lec21

# lec21 - Scott Hughes 3 May 2005 Massachusetts Institute of...

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Unformatted text preview: Scott Hughes 3 May 2005 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2005 Lecture 21: Polarization & Scattering 21.1 Summary: radiation so far In the last few lectures, we examined solutions of the source free Maxwell equations: ~ ∇ · ~ E = 0 ~ ∇ × ~ E =- 1 c ∂ ~ B ∂t ~ ∇ · ~ B = 0 ~ ∇ × ~ B = 1 c ∂ ~ E ∂t . With a little massaging, we discovered that these equations can be rewritten as wave equa- tions for ~ E and ~ B : ∂ 2 ~ E ∂t 2- c 2 ∇ 2 ~ E = 0 ∂ 2 ~ B ∂t 2- c 2 ∇ 2 ~ B = 0 . A particularly instructive solution to the wave equations are the plane wave forms: ~ E ( ~ r, t ) = ~ E sin( ~ k · ~ r- ωt ) ~ B ( ~ r, t ) = ~ B sin( ~ k · ~ r- ωt ) . This solution represents an electromagnetic wave propagating in the ˆ k = ~ k/k direction (where k = q ~ k · ~ k = q k 2 x + k 2 y + k 2 z ). By considering how the wave behaves at some fixed time, we learned that k is simply related to the wavelength λ : k = 2 π/λ . The requirement that this solution satisfy the wave equation tells us that ω = ck . From the definition ω = 2 πν (angular frequency is 2 π radians times “regular” frequency), we then obtain λν = c . Finally, requiring that the plane wave solution satisfy all of Maxwell’s equations leads to some important constraints on the vector amplitudes ~ E and ~ B . These constraints are: 195 • The amplitudes are orthogonal to the propagation direction: ˆ k · ~ E = 0, ˆ k · ~ B = 0. • The amplitudes are orthogonal to each otherL ~ E · ~ B = 0. • The amplitudes have the same magnitude: | ~ E | = | ~ B | . • The propagation direction is parallel to ~ E × ~ B . These are important and rather constraining conditions. Nonetheless, they leave us with a great deal of freedom in the amplitudes. This freedom is described in terms of the radiation’s polarization state . 21.2 Linear polarization Since plane waves propagate in a straight line, we might as well just define their propagation direction as something simple and be done with it. In what follows, we will take ˆ k = ˆ z , so our wave is of the form ~ E = ~ E sin( kz- ωt ) ~ B = ~ B sin( kz- ωt ) . Suppose that the wave is arranged (somehow) so that the electric field is aligned with the x axis: ~ E = E ˆ x . Then, our requirement that ~ E × ~ B be parallel to ˆ k tells us that ~ B = E ˆ y . (We’re also using | ~ E | = | ~ B | .) This configuration is known as a linearly polarized wave, as the electric (and magnetic) fields at all points align parallel to a line. Such an electromagnetic wave is quite simple to produce: we just need to take a conductor and arrange things so that it has an oscillating charge distribution. Here’s an example: + + +--- + + +--- t = 0 ϖ t = t = π /2 ϖ ϖ π 196 The top part of this figure shows an antenna : a long conductor in which we drive a very rapidly oscillating current, so that the conductor has a charge distribution . As we move to the right, time increases....
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lec21 - Scott Hughes 3 May 2005 Massachusetts Institute of...

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