbf {(} H_{2}{e}_{2})&\mathbf {(} H_{3}{e}_{3})\\{\frac {\partial }{\partial \mathbf {q} _{1}}}&{\frac {\partial }{\partial \mathbf {q} _{2}}}&{\frac {\partial }{\partial \mathbf {q} _{3}}}\\(A_{1}H_{1})&(A_{2}H_{2})&(A_{3}H_{3})\end{vmatrix}}$ ${\ displaystyle = {\ frac {1} {H_ {1} H_ {2} H_ {3}}} {\ begin {vmatrix} \ mathbf {(} H_ {1} {e} _ {1}) & \ mathbf {(} H_ {2} {e} _ {2}) & \ mathbf {(} H_ {3} {e} _ {3}) \\ {\ frac {\ partial} {\ partial \ mathbf {q } _ {1}}} & {\ frac {\ partial} {\ partial \ mathbf {q} _ {2}}} & {\ frac {\ partial} {\ partial \ mathbf {q} _ {3}} } \\ (A_ {1} H_ {1}) & (A_ {2} H_ {2}) & (A_ {3} H_ {3}) \ end {vmatrix}}}$ where H _i are the Lame coefficients .

Generalizations

A generalization of the rotor as applied to vector (and pseudovector) fields on spaces of arbitrary dimension (provided that the dimension of the space coincides with the dimension of the field vector) is an antisymmetric tensor field of valency two, whose components are equal to:

(\operatorname {rot} \mathbf {F} )_{ij}=\partial _{i}F_{j}-\partial _{j}F_{i}\equiv {\frac {\partial F_{j}}{\partial x_{i}}}-{\frac {\partial F_{i}}{\partial x_{j}}}

{\ displaystyle (\ operatorname {rot} \ mathbf {F}) _ {ij} = \ partial _ {i} F_ {j} - \ partial _ {j} F_ {i} \ equiv {\ frac {\ partial F_ {j}} {\ partial x_ {i}}} - {\ frac {\ partial F_ {i}} {\ partial x_ {j}}}}

The same formula can be written through an external product with the nabla operator:

\operatorname {rot} \mathbf {F} =\nabla \wedge \mathbf {F}

{\ displaystyle \ operatorname {rot} \ mathbf {F} = \ nabla \ wedge \ mathbf {F}}

For a two-dimensional plane, a similar formula with a pseudoscalar product can be used (such a rotor will be a pseudoscalar, and its value coincides with the projection of the traditional vector product on the normal to this plane if it is embedded in three-dimensional Euclidean space).

If on a two-dimensional real space (with coordinates $x$ ${\ displaystyle x}$ and $y$ ${\ displaystyle y}$ ) introduced the structure of the complex space (with coordinate $z=x+iy$ ${\ displaystyle z = x + iy}$ ) and two-dimensional vector fields are written as complex-valued functions $f(z)$ ${\ displaystyle f (z)}$ , then using differentiation with respect to the complex variable

{\frac {\partial {}}{\partial z}}={\frac {1}{2}}\left({\frac {\partial {}}{\partial x}}-i{\frac {\partial {}}{\partial y}}\right)

{\ displaystyle {\ frac {\ partial {}} {\ partial z}} = {\ frac {1} {2}} \ left ({\ frac {\ partial {}} {\ partial x}} - i { \ frac {\ partial {}} {\ partial y}} \ right)}

the rotor and divergence (and they will remain real numbers) can be written as follows:

\operatorname {rot} f=2\operatorname {Im} {\frac {\partial f}{\partial z}}

{\ displaystyle \ operatorname {rot} f = 2 \ operatorname {Im} {\ frac {\ partial f} {\ partial z}}}

,

\operatorname {div} f=2\operatorname {Re} {\frac {\partial f}{\partial z}}

{\ displaystyle \ operatorname {div} f = 2 \ operatorname {Re} {\ frac {\ partial f} {\ partial z}}}

.

