Some Advanced Calculus

© 2012 by  Fernando Caracena

Most advanced applications of science and engineering do not use the simpler calculus that we have covered in another blog, but rather,  use advanced algebra. The reader depending on individual interest should not overly be concerned with the details presented. The advanced calculus presented here is valuable in the development of many topics in modern physics. The following discussion can not be considered to be a thorough discussion of what is a large subject in mathematics, but is presented more in the spirit of demonstrating the use of mathematics as syntax in the language of physics.

Partial Derivatives

In an earlier blog, we considered the derivatives of a single valued function of a single variable. Because often the analyses of a three dimensional problem can be broken down into a set of problems that deal with motion along independent axis, that form of calculus has its use in physics. However, the physical world does not always allow us to see a complex problem as a set of purely one dimensional problems. For that reason , we need to extend calculus into the multidimensional realm.

We will still restrict our attention to derivatives of single-valued functions, but we will expand the range of such functions into that of functions of several variables, e. g., F(x,y,z,t). In this case, we can still define partial derivatives, in which we concentrate on changes in the function that are produced by changes in only one of the variables. For example, if we hold the variables, y, z, and t as constants, and compute the normal single variable derivative in x, then we get what is called the partial derivative of F with respect to x, which is written as ∂F(x,y,z,t)/∂x.

The total derivative of a function now involves all of the partial derivatives, which makes sense to write only for some simple cases.  Consider the change in value of the function F (δF) when there are 4 independent changes in the coordinates:

δF(x, y, z, t)=∂F(x,y,z,t)/∂x δx+∂F(x,y,z,t)/∂y δy+∂F(x,y,z,t)/∂z δz+∂F(x,y,z,t)/∂t δt. (1)

δF(x, y, z, t) = F • δr + ∂F(x,y,z,t)/∂t δt ,   (2a)


 = e1 ∂/∂x + e2 ∂/∂y +e3 ∂/∂z  (2b)

is a vector derivative operator, called the nabla (), which acting on a function produces a vector that characterizes the derivative of the function in three dimensional space,

∇ F= e1 ∂F/∂x + e2∂F/∂y + e3 ∂F/∂z.                                  (2c)


In Einstein's Theory of Relativity, both time and space are treated as components of a four-dimensional, time-space manifold. In classical physics time-space effects produced by motion at much less than the speed of light can be neglected, and most problems break down into separable spacial and temporal analysis. Later we will see how motion at near the speed of light modifies space-time so that they become mixed as four dimensional effects.

Another important quantity involving the nabla operator (∇) is called the divergence, which acts on a vector field as follows:

div V = V                                                                       (3a)


div V = ∂V1/∂x + ∂V2/∂y + ∂V3/∂z    ,                                        (3b)

where V is a function of at least all the spatial coordinates indicated above.

It is possible to grok what all these operators mean, but the subject is usually not taught that way. Instead, students are given a lot of problems to do, with the hope that in the process of solving them, they will gain some intuition about what they mean. Almost anybody working through all the problems in a vector calculus outline series, will gain some insight in what they mean.

Fig. 1. An infinitesmal cube having normals to the edges and faces alligned in the x, y, and z directions. The unit normals (red arrows) are displayed pointing outward from the centers of the faces (green Xs).

I will  show an example of grokking the divergence operator to demonstrate the validity of the Divergence theorem:

∫∫∫ dvol div V = ∫∫closed dSrf n V,    (4)

where dSrf is an infinitesimal portion of the surface that is the boundary enclosing the entire vol[ume], of which dvol is an infinitesimal element.

The plan of this demonstration uses the figure of the infinitesimal cbox (Fig. 1), where note that the unit normals to the faces are not scaled accordingly, otherwise they would point way off the screen. In Fig. 1, the lower left hand corner of the box is located at the point (x,y,z); the x-axis runs parallel to the bottom edge of the box that runs from right to left; the y-axis; the y-axis extends toward the upper right along the bottom edge of the cube; and the z-axis extends upward from the lower left hand corner. Note that the sides of the box are scaled as Δx, Δy, and Δz. Also note that the unit normal vectors point outward from each face either in the positive or negative direction of the corresponding coordinate axis. We begin by analysing one of the infinitesimal elements of the left hand side of (4).

