# Lec22 - Scott Hughes 6 May 2004 Massachusetts Institute of...

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Unformatted text preview: Scott Hughes 6 May 2004 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Lecture 22: The Poynting vector: energy and momentum in radiation. Transmission lines. 22.1 Electromagnetic energy It is intuitively obvious that electromagnetic radiation carries energy — otherwise, the sun would do a pretty lousy job keeping the earth warm. In this lecture, we will work out how to describe the flow of energy carried by electromagnetic waves. To begin, consider some volume V . Let its surface be the area A . This volume contains some mixture of electric and magnetic fields: Volume V Electric field E Magnetic field B Surface A The electromagnetic energy density in this volume is given by energy volume = u = 1 8 π ‡ ~ E · ~ E + ~ B · ~ B · . We are interested in understanding how the total energy, U = Z V udV = 1 8 π Z V dV ‡ ~ E · ~ E + ~ B · ~ B · changes as a function of time. So, we take its derivative: ∂U ∂t = ∂ ∂t 1 8 π Z V dV ‡ ~ E · ~ E + ~ B · ~ B · . Since we have fixed the integration region, we can take the ∂/∂t ender the integral: ∂U ∂t = 1 8 π Z V dV ∂ ∂t ‡ ~ E · ~ E + ~ B · ~ B · = 1 4 π Z V dV ~ E · ∂ ~ E ∂t + ~ B · ∂ ~ B ∂t . Rearranging the source free Maxwell equations, ∂ ~ E ∂t = c ~ ∇× ~ B ∂ ~ B ∂t =- c ~ ∇× ~ E we can get of the ~ E and ~ B time derivatives: ∂U ∂t = c 4 π Z V dV h ~ E · ‡ ~ ∇× ~ B ·- ~ B · ‡ ~ ∇× ~ E ·i . 22.2 The Poynting vector The expression for ∂U/∂t given above is as far as we can go without invoking a vector identity. With a little effort, you should be able to prove that ~ ∇· ‡ ~ E × ~ B · =- ~ E · ‡ ~ ∇× ~ B · + ~ B · ‡ ~ ∇× ~ E · . Comparing with our expression for ∂U/∂t , we see that this expression simplifies things: ∂U ∂t =- c 4 π Z V dV ~ ∇· ‡ ~ E × ~ B · ≡ - Z V dV ~ ∇· ~ S . On the second line we have defined ~ S = c 4 π ~ E × ~ B ; we’ll discuss this vector in greater detail very soon. Since we have a volume integral of a divergence, the obvious thing to do here is to apply Gauss’s theorem. This changes the integral over V into an integral over the surface A : ∂U ∂t =- Z A d~a · ~ S . In other words, The rate of change of the electromagnetic energy in the volume V is given by the flux of ~ S through V ’s surface. Notice the minus sign; this is easily explained. Recall that d~a points outward. If ~ S and d~a are parallel, energy is leaving the system, so U decreases. And vice versa. ~ S is called the Poynting 1 vector. It points 2 in the same direction as the electromagnetic energy flow. (This gives the physical picture promised last week that ~ E × ~ B gives the 1 It’s actually named after a person, John Henry Poynting. You’ve got to wonder if he was fated to work this quantity out....
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Lec22 - Scott Hughes 6 May 2004 Massachusetts Institute of...

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