Key Features

The rotor operation is linear over the constant field: for any vector fields $F$ ${\ displaystyle F}$ and $G$ ${\ displaystyle G}$ and for any numbers (constants) $a$ ${\ displaystyle a}$ and $b$ ${\ displaystyle b}$

\operatorname {rot} (a\mathbf {F} +b\mathbf {G} )=a\operatorname {rot} \mathbf {F} +b\operatorname {rot} \mathbf {G}

{\ displaystyle \ operatorname {rot} (a \ mathbf {F} + b \ mathbf {G}) = a \ operatorname {rot} \ mathbf {F} + b \ operatorname {rot} \ mathbf {G}}

If $\varphi$ ${\ displaystyle \ varphi}$ Is a scalar field (function), and $F$ ${\ displaystyle F}$ - vector, then:

\operatorname {rot} \varphi \mathbf {F} =\operatorname {grad} \varphi \times \mathbf {F} +\varphi \operatorname {rot} \mathbf {F}

{\ displaystyle \ operatorname {rot} \ varphi \ mathbf {F} = \ operatorname {grad} \ varphi \ times \ mathbf {F} + \ varphi \ operatorname {rot} \ mathbf {F}}

\nabla \times (\varphi \mathbf {F} )=(\nabla \varphi )\times \mathbf {F} +\varphi (\nabla \times \mathbf {F} )

{\ displaystyle \ nabla \ times (\ varphi \ mathbf {F}) = (\ nabla \ varphi) \ times \ mathbf {F} + \ varphi (\ nabla \ times \ mathbf {F})}

.

If the field $F$ ${\ displaystyle F}$ potentially , its rotor is zero (field $F$ ${\ displaystyle F}$ - irrotational):

\mathbf {F} =\operatorname {grad} ~\varphi \Rightarrow \operatorname {rot} ~\mathbf {F} =0

{\ displaystyle \ mathbf {F} = \ operatorname {grad} ~ \ varphi \ Rightarrow \ operatorname {rot} ~ \ mathbf {F} = 0}

The converse is true locally ^[8] : if the field is vortex-free, then locally (in sufficiently small areas) it is potentially (that is, there is such a scalar field $\varphi \$ ${\ displaystyle \ varphi \}$ , what $\mathbf {F}$ ${\ displaystyle \ mathbf {F}}$ will be its gradient):

\operatorname {rot} \mathbf {F} =0\Rightarrow \mathbf {F} =\operatorname {grad} \varphi

{\ displaystyle \ operatorname {rot} \ mathbf {F} = 0 \ Rightarrow \ mathbf {F} = \ operatorname {grad} \ varphi}

Thus, different vector fields can have the same rotor. At the same time, they will necessarily differ by the irrotational field (that is, locally - by the gradient of some scalar field).

The rotor divergence is equal to zero (the rotor field is divergent free):

\operatorname {div} \operatorname {rot} \mathbf {F} =0

{\ displaystyle \ operatorname {div} \ operatorname {rot} \ mathbf {F} = 0}

\nabla \cdot (\nabla \times \mathbf {F} )=0

{\ displaystyle \ nabla \ cdot (\ nabla \ times \ mathbf {F}) = 0}

The inverse property also runs locally - if the field $F$ ${\ displaystyle F}$ divergenceless, locally it is the rotor of a certain field $G$ ${\ displaystyle G}$ called its vector potential :

\operatorname {div} \mathbf {F} =0\Rightarrow \mathbf {F} =\operatorname {rot} \mathbf {G}

{\ displaystyle \ operatorname {div} \ mathbf {F} = 0 \ Rightarrow \ mathbf {F} = \ operatorname {rot} \ mathbf {G}}

.

