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Metric for the construction of Mercator map

  • #1
132
4

Homework Statement


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##

Homework 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##

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?
 

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,259
790
##g_{xx} = g_{ΘΘ} \frac{∂Θ}{∂x} + g_{φφ} \frac{∂φ}{∂x}##

##g_{yy} = g_{ΘΘ} \frac{∂Θ}{∂y} + g_{φφ} \frac{∂φ}{∂y}##
Are you sure that this is correct?
 
  • #3
132
4
Are you sure that this is correct?
##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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