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Vector calculus question

  1. May 25, 2005 #1
    A vector field is defined by [tex]\vec{A}(\vec{r})} = \rho^2 \hat{\phi} + \rho \sin \phi \hat{z}[/tex]

    Verify Stokes' theorem by explicit calcluation where S is the circle of radius a in the z = 0 plane.

    You may like to employ the ideas that [tex]\vec{dS} = \rho d\phi dz \hat{\rho} + dz d\rho \hat{\phi} + \rho d\rho d\phi \hat{z}[/tex] and [tex]\vec{dl} = d\rho \hat{\rho} + \rho d\phi \hat{\phi} + dz \hat{z}[/tex]

    Stokes' theorem: [tex]\int_{S} \vec{dS} \cdot (\nabla \times \vec{A}) = \int_{C} \vec{dl} \cdot \vec{A}[/tex]

    I found [tex]\nabla \times \vec{A} = \cos \phi \hat{\rho} - \sin \phi \hat{\phi} + 3\rho^2 \hat{z}[/tex]

    So, [tex]\vec{dS} \cdot (\nabla \times \vec{A}) = \cos \phi \rho d\phi dz - \sin \phi dz d\rho + 3\rho^2 d\rho d\phi[/tex]

    Now, I said dz = 0 because the circle is in the z = 0 plane, so z = constant and dz = 0. Is that right?

    [tex]\int_{S} \vec{dS} \cdot (\nabla \times \vec{A}) = \int_0^{2\pi} d\phi \int_0^a 3\rho^2 d\rho = 2\pi a^3[/tex]

    Now, [tex]\vec{dl} \cdot \vec{A} = \rho^3 d\phi[/tex] (since there's no [itex]\hat{\rho}[/itex] component and again dz = 0)

    [tex]\int_C \vec{dl} \cdot \vec{A} = \int_0^{2\pi} \rho^3 d\phi = [\rho^3 \phi]_{0}^{2\pi} = 2\pi \rho^3 = 2\pi a^3[/tex]

    This is what's supposed to happen, but I'm not sure if I've actually done the calculations right or got them right by luck. If someone could tell me where (if anywhere) I've gone wrong, thanks.
     
  2. jcsd
  3. May 26, 2005 #2

    Galileo

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    Be careful here! You may only evaluate a dot product by means of multiplying like components if you use cartesian coordinates.
    To illustrate this, consider the vectors:
    [tex]\vec A = \hat x, \qquad \vec B=-\hat x[/tex]
    , but in cilinder coordinates both vectors are written:
    [tex]\vec A = \hat r,\qquad \vec B = \hat r[/tex]
    The problem is that [itex]\hat r[/itex] depends on the particular point in question.

    In your case, since the normal to the disk points in the z-direction, you have:

    [tex]d\vec S = \hat{z}[/tex]
    and

    [tex](\nabla \times \vec A)\cdot d\vec S = 3\rho^2[/tex]

    You first wrote down the general expression, then used properties of your surface. Not how I'd do it, but I guess that works too.
     
    Last edited: May 26, 2005
  4. May 26, 2005 #3
    Thanks Galileo. We were never told about not being able to do the dot product if you're not using Cartesian co-ordinates. Grr :mad:.
     
  5. May 26, 2005 #4
    You can use the dot product, but you have to do it a different way.
     
  6. May 26, 2005 #5

    OlderDan

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    I don't think your statement that the dot product can only be done by multiplying like components if you use Cartesian coordinates is correct. Your vectors [tex]\vec A[/tex] and [tex]\vec B[/tex] are position vectors to two different points in space, and you are correct that the unit vectors having the same label are not the same vectors at different points in space. But in this context you are forming the dot product of two vector functions of position that are expressed in terms of a unique orthonormal basis at each point in space. The dot product of two vectors at each point is therefore the sum of the products of corresponding components. If the product somehow involved vectors at different points in space, then there would be a problem, but that is not the case. It is the uniqueness of the unit vectors at each point in space that permits the representation of the [tex]\nabla[/tex] operator in cylindrical and spherical coordinates, as well as Cartesian coordinates, is it not?
     
  7. May 26, 2005 #6
    I agree with OlderDan. Consider that regardless of whether you use cartesian or cylindrical coordinates, the vector gives a specific location in 3D space. Therefore, the name of the unit basis vectors is irrelevant when considering the dot product. The way each coordinate system relates to each other is based on our visual representation of them, but so long as each unit vector is orthogonal to the others, the dot product should be carried out in the same way.
     
  8. May 26, 2005 #7
    Jelfish,

    I agree with OlderDan too, but I think I disagree with you!

    The vector does not "give a specific location in 3D space". Each specific location in 3D space gives a vector. But that's not the same thing.

    The unit basis vectors in curvilinear coordinate systems are different in more than just name from those in cartesian coordinates. They're fundamentally different, because they're different at every point in space. OlderDan's point (if I understood correctly) was that in this case, the dot product of dl and A is formed from two vecors which by their definition are at the same point, so each has the same orthogonal basis set at that point. Thus the sum of the products of matching components is the dot product.
     
  9. May 27, 2005 #8
    All this is confusing me.
     
  10. May 27, 2005 #9

    Galileo

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    Aw'right, small correction. You cannot perform the vector operations in spherical/cilindrical coordinates associated with different points. The general idea of the statement still holds though. Don't just naively add components in A+B or multiply like components in AB in cilindrical/spherical coordinates like in cartesian coordinates.

    If you take [itex]\vec A = \hat x[/itex] and [itex]\vec B = -\hat x[/itex] at the points (1,0,0) and (-1,0,0) respectively, then in cilinder coordinates:
    [itex]\vec A = \hat r[/itex] and [itex]\vec B =\hat r[/itex].
    But [itex]\vec A + \vec B =0[/itex] and [itex]\vec A \cdot \vec B = -1[/itex].

    If the vectors are evaluated at the same point however, all is well (which, I think, was what Olderdan meant).
     
  11. May 27, 2005 #10
    In my exam today, there was a similar question but it was on the divergence theorem. When I had to work out [itex]\vec{dS} \cdot \vec{A}[/itex], I just did the dot product as I would in Cartesian and the answer still came out right :confused:.
     
  12. May 27, 2005 #11

    OlderDan

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    Yes. This is what I meant. The unit vectors in curvilinear coordinates are indeed functions of position, as you have stated. But for each coordinate system, at any one point in space there is a unique orthonormal basis. As long as two or more vector functions of position are expressed in the same orthonormal basis, all the familiar vector operations can be performed at every point. The orthonormaility of the basis vectors at each point results in the same cancellation of cross terms in the dot product, and gives rise to the same surviving vector terms in a cross product as you have for the cartesian basis.

    The exception to this is the point at the origin, or the central axis, where the curvilinear basis is ambiguous. I think the exception you have noted falls into this category. Your vectors are vectors to two points in space from the origin, expressed in terms of unit vectors defined at the termination points of the vectors because there is no unique set at the origin that can be used to represent them both. You can't perform the vector operations until you represent both vectors in the same basis. In most problems where curvilinear coordinates are preferred, the origin is not a factor in the calculations because of symmetry.
     
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