The divergence of the vector product of two vector fields is expressed through their rotors according to the formula:

\operatorname {div} (\mathbf {F} \times \mathbf {G} )=(\operatorname {rot} \mathbf {F} )\cdot \mathbf {G} -\mathbf {F} \cdot \operatorname {rot} \mathbf {G}

{\ displaystyle \ operatorname {div} (\ mathbf {F} \ times \ mathbf {G}) = (\ operatorname {rot} \ mathbf {F}) \ cdot \ mathbf {G} - \ mathbf {F} \ cdot \ operatorname {rot} \ mathbf {G}}

So if

F

{\ displaystyle F}

and

G

{\ displaystyle G}

- vortex-free vector fields, their vector product will be divergent-free and locally have a vector potential. For example, if

\mathbf {F} =\nabla f

{\ displaystyle \ mathbf {F} = \ nabla f}

, but

\mathbf {G} =\nabla g

{\ displaystyle \ mathbf {G} = \ nabla g}

easy to find vector potential for

\mathbf {F} \times \mathbf {G}

{\ displaystyle \ mathbf {F} \ times \ mathbf {G}}

:

\mathbf {F} \times \mathbf {G} =\operatorname {rot} (f\nabla g)

{\ displaystyle \ mathbf {F} \ times \ mathbf {G} = \ operatorname {rot} (f \ nabla g)}

.

Locally, each divergence-free vector field in a three-dimensional region is a vector product of two gradients.

The rotor rotor is equal to the divergence gradient minus the Laplacians:

\operatorname {rot} \operatorname {rot} \mathbf {F} =\operatorname {grad} ~\operatorname {div} \mathbf {F} -\Delta \mathbf {F}

{\ displaystyle \ operatorname {rot} \ operatorname {rot} \ mathbf {F} = \ operatorname {grad} ~ \ operatorname {div} \ mathbf {F} - \ Delta \ mathbf {F}}

The rotor of the vector product of the fields is:

\operatorname {rot} (\mathbf {F\times G} )=(\operatorname {div} \mathbf {G} )\mathbf {F} -(\operatorname {div} \mathbf {F} )\mathbf {G} +\nabla _{\mathbf {G} }\mathbf {F} -\nabla _{\mathbf {F} }\mathbf {G}

{\ displaystyle \ operatorname {rot} (\ mathbf {F \ times G}) = (\ operatorname {div} \ mathbf {G}) \ mathbf {F} - (\ operatorname {div} \ mathbf {F}) \ mathbf {G} + \ nabla _ {\ mathbf {G}} \ mathbf {F} - \ nabla _ {\ mathbf {F}} \ mathbf {G}}

Physical Interpretation

When a continuous medium moves , the distribution of its velocities (i.e., the fluid velocity field) near the point O is given by the Cauchy-Helmholtz formula:

\mathbf {v} (\mathbf {r} )=\mathbf {v} _{O}+\mathbf {\omega } \times \mathbf {r} +\nabla \varphi +o(\mathbf {r} ),

{\ displaystyle \ mathbf {v} (\ mathbf {r}) = \ mathbf {v} _ {O} + \ mathbf {\ omega} \ times \ mathbf {r} + \ nabla \ varphi + o (\ mathbf { r}),}

Where $\mathbf {\omega }$ ${\ displaystyle \ mathbf {\ omega}}$ Is the vector of the angular rotation of the medium element at the point O, and $\varphi$ ${\ displaystyle \ varphi}$ Is the quadratic form of the coordinates is the deformation potential of the medium element.

Thus, the motion of a continuous medium near point O is composed of translational motion (vector $\mathbf {v} _{O}$ ${\ displaystyle \ mathbf {v} _ {O}}$ ), rotational motion (vector $\mathbf {\omega } \times \mathbf {r}$ ${\ displaystyle \ mathbf {\ omega} \ times \ mathbf {r}}$ ) and potential motion - deformations (vector $\nabla \varphi$ ${\ displaystyle \ nabla \ varphi}$ ) Applying the rotor operation to the Cauchy – Helmholtz formula, we obtain that at the point O the equality $\operatorname {rot} ~\mathbf {v} =2\mathbf {\omega } ,$ ${\ displaystyle \ operatorname {rot} ~ \ mathbf {v} = 2 \ mathbf {\ omega},}$ and, therefore, we can conclude that when it comes to a vector field, which is the velocity field of a certain medium, the rotor of this vector field at a given point is equal to twice the angular rotation vector of a medium element centered at this point.

