Calculus/Vector calculus

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Vector calculus specifically refers to multi-variable calculus applied to scalar and vector fields. While vector calculus can be generalized to dimensions (), this chapter will specifically focus on 3 dimensions ()

Fields in vector calculus[edit]

Scalar fields[edit]

A scalar field is a function that assigns a real number to each point in space. Scalar fields typically denote densities or potentials at each specific point. For the sake of simplicity, all scalar fields considered by this chapter will be assumed to be defined at all points and differentiable at all points.

Vector fields[edit]

A vector field is a function that assigns a vector to each point in space. Vector fields typically denote flow densities or potential gradients at each specific point. For the sake of simplicity, all vector fields considered by this chapter will be assumed to be defined at all points and differentiable at all points.

Vector fields in cylindrical coordinates[edit]

The cylindrical coordinate system used here has the three parameters: . The Cartesian coordinate equivalent to the point is

Any vector field in cylindrical coordinates is a linear combination of the following 3 mutually orthogonal unit length basis vectors:

Note that these basis vectors are not constant with respect to position. The fact that the basis vectors change from position to position should always be considered. The cylindrical basis vectors change according to the following rates:

Any vector field expressed in cylindrical coordinates has the form:

Given an arbitrary position that changes with time, the velocity of the position is:

The coefficient of for the term originates from the fact that as the azimuth angle increases, the position swings around at a speed of .

Vector fields in spherical coordinates[edit]

The spherical coordinate system used here has the three parameters: . The Cartesian coordinate equivalent to the point is

Any vector field in spherical coordinates is a linear combination of the following 3 mutually orthogonal unit length basis vectors:

Note that these basis vectors are not constant with respect to position. The fact that the basis vectors change from position to position should always be considered. The spherical basis vectors change according to the following rates:

Any vector field expressed in spherical coordinates has the form:

Given an arbitrary position that changes with time, the velocity of this position is:

The coefficient of for the term arises from the fact that as the latitudinal angle changes, the position traverses a great circle at a speed of .

The coefficient of for the term arises from the fact that as the longitudinal angle changes, the position traverses a latitude circle at a speed of .

Volume, path, and surface integrals[edit]

Volume Integrals[edit]

Volume integrals have already been discussed in the chapter Multivariable calculus, but a brief review is given here for completeness.

Given a scalar field that denotes a density at each specific point, and an arbitrary volume , the total "mass" inside of can be determined by partitioning into infinitesimal volumes. At each position , the volume of the infinitesimal volume is denoted by the infinitesimal . This gives rise to the following integral:

Path Integrals[edit]

Given any oriented path (oriented means that there is a preferred direction), the differential denotes an infinitesimal displacement along in the preferred direction. This differential can be used in various path integrals. Letting denote an arbitrary scalar field, and denote an arbitrary vector field, various path integrals include:

, , , , and many more.

denotes the total displacement along , and denotes the total length of .

Calculating Path Integrals[edit]

To compute a path integral, the continuous oriented curve must be parameterized. will denote the point along indexed by from the range . must be the starting point of and must be the ending point of . As increases, must proceed along in the preferred direction. An infinitesimal change in , , results in the infinitesimal displacement along . In the path integral , the differential can be replaced with to get

Example 1

As an example, consider the vector field and the straight line curve that starts at and ends at . can be parameterized by where . . We can then evaluate the path integral as follows:


If a vector field denotes a "force field", which returns the force on an object as a function of position, the work performed on a point mass that traverses the oriented curve is

Example 2

Consider the gravitational field that surrounds a point mass of located at the origin: using Newton's inverse square law. The force acting on a point mass of at position is . In spherical coordinates the force is (note that are the unit length mutually orthogonal basis vectors for spherical coordinates).

Consider an arbitrary path that traverses that starts at an altitude of and ends at an altitude of . The work done by the gravitational field is:

The infinitesimal displacement is equivalent to the displacement expressed in spherical coordinates: .

