PHYS 3343 Lesson 1 - Tarleton State University



PHYS 3323 Lesson 3

Coordinate Systems, Delta Functions and Integration

A. Why So Many Coordinate Systems?

Solving a physics problem usually involves solving a differential equation under some constraint (i.e. The electric potential is constant along a sphere of radius 0.2 m.). These constraints along with the symmetry of the problem usually dictate the choice of coordinate systems in order to simplify mathematical difficulties. For instance, a sphere centered at the origin can be described as r = constant in spherical coordinates while its equation is [pic]constant in Cartesian coordinates. The trade-off for simplifying constraint equations is the need to use more complicated forms of the divergence and other vector operators.

B. Cylindrical Coordinate System

1. Coordinates

The three coordinates used in cylindrical coordinates are (, (, and z as shown below.

[pic]

2. Transforming Between Cartesian Coordinates

Using the figure above, we can obtain the following relationships

1. [pic] 2. [pic]

3. [pic] 4. z = z

In order to develop transformation equations, we need to use our knowledge of the scalar product to develop relationships between the Cartesian and Cylindrical coordinate unit vectors. These relationships can be found using the following diagram of the unit vectors in the x-y plane.

[pic]

From the diagram, we evaluate the scalar products as

[pic] [pic]

[pic] [pic]

We can use these relationships to build our substitution equations. For instance, we find the x and y components of the [pic]unit vector using the scalar product.

[pic]

Using the same technique, we find the other four relationships as

[pic]

[pic]

[pic]

[pic]

It is now easy to convert any vector between the two coordinate systems.

EXAMPLE: Convert the vector [pic] to cylindrical coordinates.

SOLUTION:

We use our knowledge of the scalar product to write the vector in cylindrical coordinates. The radial component is found by taking the scalar product of the vector and the radial unit vector.

[pic]

Following the same procedure, we find the remaining two components of the vector as

[pic]

[pic]

3. Volume and Surface Elements

In order to successfully write down up surface and volume integrals, you must be able to able to determine differential surface and volume elements for various coordinate systems. This is usually not difficult provided you understand the coordinate system that you are using and you draw a good picture.

EXAMPLE: Write the differential volume element for cylindrical coordinates.

[pic]

SOLUTION:

In the figure above, I have drawn an expanded view of a differential volume element. To make the picture easier to read, I have drawn the changes in the coordinates so large that they are no longer infinitesimal and the volume element has some curvature. However, for infinitesimal changes in the coordinates, the volume element would be a small cube of sides dS1, dS2, and dS3. The volume of any cube is found by the formula

Volume = Base x Width x Height

Using our drawing, we see that the infinitesimal volume element in cylindrical coordinates is

[pic]

You can also use the relationship Area = Base x Height and the figure to find the infinitesimal area of any of the sides.

4. Vector Operators in Cylindrical Coordinate Representation

The vector operators (Gradient, Divergence, and Curl) are more complicated in non-Cartesian coordinate systems.

EXAMPLE 1: Evaluate the volume integral [pic] for [pic]

and a sphere of radius R.

SOLUTION:

First, you must find the differential volume element dv for the sphere

dv =

So our integral in spherical coordinates is given as

[pic]

EXAMPLE 1 (CONTINUED)

We start by writing [pic]using the scalar product as

[pic]

We now use the following diagrams to evaluate the scalar products

The projection of [pic]on the z-axis is given by

The projection of [pic]on the xy plane is given by

Thus, the projection of [pic]on the x-axis and y-axis are given by

EXAMPLE 1 (CONTINUED)

Therefore, [pic]in Cartesian coordinates is given by

We substitute our results into our integral and get

[pic]

Thus, our final result is

[pic]

ANALOGY: If [pic]was a force, we would have an equal force all the way around the sphere. Thus, the net external force would be zero!! You should notice that you don’t get the correct result if you pull [pic]out of the integral!!

EXAMPLE 2: A cylinder of radius a and length 1whose bottom is centered at the origin is placed in an electric field, [pic].

[pic]

A. Evaluate the electric flux across the closed surface of the cylinder by direct integration.

SOLUTION:

The total flux integral can be broken into three surface integrals

[pic]

For the bottom integral, we have z=0 and the surface element [pic]given by

[pic]

Therefore, the flux through the bottom is given by

[pic]

EXAMPLE 2 (CONTINUED)

For the top integral, we have z = 1. Thus, the electric field has no z-component at the top of the cylinder. Since the surface element [pic]points in the [pic] direction on the top of the cylinder, electric flux through the top of the cylinder is zero!!

Thus, we are left with just the side integral where ( = a. The surface element [pic]is given by

[pic]

We now convert [pic] into its Cartesian representation using our previous work.