As an intuitive image, as described above, here one can use the idea of the rotation of a small dust particle thrown into the stream (carried away by the stream with itself, without its noticeable perturbation) or of the rotation of a small particle placed in the stream with a fixed axis (without inertia rotated by the stream, without distorting it) wheels with straight (not screw) blades. If one or the other, when looking at it, rotates counterclockwise, then this means that the rotor vector of the flow velocity field at a given point has a positive projection towards us.

Stokes Theorem

The circulation of a vector in a closed loop, which is the boundary of a surface, is equal to the flow of the rotor of this vector through this surface:

\oint \limits _{\partial S}\mathbf {F} \cdot \,\mathbf {dl} =\int \limits _{S}(\operatorname {rot} ~\mathbf {F} )\cdot \mathbf {dS}

{\ displaystyle \ oint \ limits _ {\ partial S} \ mathbf {F} \ cdot \, \ mathbf {dl} = \ int \ limits _ {S} (\ operatorname {rot} ~ \ mathbf {F}) \ cdot \ mathbf {dS}}

A special case of the Stokes theorem for a flat surface is the content of Green's theorem .

Examples

In this chapter, for unit vectors along the axes of (rectangular) Cartesian coordinates, we use the notation $\mathbf {e} _{x},\mathbf {e} _{y},\mathbf {e} _{z}.$ ${\ displaystyle \ mathbf {e} _ {x}, \ mathbf {e} _ {y}, \ mathbf {e} _ {z}.}$

A simple example

Consider a vector field F depending on the coordinates x and y like this:

\mathbf {F} (x,y)=y\mathbf {e} _{x}-x\mathbf {e} _{y}

{\ displaystyle \ mathbf {F} (x, y) = y \ mathbf {e} _ {x} -x \ mathbf {e} _ {y}}

.

Regarding this example, it is easy to see that $\mathbf {F} =\mathbf {\omega } \times \mathbf {r}$ ${\ displaystyle \ mathbf {F} = \ mathbf {\ omega} \ times \ mathbf {r}}$ where r is the radius vector, and $\omega =-1\mathbf {e} _{z}$ ${\ displaystyle \ omega = -1 \ mathbf {e} _ {z}}$ , that is, the field F can be considered as the velocity field of the points of a solid body rotating with a unit angular velocity directed in the negative direction of the z axis (that is, clockwise, when viewed from above, opposite the z axis). Intuitively, it is more or less obvious that the field is twisted clockwise. If we place a wheel with blades in a fluid flowing at such speeds (that is, rotating as a whole in a clockwise direction) anywhere, we will see that it starts to rotate in a clockwise direction. (To determine the directions we use, as usual, the rule of the right hand or the right screw ).
the z- component of the field F is assumed to be zero. However, if it is nonzero, but constant (or even depending only on z ), the result for the rotor obtained below will be the same.

We calculate the rotor:

\mathbf {\nabla } \times \mathbf {F} =0\mathbf {e} _{x}+0\mathbf {e} _{y}+\left[{\frac {\partial }{\partial x}}(-x)-{\frac {\partial }{\partial y}}y\right]\mathbf {e} _{z}=-2\mathbf {e} _{z}

{\ displaystyle \ mathbf {\ nabla} \ times \ mathbf {F} = 0 \ mathbf {e} _ {x} +0 \ mathbf {e} _ {y} + \ left [{\ frac {\ partial} { \ partial x}} (- x) - {\ frac {\ partial} {\ partial y}} y \ right] \ mathbf {e} _ {z} = - 2 \ mathbf {e} _ {z}}

As suggested, the direction coincided with the negative direction of the z axis. In this case, the rotor turned out to be constant, i.e. the field $\mathbf {\nabla } \times \mathbf {F}$ ${\ displaystyle \ mathbf {\ nabla} \ times \ mathbf {F}}$ It turned out to be homogeneous, independent of coordinates (which is natural for rotation of a solid). What a wonderful thing

the angular velocity of rotation of the fluid, calculated from the rotor and turned out to be exactly equal $\operatorname {rot} \mathbf {F} /2$ ${\ displaystyle \ operatorname {rot} \ mathbf {F} / 2}$ , exactly coincided with what is indicated in the paragraph Physical interpretation , that is, this example is a good illustration of the fact presented there. (Of course, calculations that completely repeat the above, but only for a non-unit angular velocity, give the same result $\mathbf {\nabla } \times \mathbf {F} =2\omega$ ${\ displaystyle \ mathbf {\ nabla} \ times \ mathbf {F} = 2 \ omega}$ )

The angular velocity of rotation in this example is the same at any point in space (the angle of rotation of a dust particle glued to a solid does not depend on the place where it is to stick the dust particle). The rotor graph F is therefore not very interesting:

More complex example.

Now consider a slightly more complex vector field ^[9] :

F(x,y)=-x^{2}\mathbf {e} _{y}

{\ displaystyle F (x, y) = - x ^ {2} \ mathbf {e} _ {y}}

.

His schedule:

We may not see any rotation, but, looking more closely to the right, we see a larger field at, for example, the point x = 4 than at the point x = 3. If we installed a small wheel with blades there, a larger flow on the right side would cause the wheel to rotate clockwise, which corresponds to screwing in the - z direction. If we placed the wheel on the left side of the field, a larger flow on its left side would cause the wheel to rotate counterclockwise, which corresponds to screwing in the + z direction. Let's check our guess by calculating:

\mathbf {\nabla } \times \mathbf {F} =0\mathbf {e} _{x}+0\mathbf {e} _{y}+{\frac {\partial }{\partial x}}(-x^{2})\mathbf {e} _{z}=-2x\mathbf {e} _{z}

{\ displaystyle \ mathbf {\ nabla} \ times \ mathbf {F} = 0 \ mathbf {e} _ {x} +0 \ mathbf {e} _ {y} + {\ frac {\ partial} {\ partial x }} (- x ^ {2}) \ mathbf {e} _ {z} = - 2x \ mathbf {e} _ {z}}

Indeed, screwing occurs in the + z direction for negative x and - z for positive x , as expected. Since this rotor is not the same at every point, its graph looks a bit more interesting:

Rotor F with plane x = 0 highlighted in dark blue

You can see that the graph of this rotor is independent of y or z (as it should be) and is directed along - z for positive x and in the + z direction for negative x .

Explanatory Examples

In a tornado, winds rotate around the center, and the vector field of wind speeds has a nonzero rotor (somewhere) in the central region. (see Vortex motion ). (True, closer to the edge somewhere the rotor may take a zero value, see below ).
For the vector field v of the velocities of the points of rotation of a rotating solid (absolutely solid) body, rot v is the same throughout the volume of this body and is equal to (the vector) twice the angular velocity of rotation ( for more details, see above ). In the particular case of purely translational motion or rest, this rotor can be equal to zero, like the angular velocity, also for all points of the body.
If the speeds of the cars on the track were described by a vector field and the different lanes had different speed limits, the rotor at the border between the lanes would be non-zero.
The Faraday law of electromagnetic induction , one of Maxwell's equations , is simply written (in differential form) through the rotor: the rotor of the electric field is equal to the rate of change of the magnetic field (with time), taken with the opposite sign.
The fourth Maxwell equation — the Ampere – Maxwell law — is also written in differential form using a rotor: the rotor of the magnetic field is equal to the sum of the normal current densities and the bias current ^[10] .

Important counterintuitive example

It should be borne in mind that in principle (although not always) the direction of the rotor of the vector field may not correspond to the direction of rotation ^[11] , which is obvious even in the direction of curvature of the streamlines (we will speak for concreteness about the velocity field of a liquid or gas and rotation of this medium). It may even have the opposite direction (and in the particular case, the rotor may turn out to be zero, although the streamlines are bent or, in particular, even represent exact circles).

