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PHYSICS 131 hw-1-soln

PHYSICS 131 hw-1-soln - HOMEWORK SET 1 PHYS 131 Solutions...

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HOMEWORK SET 1, PHYS. 131 Solutions originally by Matt Pillsbury, modifed by Kevin Moore Instructor: Anthony Zee Email: [email protected] TA: Kevin Moore Email: [email protected] 2.7. Consider the Following Familiar rectangular coordinates ( x,y ), labelling points in the plane to a new set oF coordinates ( μ,ν ): x = μν, y = 1 2 ( μ 2 - ν 2 ) (a) Sketch the curves oF constant μ and constant ν in the xy plane. IF we parametricly plot x and y with respect to ν while holding μ constant, we get the red curves; iF we plot with respect to μ holding ν constant, we get the curves shown in blue. Only curves with ν 0 are shown and the entire xy plane is covered. ±igure 1: Lines oF constant μ and ν in the xy plane. (b) TransForm the line element dS 2 = dx 2 + dy 2 into ( ) coordinates. 1

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We have dx = ν dμ + μdν dy = μdμ - ν dν so dS 2 =( ν dμ + ) 2 +( - ν dν ) 2 μ 2 + ν 2 )[ 2 + 2 ] (c) Do the curves of constant μ and ν intersect at right angles? Yes. There are two ways to show this: 1. Complex analysis. DeFne complex coordinates z = x + iy and w = μ - so the coordinate transformation can be written as an analytic function, z = i 2 w 2 . Analytic functions are conformal mappings of the complex plane into itself, preserving the angles of intersecting curves! Curves of constant μ and ν intersect at right angles in the μν plane, so they intersect at right angles in the xy plane as well. 2. Explicit calculation. Tangent vectors to curves of constant μ and ν in the xy plane are given by ! T μ = ± ∂x ∂ν ∂y ² = ³ μ - ν ´ , ! T ν = ± ∂μ ² = ³ ν μ ´ Then ! T μ · ! T ν = μν - νμ = 0, so curves of constant μ must intersect curves of constant ν at right angles. (d) ±ind the equation of a circle of constant radius r centered at the origin. The circle satisFes r = ( x 2 + y 2 ) 1 / 2 = µ μ 2 ν 2 + 1 4 ( μ 4 - 2 μ 2 ν 2 + ν 2 1 / 2 = 1 2 ( μ 2 + ν 2 ) (e) Calculate the ratio of the circumference to the radius of the circle using ( μ,ν ) coordinates. Do you get the right answer? We can Fnd the circumference by integrating the line element dS around the circle, C = · ¸ ( μ 2 + ν 2 )( 2 + 2 ) 2
Using the equation from part (d) on the second line, and the fact that the ν 0 half-plane covers the entire xy plane, ν = ± 2 r - μ 2 , = μ ± 2 r - μ 2 Thus C = 2 r ² 2 R - 2 R ± 1 - μ 2 / 2 r =2 r ² 1 - 1 ± 1 - ξ 2 πr where the integral is identical to (2.11) from p. 22 of the book. Thus, C/r π , as desired.

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PHYSICS 131 hw-1-soln - HOMEWORK SET 1 PHYS 131 Solutions...

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