1. 
   

   $\parbox{2in}{ We want the area (as measured in the surface) of the p... ...theta \le \alpha$. Using the infinitesimal surface area element, \vspace{1in} }$
   
   @25<@@25 setsize @25<@@25 setsizeSetFigFont>@ pt@@ pt xxxxxxsplain
   
   

   $$A = \int_0^{\alpha} \ R\, d\theta\ \int_0^{2\pi} \ R\,\sin\theta\... ...^2\ \int_0^{\alpha} \ d\theta\ \sin\theta = 2\pi \, R^2\ (\ 1 - \cos\alpha\ ).$$

   Noting that $\alpha = a/R$, and Taylor expanding in $\alpha$ gives

   $$A = 2\pi \, R^2\ \left(\ \frac{\alpha^2}{2} - \frac{\alpha^4}{4!}... ...ha^6)\ \right) = \pi \, a^2\ \left(\ 1 - \frac{a^2}{12R^2} + O(a^4)\ \right)\ .$$
   Thus
   $$\frac{\pi\, a^4}{12R^2} = \pi \, a^2 - A + O(a^6)$$
   which implies
   $$K = \frac{1}{R^2} = (12/\pi)\ \ \frac{\pi \, a^2 - A + O(a^6) }{a^4} \ .$$
   If we take the limit as $a$ goes to zero the terms of order $a^6/a^4$ vanish and we get the desired identity:
   $$K = (12/\pi)\ \lim_{a\rightarrow 0} \ \frac{\pi \, a^2 - A}{a^4} \ \ .$$

2. 1. Yes, Euclid's 5th postulate is true, for 2-dimensional geometry on the cylinder.

   2. To show this, we must show that for any type of geodesic $L$, and point $P$ not on $L$, there is exactly one geodesic $L'$ through $P$ that never intersects $L$. There are three cases for $L$:
      1. $L$ is a vertical line, which we parametrize by $\theta = \theta_L = 0$. $P$ is the point $(z_p, \theta_p \ne 0)$ not on $L$. Geodesics $L'$ through $P$ include

         $\parbox{4in}{\vspace{-1.5in} \begin{enumerate}\item Helices $\theta ... ...P$\ that never intersects the vertical line $L$, the vertical line $L'$. \qed }$

      2. $L$ is a circle, which we parametrize by $z = z_L = 0$. $P$ is the point $(z_p \ne 0, \theta_p)$ not on $L$. Geodesics $L'$ through $P$ include

         $\parbox{4in}{\vspace{-1.5in} \begin{enumerate}\item Helices $z = m\,... ...esic through $P$\ that never intersects the circle $L$, the circle $L'$. \qed }$

      3. $L$ is the helix $\theta = nz + \theta_L \equiv nz$, for nonzero $n$. $P$ is the point $(z_p, \theta_p),$ where $\theta_p \ne nz_p$ (mod $2\pi$ ) because $P$ is not on $L$. Geodesics $L'$ through $P$ include

         $\parbox{4in}{\vspace{-2.5in} \begin{enumerate}\par \item The line $\... ... intersects the helix $L$, the helix $L'$\ with the same slope $n' = n$. \qed }$

      (Note that I presented each case fully for clarity, but didn't expect you to be so elaborate or redundant.)

   3. $K = 0$, since Euclid's 5th postulate is obeyed. This makes sense because $K^{-1}$ is the product of the two radii of curvature, defined as the radii of the largest and smallest circles which locally approximate any curve through a point $P$ on the surface. Here those radii of curvature are the cylinder radius $R$, which approximates the horizontal circular geodesic , and infinity, as straight vertical lines on the cylinder are approximated by circles of infinite radius.

3. 1. We must calculate the Christoffel symbols
      $$\Gamma^i_{\ \ jk} = \frac{1}{2}\ g^{il}\ \left(\ \partial_j\, g_{lk} + \partial_k\, g_{lj} - \partial_l\, g_{jk} \ \right)$$
      for the metric
      $$g_{ij} = \left(\ \begin{array}{cc} 1&\\ &r^2 \end{array}\ \right)\ \ ,$$
      which is diagonal, and has only one nontrivial derivative, $\partial_r\, g_{\theta\theta} = 2r$. Note that the inverse metric is just
      $$g^{ij} = \left(\ \begin{array}{cc} 1&\\ &r^{-2} \end{array}\ \right)\ \ .$$