Consider this example. Let the fluid rotate (the movement is two-dimensional for simplicity, nothing depends on the z coordinate — let it be vertical — nothing) around the origin (streamlines — circles with a center at the origin) so that the fluid velocity field v is defined on the x axis by the formula:

\mathbf {v} =f(x)\mathbf {e} _{y}

{\ displaystyle \ mathbf {v} = f (x) \ mathbf {e} _ {y}}

(f is a positive definite function, that is, counterclockwise rotation, viewed from above, against the direction of the z axis), then the vertical component of the speed rotor on the x axis

(\operatorname {rot} \mathbf {v} )_{z}=\partial _{x}v_{y}-\partial _{y}v_{x}\equiv {\frac {\partial v_{y}}{\partial x}}-{\frac {\partial v_{x}}{\partial y}}=

{\ displaystyle (\ operatorname {rot} \ mathbf {v}) _ {z} = \ partial _ {x} v_ {y} - \ partial _ {y} v_ {x} \ equiv {\ frac {\ partial v_ {y}} {\ partial x}} - {\ frac {\ partial v_ {x}} {\ partial y}} =}

f'(x)-f(x)/x

{\ displaystyle f '(x) -f (x) / x}

.

Then if $f'(x)-f(x)/x<0$ ${\ displaystyle f '(x) -f (x) / x <0}$ , which, for example, is obviously realized when $f'(x)\leq 0$ ${\ displaystyle f '(x) \ leq 0}$ (hereafter, for simplicity, we will only talk about this case), then the rotor everywhere (except, however, the origin) is directed downwards, which means that every small piece of liquid that is not at the origin is twisted clockwise (and is deformed, t .k. rotation is not like a solid). In other words, if we place dust particles in the liquid, the centers of these dust particles will move counterclockwise, while each dust particle itself will rotate counterclockwise (with the exception of one dust particle at the origin, which will still rotate counterclockwise , moreover, infinitely fast, since f (0) cannot turn out to be zero in our case).

In fact, what has been said (with reference to a liquid or gas) means that the medium can rotate as a whole in one direction, and each small volume can rotate in the opposite direction or not rotate at all (although not necessarily).

As applied to abstract geometry, this means that the direction of curvature of the vector lines of a vector field is generally not directly related to the direction of the rotor vector of this field, i.e. generally speaking, they do not condition each other. In particular, vector lines can be straight (forward flow), and the rotor is nonzero, or vice versa vector lines, for example, circles, and the rotor is zero or directed in any direction (including the opposite of the “expected” or orthogonal).

The only restriction on this independence is imposed by the Stokes theorem , and it does not have a local character, that is, it does not limit the possible magnitude and direction of the rotor at any particular specific point, but, like the theorem itself, is integral. More specifically, if the circulation of the field along some contour has a certain sign, then somewhere inside this contour the field rotor will necessarily have a projection of the same sign on the surface stretched over the contour. For example, if the fluid simply rotates (the streamlines are then closed and have the shape of circles, and the circulation of the velocity field along each such line has a sign corresponding to the direction of rotation), then at least somewhere (not necessarily everywhere!) Inside such a circle on the plane, in which the circle lies, the rotor of the fluid velocity must have a projection on this plane of the same sign as the circulation, that is, at least somewhere inside the rotating fluid the direction of the rotor of the velocity must coincide with the direction of rotation of the fluid. In an abstract field, this can be just one point (then this is a special point where the field tends to infinity), but if we are talking about real physical fields that always have finite values, then the region in question will have a finite size (although it may turn out to be very small).

On the other hand, it also follows from the Stokes theorem that if the rotation is local in character ^[12] , then there will be another region outside the region of rotation mentioned by the first one, in which the rotor will have the opposite sign.

