Prof. David R. Jackson ECE Dept. Spring 2014 Notes 21 ECE 6341 1

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Prof. David R. JacksonECE Dept.

Spring 2014

Notes 21

ECE 6341

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Vector Potentials

If we choose

However, the E and H fields in spherical coordinates are complicated! (see Prob. 6.1 in Harrington).

, z zA F

2 2 0 k

cos cos

sincos

mn

n mn

P mb kr

mQ

then

and

ˆˆ cos sinz zA r A A

This is because we have two components in spherical coordinates:

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Vector Potentials (cont.)A better choice is

ˆ

ˆr

r

A r A

F r F

“Debye Potentials”

The electric and magnetic fields are given in terms of these potentials in a fairly simple manner (please see the next slide).

Peter Joseph William Debye (March 24, 1884 – November 2, 1966) was a Dutch physicist and physical chemist, and Nobel laureate in Chemistry.

http://en.wikipedia.org/wiki/Peter_Debye

“…he studied under the theoretical physicist Arnold Sommerfeld, who later claimed that his most important discovery was Peter Debye.”

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Vector Potentials (cont.)

1 1

1 1

E A Fj

H A Fj

Use this together with the basic field equations:

ˆ

ˆr

r

A r A

F r F

5

Vector Potentials (cont.)

rA

22

2

1rE k

j r

0rH

21E

j r r

1

sinH

r

21

sinE

j r r

1H

r

rF

0rE

22

2

1rH k

j r

1

sinE

r

21H

j r r

1E

r

21

sinH

j r r

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Vector Potentials (cont.)

2 2 0A k A

First, let’s assume that we have the “usual” solution which has enforced the Lorenz Gauge:

How do we represent the solution for Ar and Fr in spherical coordinates?

2 2

2 2

2 2

0

0

0

x x

y y

z z

A k A

A k A

A k A

2 2 2 2ˆ ˆ ˆx y zA x A y A z A since

(vector Helmholtz equation)

A j

We then have

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Vector Potentials (cont.)

since

2 2rr

A A

2 2 2 2ˆ ˆˆ rA r A A A

However,

2 2 0A k A

Hence 2 2 0 r rA k A

cos cos

sincos

mn

r n mn

P mA b kr

mQ

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Vector Potentials (cont.)

When using the Lorenz gauge (so that we have the vector Helmholtz equation), Ar and Fr do not satisfy the scalar Helmholtz equation.

We can show from Maxwell’s equations that using Ar and Fr implies that we cannot not have the Lorenz gauge (proof given next).

Observations

9

Vector Potentials (cont.)

From Maxwell’s Equations:

0

1

H

H A

0

E j H

E j A

E j A

E j A

10

Vector Potentials (cont.)

H j E

2

1A j j A

A k A j

2A k A j

(vector wave equation in mixed-potential form)

Note: We have not assumed any “gauge” here.

(source-free region)

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Vector Potentials (cont.)

Take components of the vector wave equation:

Assume ˆ rA r A

ˆ ˆ, 2

2

1 1

1 1

sin sin

r

r

Aj

r r r

Aj

r r r

Both are satisfied if we choose

rA jr

“Debye Gauge”

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Compare with Lorenz Gauge:

Not the same as the Debye Gauge !

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ˆ

1

2

r

r

rr

A j

rA j

r A jr rA

A jr r

2 2 0 A k AHence when we use the Debye Gauge

Vector Potentials (cont.)

If we wish to use the Debye potentials, we must have the Debye gauge, and therefore we do not have the vector Helmholtz equation.

Hence, A is different in spherical coordinates (using the Debye gauge) than in rectangular coordinates (using the Lorenz gauge).

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Vector Potentials (cont.)

Take the radial component of the vector wave equation, to obtain

a differential equation for Ar.

Next step:

14

Vector Potentials (cont.)

(using Debye gauge)

2

2

2

ˆ ˆ r r

r

r rA k A jr

A

r

2

22

ˆ ˆ 0rr r

Ar rA k A

r

2 2

2 2 2 2 2

2

1 1sin

sin sin

0

r r r

r

A A A

r r r

k A

Hence

Expanding, we have

2A k A j so

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Vector Potentials (cont.)The potential therefore satisfies

Not the same!

2 22

2 2 2 2 2

1 1sin 0

sin sinr r r

r

A A Ak A

r r r

2 2 0r rA k A

Even when using the Debye Gauge, we don’t get the Helmholtz equation!

22 2

2 2 2 2 2

1 1 1sin 0

sin sinr k

r r r r r

2 2 0k Compare with

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Vector Potentials (cont.)

Try this: rA r

2 2

2 2

2

2

2

2

22

2

2

1

r

r

Ar

r r

Ar

r r r r

rr r

rr r r

r rr r r

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Vector Potentials (cont.)

Hence

22

2 2 2 2 2

2

1 1 1sin

sin sin

0

r r rr r r r r

k r

2 2

2 2 2 2 2

2

1 1sin

sin sin

0

r r r

r

A A A

r r r

k A

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Vector Potentials (cont.)

22

2 2 2 2 2

2

1 1 1sin

sin sin

0

r r rr r r r r

k r

22 2

2 2 2 2 2

1 1 1sin 0

sin sinr k

r r r r r

Now compare with 2 2 0k

They are the same!

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Vector Potentials (cont.)

Hence

2 2 0 k

cos cos

sincos

mn

n mn

P mb kr

mQ

rA r

(The same holds for Fr.)

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Vector Potentials (cont.)

Define “Schelkunoff Spherical Bessel function”

cos cos

sincos

mn

r n mn

P mA kr b kr

mQ

1/2ˆ ( ) ( )

2n n nB x x b x x B xx

cos cosˆ

sincos

mn

r n mn

P mA B kr

mQ

Hence

Then we have

Note: The Schelkunoff Bessel Functions are all given in closed form (for n an integer).

(The same holds for Fr.)

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Vector Potentials (cont.)

1/2ˆ ( ) ( )

2n n nB x x b x x B xx

cos cosˆ

sincos

mn

r n mn

P mA B kr

mQ

Summary

(The same holds for Fr.)

m w

n

In general,

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Vector Potentials (cont.)

Note:

The Schelkunoff Bessel functions do not go to zero as x !

2 (2)1/2

1

2

1

ˆ ( ) ( )2

2~

2

n n

njx

n jx

H x x H xx

jx e j

x x

j e

Hence Ar and Fr do not go to zero at infinity.

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Vector Potentials (cont.)

20ˆ ( )H xExample: Calculate

1/2

2( ) sinJ x x

x

1/2

2( ) cosJ x x

x

2 (2) (2)0 0 1/2ˆ ( ) ( )

2H x xh x x H x

x

( ) cos( ) ( )( )

sin( )

J x J xY x

1/2 1/2( )Y x J x

(2)1/2 1/2 1/2H x J x jY x

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Vector Potentials (cont.)

21/2

2 2( ) sin cos jxH x x j x j e

x x

Hence

10ˆ ( ) jxH x j e

Similarly,

20ˆ ( ) jxH x j e

so

2 (2)0 1/2

2ˆ ( )2 2

jxH x x H x x j ex x x