In the last lecture on the "Mathematical Preliminaries" we have covered the expressions for Gradient, Divergence and Curl for scalar and vector fields
in the General Orthogonal Curvilinear Co-ordinate Systems and then we introduced
the Dirac Delta function ( which is not really a function but a generalized function or a Distribution.)
The expressions for the operators in Curvilinear co-ordinate systems may be obtained by the tedious method of expressing the partial derivatives with respect to (x, y, z)
in terms of the desired coordinates by the chain rule for partial derivatives and then use the transformation equations between the two systems. However an easier more direct method is through the use of the Boundary Theorems. This is given in Appendix A, Griffith.
The procedure for obtaining the Gradient operator expression was discussed in the lecture. This invloved the fact that the boundary theorem for gradient involves
the line integral of the scalar product of the gradient of a scalar field ∇ φ. dl over some curve C between two points A and B. The RHS is simply the difference in the value of φ (B) - φ (A). Now the line element in the general curvilinear coordinate system (u, v, w) is given by the sum of the differentials du, dv and dw multiplied by their respective scale factor functions (f, g, h). On the RHS dφ (u, v, w) may be expressed in terms of the partial derivatives of φ with respect to (u, v, w) and the differentials du, dv, dw. Now comparing the LHS with the RHS we get the expression for ∇ φ in the general coordinates u, v, w. From this expression we can now specialise to any of the specific curvilinear
system by writing down the specific coordinates like (r, θ, φ) and the scale factors.
We also discussed the Dirac delta function. First thing is that it is NOT An ordinary function as no ordinary function can be zero everywhere except at x=0 and still give an integral over all x to be equal to 1. It is something called a Generalized Function or a Distribution and the theory of this is an advanced mathematical issue beyond the scope of the present course. However we will consider the delta function to be the limit of a sequence of suitable peaked functions. One example of such a sequence of functions is a sequence of Gaussian
f_a(x) =(1/a π1/2) exp {- (x2/a2)} as the Lim a goes to ZERO. As a goes to small values the Gaussian becomes more peaked and less wide and in the limit a goes to zero it is zero everywhere except a zero where it is infinite and is hence a delta function.
But this function is NOT UNIQUE that is a sequence any peaked function of certain width witha parameter that controls the peak and the width will provide a delta function in the limit of that parameter going to zero. The correct unique way to approach delta functions is through the "Theory of Generalized Functions or Distributions" which is beyond the scope of the course.
A naive way of justifying that integral of delta ( x) over all x is zero is to integrate such a sequence of functions as the limit a goes to zero and this will give 1 if the integration is done before taking the limit in careful way. It can also be seen graphically from the link provided in this post. We will treat the delta function as a normal function with the special properties mentioned but always under the integral sign. Under an integral the delta function when multiplied by another function always gives the result of the function at the singularity
of the delta function that is where the delta function is infinite if the range of integration includes that point.
Some properties of the delta function is given in the Image at the beginning of the post. The way to prove these properties is to multiply both sides by a smooth function which goes to zero at plus minus infinity and integrate both sides over all
space ( real line for single x).
The link below provides extra reference for the delta function but some aspects are very advanced
which are not required for this course
http://en.wikipedia.org/wiki/Dirac_delta_function
we define dirac delta function as 0 for all values except x=zero while infinity for x=0. Why could it not be defined as negative infinity for x=0 and 0 for all other x??
ReplyDeleteFirstly the delta function is even so delta (x)=delta(-x) in the sense of an identity when multiplied by a smooth function and integrated.
ReplyDeleteSecondly it really does not matter because the delta function arises when we have singularities in physical quantities which means that they assume large values, so negative infinity or positive infinity does not make much difference. Thirdly the definition of delta function ( at our
level ) is the limit of a sequence of peaked functions which are all positive. So except for maybe change certain signs the math and physics will remain the same.
someone had asked me about a proof of integral of delta x is 1. I have provided a justification of this through the intergral of the limit of the sequence of functions which define a delta function.
ReplyDeleteOf course this is not a proof because the delta function as the limit of sequence of functions is not unique.