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Orthogonality of a curvilinear coordinate system

  1. Aug 7, 2015 #1
    1. The problem statement, all variables and given/known data

    Show that the uvw-system is orthogonal.

    [itex] r, \theta, \varphi[/itex] are spherical coordinates.

    $$u=r(1-\cos\theta)$$
    $$v=r(1+\cos\theta)$$
    $$w=\varphi$$

    3. The attempt at a solution

    So basically I want to show that the scalar products between [itex]\frac{\partial \vec{r}}{\partial u} [/itex] [itex] \frac{\partial \vec{r}}{\partial v} [/itex] [itex] \frac{\partial \vec{r}}{\partial w}[/itex] amount to zero.

    Which means that I can't avoid finding [itex] \vec{r}[/itex]. However, as transforming all the way to cartesian coordinates seems to be a minor nightmare I hope to show that uvw is orthogonal in spherical space, which (?? got no proof) implies it's orthogonal also in cartesian space.

    I call the position vector in spherical space [itex]\vec{q}[/itex], so I want to show that these ones are orthogonal [itex]\frac{\partial \vec{q}}{\partial u} [/itex] [itex] \frac{\partial \vec{q}}{\partial v} [/itex] [itex] \frac{\partial \vec{q}}{\partial w}[/itex]

    Which means I need uvw in [itex]q=q(r, \theta, \varphi)[/itex].

    I find that (by adding or subtracting the expressions for u and v given in the problem statement)

    $$r=\frac{u+v}{2}$$
    $$\theta=arccos(\frac{u-v}{2})$$
    $$\varphi = w$$

    Proceeding by determining the tangent vectors I get

    $$\frac{\partial \vec{q}}{\partial u} = \frac{1}{2} ( 1 \hat{r} - (1-(\frac{u-v}{2})^2)^{-1/2} \hat{\theta})$$
    $$\frac{\partial \vec{q}}{\partial v} = \frac{1}{2} ( 1 \hat{r} + (1-(\frac{u-v}{2})^2)^{-1/2} \hat{\theta})$$
    $$\frac{\partial \vec{q}}{\partial w} = 1 \hat{\varphi}$$

    The last vector is clearly orthogonal to the other ones, but the first two aren't orthogonal as far as I can tell, which means there's something I'm missing. Maybe I got the idea wrong, but at the moment I can't put my finger on it.

    Any thoughts?
     
    Last edited: Aug 7, 2015
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  3. Aug 7, 2015 #2

    Orodruin

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    Yes, you can easily avoid finding ##\vec r##. Have you tried using the chain rule and what you already know about spherical coordinates?
     
  4. Aug 7, 2015 #3
    I'll think about what you said, but I'm under the impression that I use the chain rule above. There must be something wrong with my reasoning there, and that's what bothers me.

    Doing a [itex]\frac{\partial \vec{r}}{\partial u} = \frac{\partial \vec{r}}{\partial x} \frac{\partial x}{\partial u } + \frac{\partial \vec{r}}{\partial y} \frac{\partial y}{\partial u } + \frac{\partial \vec{r}}{\partial z} \frac{\partial z}{\partial u } [/itex] (for each) is doable but seems messy, perhaps even if I expand [itex]\frac{\partial x}{\partial u} = \frac{\partial x}{\partial r} \frac{\partial r}{\partial u} + ... [/itex] or is that the idea?
     
  5. Aug 7, 2015 #4

    Orodruin

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    That would be the general idea yes. The easiest way would be if you are familiar with how to express the metric in different coordinate systems and how to transform it between coordinate systems. In that case you only need to look at the metric transformation to the new coordinate system.
     
  6. Aug 7, 2015 #5

    Ray Vickson

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    Since there are two conventions about spherical coordinates, which one are you using? Is ##\theta## the latitude (measured down from the north pole) and ##\varphi## the longitude (common in physics), or is ##\theta## the longitude and ##\varphi## the latitude down from N. pole (quite common in mathematics)? See, eg., https://en.wikipedia.org/wiki/Spherical_coordinate_system .
     
  7. Aug 8, 2015 #6
    Yes, it can be confusing, here theta is meant to be the latitude as measured from the +z axis while phi denotes the longitude.

