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Jacobian in spherical coordinates?

  1. Aug 24, 2013 #1


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    Started to learn about Jacobians recently and found something I do not understand.

    Say there is a vector field F(r, phi, theta), and I want to find the flux across the surface of a sphere. eg:


    Do I need to use the Jacobian if the function is already in spherical coordinates?

    My notes has an similar example that show you do use the Jacobian but I do not understand why. My understanding is you only use the Jacobian when there is a change in coordinates, but the function F is already in the desired coordinate system.

    Last edited: Aug 24, 2013
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  3. Aug 24, 2013 #2


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    Hi Uan! Welcome to PF! :smile:
    Suppose you want to find a volume (or area) by integrating, and everything is already in spherical coordinates …

    you still need to use the jacobian (instead of just drdθdφ) because volume (or area) is defined in terms of cartesian (x,y,z) coordinates, so you have made a transformation!

    Similarly, flux is defined in terms of cartesian coordinates. :wink:
  4. Aug 24, 2013 #3


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    Ok that makes sense.

    One other question...

    How do they get the side lengths r*d(theta) and r*sin(theta)*d(phi) of element dA in the diagram below?

  5. Aug 24, 2013 #4


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    Hi Uan! :smile:
    rdθ (the length of the side of A) is length of the side of that triangle with two sides r and angle dθ …

    so it's 2rsin(dθ/2), = 2r(dθ/2), = rdθ :wink:

    (alternatively, if you're happy using arc-length instead of "straight" length, then the arc-length is rdθ by definition)

    and the other one is calculated the same way, except that the side of the triangle is rsinθ instead of r
  6. Aug 24, 2013 #5


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    Ohh yeaahhh!! Small angle approximation duh! Thanks tiny-tim! I really appreciate your help!

    By the way, I derived it as rsin(dθ), = rdθ, as when the angle dθ goes infinitely small in the triangle with sidelengths r, r and dθ (I've redrawn it), its like it has 2 right angles (180° ~= 90° + 90° + ~0°) so hence rsin(dθ), = rdθ


    My way is correct yes?
    How did you get your 2s in 2rsin(dθ/2)?

    (Time for F1 break)
    Last edited by a moderator: May 6, 2017
  7. Aug 24, 2013 #6


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    (what's "F1 break"?)
    i used a triangle with two equal sides, and split it in two to make two right-angled triangles, so that my result of 2rsin(dθ/2) was precise

    your triangle is not precise :wink:
  8. Aug 24, 2013 #7


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    Ah yes! That works out beautifully! Cheers! :thumbs:

    (Formula 1 Qualifying at Spa! :wink:)
  9. Aug 27, 2013 #8


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    Of course, for the spherical shell it's pretty easy to get the surface vector element [itex]\mathrm{d}^2 \vec{A}[/itex] in this geometrical way, but it can become difficult for more complicated shapes. Thus, here the general way to get it.

    The surface-element vector is defined as a vector perpendicular to the surface and of the length of the area element. If you have given the surface [itex]S[/itex] in parameter form
    [tex]S: \quad \vec{r}=\vec{r}(u,v),[/tex]
    where [itex]u[/itex] and [itex]v[/itex] are real parameters, then the surface-element vector is given by
    [tex]\mathrm{d} \vec{A}=\mathrm{d} u \mathrm{d} v \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}.[/tex]
    That's so, because obviously this vector is perpendicular to the two (by assumption linearly independent) tangent vectors [itex]\partial_u \vec{r}[/itex] and [itex]\partial_v \vec{r}[/itex] and thus to all tangent vectors of the surface in the point under consideration, and the cross product has the magnitude corresponding to the infinitesimal parallelogram spanned by the vectors [itex]\mathrm{d} u \partial_u \vec{r}[/itex] and [itex]\mathrm{d} v \partial_v \vec{r}[/itex]. You only must be careful concerning the orientation of the surface element, beause obviously it switches sign when you change the order of the parameters, because the cross product in skew symmetric.

    For the sphere you take
    [tex]\vec{r}=R \begin{pmatrix}
    \cos \varphi \sin \vartheta \\ \sin \varphi \sin \vartheta \\ \cos \vartheta
    with [itex]u=\vartheta[/itex] and [itex]v=\varphi[/itex]. This gives
    [tex]\mathrm{d} \vec{A} =\mathrm{d} \vartheta \mathrm{d} \varphi R^2 \begin{pmatrix} \cos \varphi \cos \vartheta \\ \sin \varphi \cos \vartheta \\ -\sin \vartheta \end{pmatrix} \times
    -\sin \varphi \sin \vartheta \\ -\cos \varphi \sin \vartheta \\ 0
    \end{pmatrix}=R^2 \mathrm{d} \vartheta \mathrm{d} \varphi \begin{pmatrix}
    \cos \varphi \sin \vartheta \\ \sin \varphi \sin \vartheta \\ \cos \vartheta
    For the full sphere the parameters run over the ranges [itex]\vartheta=(0,\pi)[/itex] and [itex]\varphi \in [0,2 \pi)[/itex].
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