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Some partial derivative questions

  1. May 19, 2006 #1
    I've have met partial derivatives and the [tex]\nabla[/tex] symbol, however, I was asked today what was the geometrical representation and meaning of [tex] \nabla \times r [/tex] and [tex] \nabla \cdot r [/tex] where r was a surface in 3D (i.e. r(x,y,z) = ...).

    For the first one, I think that the answer might be:
    [tex] \left( \frac{\partial{r}}{\partial{y}} - \frac{\partial{r}}{\partial{z}} \right)i + \left( \frac{\partial{r}}{\partial{z}} - \frac{\partial{r}}{\partial{x}} \right)j + \left( \frac{\partial{r}}{\partial{x}} - \frac{\partial{r}}{\partial{y}} \right)k [/tex]

    Where i, j and k are the unit vectors in the x,y and z directions. However, assuming that all that is right, what does it mean geometrically?

    Would the second one simply be:
    [tex] \frac{\partial{r}}{\partial{x}} + \frac{\partial{r}}{\partial{y}} + \frac{\partial{r}}{\partial{z}} [/tex]

    What does this scalar mean, if anything?
     
  2. jcsd
  3. May 19, 2006 #2
  4. May 19, 2006 #3
    Are the expressions I gave in my first post correct?

    So, in a vector field, if [tex] \nabla \cdot r = 0[/tex], it means that the vector density is constant, right?
     
  5. May 19, 2006 #4

    nrqed

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    Your initial post is a bit confusing because there are two types of operation related to the nabla operator. You can have the deivergence of a vector field, [itex] \nabla \cdot {\vec V} [/itex] which gives a scalar or you may have the gradient of a *scalar* field [/itex] \nabla f(x,y,z) [/itex]which gives a vector.

    In your first post you seem to write an expression for the divergence but then talked about you r as being a scalar function.

    In any case, the fact that the divergenceof a vector field is zero does not imply that the vector field is constant. The geometrical meaning of the divergence of a vector field at a point is the net "flux" of the electric field going through an infinitesimal cube centered at that point. The geometrical meaning of the curl at a point is the net circulation around an infinitesimal loop around that point.

    Patrick
    EDIT: As an example, the vector field [itex] y {\vec i} + x {\vec j} [/itex] is not constant but its divergence is zero.
     
  6. May 19, 2006 #5
    Sorry about the confusion of my first post, I understand now where the flaw was in my understanding, although, the flux through an infinitesimal cube seems a rather strange concept to grasp!

    Just to reinforce the idea, could someone go through a quick example of calculating the divergence and the curl of the vector field [tex] yi + xj [/tex] please?
     
  7. May 19, 2006 #6

    nrqed

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    Consider a vector field of the form [itex] V_x {\vec i} + V_y {\vec j} + V_z {\vec k} [/itex], then the divergence is

    [tex] \nabla \cdot {\vec V} = { \partial V_x \over \partial x} + { \partial V_y \over \partial y} + { \partial V_z \over \partial z} [/tex]
    as you wrote in your post,
    whereas the curl is
    [tex] ( { \partial_y V_z - \partial_z V_y} ){\vec i} \ldots [/tex]
    I won't write the whole thing here since you already wrote it in a previous post.

    Just try it out with the vector field y i + x j and I will tell you if I agree with your answer.

    By the way, the flux through a small volume is a very useful concept in physics!

    Regards

    Pat
     
  8. May 20, 2006 #7
    So, [tex] \nabla \cdot r = \frac{\partial y}{\partial x} + \frac{\partial x}{\partial y} [/tex], if r = yi + xj, however, is [tex] \frac{\partial y}{\partial x} = 0 [/tex] and [tex] \frac{\partial x}{\partial y} = 0 [/tex] true, because it seems a rather simple solution, so [tex] \nabla \cdot r = 0[/tex].

    For the curl, is it zero because all the partial derivatives equal zero, I think, is this right?
     