Notes

↑ Also in German, from where, apparently, this designation fell into Russian, and almost everywhere in Europe, except England, where such a designation is considered “alternative” (possibly due to the lack of sound: English rot - rot, rot) .
↑ O. Heaviside . The relations between magnetic force and electric current . // The Electrician, 1882.
↑ More precisely, if F is a pseudovector field, then rot F is an ordinary vector field (vector rot F is polar), and vice versa, if F is a field of ordinary (polar) vector, then rot F is a pseudovector field.
↑ Pulling to a point is a prerequisite, just tending Δ S to zero is not enough, because we want to get the field characteristics at one specific point.
↑ The usual agreement, consistent with the definition through a vector product with the nabla operator.
↑ The equivalence of these definitions, if the limit exists and does not depend on the way the point is contracted, is visible if we select the surface S of the second definition in the form of a cylindrical surface with bases obtained by parallel transfer of the site of the first definition Δ S to a very small distance in two opposite directions orthogonal to Δ S. In the limit, they should approach Δ S faster than the size of Δ S itself decreases. Then the expression of the second definition is divided into two terms, one containing the integral over the lateral surface coincides with the first definition, and the second gives zero in the projection onto the normal to the bases, since $\mathbf {dS}$ ${\ displaystyle \ mathbf {dS}}$ on the grounds itself orthogonal to him. Instead, you can consider just a small box as a surface, then it is not so easy right away strictly, but in general a similar one is clear.
↑ Formally similar to the definition of divergence through a stream through a surface:
$\mathrm {rot} \ \mathbf {a} {\Big |}_{O}=\lim _{S\rightarrow O}{\frac {\oint \limits _{S}(\mathbf {a} \cdot \mathbf {dS} )}{V}}.$ ${\ displaystyle \ mathrm {rot} \ \ mathbf {a} {\ Big |} _ {O} = \ lim _ {S \ rightarrow O} {\ frac {\ oint \ limits _ {S} (\ mathbf {a } \ cdot \ mathbf {dS})} {V}}.}$
↑ The locality clause is important for the general case when the fields considered here $\mathbf {F}$ ${\ displaystyle \ mathbf {F}}$ and $\varphi$ ${\ displaystyle \ varphi}$ can be defined on a space (manifold) or domain of a nontrivial topology, and when the conditions $\operatorname {rot} \mathbf {F} =0$ ${\ displaystyle \ operatorname {rot} \ mathbf {F} = 0}$ also holds generally speaking on a space or domain of a nontrivial topology. For the case of Euclidean space or its simply connected region, the locality clause is not needed, a field whose rotor is zero on the whole such space or simply connected region will be potential on this whole space or this region. That is, then there is such a scalar field $\varphi$ ${\ displaystyle \ varphi}$ , what $\mathbf {F} =\operatorname {grad} \varphi$ ${\ displaystyle \ mathbf {F} = \ operatorname {grad} \ varphi}$ will be true everywhere in this space or this area.
↑ The simplest physical realization of such a field (accurate to an additive constant, which does not affect the calculation of the rotor, since rot const = 0 ; in addition, if desired, this constant can be reset by transition to a reference system associated with the fastest flowing water in the center of the jet ) is the laminar flow of a (viscous) fluid between two parallel solid planes perpendicular to the x axis under the influence of a uniform force field (gravity) or pressure difference. The fluid flow in a round pipe gives the same relationship $v_{y}(x)$ ${\ displaystyle v_ {y} (x)}$ , therefore, the calculation of the rotor given below is also applicable to this case (it is easiest to take the y axis coinciding with the axis of the pipe, and although the dependence $\mathbf {v} (z)$ ${\ displaystyle \ mathbf {v} (z)}$ will not be constant anymore however $\partial v_{y}/\partial z$ ${\ displaystyle \ partial v_ {y} / \ partial z}$ will be zero at z = 0, as in the main example, i.e. the calculation and answer for any plane passing through the axis of the pipe is the same, and this solves the problem).
↑ Mathematical Dictionary of Higher Education. V.T. Vodnev, A.F. Naumovich, N.F. Naumovich
↑ This refers, for example, to the rotation of a fluid, if we are talking about the field of the velocity of the fluid (or gas), or, if we are talking about another (for example, purely abstract) field, then we are talking about a formal analogy with the rotation of a fluid (gas).
↑ That is, if there is no movement outside a certain finite region, or if it decays quickly enough when moving away from this region, so that in the limit of an infinitely large contour the circulation tends to zero