Thus, the surface element [pic]is given by

[pic]

Therefore, the electric flux through the side of the cylinder is given by

[pic]

[pic]

Thus, the total electric flux through the cylinder is

[pic]

EXAMPLE 2 (CONTINUED)

B. Evaluate the electric flux across the closed surface of the cylinder by applying Gauss’s Theorem.

SOLUTION:

Since the electric field is given in Cartesian coordinates, we use the Cartesian representation for the divergence. Thus, the divergence of the electric field is given by

[pic]

The volume element in cylindrical coordinates is given by

[pic]

Using Gauss’ Theorem, we find the electric flux by

[pic]

[pic]

C. According to Gauss’ Law of Electrostatics, what is the interpretation of your results from parts A&B.

EXAMPLE 3: The electric potential V is given by [pic]. Compute

the gradient of the electric potential (negative of the electrostatic field).

SOLUTION:

EXAMPLE 4: Starting with the Cartesian representation for the Laplacian

in two dimensions derive the Laplacian’s Polar representation.

SOLUTION:

The Cartesian representation of the Laplacian is

[pic]

By the chain rule of Calculus, we have

[pic]

[pic]

We need to express the Cartesian coordinates x and y and their differentials

dx and dy in their polar representations

x = y =

dx =

dy =

We now perform some algebra to obtain the polar coordinate differentials d(

and d(.

[pic]

[pic]

EXAMPLE 4 (CONTINUED)

Multiplying the first equation by cos( and the second one by sin(, we get

[pic]

[pic]

We now add the equations together to obtain d(.

[pic]

We now do a similar algebraic set of tricks to get d(.

[pic]

[pic]

[pic]

[pic]

[pic]

From the chain rule and our equations for d( and d(, we find

[pic] [pic]

[pic] [pic]

EXAMPLE 4 (CONTINUED)

Substituting our results back into our equations for the partials with respect

To x and y, we get

[pic]

[pic]

We now find the second derivatives by repeating our process

[pic]

[pic]

[pic]

[pic]

EXAMPLE 4 (CONTINUED)

[pic]

[pic]

Thus, our final result for the Laplacian in polar coordinates is

[pic]

[pic]

C. Generalized Orthogonal Coordinates

You need to become familiar with the mathematics of generalized orthogonal coordinate systems. This material is covered in Chapter 2 of Mathematical Methods of Physics by Arfkin and Weber and my class notes for Mathematical Methods. I strongly suggest that you become familiar with the material available on vector analysis and coordinate systems in your Math Handbook. If you are using the Schaum’s Outline Mathematical Handbook by Spiegel, the material is on pages 116-130.

D. Dirac Delta Function

The Dirac delta function is not actually a function since it isn’t finite. It is defined under the theory of distributions (see Mathematical Physics by Butkov for more information).

It is important to remember that the Dirac Delta function is only defined under an integral!! Thus, you usually derive its properties by showing that two integrals are equal.

Another way of deriving properties of the Dirac delta function are to represent it in terms of well behaved functions that approach the delta function under a suitable limit (See my Math Methods notes and Chapter 1 of Mathematical Methods for Physicists by Arfkin and Weber).

The Dirac Delta function’s practical purpose is to represent point sources. For example, we know from PHYS2424 that we can obtain the net charge in an interval by integrating the linear charge density over the interval.

[pic]

We now consider a problem in which three point charges each of charge q exist along the x-axis as shown below:

How are we going to write the linear charge density for this charge distribution?

ANSWERE:

EXAMPLE PROBLEM:

For the charge distribution shown above, find the charge in the interval between x=-5 and x=3.

SOLUTION:

[pic]

The first two integrals are equal to one as the Dirac delta function’s argument goes to zero over the interval. For the same reason, the last integral is zero!! Thus, we have

We can also write three dimensional Dirac delta functions by taking the product of three 1-dimensional Dirac delta functions.

[pic]

EXAMPLE: Evaluate the following integral

[pic]

SOLUTION:

-----------------------

x

3-Dimensional Dirac Delta Function in Cartesian Coordinates

[pic]

[pic]

[pic]

x = -3

x = 4

x = 2

The Dirac delta function is defined by its assigned properties:

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

1

a

(

[pic]

(

rxy

rxy

x

y

z

y

x

Trick: Replace the unit vectors with their representations in Cartesian coordinates. The Cartesian unit vectors ( [pic], [pic], and [pic]) are constants and can be removed from the integral.

dS1 = d(

Important Point: The unit vectors in spherical coordinates are not constant (i.e. They change direction) so you can pull them out of the integral.

dS3 = dz

dS2 = (d(

(

d(

z

y

[pic]

[pic]

[pic]

[pic]

z

(

(

z

y

x

If you have really mastered the vector material in lesson 1, you should have no problem converting vectors between different coordinate systems. Thus, I will leave the remaining transformations for you to work out.

Learn the technique and don’t just try and memorize the results!

[pic]

[pic]

[pic]

y

x

[pic]

[pic]

[pic]

[pic]

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