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Physics LibreTexts

S01. Euclidean Geometry - SOLUTIONS

  • Page ID
    5056
  • Exercise 1.1.1

    Answer

    Solution not available yet.

    Exercise 1.2.1

    Answer

    \[\begin{equation*}
      \begin{aligned}
            dx' &= \cos\theta dx - \sin\theta dy \\ dy' &= \sin\theta dx + \cos\theta dy \\ dz' &= dz
      \end{aligned}
    \end{equation*}\]

    \[\begin{equation*}
      \begin{aligned}
            (dx')^2 &= \cos^2\theta dx^2 - 2\cos\theta\sin\theta dxdy + \sin^2\theta dy^2 \\ (dy')^2 &= \sin^2\theta dx^2 + 2\cos\theta\sin\theta dxdy + \cos^2\theta dy^2 \\ (dz')^2 &= dz^2
      \end{aligned}
    \end{equation*}\]

    \[\begin{equation*}
      \begin{aligned}
            (d\ell')^2 &= dx^2 + dy^2 + dz^2 \\ d\ell' &= d\ell
      \end{aligned}
    \end{equation*}\]

    Exercise 1.3.1

    Answer

    \[\begin{equation*}
      \begin{aligned}
            dx &= \cos\theta dr - r\sin\theta d\theta \\ dy &= \sin\theta dr + r\cos\theta d\theta
      \end{aligned}
    \end{equation*}\]

    \[\begin{equation*}
      \begin{aligned}
            dx^2 &= \cos^2\theta dr^2 - 2r\cos\theta\sin\theta drd\theta + r^2sin^2\theta d\theta^2 \\ dy^2 &= \sin^2\theta dr^2 + 2r\cos\theta\sin\theta drd\theta + r^2\cos^2\theta d\theta^2
      \end{aligned}
    \end{equation*}\]

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 &= dx^2 + dy^2 \\ d\ell^2 &= dr^2 + r^2d\theta^2
      \end{aligned}
    \end{equation*}\]

    Exercise 1.4.1

    Answer

    Using \(d\ell^2 = dx^2 + dy^2 + dz^2\) we get

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 = r_1^2sin^2\lambda d\lambda^2 + r_1^2cos^2\lambda d\lambda^2
      \end{aligned}
    \end{equation*}\]

    which simplifies to

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 = r_1^2 d\lambda^2 \quad \rightarrow \quad d\ell = r_1 d\lambda
      \end{aligned}
    \end{equation*}\]

    Now integrate both sides, with \(\lambda\) from \(0\) to \(2\pi\) (one period)

    \[\begin{equation*}
      \begin{aligned}
            \ell = \int_{0}^{2\pi} r_1 d\lambda \quad \rightarrow \quad \ell = 2\pi r_1
      \end{aligned}
    \end{equation*}\]

    This is the circles circumference.

    Exercise 1.4.2

    Answer

    To Cartesian coordinates to Spherical:

    \[\begin{equation*}
      \begin{aligned}
            x &= rcos\phi sin\theta \\ y &= rsin\phi sin\theta \\ z &= rcos\theta
      \end{aligned}
    \end{equation*}\]

    substituting in \(\phi = \lambda\), \(\theta = \pi/2\), and \(r = r_1\), we get

    \[\begin{equation*}
      \begin{aligned}
            x &= r_1cos\lambda \\ y &= r_1sin\lambda  \\ z &= 0
      \end{aligned}
    \end{equation*}\]

    which is indeed the same path given in Exercise 1.3.1.

    Exercise 1.4.3

    Answer

    Substituting in the above to Equation 1.7 we get

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 = dr^2 + r_1^2(d\theta^2 + sin^2(\frac{\pi}{2})d\lambda^2)
      \end{aligned}
    \end{equation*}\]

    and since \(r\) and \(\lambda\) are constants the whole thing simplifies to

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 = r_1^2 d\lambda^2 \quad \rightarrow \quad  d\ell = r_1 d\lambda
      \end{aligned}
    \end{equation*}\]

    Now integrate both sides, again with \(\lambda\) from \(0\) to \(2\pi\)

    \[\begin{equation*}
      \begin{aligned}
            \ell = \int_{0}^{2\pi} r_1 d\lambda \quad \rightarrow \quad \ell = 2\pi r_1
      \end{aligned}
    \end{equation*}\]

    This is the same result as we got in Exercise 1.3.1.

    Exercise 1.4.4

    Answer

    Substituting in the above to Equation 1.7 we get

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 = d\mu^2 + \mu^2(d\theta^2 + sin^2(\frac{\pi}{2})d\lambda^2)
      \end{aligned}
    \end{equation*}\]

    and since \(\theta\) and \(\lambda\) are fixed the whole thing simplifies to

    \[\begin{equation*}
      \begin{aligned}
            d\ell^2 = d\mu^2 \quad \rightarrow \quad  d\ell = d\mu
      \end{aligned}
    \end{equation*}\]

    It's now easy to see that  \(\int_0^{r_1} (d\ell/d\mu) d\mu\) does indeed equal \(r_1\). This is the radius.

    Exercise 1.5.1

    Answer

    Solution not available yet.