      First consider $\Gamma^r_{\ \ jk}$. $g^{rl}$ is nonzero only when $l= r$, giving for the sum over $l$ only the contribution from $l= r$:

      $$\Gamma^r_{\ \ jk} = \frac{1}{2}\ \left(\ \partial_j\, g_{rk} + \partial_k\, g_{rj} - \partial_r\, g_{jk} \ \right)\ ,$$
      where we have used the fact that $g^{rr} = 1$. There is no component $g_{rk}$ of the metric with nontrivial coordinate derivatives, so the first 2 terms always vanish. The last term vanishes unless $j=k= \theta$, for which we get the only nonzero Christoffel symbol of the form $\Gamma^r_{\ \ jk}$,
      $$\Gamma^r_{\ \ \theta\theta} = -\ \frac{1}{2}\ \partial_r\, g_{\theta\theta} = -\ r\ \ .$$

      We then consider $\Gamma^\theta_{\ \ jk}$. $g^{\theta l}$ is nonzero only when $l= \theta$, giving for the sum over $l$ only the contribution from $l= \theta$:

      $$\Gamma^\theta_{\ \ jk} = \frac{1}{2\, r^2}\ \left(\ \partial_j\, g_{\theta k} + \partial_k\, g_{\theta j} - \partial_\theta\, g_{jk} \ \right)\ ,$$
      where we have used the fact that $g^{\theta\theta} = r^{-2}$. There is no component $g_{jk}$ of the metric with nontrivial $\theta$-derivative, so the last term always vanishes. The first term vanishes unless $j=r$ and $k= \theta$; the second unless $j=\theta$ and $k=r$. Thus we get the only nonzero Christoffel symbols of the form $\Gamma^\theta_{\ \ jk}$ as being
      $$\Gamma^\theta_{\ \ r\theta} =\Gamma^\theta_{\ \ \theta r} = \frac{1}{2\, r^2}\ \partial_r\, g_{\theta\theta} = \frac{1}{r}\ \ .$$

      Summarizing, we get the Christoffel symbols

      $$\Gamma^r_{\ \ \theta\theta} = -\ r \hspace{1in} \Gamma^\theta_{\ \ r\theta} =\Gamma^\theta_{\ \ \theta r} = \frac{1}{r}\ \ ,$$
      with all remaining components zero.

   2. First, note that $R_{ijkl}$ is antisymmetric under the exchanges $i \leftrightarrow j$ and $k \leftrightarrow l$. Thus
      $$R_{rrkl} = R_{\theta\theta kl} = R_{ijrr} = R_{ij\theta\theta} = 0\ \ ,$$
      with all nonzero $R_{ijkl}$ determined by symmetry from $R_{r\theta r\theta}$, the lowered form of
      $$R_{r\theta r}^{\ \ \ \ \theta} = \frac{\partial\ }{\partial \thet... ...ma^\theta_{\ \ m\theta} - \Gamma^m_{\ \ \theta r}\ \Gamma^\theta_{\ \ mr} \ \ ,$$
      plugging in $i = k = r$ and $j = l = \theta$ to our expression for $R_{ijk}^{\ \ \ \ l}$. Term 1 vanishes, as $\Gamma^\theta_{\ \ \theta r}$ is independent of $\theta$. Term 3 vanishes, as $\Gamma^m_{\ \ rr} = 0$ for all $m$. Term 4 is nonzero only for $m = \theta$. Thus
      $$R_{r\theta r}^{\ \ \ \ \theta} = - \ \frac{\partial\ }{\partial ... ...heta_{\ \ \theta r} \ \ =\ \ 1/r^{2} - \left(\ 1/r\ \right)^2 \ \ =\ \ 0 \ \ .$$
      All $R_{ijkl}$ related to $R_{r\theta r}^{\ \ \ \ \theta}$ by symmetry are thus also zero, giving $R_{ijkl} = 0$ for all values of $ijkl$.

4. 1. We use the coordinate transformation
      $$r = \sqrt{x^2 + y^2} \hspace{1.5in} \theta = \tan^{-1} \ \frac{y}{x}$$
      to find the Jacobian matrix
      $$J^{\ i}_{\ \ \ j} \equiv \frac{\partial{\tilde{x}^i}}{\partia... ...c{\partial{\theta}}{\partial{y}} \end{array}\ \ \right)\ \ \ ,$$
      remembering that a partial derivative with respect to $x$ involves holding its orthogonal variable $y$ constant (which is why we need the functions $r(x,y), \theta(x,y)$.) Evaluating partial derivatives gives
      $$J^{\ i}_{\ \ \ j} = \quad \left(\ \ \begin{array}{cc} \frac{... ...theta}{r} & \frac{\cos\theta}{r} \end{array}\ \ \right)\ \ \ .$$

