lec22

lec22 - Scott Hughes 6 May 2005 Massachusetts Institute of...

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Unformatted text preview: Scott Hughes 6 May 2005 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2005 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 parenleftBig vector E vector E + vector B vector B parenrightBig . We are interested in understanding how the total energy, U = integraldisplay V u dV = 1 8 integraldisplay V dV parenleftBig vector E vector E + vector B vector B parenrightBig changes as a function of time. So, we take its derivative: U t = t 1 8 integraldisplay V dV parenleftBig vector E vector E + vector B vector B parenrightBig . 205 Since we have fixed the integration region, we can take the /t ender the integral: U t = 1 8 integraldisplay V dV t parenleftBig vector E vector E + vector B vector B parenrightBig = 1 4 integraldisplay V dV vector E vector E t + vector B vector B t . Rearranging the source free Maxwell equations, vector E t = c vector vector B vector B t = c vector vector E we can get rid of the vector E and vector B time derivatives: U t = c 4 integraldisplay V dV bracketleftBig vector E parenleftBig vector vector B parenrightBig vector B parenleftBig vector vector E parenrightBigbracketrightBig . 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 vector parenleftBig vector E vector B parenrightBig = vector E parenleftBig vector vector B parenrightBig + vector B parenleftBig vector vector E parenrightBig . Comparing with our expression for U/t , we see that this expression simplifies things: U t = c 4 integraldisplay V dV vector parenleftBig vector E vector B parenrightBig integraldisplay V dV vector vector S . On the second line we have defined vector S = c 4 vector E vector B ; well 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 Gausss theorem. This changes the integral over V into an integral over the surface A : U t = integraldisplay A dvectora vector S ....
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This note was uploaded on 07/19/2010 for the course 8 8.022 taught by Professor Scotthughes during the Spring '10 term at MIT.

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lec22 - Scott Hughes 6 May 2005 Massachusetts Institute of...

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