In the chapter on vector calculus, the differential operator of the gradient (), the divergence (), and the curl () were introduced. This chapter will focus on inverting these differential operators.
The gradient, divergence, and curl operators are all "linear", meaning that given arbitrary scalar fields , vector fields , and scalars , that:
More generally given a family of scalar fields , vector fields , and coefficients (each corresponds to a scalar field, a vector field, and a scalar coefficient),
Inverting Linear Operators
Recall from linear algebra that when given a bijective linear operator , that an inverse can be created by computing a solution to each of where is the elementary basis vector for each . When attempting to find an that solves for an arbitrary , the solutions to can be stacked in a linear manner to get as a possible solution. This yields an approach to inverting .
This same approach will be used to compute "Green's functions" for each vector calculus differential operator.
The Dirac delta function
The Dirac delta function is a hypothetical function that returns for all and for . The Dirac delta is not meant to be evaluated at , but instead assumed to satisfy the following integral property: . More generally, given any interval that strictly contains 0: , then . is effectively a density function that describes an infinitely dense total mass of 1 at . For an arbitrary , describes a density function that describes an infinitely dense total mass of 1 at .
Even though , it is not the case that , or . The integrals are: and . In general, for any , .
Consider the piece-wise function defined by: . It is common to accept that is not differentiable at and that . With the Dirac delta function, the derivative of can be expressed as . With this derivative, part II of the fundamental theorem of calculus holds even for intervals that contain .
In this chapter, it will generally be assumed that all scalar fields and vector fields are continuous and differentiable everywhere. However if this is not the case, the Dirac delta function will be used to model the derivative operators at points of non-differentiability.
Inverting linear differential operators
Given an arbitrary function , can be expressed as the linear combination of Dirac delta functions. Let be a linear differential operator that takes a real valued single variable function , and returns another real valued single variable function . If a solution exists to for all , then given an arbitrary , a solution exists to which is . The family of functions are referred to as "Green's functions".
Dirac delta function variants
In this chapter, given an arbitrary point , the function will denote a 3-dimensional variant of the Dirac delta function. The key integral property is that for any volume , that . Note that .
Given an arbitrary oriented curve , will denote another variant of the Dirac delta function which returns vectors. for all , and is infinite in the direction of for all points from . The key integral property is that for any oriented surface , that where is the net number of times passes through in the preferred direction ( passing through in the reverse direction reduces by 1).
Given an arbitrary oriented surface , will denote another variant of the Dirac delta function which returns vectors. for all , and is infinite in the direction of the oriented normals of for all points from . The key integral property is that for any oriented curve , that where is the net number of times passes through in the preferred direction ( passing through in the reverse direction reduces by 1).
Multi-paths and multi-surfaces
Given an oriented curve , then can be denoted by the Dirac delta vector field . If is continuous and starts at and ends at , then it can be proven using Gauss's Divergence Theorem that .
Given a collection of oriented paths, then the vector field effectively denotes the "superposition" of . This superposition is referred to as a "multi-path". Not all paths have to have a weight of 1. With multi-path , the weights on and are both . This multi-path is an even 50%/50% superposition between and .
Any vector field can be envisioned as a superposition of a possibly infinite number of paths. Each path may have an infinitesimal weight. When a vector field is envisioned as a multi-path, the decomposition into individual paths is not unique. When vector field denotes a multi-path, is the net density of path origin points minus the density of destination points: the path starting point density.
When everywhere, can be envisioned as a superposition of a possibly infinite number of paths that are either closed or extend to infinity. If it is also the case that is for some , then all of the paths have to close and is effectively a "multi-loop". ( is if and only if there exists some threshold and factor such that )
A demonstration of how a divergence free vector field can be envisioned as the superposition of several simple loops.
In the image to the right, a divergence free vector field that denotes flow density is decomposed into the superposition of multiple simple loops. 2 dimensional space is depicted as a lattice of infinitely small squares. The vector field is the top-left section. The flow along each horizontal edge, denoted by the direction and number of arrows, is the horizontal component of the vector field at the current edge (or neighboring vertex). The flow along each vertical edge, denoted by the direction and number of arrows, is the vertical component of the vector field at the current edge (or neighboring vertex). The remaining 3 sections show 3 simple loops whose superposition forms the vector field.
