# Equations i 10 and i 11 represents in some sense a

• Notes
• 100000464160110_ch
• 12

This preview shows 8 out of 12 pages.

Equations [ I-10 ] and [ I-11 ] represents in some sense a complete formal solution of the wave propagation problem. However, from a practical point of view it is essential to consider how energy propagates in anisotropic

Subscribe to view the full document.

O N C LASSICAL E LECTROMAGNETIC F IELDS P AGE 46 R. Victor Jones, December 19, 2000 media. To that end, we note that the time averaged Poynting vector associated with a given plane wave-- viz. v S r k , ϖ ( 29 = 1 2 r E r k , ϖ ( 29 × r H * r k , ϖ ( 29 [ V-26 ] propagates in a direction ˆ s r k , ϖ ( 29 -- conventionally designated the “ray” direction -- which is orthogonal to both r E r k , ϖ ( 29 and r H r k , ϖ ( 29 as shown below. The time averaged total stored energy is given by U r k , ϖ ( 29 = 1 4 r E r k , ϖ ( 29 r D * r k , ϖ ( 29 + r B r k , ϖ ( 29 r H * r k , ϖ ( 29 [ ] = r k 2 ϖ ˆ k r E r k , ϖ ( 29 × r H * r k , ϖ ( 29 = r k ϖ ˆ k r S r k , ϖ ( 29 [ V-27 ]
O N C LASSICAL E LECTROMAGNETIC F IELDS P AGE 47 R. Victor Jones, December 19, 2000 and, thus, we see that the “ray” or “energy flow” velocity, v ray , for a given ˆ s r k , ϖ ( 29 is given by 1 v ray = r k ϖ ˆ k ˆ s ˆ k , ϖ ( 29 = 1 v phase ˆ k ˆ s ˆ k , ϖ ( 29 [ V-28 ] We write the time averaged Poynting vector associated with a given eigenmode as v S σ ( 29 r k ( 29 = 1 2 r E σ ( 29 r k ( 29 × r H σ ( 29 * r k ( 29 = 1 2 ϖ μ 0 r E σ ( 29 r k ( 29 × r k × r E σ ( 29 * r k ( 29 = 1 2 ϖ μ 0 r k r E σ ( 29 r k ( 29 2 - r E σ ( 29 * r k , ϖ ( 29 r k r E σ ( 29 r k ( 29 [ ] [ V-29 ] where r E σ ( 29 r k ( 29 is the electric field associated with the eigenmode. Using Equations [ V-3 ], [ V-10 ] and [ V-11 ] this field can be expressed as r E ( σ ) r k ( 29 = ( t ( σ ) ( ) k ) t ε - 1 ) t ( σ ) ( ( k ) [ ] ) t ( σ ) ( ( k ) + ) k t ε - 1 ) t ( σ ) ( ( k ) [ ] ) k { } D ( σ ) ( r k ) = 1 ε 0 n ( σ ) ( r k ) [ ] 2 ) t ( σ ) ( ( k ) + ε 0 n ( σ ) ( r k ) [ ] 2 ) k t ε - 1 ) t ( σ ) ( ( k ) [ ] ) k { } D ( σ ) ( r k ) = 1 ε 0 n ( σ ) ( r k ) [ ] 2 ) t ( σ ) ( ( k ) + g ( σ ) ( ( k ) ) k { } D ( σ ) ( r k ) [ V-30 ] where g ( σ ) ( ( k ) = ε 0 n ( σ ) ( r k ) [ ] 2 ) k t ε - 1 ) t ( σ ) ( ( k ) [ ] = ) k t ε - 1 ) t ( σ ) ( ( k ) [ ] ( t ( σ ) ( ) k ) t ε - 1 ) t ( σ ) ( ( k ) [ ] . Using this parameterization, the modal Poynting vector can be expressed as v S σ ( 29 r k ( 29 = 1 2 ε 0 μ 0 D ( σ ) ( r k ) ε 0 2 n ( σ ) ( r k ) [ ] 3 ) k - ) t ( σ ) ( ( k ) g ( σ ) ( ( k ) { } [ V-31 ]

Subscribe to view the full document.

O N C LASSICAL E LECTROMAGNETIC F IELDS P AGE 48 R. Victor Jones, December 19, 2000 and the associated ray vector as ˆ s σ ( 29 r k ( 29 = ) k - ) t ( σ ) ( ( k ) g ( σ ) ( ( k ) ) k - ) t ( σ ) ( ( k ) g ( σ ) ( ( k ) = ) k - ) t ( σ ) ( ( k ) g ( σ ) ( ( k ) 1 + g ( σ ) ( ( k ) 2 [ V-32 ] If we take ) k and ) t ( σ ) ( ( k ) as reference directions, the ray vector, ˆ s σ ( 29 r k ( 29 , lies in the plane containing ) k and ) t ( σ ) ( ( k ) at an angle η ( σ ) ( ( k ) = - tan - 1 g ( σ ) ( ( k ) [ ] with respect to the direction ) k . For an optical material with uniaxial symmetry, we may use Equations [ V-16 ], [ V-19 ] and [ V-20 ] to evaluate g ( σ ) ( ( k ). In particular, we may easily see that for the ordinary mode tan η (o) ( ( k ) = - g (o) ( ( k ) = 0 [ V-33a ] or ˆ s (o) r k ( 29 = ) k and for the extraordinary mode tan η (e) ( ( k ) = - g (e) ( ( k ) = sin ϑ cos ϑ ε - 1 - ε || - 1 ( 29 ε - 1 cos 2 ϑ + ε || - 1 sin 2 ϑ . [ V-33b ]
O N C LASSICAL E LECTROMAGNETIC F IELDS P AGE 49 R. Victor Jones, December 19, 2000
You've reached the end of this preview.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern