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Curl in spherical coords with seeming cartesian unit vector

  1. Nov 9, 2013 #1
    1. The problem statement, all variables and given/known data
    I have a problem that is the curl of jln(rsinθ)

    Since this is in spherical, why is there a bold j in the problem? Doesn't that indicate it's a unit vector in cartesian coordinates? Can I do the curl in spherical coordinates when I have a cartesian unit vector in the problem?

    Thanks.
     
  2. jcsd
  3. Nov 9, 2013 #2

    UltrafastPED

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    You will need to provide the details of the problem; perhaps an image of the actual problem.
     
  4. Nov 9, 2013 #3
    Actually, I made a mistake. It's not spherical, it's cylindrical. But I'm supposed to do this problem in cylindrical coordinates:

    ∇x jln(rsinθ)

    That's all the information I'm given.
     
  5. Nov 9, 2013 #4

    vanhees71

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    I'd do the problem in Cartesian coordinates. Just think, what [itex]r \cos theta[/itex] is!
     
  6. Nov 9, 2013 #5

    UltrafastPED

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    Just do it with constant j; it is a valid direction.
     
  7. Nov 9, 2013 #6
    You mean I can treat j as a constant instead of a unit vector?
     
  8. Nov 9, 2013 #7

    UltrafastPED

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    Yes, j is a constant in this case, but the coefficients are not.
     
  9. Nov 9, 2013 #8
    Cylindrical coordinates use (r, θ, z), so to input this into the cylindrical coordinates formula for curl, I need the function in one of those directions. Like the first part of it is ([itex]\frac{1}{r}\frac{∂V_{z}}{∂θ}-\frac{∂V_{θ}}{∂z}[/itex])r. How would I input this function I'm given into that without having it along any of the (r, θ, z) vectors? The only vector it's in the direction of is k, but since this is cylindrical coordinates, that information is apparently useless. Or is it?

    Thanks.

    edit: How about changing that j to cylindrical? Like y = rsinθ? Would that work?
     
    Last edited: Nov 9, 2013
  10. Nov 9, 2013 #9

    UltrafastPED

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    You could convert j to polar coordinates .... in that case theta = pi/2.
     
  11. Nov 9, 2013 #10
    Yeah, I meant j. So j = rsin(pi/2)? I know y = rsinθ, but y and j are different. y is a vector in the y direction, but j is a unit vector in the y direction. Does that matter?

    Thanks
     
  12. Nov 9, 2013 #11

    UltrafastPED

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    You could convert F1=j ln(r sin(θ)) to rectangular coordinate: F2= j ln(y).

    Then take the curl; then convert back to cylindrical coordinates.
     
  13. Nov 10, 2013 #12

    vanhees71

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    I think, it would be good to rethink, what cylinder coordinates are, before going on. There seem to be a lot of misconceptions in your mind.

    First of all you should clearly distinguish between vectors and components of vectors with respect to a basis. To introduce cylinder coordinates one usually starts with cartesian coordinates. The position vector is given by
    [tex]\vec{r}=x \vec{i} + y \vec{j} + z \vec{k}.[/tex]
    Here, [itex](\vec{i},\vec{j},\vec{k})[/itex] are a set of three constant unit vectors that are perpendicular to each other and build a right-handed dreibein, i.e., they fulfill
    [tex]\vec{k}=\vec{i} \times \vec{j}.[/tex]
    Then you introduce cylinder coordinates by setting
    [tex]\vec{r}=\rho \cos \theta \vec{i} + \rho \sin \theta \vec{j} + z \vec{k}.[/tex]
    I use [itex]\rho[/itex] instead of [itex]r[/itex], because usually [itex]r=|\vec{r}|=\sqrt{\rho^2+z^2}\neq \rho[/itex].

    The ranges of the coordinates are [itex]\rho>0[/itex], [itex]\theta \in [0,2\pi)[/itex], [itex]z \in \mathbb{R}[/itex]. This covers the entire space except the [itex]z[/itex] axis, where the cylindrical coordinates are singular.

    Now, you introduce also a new basis, adapted to the new coordinates. The new basis is defined by the tangent vectors to the coordinate lines. If these tangent vectors are perpendicular at any point in the domain of the new coordinates, you call the coordinates "orthogonal curvilinear coordinates" and then you use a normalized basis. Then you have again a orthonormal basis, but in general these basis vectors will not be constant.

    In the case of cylinder coordinates the new basis vectors are
    [tex]\hat{\rho}=\frac{\partial_{\rho} \vec{r}}{|\partial_{\rho} \vec{r}|}=\cos \theta \vec{i} + \sin \theta \vec{j}, \\
    \hat{\theta}=\hat{\rho}=\frac{\partial_{\theta} \vec{r}}{|\partial_{\theta} \vec{r}|}=-\sin \theta \vec{i}+\cos \theta \vec{j},\\
    \hat{z}=\frac{\partial_{z} \vec{r}}{|\partial_{z} \vec{r}|}=\vec{k}.[/tex]
    Obviously these vectors are perpendicular to each other, and also build a right-handed system at each point (except along the [itex]z[/itex] axis where the cylinder coordinates are singular).

    Now you write the vector fields in terms of the components wrt. the new basis, i.e.,
    [tex]\vec{V}=V_r \hat{r} + V_{\theta} \hat{\theta} + V_z \hat{z}.[/tex]
    For your case you need [itex]\vec{j}[/itex] in terms of the new coordinates. Since you have orthonormal coordinates, the components are easily calculated by taking the dot products with the new basis, i.e., you have
    [tex]\vec{j} \cdot \hat{r}=\sin \theta, \quad \vec{j} \cdot \hat{\theta}=\cos \theta, \quad \vec{j} \cdot \hat{z}=0.[/tex]
    Thus you have
    [tex]\vec{j}=\sin \theta \hat{r} + \cos \theta \hat{\theta},[/tex]
    and you vector field, expressed in cylinder coordinates, is
    [tex]\vec{V}=\ln(\rho \sin \theta) (\sin \theta \hat{r} + \cos \theta \hat{\theta}).[/tex]
    Now you can take the curl directly in cylinder coordinates, using the corresponding formula from your textbook.

    That's, however, not very convenient, because it's easier to take the curl in Cartesian coordinates in this case, because obviously
    [tex]\vec{V}=\vec{j} \ln y.[/tex]
    Further, since [itex]\vec{j}=\text{const}[/itex] you have
    [tex]\vec{\nabla} \times \vec{V} =-\vec{j} \times \vec{\nabla} \ln y.[/tex]
    You should calculate the curl in both ways and prove that you indeed get the same result!
     
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