Given an oriented surface , then can be denoted by the Dirac delta vector field . If has the counter-clockwise oriented boundary , then it can be proven using Stokes' Theorem that .
Given a collection of oriented surfaces, then the vector field effectively denotes the "superposition" of . This superposition is referred to as a "multi-surface". Not all surfaces have to have a weight of 1. With multi-surface , the weights on and are both . This multi-surface is an even 50%/50% superposition between and .
Any vector field can be envisioned as a superposition of a possibly infinite number of surfaces. Each surface may have an infinitesimal weight. When a vector field is envisioned as a multi-surface, the decomposition into individual surfaces is not unique. When vector field denotes a multi-surface, is the multi-loop that is the counter-clockwise oriented boundary of the multi-surface denoted by .
When everywhere, can be envisioned as a superposition of a possibly infinite number of surfaces that are either closed with no boundaries or extend to infinity. If it is also the case that is for some , then all of the surfaces have to close without extending to infinity and is effectively a "multi closed surface".
Given an oriented curve and a vector field , then the path integral is equivalent to the volume integral . This statement, while intuitive to a certain degree given the definition of , is proven in the box below:
To start, note that for all that . Positions that do not lie in have no contribution to the integral nor have any contribution to the integral .
Vector field denotes a flow density field where a total flow of 1 is moving along curve . Imagining as an infinitesimally narrow "pipe", since a flow of 1 is moving through this pipe, the density of the flow is infinite. To properly model , path will be assumed to have a non-zero infinitesimally small cross-sectional area of so the flow density is . Path with a non-zero thickness of will be denoted by . For any point , will denote a unit length vector that is parallel to the direction of . will be approximated by . The volume integral becomes:
Now consider an arbitrary infinitesimal displacement along that is located at . The corresponding volume segment from has a volume of . The contribution to the volume integral is approximately which is equivalent to the contribution to the path integral . Integrating over all path segments and letting yields:
In a general sense, the infinitesimal displacement from a path integral along can be replaced by when the integral is converted to a volume integral over . For example, given a continuous curve which begins at and ends at , it can be derived that: .
Given an oriented surface and a vector field , then the surface integral is equivalent to the volume integral . This statement, while intuitive to a certain degree given the definition of , is proven in the box below:
To start, note that for all that . Positions that do not lie in have no contribution to the integral nor have any contribution to the integral .
Vector field denotes a vector field where a total gain of 1 is acquired when a path integral crosses in the preferred direction. Imagining as an infinitesimally thin "sheet", since a gain of 1 is acquired moving through this sheet in the preferred direction, the rate of gain is infinite. To properly model , surface will be assumed to have a non-zero infinitesimally small thickness of so the rate of gain is . Surface with a non-zero thickness of will be denoted by . For any point , will denote a unit length vector that is normal to and points in the preferred direction. will be approximated by . The volume integral becomes:
Now consider an arbitrary surface vector that denotes an infinitesimal portion of that is located at . The corresponding volume segment from has a volume of . The contribution to the volume integral is approximately which is equivalent to the contribution to the surface integral . Integrating over all surface segments and letting yields:
In a general sense, the infinitesimal surface vector from a surface integral over can be replaced by when the integral is converted to a volume integral over . For example, given an oriented surface which has a total surface vector of , it can be derived that: .
Given a curve and a surface , the net number of times passes through in the preferred direction, denoted by , is given by:
Given vector fields and where denotes a multi-path and denotes a multi-surface, then denotes the total "flux" of multi-path through multi-surface .
If is closed loop and is a closed surface then since every time passes through , must pass through in the opposite direction in order to close itself. More generally, if vector field denotes a multi-loop, which means that and is for some , and vector field denotes a multi closed surface, which means that and is for some , then . An algebraic proof is given in the box below:
To start, let and .
It is important to note that:
- As , , or , that .
- As , , or , that .
- Since as , then as .
