Homework11-12 - Homework 11 and 12 Answers, 95.658 Spring...

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Homework 11 and 12 Answers, 95.658 Spring 2011, Electromagnetic Theory II Dr. Christopher S. Baird, UMass Lowell Problem 1 Jackson 11.1 Two equivalent inertial frames K and K ' are such that K ' moves in the positive x direction with speed v as seen from K . The spatial coordinate axes in K ' are parallel to those in K and the two origins are coincident at times t = t ' = 0. (a) Show that the isotropy and homogeneity of space-time and equivalence of different inertial frames (first postulate of relativity) require that the most general transformation between the space-time coordinates ( x , y , z , t ) and ( x ', y ', z ', t ') is the linear transformation, x ' = f v 2 x v f v 2 t ; t ' = g v 2 t v h v 2 x ; y ' = y ; z ' = z and the inverse, x = f v 2 x ' v f v 2 t ' ; t = g v 2 t ' v h v 2 x ' ; y = y ' ; z = z ' where f , g , and h are functions of v 2 , the structures of the x ' and x equations are determined by the definition of the inertial frames in relative motion, and the signs of the inverse equation are a reflection of a reversal of roles of the two frames. SOLUTION: First of all, there is no motion in the y or z directions, so that these coordinates must be identical: y ' = y ; z ' = z Homogeneity of space-time means that points in space-time have the same density at all points as observed in one particular frame. Mathematically, this requires linear relationships between all space and time coordinates. (If, for instance, x ' = x 3 , than at locations marked by higher x , successive points in x ' would be further and further spaced apart.) This gives: x ' = f x f 2 t ; t ' = g t h x ; y ' = y ; z ' = z At this point, f , f 2 , g , and h are arbitrary functions that cannot depend on x or t . The only thing left that they could depend is the frame's velocity v and any universal constants that fall out. We have inertial frames in relative motion, so that the shifting of the origins must be taken into account. In Galilean relativity, this was taken into account by adding - vt to the spatial relation. We can get this in the same general form by taking a - v out of the arbitrary f 2 and h : x ' = f x v f 2 t ; t ' = g t v h x ; y ' = y ; z ' = z Again, the homogenous nature of space-time means that as time marches on, the points cannot spread out. This forces upon us f 2 = f : x ' = f x v f t ; t ' = g t v h x ; y ' = y ; z ' = z
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If the functions f , g , h are functions of v , they should be functions of the magnitude of v and not the sign of v , so that the signs we have chosen above are preserved to keep the origins shifting in the right way. The most natural way to be the function of a variable's magnitude but not sign is to square it. We
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This note was uploaded on 02/13/2012 for the course PHYSICS 95.658 taught by Professor Staff during the Spring '11 term at UMass Lowell.

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Homework11-12 - Homework 11 and 12 Answers, 95.658 Spring...

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