The work is equal to the amount of gravitational potential energy lost, so one possible function for the gravitational potential energy is or equivalently, .


Example 3

Consider the spiral parameterized with respect to in cylindrical coordinates by . Consider the problem of determining the spiral's length with restricted to the range . An infinitesimal change of in results in the infinitesimal displacement:

The length of the infinitesimal displacement is .

The length of the spiral is therefore:



Surface Integrals[edit]

Given any oriented surface (oriented means that the there is a preferred direction to pass through the surface), an infinitesimal portion of the surface is defined by an infinitesimal area , and a unit length outwards oriented normal vector . has a length of 1 and is perpendicular to the surface of , while penetrating in the preferred direction. The infinitesimal portion of the surface is denoted by the infinitesimal "surface vector": . If a vector field denotes a flow density, then the flow through the infinitesimal surface portion in the preferred direction is .

The infinitesimal "surface vector" describes the infinitesimal surface element in a manner similar to how the infinitesimal displacement describes an infinitesimal portion of a path. More specifically, similar to how the interior points on a path do not affect the total displacement, the interior points on a surface to not affect the total surface vector.

The displacement between two points is independent of the path that connects them.

Consider for instance two paths and that both start at point , and end at point . The total displacements, and , are both equivalent and equal to the displacement between and . Note however that the total lengths and are not necessarily equivalent.

Similarly, given two surfaces and that both share the same counter-clockwise oriented boundary , the total surface vectors and are both equivalent and are a function of the boundary . This implies that a surface can be freely deformed within its boundaries without changing the total surface vector. Note however that the surface areas and are not necessarily equivalent.

The fact that the total surface vectors of and are equivalent is not immediately obvious. To prove this fact, let be a constant vector field. and share the same boundary, so the flux/flow of through and is equivalent. The flux through is , and similarly for is . Since for every choice of , it follows that .

The geometric significance of the total surface vector is that each component measures the area of the projection of the surface onto the plane formed by the other two dimensions. Let be a surface with surface vector . It is then the case that: is the area of the projection of onto the yz-plane; is the area of the projection of onto the xz-plane; and is the area of the projection of onto the xy-plane.

The boundary of is counter-clockwise oriented.

Given an oriented surface , another important concept is the oriented boundary. The boundary of is an oriented curve but how is the orientation chosen? If the boundary is "counter-clockwise" oriented, then the boundary must follow a counter-clockwise direction when the oriented surface normal vectors point towards the viewer. The counter-clockwise boundary also obeys the "right-hand rule": If you hold your right hand with your thumb in the direction of the surface normals (penetrating the surface in the "preferred" direction), then your fingers will wrap around in the direction of the counter-clockwise oriented boundary.

Example 1

Consider the Cartesian points ; ; ; and .

Let be the surface formed by the triangular planes ; ; and where the vertices are listed in a counterclockwise direction relative to the surface normal directions. The surface vectors of each plane are respectively ; ; and respectively which add to a total surface vector of .

Let be the surface formed by the single triangular plane where the vertices are listed in a counterclockwise direction relative to the normal direction. It can be seen that and share a the common counter clockwise boundary The surface vector is which is equivalent to .


Example 2

This example will show how moving a point that is in the interior of a "triangular mesh" does not affect the total surface vector. Consider the points where . Let the closed path be defined by the cycle . For simplicity, . For each , will denote the displacement of relative to . Like with , .

Let denote a surface that is a "triangular mesh" comprised of the closed fan of triangles: ; ; ...; ; where the vertices of each triangle are listed in a counterclockwise direction. It can be seen that the counterclockwise boundary of is and does not depend on the location of . The total surface vector for is:

Now displace by to get . The displacement vector of relative to becomes . The counterclockwise boundary is unaffected. The total surface vector is:

Therefore moving the interior point neither affects the boundary, nor the total surface vector.