Evaluating the integral gives:
An example application of multi-paths and multi-surfaces is given in the box below:
Given a closed loop that is carrying an electric current of , the magnetic field generated by this closed loop obeys the following equations:
(Ampere's Law in magentostatics)
where is magnetic permeability of free space . It is also required that as .
Let and be two closed loops which are the counterclockwise boundaries of surfaces and . Let denote the magnetic field generated by running an electric current of around . Note that is proportional to . Now let denote the total flux of through in the preferred direction: . Note that since , that is a function of instead of . It is the case that is proportional to , and the constant of proportionality is the "mutual inductance" from to .
The mutual inductance is purely a function of and . It is also the case that the mutual inductance is symmetric: . The symmetry of the mutual inductance is not obvious, and while the symmetry is apparent from explicit formulas for the mutual inductance such as the "Neumann formula" , the symmetry can be made clear by interpreting the magnetic field as both a multi-loop and a multi-surface.
The vector field satisfies both and . While not proven here, the magnetic field generated by a closed loop of current that does not extend to infinity is so is both a multi-loop and a multi-surface with counterclockwise boundary . Doing the same for loop gives where is the magnetic field generated solely by running a current of around . is both a multi-loop and a multi-surface with counterclockwise boundary .
The flux of through is:
Since , the vector field is a multi closed surface, and the total flux of multi-loop through this multi closed surface is 0: .
Therefore so the mutual inductance is . The symmetry of the mutual inductance is now apparent from this formula: .
Given a continuous oriented curve that originates from and terminates at , it has been noted that . It can be derived that: . Generalizing to a multi-path vector field , it is the case that: . Recall that is a measure of the "starting point density" of .
Given an oriented surface with counterclockwise boundary , it has been noted that . It can be derived that: . Generalizing to a multi-surface vector field , it is the case that: . Recall that is the counterclockwise boundary density of .
Inverting the divergence operator (the inverse square law)
Given an arbitrary scalar field , the problem of interest is that of finding a vector field that satisfies and . In other words, we want to find a vector field whose divergence is given by and is irrotational. For reasons that will soon become apparent, it will be assumed that is as for some . The "big O" means that .
can be expressed as a linear combination of Dirac delta functions: .
If an irrotational vector field can be determined such that (that is divergence-free everywhere except the origin where the divergence is infinite), then for all : . Since the divergence operator distributes over linear combinations, is an irrotational vector field that satisfies .
An intuitive candidate for an irrotational vector field that is divergence free everywhere except the origin is a radially symmetric vector field where is the distance from the origin, and is the unit vector that points away from the origin (see spherical coordinates). is unknown at this point. The inverse square indicates that the flow diffuses out over a larger area as the distance from the origin increases. It can easily be checked that is irrotational everywhere (including the origin), and is divergence free everywhere except the origin. All that remains is to determine such that . Consider a sphere of radius centered on the origin. The total outwards flow/flux through the surface of this sphere is , so the total flow generated inside the sphere is . Since , the total flow generated inside the sphere is . This gives . Therefore .
In total, is an irrotational vector field that satisfies . The assumption that is as for some implies that the volume integral does not diverge to infinity. Also note that the apparent singularity at does not impact the integral (to be discussed below).
An important question is if the vector field is the only irrotational vector field that satisfies . If is another possible solution, then is a vector field that is both irrotational and divergence free at all points: and . There is then the following theorem:
- Divergence free and irrotational vector fields
Given a vector field that is divergence free (meaning: For all closed oriented surfaces : )
and irrotational, (meaning: For all closed oriented curves : )
it must then be the case that is everywhere constant:
is a constant vector , and therefore an irrotational vector field that satisfies is unique up to the addition of a constant vector field.
About Improper Integrals
The volume integral has a pole/singularity when . Its range also extends to infinity. Both of these irregularities have the potential to result in a divergent (infinite) integral.
To analyse whether or not the integral diverges due to the pole/singularity or infinite range, the volume integral will be expressed as the integral of concentric spherical shells centered on :
where denotes the outwards oriented surface of a sphere centered on with radius . denotes a solid sphere centered on with radius .
The inner surface integral does not present any irregularities. The lower bound of denotes the pole, and the upper bound of denotes the infinite range.
The integral where the inner radius is small and the outer radius is large is regular. The goal is to analyse the integral's behavior as and .
