Harald Kirsch

genug Unfug.

2013-08-17

Mercator Map (I.5.3, p. 79)*

Contents

Exercise

Given the coordinate transformation \begin{align} \label{trafo} x &= \frac{W}{2\pi}\phi\\ y &= - \frac{W}{2\pi} \log\tan(\theta/2), \label{eqy} \end{align} we are asked to prove that $$ ds^2 = \Omega^2(x,y)(dx^2+dy^2)$$ and to find out $\Omega$. Further we know how $ds$ looks like in terms of $\theta$ and $\phi$: $$ds^2=d\theta^2 + \sin^2(\theta) d\phi^2$$

Solution

We will solve for $d\theta$ and $d\phi$ and insert the result into this last equation. The easy one is $$d\phi=\frac{2\pi}{W}dx.$$ Now we differentiate the tangent first as the quotient of sine and cosine. \begin{align} \frac{d}{dq}\tan(q) &= \frac{d}{dq}\frac{\sin(q)}{\cos(q)}\\ &= \frac{\cos^2(q) + \sin^2(q)}{\cos^2(q)}\\ &= \frac{1}{\cos^2(q)} \end{align} With the logarithm in front, the whole thing looks like \begin{align} \frac{d}{dq} \log(\tan(q)) &= \frac{1}{\tan(q)}\frac{1}{\cos^2(q)}\\ &= \frac{\cos(q)}{\sin(q)}\frac{1}{\cos^2(q)}\\ &= \frac{1}{\sin(q)\cos(q)} \end{align} But we actually have $q=\theta/2$, which requires a factor $1/2$ that we shall not forget when we now differentiate the coordinate transform for $y$ of equation \ref{eqy}: \begin{align} dy &=-\frac{W}{2\pi}\frac{1}{\sin(\theta/2)\cos(\theta/2)}\frac{1}{2} d\theta \end{align} Looking up the expression involving the half angle in a collection of formulas, we find $$ 2 \sin(\theta/2)\cos(\theta/2) = \sin(\theta).$$ Inserting into the last equation and solving for $d\theta$ resuls in $$d\theta = -\frac{2\pi}{W}\sin(\theta) dy.$$ Plugging all this into the given equation for $ds^2$ gets us: \begin{align} ds^2 &=d\theta^2 + \sin^2(\theta) d\phi^2\\ &= (\frac{2\pi}{W}\sin(\theta) dy)^2 + \sin(\theta)^2 (\frac{2\pi}{W}dx)^2\\ &= (\frac{2\pi}{W}\sin(\theta))^2 (dx^2 + dy^2). \end{align} From the last line we see that $\Omega = \frac{2\pi}{W}\sin(\theta)$, so we are nearly there, except that we need to express the $\theta$ in terms of $x$ and $y$.

So far we found $\Omega = \frac{2\pi}{W} \sin(\theta)$ and we also know that \begin{equation} y=-\frac{W}{2\pi} \log\tan(\theta/2)\,. \label{eqy2} \end{equation} With the shortcut $b=\frac{2\pi}{W}$, \ref{eqy2} can be rearranged into $$\theta = 2 \arctan(e^{-y b}).$$ Interestingly, when I ask Wolframalpha for the sine of this, I get $$\sin(\theta)=\frac{2 e^{-y b}}{e^{-2y b}+1}.$$ For now I believe it, so that we end up with $$\Omega(x,y)=b \frac{2 e^{-y b}}{e^{-2y b}+1} $$ with $b$ as above. Why Zee keeps the dependency on $x$, I don't know. Too bad this does not look like the rather terse solution $$\Omega = \frac{b}{\cosh(b y)}$$ given in the solution section of the book. But lets insert the definition $\cosh(x) = 1/2(e^x+e^{-x})$ to get \begin{align} \Omega &= \frac{2b}{e^{by}+e^{-by}}\\ &=\frac{2b}{e^{by}+e^{-by}} \frac{e^{-by}}{e^{-by}}\\ &=\frac{2b e^{-by}}{e^{-2by}+1}, \end{align} which is indeed the solution we got. (sigh)