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Converting one coordinate to full Cartesian/Cylindrical/Spherical

  1. Sep 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Convert to the two other coordinate systems: θ = π/4


    2. Relevant equations
    tan = x/y
    r = √(x[itex]^{}2[/itex] + (y[itex]^{}2[/itex])


    3. The attempt at a solution
    The tangent equals one, so doesn't this make just the line x = y? But the cylinder (I'm assuming since the second part is the same exact problem but with [itex]\phi[/itex]). But that doesn't make sense because isn't this just a cylinder slice? And how do I convert it to spherical afterwards?

    I can't find any help anywhere (book or online) about being given a single coordinate.
     
  2. jcsd
  3. Sep 2, 2014 #2

    vanhees71

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    You have to tell us a bit more. Particularly you should give us the definitions of the cylinder and spherical coordinates you use. Most probably your equation for the angle in terms of cartesian coordinates of the plane is wrong. Usually [itex]\theta[/itex] is the angle of the radius vector wrt. to the positive [itex]x[/itex] axis. It takes values along an interval of length [itex]2 \pi[/itex]. Usually one chooses [itex][0,2 \pi[[/itex] or [itex]]-\pi,\pi][/itex]. In the latter case the correct formula is
    [tex]\theta=\mathrm{atan2}(y,x)=\mathrm{sign} \, x \; \arccos \left (\frac{x}{\sqrt{x^2+y^2}} \right ).[/tex]
    The formulae for the Cartesian components in polar coordinates read
    [tex]x=r \cos \theta, \quad y=r \sin \theta.[/tex]
     
  4. Sep 2, 2014 #3

    Simon Bridge

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    Since there is a ##\phi## right after, I'm guessing the angle comes from the spherical coordinate system ... is that correct?

    If "yes" then there's a problem:
    In spherical polar coordinates, ##\theta## is often the azimuthal angle (measured from the z axis), but may be the polar angle (from the x axis). We cannot tell which convention you are expected to use just from the number.

    Even if we knew which angle it was, that is not enough to specify a unique position.

    The question is trying to get you to explore what those angles mean ... find a picture, a diagram, of what the different coordinates mean and see if you can work out what the set of all points with the same angle looks like - hint: it's a surface.

    eg. If you are given just x=1 in cartesian coordinates, then the set of all points with x=1 will be the plane perpendicular to the x-axis passing through x=1. You need a similar kind of thinking. This can test your 3D sketching skills too, but it is the best way.

    Of course, the problem statement may have other useful information in it that you have not told us yet.
     
  5. Sep 2, 2014 #4
    I think the first is a cylinder and the second a sphere, but it's very very vague. For cylinders, we use:

    r[itex]^{2}[/itex]= x[itex]^{2}[/itex] + y[itex]^{2}[/itex]
    tan[itex]\theta[/itex] = y/x
    z = z

    For spheres:

    [itex]\rho[/itex][itex]^{2}[/itex] = x[itex]^{2}[/itex] + y[itex]^{2}[/itex] + z[itex]^{2}[/itex]
    tan[itex]\theta[/itex] = y/x
    [itex]\phi[/itex] = z/[itex]\rho[/itex]

    Very ambiguous and I'm not sure what they want exactly . . . I know it's a surface but how do you "convert" it to each coordinate system without just "y=x with no z bounds". That doesn't sound right, but mathematically it seems to work . . .
     
  6. Sep 2, 2014 #5

    Ray Vickson

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    You still have not answered the basic questions several people have asked you: what are the new coordinates in general? Are you dealing with a "cylindrical polar" system or a "spherical polar" system, and if the latter, what are the definitions of ##\theta## and ##\phi##?
     
  7. Sep 2, 2014 #6
    Uh, so you're asking for this?:

    (r, θ, z) - cylindrical
    ([itex]\rho[/itex], θ, [itex]\phi[/itex])

    Because that's all I have to give. You now are 100% up to speed on what I'm trying to solve and the tools I have to solve it.
     
  8. Sep 2, 2014 #7

    Ray Vickson

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    Well, I know the usual convention for what is ##\theta## and what is ##\phi##, but I have seen some writers use the opposite convention! So, the question is not stupid: what is ##\theta## and what is ##\phi## in your case?
     
  9. Sep 3, 2014 #8

    Simon Bridge

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    If ##\tan\theta = x/y## then ##\theta## is the polar angle.
    But I'm only guessing - I don't like guessing - don't make me guess.
    If you don't answer questions it is very difficult to hep you.

    For cartesian coordinates, the plane where y=x is almost there.
    Consider: what is the difference between ##\theta = \pi/4## and ##\theta = 5\pi/4##?

    The trick is to sketch out the cartesian axes and draw in the line in the x-y plane for the polar angle.
    Where does it start? Where does it stop?
     
  10. Sep 3, 2014 #9

    vanhees71

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    If you want to use the tan, which has to be used however with care, then it's
    [tex]\tan \theta=\frac{y}{x}[/tex]
    for the usual polar/cylinder coordinates, where the angle is with respect to the x-axis of the Cartesian coordinate system.

    A save formula, giving angles in the full range [itex]\theta \in ]-\pi,\pi][/itex] is
    [tex]\theta=\mathrm{sign} \, y \; \arccos \left (\frac{x}{r} \right ) \quad \text{with} \quad r=\sqrt{x^2+y^2}.[/tex]
     
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