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# Electromagnetic Waves

In a vacuum where there are no currents and no charges, so that Maxwell's equations become (6.12) (6.13) (6.14) (6.15)

Consider equations (6.13) and (6.15). If we take, of (6.13) we obtain Now we use the curl of (6.15) to rewrite the right hand side in terms of alone. Hence, Finally, using (6.14) we obtain (6.16)

where is the speed of light squared. Thus, the magnetic field and the electric field support electromagnetic waves that propagate at the speed of light. Note, that this represents very rapid time variations so that the usual MHD approximation does not apply.

In Cartesian coordinates, the wave equation has constant coefficients and solutions can be represented in the form  where and the dispersion relation is or (6.17)

Electromagnetic waves, such as gamma rays, x-rays, UV, visible light, infra-red waves and radio waves, have wavelengths ranging from to , all propagate at the speed of light. (6.12) and (6.14) imply that and so the electromagnetic waves are transverse since they propagate in a direction that is at right angles to the direction of the electric and magnetic field vectors. In addition, (6.13) gives so that and the electric and magnetic fields associated with the waves are perpendicular to each other. Finally, (6.15) implies where is a unit vector in the direction of the magnetic field, and so Example 6. .2Suppose that , then (6.12) and (6.14) are satisfied identically and (6.13) and (6.15) imply (6.18) (6.19)

and so (6.20)

One possible solution is and    Next: MHD Waves Up: Waves Previous: Sound Waves
Prof. Alan Hood
2000-11-06