Calculating Surface Integrals[edit]

To calculate a surface integral, the oriented surface must be parameterized. Let be a continuous function that maps each point from a two-dimensional domain to a point in . must be continuous and onto. While does not necessarily have to be one to one, the parameterization should never "fold back" on itself. The infinitesimal increases in and are respectively and . These respectively give rise to the displacements and . Assuming that the surface's orientation follows the right hand rule with respect to the displacements and , the surface vector that arises is .

In the surface integral , the differential can be replaced with to get .

Example 3

Consider the problem of computing the surface area of a sphere of radius .

Center the sphere on the origin, and using and as the parameter variables, the sphere can be parameterized in spherical coordinates via where and . The infinitesimal displacements from small changes in the parameters are:

causes

causes

The infinitesimal surface vector is hence . While not important to this example, note how the parameterization was chosen so that the surface vector points outwards. The area is .

The total surface area is hence:


The Gradient and Directional Derivatives[edit]

Given a scalar field that denotes a potential, and given a curve , a commonly sought after quantity is the rate of change in as is being traversed. Let be an arbitrary parameter for , and let denote the point indexed by . Given an arbitrary which corresponds to the point , then using the chain rule gives the following expression for the rate of increase of at , :

where is a vector field that denotes the "gradient" of , and is the unnormalized tangent of .

If is infact an arc-length parameter, i.e. , then the direction of the gradient is the direction of maximum gain: Given any unit length tangent , the direction will maximize the rate of increase in . This maximum rate of increase is .

Calculating total gain[edit]

Given the gradient of a scalar field : , the difference between at two different points can be calculated, provided that there is a continuous path that links the two points. Let denote an arbitrary continuous path that starts at point and ends at point . Given an infinitesimal path segment with endpoints and , let be an arbitrary point in . denotes the infinitesimal displacement denoted by . The increase in along is:

The relative error in the approximations vanish as . Adding together the above equation over all infinitesimal path segments of yields the following path integral equation:

This is the path integral analog of the fundamental theorem of calculus.

The gradient in cylindrical coordinates[edit]

Let be a scalar field that denotes a potential and a curve that is parameterized by : . Let the rate of change in be quantified by the vector . The rate of change in is:

Therefore in cylindrical coordinates, the gradient is:

The gradient in spherical coordinates[edit]

Let be a scalar field that denotes a potential and a curve that is parameterized by : . Let the rate of change in be quantified by the vector . The rate of change in is:

Therefore in spherical coordinates, the gradient is:

The Directional Derivative[edit]

Given a scalar field and a vector , scalar field computes the rate of change in at each position where the velocity of is . Scalar field can also be expressed as . Velocity can also be a vector field so depends on the position . Scalar field becomes .

In Cartesian coordinates where the directional derivative is:

In cylindrical coordinates where the directional derivative is:

In spherical coordinates where the directional derivative is:

What makes the discussion of directional derivatives nontrivial is the fact that can instead be a vector field . Vector field computes at each position where .

In cylindrical coordinates, basis vectors and are not fixed, and in spherical coordinates, all of the basis vectors , , and are not fixed. This makes determining the directional derivative of a vector field that is expressed using the cylindrical or spherical basis vectors non-trivial. To directly compute the directional derivative, the rates of change of each basis vector with respect to each coordinate should be used. Alternatively, the following identities related to the directional derivative can be used (proofs can be found here):

Given vector fields , , and , then

Given vector fields and , and scalar field , then

In cylindrical coordinates, and

In spherical coordinates, , and , and

Full Expansions

In Cartesian coordinates where and the directional derivative is:

In cylindrical coordinates where and the directional derivative is:

In spherical coordinates where and the directional derivative is:


The Divergence and Gauss's Divergence Theorem[edit]

Let denote a vector field that denotes "flow density". For any infinitesimal surface vector at position , the flow through in the preferred direction is . is the flow density parallel to the x-axis etc.

Given a volume with a closed surface boundary with an outwards orientation, the the total outwards flow/flux through is given by the surface integral . This outwards flow is equal to the total flow that is being generated in the interior of .

For an infinitesimal rectangular prism (, , and ) that is centered on position , the outwards flow through the surface is:

All relative errors vanish as .

is the "divergence" of and is the density of "flow generation" at . As noted above, the total outwards flow through is the total flow generated inside of , which gives Gauss's divergence theorem:

This image depicts an example of the total flow across a closed boundary being the total flow generated inside the boundary.

In the image to the right, an example of the total flow across a closed boundary being the total flow generated in the interior of the boundary is given. The direction of the flow across each edge is denoted by the direction of the arrows, and the rate is denoted by the number of arrows. Each node inside the boundary is labelled with the rate of flow generation at the current node. It can be checked that a net total of 2 units of flow is being drawn into the boundary, and the total rate of flow generation across all interior nodes is a net consumption of 2 units.


The divergence in cylindrical coordinates[edit]

Let denote a vector field that denotes "flow density". In order to compute the divergence (flow generation density) of , consider an infinitesimal volume defined by all points where , , and . Note that is not a rectangular prism. Let , , and . Let be an arbitrary point from .

The volume of is approximately . The 6 surfaces bounding are described in the following table:

Surface approximate area direction approximate flow density
, ,
, ,
, ,
, ,
, ,
, ,

The total outwards flow through the surface of is:

All relative errors vanish as .

The divergence (flow generation density) is therefore:

a note about the approximations

A reader may wonder why the area of surface , , is approximated by instead of since the difference between and approaches 0 as . While the absolute difference between and approaches 0 as , the difference relative to the infinitesimal does not approach 0: .

With respect to the surface , , , the area can be approximated by , , or since is already a factor, and the differences between , , and relative to the infinitesimal do approach 0 as .


The divergence in spherical coordinates[edit]

Let denote a vector field that denotes "flow density". In order to compute the divergence (flow generation density) of , consider an infinitesimal volume defined by all points where , , and . Note that is not a rectangular prism. Let , , and . Let be an arbitrary point from .

The volume of is approximately . The 6 surfaces bounding are shown in the following table:

Surface approximate area direction approximate flow density
, ,
, ,
, ,
, ,
, ,
, ,

The total outwards flow through the surface of is:

All relative errors vanish as

The divergence (flow generation density) is therefore:


Divergence free vector fields[edit]

A vector field for which is a "divergence free" vector field. can also be referred to as "incompressible" (since the flow density of an incompressible fluid is divergence free) or "solenoidal" (since magnetic fields are divergence free).

A key property of a divergence free vector field is that the flux of through a surface is purely a function of the surface's boundary. If and are two surfaces which share the same counterclockwise oriented boundary , then . In other words, the flux is purely a function of . This property can be derived from Gauss's divergence theorem as follows:

To begin, it will be assumed that and do not intersect each other, except for at the common boundary . The argument presented here easily generalizes to cases where and do intersect each other. Invert the orientation of to get and combine and to get a closed surface , stitching the surfaces together along the seam . Let denote the volume which is the interior of , and it will also be assumed that is oriented outwards (which is the case if is "in front" of ).

Gauss's divergence theorem states that . The flux through is the flux through minus the flux through : . Therefore:

The Laplacian Operator[edit]

Given a scalar field , if the gradient is interpreted as denoting flow density, the the rate of flow generation at each point is which is referred to as the "Laplacian" of and is denoted by (or alternately ).

The laplacian is effectively a measure of the "convexity" of at each point . When there is a net flow of the gradient away from , this means that is "low" compared to its neighboring points and that the convexity is positive. When there is a net flow of the gradient towards , this means that is "high" compared to its neighboring points and that the convexity is negative.

In Cartesian coordinates, the Laplacian is: