20 December 2002

Yet Another Tensor... (Well, next week)

by George E. Hrabovsky, President, MAST

News from MAST

I want to wish all of you a Happy Thanskgiving from the people of MAST! Sorry about being so late with this column, but I have just completed the first vacation I have ever had. I would also like to announce that I am now engaged to a wonderful lady from Michigan named Diana. We plan to be wed sometime in May and will be taking a stormchasing honeymoon.

It has just come to my attention (thanks Diana) that I made two mistakes in the column entitled, "Close Enough for Scientific Work." The value quoted is not for the Z boson as stated, it is for the W boson. I also made the stupid mistake of writing the mass as eighty one hundred MeV instead of the actual value eighty one thousand MeV (the picture is correct). Finally, there is the word websight, it should read web site. My apologies.

Momentum

We have already discussed the concept of momentum ("The Energy of Motion") and the idea of the center of mass ("Getting to the Center of Things"). With a rigid body we have a large number of particles so using the summation notation will be unwieldy. Instead we recall that integration is a way of summing up all of the little pieces of what we are integration. So,

P = Underoverscript[∑, i = 1, arg3] m _ i v _ i =    Underoverscript[∑, i = 1, arg3] m _ i Overscript[r, ·] _ i

becomes,

P = ∫ m _ i Overscript[r, ·] _ i d V .

This tells us to add up all of the little boxes that make up the volume of the rigid body.

Multiple Integrals

Let's examine this last expression a little more carefully. We just did something that looks normal, but it really isn't. We are integrating with respect to a volume. We don't know how to do that yet. We know how to integrate with respect to lengths, but not areas or volumes. How do we do this?

The answer comes to us if we think about the relationship between length, area, and volume. If we integrate over a distance, we have

P = ∫ m _ i Overscript[r, ·] _ i d x .

If we want to integrate with respect to area we have to think about how area and length are related. The area of a square of side x is x^2 of x x. Let us say we have the x direction and the y direction, then we would integrate first with respect to x then y.

P = ∫ ∫ m _ i Overscript[r, ·] _ i d x d y      = ͪ ... middot;] _ i d x ) d y      = ∫ m _ i Overscript[r, ·] _ i d A .

This is called, reasonably enough, a double integral.

This works in a similar way for volumes. Assume we have an x, y, z coordinate system; then we start by integrating with respect to x, then y, and then z;

P = ∫ ∫ ∫ m _ i Overscript[r, ·] _ i d x d y d z    &n ... ;] _ i d x ) d y ] d z      = ∫ m _ i Overscript[r, ·] _ i d V .

The Frame of a Curve

There is an interesting fact that any point on a curve C can have its own coordinate frame. We begin by defining the rate at which the position vector r to a point ÷ changes with respect to the length of the curve s to the same point,

T = (d r)/(d s) .

This is called the unit tangent vector. This forms one axis of a three dimensional frame.
    If we define the curvature of C as,

κ = | (d T)/(d s) |

and the radius of curvature as,

R = 1/κ

then we can define another axis of the three dimensional frame as

N = R (d T)/(d s) .

This is called the unit normal vector.
    The final axis of the frame is the outer product of the unit tangent vector and the unit norma vector,

B = T × N .

These three axes form the components of a special frame called the Frenet-Serret frame.

Vector Integrals

There are basically three kinds of integrals for vectors; line, surface, and volume integrals. A line integral is the sum of all of the tiny bits of the vector (or even a scalar field) along a line. Given a vector A and a line element d r, the line integral is

FormBox[RowBox[{∫ _ C A · d r, , =, , RowBox[{RowBox[{∫ A _ 1 d x^1 + & ... p;           ]}], (1)}]}], TraditionalForm]

    The surface integral is the sum of all of the tiny bits of the vector (or scalar field) on a surface. Given a vector A the surface is divided into n individual elements of ΔS each having its own unit normal vector N. So we have,

∫ _ S A · N d S = Underscript[lim, n -> ∞] Underoverscript[∑, i = 1, arg3] A _ i · N _ i ΔS _ i .

The surface element can be defined,

d S = (d A)/(| N · Overscript[e,^] |)

where Overscript[e,^] is the unit vector of A. We can thus write,

∫ _ S A · N d S = ∫ ∫ A · N (d x^1 d x^2)/(| N · Overscript[e,^] |) .

There is a relationship between the surface integral and the line integral,

∫ _ C A · d r = ∫ _ S (∇ × A ) · N d S .

This result is called Stokes' theorem.
    It seems strange to me that the volume integral is the simplest of all of the vector integrals. It is simply the triple integral over the volume,

∫ _ V A d V = ∫ ∫ ∫ A d x^1 d x^2 d x^3 .

There is a relationship between the volume and surface integrals,

∫ _ V ∇ · A    d V    = ∫ _ S A · N d S

this is called the Divergence Theorem.
    I had planned to get to more material, but this is getting long and complicated, so i think that I will pick it up from here next time.

Answer to Last Column's Theory Challenge

See The Frame of a Curve given above.

Theory Challenge

Have a happy holiday season!

Books That I Like

Murray R. Speigel (1968), Schaum's Outline Series Mathematical Handbook, McGraw Hill, Inc. (34th Printing, 1995). Even with Mathematica I still look stuff up in this book. At $15 it can't be beat as a quick reference.

K. F. Riley, M. P. Hobson, S. J. Bence, (1997), Mathematical Methods for Physics and Engineering, Cambridge University Press. This book has several excellent chapters on vector analysis.

Converted by Mathematica  (December 19, 2002)