Δx Δy Δz (∂V1/∂x + ∂V2/∂y + ∂V3/∂z )=ΔV1 Δy Δz +ΔV2 Δx Δz + ΔV3 Δx Δy

"  =  [V1(x+Δx, y, z) -   V1(x, y, z)] Δy Δz + [V2 (x, y+Δy, z) - V2 (x, y, z)]  Δx Δz

                                                           +[V3 (x, y, z+Δz) - V3 (x, y, z)]  Δx Δy

" = [e1 V(x+Δx, y, z) - e1 V(x, y, z)] Δy Δz

   +[e2 V (x, y+Δy, z) - e2 V (x, y, z)]  Δx Δz

   +[e3 V (x, y, z+Δz) - e3 V (x, y, z)]  Δx Δy

Now note that we can number all the bounding faces of the box, values of V and their normals as follows:

V1 = V(x+Δx, y, z)

V2= V(x, y, z)

V3 = V(x, y+Δy, z)

V4= V(x, y, z)

V5 = V(x, y, z+Δz)

V6 = V(x, y, z)

dS1 =Δy Δz

dS2 =Δy Δz

dS3 =Δx Δz

dS4=Δx Δz

dS4 =Δx Δz

dS5 = Δx Δy

dS6 = Δx Δy

dS5 = Δx Δy

dS6 = Δx Δy

n1 = -e1

n2 = e1

n3 = -e2

n4 = e2

n5 = -e3

n6 = e3

Using the above notation we can write the divergence of V over the volume Δx Δy Δz as the following sum over all the six faces that bound the cube:

Δx Δy Δz (∂V1/∂x + ∂V2/∂y + ∂V3/∂z )= ∑i=1,..,6   n V dSi .      (5)

OK, so we have shown that the Divergence Theorem applies to an infinitesimal box of dimensions Δx Δy Δz; but what about a sum over a finite volume, which is expressed in terms of infinitesimal boxes? To do this, we imagine pushing two infinitesimal boxes together side by side and note what happens at the common face. The outward normal of one box is in the opposite direction of the other along the common face, as a result of which the contributions of the common faces to the volume sum cancel and the sum collapses to that of all the outward faces. Repeating this process many times results in a sum similar to (5) but over all of the outward faces of the bounding volume that is approximated by a very large number of infinitesimal cubes stacked together. As we let the volumes of the infinitesimal boxes shrink to zero and their number grow to infinity to progressively approach the volume of the integral, we get (4) as the limit.

An additional operator, which is important in advanced applications of calculus involving the cross product of a nabla with the vector function, is the curl of a vector function [V(x,y,z)],

curl V = x V                                              (7a)

curl V = e1 (∂V3/∂y-∂V2/∂z) + e2 (∂V1/∂z-∂V3/∂x) + e3 (∂V2/∂x-∂V1/∂y).     (7b)

There is another theorem involving the curl of a vector function, called Stoke's Theorem, which is valuable in advanced applications. It runs as follows

∫∫ (∇ x V)•n dS = ∫c V dl   ,                      (8)

Fig. 2. Depicts Stoke's Theorem.

where the surface integral [LHS (8)] extends over a patch that has a bounding closed surve (c) (Fig. 2). Each area element (dS) has a unit normal that extends outward from its surface. Stokes theorem states that the normal component of the curl of a vector field integrated over an open surface is just equivalent to a line integral of the tangential component of the vector field around a closed loop around the entire curve (c) that is the boundary of the surface. Note that this result can be demonstrated in a way similar to that of the Divergence Theorem, which is left as an exercise for the reader.

One final identity is that the divergence of the curl of a vector function vanishes,

x V =0   .                             (9a)

Applying the Divergence Theorem (4) to (9) results in the following:

∫∫∫ dvol x V = ∫∫closed dS n x V ,


∫∫closed dS n x V =0 .              (9b)

The result of the application of the Divergence theorem to the identity in (9a) has the result (9b) that the areal sum of the normal component of the curl of a vector function vanishes when the integral is taken over an entire closed surface, i.e., when it has no bounding curve.



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