    Expressing xyz in uvw is no problem. I tried the suggested chain rule approach but can't seem to avoid ending up with pages of computations. If there is an easier way I can't see it falling out mathematically from the above momentarily. My initial attempt in the first post is an attempt to avoid this, but seems flawed - for a reason I can't detect. Finding what's wrong there would lift a lot of my confusion around this problem.
     
  8. Aug 8, 2015 #7

    Orodruin

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    Did you try transforming the metric tensor?
     
  9. Aug 10, 2015 #8
    I'm not accustomed to the tensor method, it's described in full in a later chapter (however, using scale factors and sums should be equivalent at this stage anyway) but I tried the following


    [tex] g_{uv}^{'} = g_{r r} \frac{\partial r}{\partial u} \frac{\partial r}{\partial v} + g_{\theta \theta} \frac{\partial \theta}{\partial u} \frac{\partial \theta}{\partial v} + g_{\varphi \varphi} \frac{\partial \varphi}{\partial u} \frac{\partial \varphi}{\partial v} \stackrel{?}{=} 0 [/tex]

    As the spherical coordinate system is orthogonal [itex]g_{ij}=0[/itex] when [itex] i \neq j [/itex].

    This computation is somewhat easier, but does not return the wanted result. The last g-term vanishes, but the other two don't form a zero (using [itex]g_{\theta \theta} = r^2[/itex] ).

    Edit: Nevermind, I think I see the error now. The expression for theta is wrong. That's what you get for not drinking coffee for a week.

    It should be
    $$ \theta = arccos( \frac{v-u}{u+v} )$$
     
    Last edited: Aug 10, 2015
  10. Aug 10, 2015 #9

    Orodruin

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    I just noticed your expression for ##\theta## in terms of u and v is not correct. Subtracting the expressions for u and v gives ##2r\cos\theta = u -v##, not ##2\cos\theta = u-v##...
     
  11. Aug 10, 2015 #10
    Now it all works out nicely. Thanks for your replies, they helped and I learnt something.

    The theta was the first mistake I made, it's more of a "book-keeping" issue, while the most important error seems to have been to not include the metric. I must be doing some illegal operation when computing the scalar product of the tangent vectors there (or when I compute the tangent vectors themselves), because the scale factors don't seem to appear by (mindlessly) just doing the maths. I'll have to review that.
     
  12. Sep 8, 2015 #11
    I finally had the opportunity to revisit this problem, thinking about if orthogonality expressed in spherical coordinates implies orthogonality in cartesian such. It seemed intuitive, but it doesn't seem to be the case judging from the problem above?

    [itex]\vec{o_i} \cdot \vec{o_j} = \frac{1}{k_i} \frac{1}{k_j} \frac{\partial \vec{q}}{\partial u_i} \frac{\partial \vec{q}}{\partial u_j} = 0 \stackrel{?}{\implies} \vec{e_i} \cdot \vec{e_j} = \frac{1}{h_i} \frac{1}{h_j} \frac{\partial \vec{r}}{\partial u_i} \frac{\partial \vec{r}}{\partial u_j} = 0[/itex]

    or

    [itex]\vec{o_i} \cdot \vec{o_j} = \frac{\partial \vec{q}}{\partial u_i} \frac{\partial \vec{q}}{\partial u_j} = 0
    \stackrel{?}{\implies} \vec{e_i} \cdot \vec{e_j} = \frac{\partial \vec{r}}{\partial u_i} \frac{\partial \vec{r}}{\partial u_j} = 0[/itex]

    where [itex]\vec{o_i}[/itex] is a base vector expressed in spherical coordinates.
     
  13. Sep 10, 2015 #12
    I also calculated the scale factors and nabla expressed in u,v,w , which was pretty straightforward. However, determining the position vector turned out to be a mess, but it could be that I use my naive book-keeping approach to compute it.

    I start off by expressing the position vector in cartesian coordinates, then I compute the position vector in spherical coordinates by inverting the transformation matrix using gaussian elimination and arrive at [itex]\vec{r} = r \hat{r}[/itex]. From that point I use the nabla computed earlier and apply it like this to get [itex] \hat{r} = \frac{\nabla r}{\mid \nabla r \mid}[/itex] in u, v and w, which gives me the right answer combined with the expression for the spherical position vector.

    However, is there an easier way? Doing the gaussian elimination is a major time sink.
     
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