Fundamental Theorems & Dirac Delta Function | Mathematical Methods - Physics PDF Download

The Fundamental Theorem of Calculus

Suppose f (x) is a function of one variable. The fundamental theorem of calculus states:

where df /dx = F(x).

Geometrical Interpretation

According to equation df = (df /dx) dx is the infinitesimal change in f when one goes from (x) to (x + dx). The fundamental theorem says that if you chop the interval from a to b into many tiny pieces, dx, and add up the increments df from each little piece, the result is equal to the total change in f  is  f (b) – f (a).  In other words, there are two ways to determine the total change in the function: either subtract the values at the ends or go step-by-step, adding up all the tiny increments as you go. You’ll get the same answer either way.

The Fundamental Theorem for Gradients

Suppose we have a scalar function of three variables V(x, y, z). Starting at point a, we moves a small distance  Then

Now we move a little further, by an additional small displacementthe incremental change in V will be In this manner, proceeding by infinitesimal steps, we make the journey to point b. At each step we compute the gradient of V (at that point) and dot it into the displacement  …this gives us the change in V. Evidently the total change in V in going from a to b along the path selected is

This is called the fundamental theorem for gradients; like the “ordinary” fundamental theorem, it says that the integral (here a line integral) of a derivative (here the gradient) is given by the value of the function at the boundaries (a and b).
Geometrical Interpretation
Suppose you wanted to determine the height of the Eiffel Tower. You could climb the stairs, using a ruler to measure the rise at each step, and adding them all up or you could place altimeters at the top and the bottom, and subtract the two readings; you should get the same answer either way (that's the fundamental theorem).
Corollary 1:is independent of path taken from a to b.
Corollary 2:since the beginning and end points are identical, and hence
V(b) – V(a) = 0.

Example 22: Let V = xy², and take point a to be the origin (0, 0, 0) and b the point (2, 1, 0). Check the fundamental theorem for gradients.

Although the integral is independent of path, we must pick a specific path in order to evaluate it. Let's go out along the x axis (step i) and then up (step ii). As always,


(i)
(ii) 

Evidently the total line integral is 2.  This consistent with the fundamental theorem: T(b) – T(a) =2 – 0 = 2.
Calculate the same integral along path (iii) (the straight line from a to b):
(iii)
Thus the integral is independent of path.

The Fundamental Theorem for Divergences

The fundamental theorem for divergences states that:

This theorem has at least three special names: Gauss’s theorem, Green’s theorem, or, simply, the divergence theorem. Like the other “fundamental theorems,” it says that the integral of a derivative (in this case the divergence) over a region (in this case a volume) is equal to the value of the function at the boundary (in this case the surface that bounds the volume). Notice that the boundary term is itself an integral (specifically, a surface integral). This is reasonable: the “boundary” of a line is just two end points, but the boundary of a volume is a (closed) surface.

Geometrical Interpretation
If  represents the flow of an incompressible fluid, then “the flux of (the right side of equation) is the total amount of fluid passing out through the surface, per unit time and the left side of equation shows an equal amount of liquid will be forced out through the boundaries of the region.

Example 23: Check the divergence theorem using the function  and the unit cube situated at the origin.

In this case
 
and


Evidently,
To evaluate the surface integral we must consider separately the six sides of the cube:
(i)
(ii) 
(iii)
(iv)
(v)
(vi)
So the total flux is:

Example 24:  A vector field is given in spherical coordinates. Evaluate both sides of Divergence Theorem for the volume enclosed between
(i) r = 1 and r = 2, and
(ii)

Divergence theorem states that
Since

(i)
 



So L.H.S. = R.H.S. = 75π
Divergence theorem proved.

(ii) L.H.S. of Divergence Theorem

R.H.S. of Divergence Theorem


L.H.S. = R.H.S. = 588.91. Divergence theorem proved.

The Fundamental Theorem for Curls

The fundamental theorem for curls, which goes by the special name of Stokes’ theorem, states that

As always, the integral of a derivative (here, the curl) over a region (here, a patch of surface) is equal to the value of the function at the boundary (here, the perimeter of the patch). As in the case of the divergence theorem, the boundary term is itself an integralspecifically, a closed line integral.

Geometrical Interpretation

The integral of the curl over some surface (or, more precisely, the flux of the curl through that surface) represents the “total amount of swirl,” and we can determine that swirl just as well by going around the edge and finding how much the flow is following the boundary (as shown in figure).

Corollary 1:depends only on the boundary line, not on the particular surface used.
Corollary 2: for any closed surface, since the boundary line, like the mouth of a balloon, shrinks down to a point, and hence the right side of equation vanishes.

