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Metric for the construction of Mercator map

  1. Feb 3, 2016 #1
    1. The problem statement, all variables and given/known data
    The familiar Mercator map of the world is obtained by transforming spherical coordinates θ , ϕ to coordinates x , y given by
    ##x = \frac{W}{2π} φ,
    y = -\frac{W}{2π} log (tan (\frac{Θ}{2}))##

    Show that ##ds^2 = Ω^2(x,y) (dx^2 + dy^2)## and find ##Ω##

    2. Relevant equations
    ##g'_{ρσ} = g_{μν} (\frac{∂x^μ}{∂x'^ρ}) (\frac{∂x^ν}{∂x'^σ})##

    In spherical coordinates ##ds^2 = dΘ^2 + \sin^2(Θ)dφ^2## with ##r=1##

    3. The attempt at a solution
    So in this case, I have to find ##g_{xx}## and ##g_{yy}##, they should be the same since ##Ω^2(x,y)## is factored out.

    By isolating ##Θ## and ##φ##, I got

    ##Θ=2\arctan(e^\frac{-2πy}{W})## and ##φ = \frac{2πx}{W}##

    ##g_{xx} = g_{ΘΘ} \frac{∂Θ}{∂x} + g_{φφ} \frac{∂φ}{∂x} = \frac{2π}{W}\sin^2(2\arctan(e^\frac{-2πy}{W}))##

    ##g_{yy} = g_{ΘΘ} \frac{∂Θ}{∂y} + g_{φφ} \frac{∂φ}{∂y}##

    For this part, ##y = -\frac{W}{2π} log (tan (\frac{Θ}{2}))##, manipulate so that ##tan (\frac{Θ}{2}) = e^\frac{-2πy}{W}##, then differentiate with respect to

    y so that ##\frac{∂Θ}{∂y} = \frac{-4πe^\frac{-2πy}{W}}{Wsec^2(\frac{Θ}{2})}##

    After some manipulation I ended up

    ##g_{yy} = g_{ΘΘ} \frac{∂Θ}{∂y} + g_{φφ} \frac{∂φ}{∂y} = \frac{-2π}{Wcosh(\frac{2πy}{W})}##

    According to the book I'm reading the answer should be

    ##Ω = \frac{2π}{Wcosh(\frac{2πy}{W})}##

    Can anyone help me find out what is wrong?
     
  2. jcsd
  3. Feb 4, 2016 #2

    George Jones

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    Are you sure that this is correct?
     
  4. Feb 5, 2016 #3
    ##g_{xx} = g_{ΘΘ} (\frac{∂Θ}{∂x})^2 + g_{φφ} (\frac{∂φ}{∂x})^2##
    ##g_{yy} = g_{ΘΘ} (\frac{∂Θ}{∂y})^2 + g_{φφ} (\frac{∂φ}{∂y})^2##

    Forgot the squared!!!!!!!!!!! How dumb. Thanks!!!
     
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