Assume that for all for each radius . The magnitude of the inner surface integral is now bounded from above by:
As can be seen, the surface area of cancels out .
The magnitude of is bounded from above by:
Assuming that is bounded everywhere, is bounded everywhere, so it is clear that does not approach infinity as . This settles the pole/singularity at . For the integral to not diverge as , the condition that is as for some is sufficient. This condition implies that is less than a multiple of when becomes sufficiently large. The integral of converges at infinity provided that .
Inverting the gradient operator
Given a vector field , the problem of interest is finding a scalar field such that . It should be noted that does not exist for most vector fields. The gradient theorem implies that for all closed continuous curves , that . It must be the case that for all closed continuous curves : must be conservative. Equivalently, everywhere: should be irrotational.
Given a conservative vector field , can be determined by choosing an origin point , and a constant . For each , a continuous oriented curve that starts at and ends at should be generated. is assigned: . The choice of curve is irrelevant since is conservative/irrotational.
It can be confirmed that satisfies for each point by evaluating the directional derivative along arbitrary curves that pass through . For an arbitrary curve parameterized by that passes through at , the directional derivative of at is . This confirms that .
Spherical volume integral solution
Assume that is as for some . This means that for some , that there exists such that .
This property implies that given a sufficiently large , that any path integral between any two points outside of the sphere that remains outside of the sphere is arbitrarily small. Therefore the origin point can freely shift between points at infinity.
Choose infinity as the origin point and let . Given an arbitrary point , choose a direction quantified by unit length vector . The path travels backwards along a ray that starts at and points in the direction given by . . Since this integral does not depend on the direction , the average over all directions is:
where is an infinitesimal surface portion of the unit sphere that constrains .
Letting the position vector variable be , the new volume differential is .
Lastly, letting gives:
This new formula is a volume integral that expresses the potential as a linear combination of functions that exhibit a degree of spherical symmetry. This formula is similar to the inverse square law for the inverse of the divergence.
Next, the above formula will be derived using a Green's function approach in a manner similar to the inverse square law for the divergence operator.
Green's function solution
This section will derive a formula identical to the formula above using a Green's function approach. While the derivation will be complicated and the result will not be new, the derivation itself will yield many interesting intermediate results. Again, it will be assumed that is as for some . It will also be assumed that is continuous.
Vector field can be expressed as the following linear combination of Dirac delta functions: . This linear combination however does not facilitate the inversion of the gradient since for each , the vector fields , , and are all not conservative, so there do not exist any scalar fields , , and such that , , and . This prevents a simple solution of .
It is first necessary to express as a linear combination of vector fields that are conservative, so that for each component/basis vector field, a scalar field exists where the gradient is the component vector field. In mathematical terms should be decomposed into the following linear combination: where for each , vector fields , , and are all conservative. Specifically a vector valued function is required that is linear with respect to : . In essence, is effectively a linear combination of , , and where the coefficients are the components of . The linear combination that must be decomposed into is: .
A candidate function that exhibits a degree of spherical symmetry has the form:
where is the projection of onto , and is the perpendicular component of relative to . and are real valued functions that scale the components of relative to the displacement . and have yet to be determined.
and have to be chosen such that provided that satisfies all of the required conditions (most importantly is conservative) and that is always a conservative vector field with as the position parameter.
Using the assumption that is continuous, for an arbitrarily small , it can be assumed that for . can be added as an additional parameter to , , and to get:
The condition that becomes: .
It can be proven with difficulty that if where is an arbitrary monotone decreasing function and and , then choosing and will satisfy the condition .
being conservative, continuous, and vanishing at infinity are all important.
Decomposition of Conservative Vector Fields
Let be an arbitrary vector field that is:
- as for some
Let arbitrary function be such that:
- is monotone decreasing
Let be an arbitrary real number.
If , , and then
To begin, fix . To simplify notation let:
- denotes the unit length vector at position that points away from
- denote the angle between and
When , . is continuous, so approximating with gives: . The error relative to in the approximation vanishes as .
The challenge is to evaluate the integral when : where can no longer be approximated by . To start, hollow out the vector field around