   2. For the inverse coordinate transformation
      $$x = r\cos\theta \hspace{1.5in} y = r\sin\theta\ \ .$$
      we want the Jacobian matrix
      $$\tilde{J}^{\ i}_{\ \ \ j} \equiv\quad \frac{\partial{x^i}}{\... ...c{\partial{y}}{\partial{\theta}} \end{array}\ \ \right)\ \ \ .$$
      Evaluating partial derivatives gives
      $$\tilde{J}^{\ i}_{\ \ \ j} = \quad \left(\ \ \begin{array}{cc... ...4pt] \sin\theta & r\, \cos\theta \end{array}\ \ \right)\ \ \ .$$

      Note that $\vec{\tilde{J}} =\vec{J^{-1}}$, since

      $$\vec{\tilde{J}} \ \vec{J} = \left(\ \ \begin{array}{cc} \cos... ...\ \begin{array}{cc} 1& 0\\ 0&1 \end{array}\ \ \right)\ \ \ .$$

   3. Under our coordinate transformation,
      $$\left(\ \begin{array}{c} v^x\\ v^y \end{array}\ \right) \qu... ...}\ \left(\ \begin{array}{c} v^x\\ v^y \end{array}\ \right)\ .$$
      Using our calculated Jacobian matrix $\vec{J}$ from part (a),
      $$\left(\ \begin{array}{c} v^r\\ v^\theta \end{array}\ \right)... ...-\sin\theta\ v^x + \cos\theta\ v^y \ )\end{array}\ \right)\ \ .$$
      Note that $v_r$ is the usual radial component of a vector; $v_\theta$ is $1/r$ times the usual angular component, because we are using a metric which has explicit factors of $r$ in converting angular components to length via the dot product (instead of the usual approach normalizing ${\boldsymbol {\hat{\theta}}}$ with a trivial metric).

   4. 1. The question showed that the matrix multiplication
         $$\left(\ \begin{array}{c} \partial_r \\ \partial_\theta \end{arra... ...)^T \ \left(\ \begin{array}{c} \partial_x \\ \partial_y \end{array}\ \right)$$
         implements the chain rule,
         $$\partial_r = \frac{\partial{x}}{\partial{r}} \ \partial_x + \frac{\partial{y}}{\partial{r}} \ \partial_y\ ,$$
         etcetera. Performing this multiplication gives (using the result for $\vec{J^{-1}}$ from part (b))
         $$\left(\ \begin{array}{c} \partial_r \\ \partial_\theta \end{... ...partial_x + \cos\theta\ \partial_y\ ) \end{array}\ \right).$$
         Looking up the partial derivatives $\frac{\partial{x}}{\partial{r}}, \frac{\partial{y}}{\partial{r}},$ etc from part (b) confirms that our result does indeed agree with the chain rule.

      2. By a similar argument to that posed in the question,
         $$\frac{\partial{\ }}{\partial{{x}^j}} = \frac{\partial{\tilde{x}^i... ...\left(\ \vec{{J}}\ \right)^T \ \frac{\partial{\ }}{\partial{\tilde{x}^i}} \ \ ,$$
         if we think of $\frac{\partial{\ \ }}{\partial{\tilde{x}^i}}$ as a column vector. Performing this multiplication gives (using the result for $\vec{J}$ from part (a))
         $$\left(\ \begin{array}{c} \partial_x \\ \partial_y \end{array... ...rac{\cos\theta}{r}\ \partial_\theta\ \end{array}\ \right).$$
         Again, looking up the partial derivatives $\frac{\partial{r}}{\partial{x}}, \frac{\partial{\theta}}{\partial{x}},$ etc from part (a) confirms that our result does indeed agree with the chain rule.

   5. In Cartesian coordinates,
      $$\partial_i\ v^i = \partial_x\ v^x + \partial_y\ v^y\ \ .$$
      In polar coordinates,
      

      $$\partial_i\ v^i = \partial_r\ v^r + \partial_\theta\ v^\theta$$

      $$= \partial_r\ (\ \cos\theta\ v^x + \sin\theta\ v^y\ ) + \partial_\theta\ \left(\ \frac{1}{r} \ ( -\sin\theta\ v^x + \cos\theta\ v^y )\ \right)\ \ ,$$

      
      
      using our result for $v_r, v_\theta$ from (c). Performing the derivatives gives
      

      $$\partial_i\ v^i = (\ \cos\theta\ (\ \partial_r\ v^x\ ) + \sin\theta\ (\ \partial_r\ v^y\ )\ ) + \frac{1}{r} \ ( -\cos\theta\ v^x - \sin\theta\ v^y )$$

      $$\quad \quad + \frac{1}{r} \ ( -\sin\theta\ (\ \partial_\theta\ v^x\ ) + \cos\theta\ (\ \partial_\theta\ v^y \ )\ )\$$

      $$= (\ \cos\theta\ (\ \partial_r\ v^x\ ) - \ \frac{\sin\theta}{r}\ (\... ...ial_\theta v^y\ )\ ) - \frac{1}{r} \ ( \cos\theta\ v^x + \sin\theta\ v^y )\ \ ,$$

      
      
      after regrouping terms. From part (d-ii), we recognize the derivative operators as
      

      $$\partial_i\ v^i = \partial_x\ v^x\ + \partial_y\ v^y - \frac{1}{r} \ ( \cos\theta\ v^x + \sin\theta\ v^y )$$

      $$= \partial_x\ v^x\ + \partial_y\ v^y - \frac{1}{r} \ v^r \ \ ,$$

      
      
      again using our result for $v_r$ from (c).

   6. In polar coordinates,
      

      $${\boldsymbol {\nabla}}_i\ v^i = \partial_i \ v^i + \Gamma^i_{\ \ ij}\ v^j$$

      $$= \partial_r \ v^r + \partial_\theta \ v^\theta + \Gamma^r_{\ \ rj}\ v^j + \Gamma^\theta_{\ \ \theta j}\ v^j \ \ .$$

      
      
      $\Gamma^r_{\ \ rj}$ vanishes for all $j$, while $\Gamma^\theta_{\ \ \theta j}$ contributes only when $j=r$. Thus we get
      $${\boldsymbol {\nabla}}_i\ v^i = \partial_r \ v^r + \partial_\theta \ v^\theta + \frac{1}{r}\ v^r \ \ ,$$
      plugging in $\Gamma^\theta_{\ \ \theta r} = 1/r$ from problem 1(b). From our answer to 2(e), this just gives
      $${\boldsymbol {\nabla}}_i\ v^i = \partial_x\ v^x\ + \partial_y\ v^y \ \ ,$$
      which is what we obtain for ${\boldsymbol {\nabla}}_i\ v^i$ in Cartesian coordinates, where all Christoffel symbols vanish.

5. 1. See attached mathematica printout.
   2. See attached mathematica printout.

   3. By inspection, we have nonzero Riemann tensor entries
      $$\begin{array}{rcrcccrcrcc} R_{r\theta r}^{\ \ \ \ \theta} &=&... ...\phi}^{\ \ \ \ \theta} &=& {kr^2\sin^2 \theta}\ \ , \end{array}$$
      where we have always grouped $R_{ijk}^{\ \ \ \ l} = -R_{jik}^{\ \ \ \ l}$ together to show the antisymmetry.

   4. To find the Ricci tensor, we must perform the contraction $R_{ik} = R_{ilk}^{\ \ \ \ l}$. Note for the Riemann tensor entries above, the entry $R_{ijk}^{\ \ \ \ l}$ is nonzero only if the set of indices ${ij}$ equals the set ${kl}$. In our contraction we further require $j=l$, so we get nonzero Ricci tensor entries only when $i=k$. The possibilities are
      

      $$R_{rr} \ \ =\ \ R_{r\theta r}^{\ \ \ \ \theta} + R_{r\phi r}^{\ \ \ \ \phi} = \frac{2k}{1-kr^2}$$

      $$R_{\theta\theta} \ \ =\ \ R_{\theta r\theta}^{\ \ \ \ r} + R_{\theta\phi \theta}^{\ \ \ \ \phi} = {2kr^2}$$

      $$R_{\phi\phi} \ \ =\ \ R_{\phi \theta\phi}^{\ \ \ \ \theta} + R_{\phi r\phi}^{\ \ \ \ r} = {2kr^2\sin^2 \theta}\ \ .$$

      
      

   5. The Ricci tensor ${\cal R}$ is given by
      $${\cal R} \ \ =\ \ R_{i}^{\ i}\ \ =\ \ g^{ij} R_{ij} \ \ =\ \ \sum_i\ g^{ii}R_{ii}\ \ ,$$
      the last step because the metric $g_{ij}$ is diagonal, nonzero only if $i=j$. Performing the sum,
      

      $${\cal R} = g^{rr}R_{rr} + g^{\theta\theta}R_{\theta\theta} + g^{\phi\phi}R_{\phi\phi}$$

      $$= \left(\ \frac{R^2}{1-kr^2}\ \right)^{-1}\ \ \frac{2k}{1-kr^2} \ +... ...R^2r^2\sin^2\theta} \ \ (\ {2kr^2\sin^2 \theta}\ ) \ \ =\ \ \frac{6k}{R^2}\ \ .$$

      
      

   6. Note that our metric $g_{ij}$ is diagonal, with entries
      

      $$g_{rr} = \frac{R^2}{1-kr^2}$$

      $$g_{\theta\theta} = R^2 \, r^2$$

      $$g_{\phi\phi} = R^2\, r^2\, \sin^2\theta \ \ \ .$$

      
      
      By inspection from part (d), this is a space of constant curvature, with $R_{ik} = (2k/R^2)\ g_{ij}$.

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