hw2sol - Physics 570 Homework No. 2 1. Solutions: due...

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Physics 570 Homework No. 2 Solutions: due Wednesday, 3 February, 2010 1. On a 2-dimensional manifold, using Cartesian coordinates { x,y } , we are given the local vector fields e V and e C , with components relative to the standard coordinate basis as V x = xy , V y = - y 2 , C x = y , C y = - x . a. For each of these vector fields, first determine the curve Γ( η ) for which this vector field is the tangent vector, and normalize constants of integration so that for η = 0 the curve is at the initial point on the manifold, x = 2 ,y = 1. Then produce a sketch of the curve and also a few other adjacent curves, i.e., ones with the same tangent vector field but having an initial point somewhere nearby. Such a set of curves is referred to as a congruence of curves , because they are all “congruent” to each other in the sense of the use of the word in plane geometry. b. If the initial point is not particularly nearby, do the curves still look basically the same? Test this by now choosing an initial point as x = - 2 ,y = - 1. [5 pts] ........................................................................................... First we consider the curve for which e V is the tangent vector. We know that a tangent vector has components that are just the derivatives of the components along the curve; therefore, we may write the following: dx = V x = xy , dy = V y = - y 2 . We easily integrate the second of these integrations, to give y = 1 / ( η - c ), where c is a constant of integration. Inserting this information into the first equation it is also, now, easily integrated to give x = α ( η - c ), with α a second constant of integration. We may then simply specify the curve in the way shown on the first line of the equations just below, which is satisfactory for the requested solution to the problem; however, an optimal presentation is, instead, shown by determining some “meaning” for the constants of integration, by re-writing them in terms of the
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beginning point for the curve, i.e., the point with coordinates ( x 0 ,y 0 ) that corresponds to η = 0, which is shown in the equations on the second line: x { Γ( η ) } = α ( η - c ) , y { Γ( η ) } = 1 η - c , or x { Γ( η ) } = x 0 (1 + y 0 η ) , y { Γ( η ) } = y 0 1 + y 0 η . We are asked to look at the particular curve that has ( x 0 ,y 0 ) = (2 , 1), and nearby curves. That curve is specified by the following: x { Γ( η ) } = 2(1 + η ) , y { Γ( η ) } = 1 1 + η , while the figure on the left just below shows that curve and 4 other ones that are quite nearby: We are also asked to compare that congruence of nearby curves with one that goes through a rather more distance point on the manifold, namely ( x 0 ,y 0 ) = ( - 2 , - 1), which gives x { Γ( η ) } = 2( η - 1) , y { Γ( η ) } = 1 η - 1 . This can be seen on the graph on the right above, where it is also compared with a curve going
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This note was uploaded on 03/08/2010 for the course PHYSICS AN 570 taught by Professor Davids.king during the Spring '10 term at Caltech.

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hw2sol - Physics 570 Homework No. 2 1. Solutions: due...

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