Example 25: Suppose Check Stokes’ theorem for the square surface shown in figure.

Here
 
(In saying thatpoints in the x direction, we are chosen to a counterclockwise line integral. We could as well writebut then we have to go clockwise.) Since x = 0 for this surface,

Now, what about the line integral? We must break this up into four segments:
(i)
(ii) 
(iii)
(iv)
So

Example 26: Given in cylindrical coordinates. For the contour shown in figure, verify the Stokes’ Theorem.

Stokes’ Theorem
In cylindrical coordinates,
Ar = 2r cos∅ , A_{∅ =} r, Az = 0

L.H.S. = R.H.S. =

Example 27: Given a vector field Verify stokes,’ theorem over the path shown in figure.

Stokes’ theorem







(Equation of quarter circle x² + y² = 9; 0 < x, y < 3 )

The Dirac Delta Function

The Divergence of

The plot thickens if you apply the divergence theorem to this function. Suppose we integrate over a sphere of radius R, centered at the origin; the surface integral is

But the volume integral,is zero. Does this mean that the divergence theorem is false?
The source of the problem is the point r = 0, where  blows up. It is quite true that=0 everywhere except the origin, but right at the origin the situation is more complicated.
Notice that the surface integral is independent of R; if the divergence theorem is right (and it is), we should get for any sphere centered at the origin, no matter how small. Evidently the entire contribution must be coming from the point r = 0!
Thus,has the bizarre property that it vanishes everywhere except at one point, and yet its integral (over any volume containing that point) is 4π . No ordinary function behaves like that. (On the other hand, a physical example does come to mind: the density (mass per unit volume) of a point particle. It's zero except at the exact location of the particle, and yet its integral is finite namely, the mass of the particle.) What we have stumbled on is a mathematical object known to physicists as the Dirac delta function. 

The One- Dimensional Dirac Delta Function

The one dimensional Dirac delta function, δ(x) , can be pictured as an infinitely high, infinitesimally narrow "spike," with area 1 (as shown in figure).

That is to say:

and
If f (x) is some “ordinary” function then the product f(x)δ(x) is zero everywhere except at x = 0 . It follows that
f(x)δ(x)= f(0)δ(x)

Of course, we can shift the spike from x = 0 to some other point, x = a :
 
also
and


Example 28: Show that
 
where k is any (nonzero constant). (In particular

For an arbitrary test function f(x), consider the integral
Changing variables, we let y = kx , so that x = y /kx, and If k is positive, the integration still runs from -∞ to + ∞  but if k is negative, then x = ∞ implies y = -∞ , and vice versa, so the order of limit is reversed. Restoring the "proper" order costs a minus sign. Thus

(The lower signs apply when k is negative, and we account for this neatly by putting absolute value bars around the final k, as indicated.) Under the integral sign, then, δk(x) serves the same purpose as

Example 29: Evaluate the integral

Answer would be 0, because the spike would then be outside the domain of integration.

Example 30: Evaluate the integral

The delta function picks out the value of x³ at the point x = 2 so the integral is 2³ = 8 .

Example 31: Show that

Similarly

Example 32: Let θ (x) be the step function:

Show that 

The Three-Dimensional Delta Function

It is an easy matter to generalize the delta function to three dimensions:

(As always, is the position vector, extending from the origin to the point (x, y, z)). This three-dimensional delta function is zero everywhere except at (0, 0, 0), where it blows up. Its volume integral is 1

and

Since the divergence of  is zero everywhere except at the origin, and yet its integral over any volume containing the origin is a constant ( 4π). These are precisely the defining conditions for the Dirac delta function; evidently

Example 33: Evaluate the integral where v is a sphere of radius R centered  at the origin.

The Theory of Vector Fields

If the curl of a vector field vanishes (everywhere), then can be written as the gradient of a scalar potential (V):
(The minus sign is purely conventional.)
Theorem 1: Curl-less (or "irrotational") fields. The following conditions are equivalent (that is,  satisfies one if and only if it satisfies all the others):
(a) everywhere.
(b) is independent of path, for any given end points.
(c) for any closed loop.

(d)  is the gradient of some scalar,
The scalar potential is not unique-any constant can be added to V with impunity, since this will not affect its gradient.
If the divergence of a vector field vanishes (everywhere), then can be expressed as the curl of a vector potential

That’s the main conclusion of the following theorem:
Theorem 2: Divergence-less (or “solenoidal”) fields. The following conditions are equivalent:
(a)everywhere.
(b) is independent of surface, for any given boundary line.
(c)for any closed surface.

(d) is the curl of some vector,
The vector potential is not unique-the gradient of any scalar function can be added to without affecting the curl, since the curl of a gradient is zero.