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Unformatted text preview: Vector operators in curvilinear coordinate systems In a Cartesian system, take x 1 = x , x 2 = y , and x 3 = z , then an element of arc length ds 2 is, ds 2 = dx 2 1 + dx 2 2 + dx 2 3 In a general system of coordinates, we still have x 1 , x 2 , and x 3 For example, in cylindrical coordinates, we have x 1 = r , x 2 = , and x 3 = z We have already shown how we can write ds 2 in cylindrical coordinates, ds 2 = dr 2 + r 2 d + dz 2 = dx 2 1 + x 2 1 dx 2 2 + dx 2 3 We write this in a general form, with h i being the scale factors ds 2 = h 2 1 dx 2 1 + h 2 2 dx 2 2 + h 2 3 dx 2 3 We see then for cylindrical coordinates, h 1 = 1, h 2 = r , and h 3 = 1 Curvilinear coordinates For an vector displacement ~ ds ~ ds = e 1 h 1 dx 1 + e 2 h 2 dx 2 + e 3 h 3 dx 3 Back to our example of cylindrical coordiantes, e 1 = e r , e 2 = e , and e 3 = e z , and ~ ds = e r dr + e rd + e z dz These are orthogonal systems, but it would not have to be! ds 2 = 3 X i =1 3 X j =1 g ij dx i dx i The g ij is the metric tensor, and for an orthogonal system it is diagonal with g i = h 2 i Vector operators in general curvilinear coordinates Recall the directional derivative d ds along ~ u , where ~ u was a unit vector d ds = ~ u Now the ~ u becomes the unit vectors in an orthogonal system, for example in cylindrical coordinates Now we recall that ds 2 = ds 2 = h 2 1 dx 2 1 + h 2 2 dx 2 2 + h 2 3 dx 2 3 Lets take a cylindrical system, first consider ~ u = e r , then ds = dr ~ ( r ,, z ) e r = r Vector operators in general curvilinear coordinates Next ~ u = e r , then ds = rd ( h 2 = r ) ~ ( r ,, z ) e = 1 r It is also easy to show, ~ ( r ,, z ) e z = z Now that we have the projections, we can find ~ in cylindrical coordinates, ~ = r e r + 1 r e + z e z Gradient in curvilinear (orthogonal) coordinate system...
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This note was uploaded on 07/30/2011 for the course PHZ 3113 taught by Professor Staff during the Spring '03 term at University of Central Florida.
 Spring '03
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