  9. May 20, 2006 #8
    Is anything that I said in my previous post right?
     
  10. May 20, 2006 #9

    nrqed

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    The divergence is indeed zero. However the curl is not zero (the partial derivatives are not all zero ...for example , [itex] {\partial V_x \over \partial y}[/itex] is not zero, and so on...)

    EDIT: The curl *is* zero but not because all the partial derivatives are zero
     
    Last edited: May 20, 2006
  11. May 20, 2006 #10

    nrqed

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    I will write the curl explicitly:

    [tex] ( { \partial_y V_z - \partial_z V_y} ){\vec i}
    + ( { \partial_z V_x - \partial_x V_z} ){\vec j}
    + ( { \partial_x V_y - \partial_y V_x} ){\vec k}
    [/tex]
     
  12. May 20, 2006 #11

    nrqed

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    If you want to practice more, try the vector field [itex] y {\vec i} + x^2 {\vec j} + xz {\vec k} [/itex]
     
  13. May 20, 2006 #12
    Thank you for spending the time answering, I will try finding the curl and divergence of that vector field tommorrow (its quite late where I am), and will post my answer tommorrow.

    Sorry for taking so long to understand this, I have only a feeble mind!:smile:
     
  14. May 20, 2006 #13

    nrqed

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    No problem! Take your time. I will check in tomorrow.
    After you have done a couple of examples you will find it easy.

    Regards

    Patrick
     
  15. May 21, 2006 #14
    ok, had a little think, and do not have complete confidence in my answers, however, this is what I got:
    [tex] \nabla \cdot V = \frac{\partial y}{\partial x} + \frac{\partial (x^2)}{\partial y} + \frac{\partial (xz)}{\partial x} = \frac{\partial y}{\partial x} + \frac{\partial (x^2)}{\partial y} + z [/tex]

    Where [tex] V = yi + x^2j + xzk [/tex]

    [tex] \nabla \times V = \left( \frac{\partial (xy)}{\partial{y}} - \frac{\partial y}{\partial{z}} \right)i + \left( \frac{\partial y}{\partial{z}} - \frac{\partial (xz)}{\partial{x}} \right)j + \left( \frac{\partial (x^2)}{\partial{x}} - \frac{\partial y}{\partial{y}} \right)k = \left( x - \frac{\partial y}{\partial{z}} \right)i + \left( \frac{\partial y}{\partial{z}} - z \right)j + \left( 2x -1 \right)k [/tex]

    I'm not sure about all the partial derivatives that I have left, that's what makes me think that it's wrong.
    Is [tex] \partial (x)}/{\partial{y}} [/tex] (for example) equal to zero?
     
    Last edited by a moderator: May 21, 2006
  16. May 21, 2006 #15

    nrqed

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    when you do a partial derivative with respect to x, let's say, it means that you vary x while keeping y and z constant so [itex] {\partial y \over \partial x } = {\partial z \over \partial x } = 0 [/itex] and so on. The same for the other partial derivatives.

    You might not realize this but you have already used this when you wrote [itex] {\partial (xy) \over \partial y} = x [/itex]. This is correct but it comes from treating x as a constant when taking a partial derivative with respect to y.

    also, in your divergence you made a mistake..the third derivative should be with respect to z, not x.
    So the divergence gives simply x in the end.

    Patrick
     
  17. May 21, 2006 #16

    nrqed

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    Your curl is incorrect. It is
    [tex] ( { \partial_y V_z - \partial_z V_y} ){\vec i}+ ( { \partial_z V_x - \partial_x V_z} ){\vec j}+ ( { \partial_x V_y - \partial_y V_x} ){\vec k} [/tex]
    and here, V_x = y, V_y = x^2 and V_z=xz, so you should have

    [tex] ( { \partial_y (xz) - \partial_z (x^2)} ){\vec i}+ ( { \partial_z y - \partial_x (xz)} ){\vec j}+ ( { \partial_x x^2 - \partial_y y} ){\vec k}=-z {\vec j} +(2x-1) {\vec k} [/tex]
     
  18. May 22, 2006 #17
    Dam it:grumpy: :mad: . I'm annoyed with myself for making such schoolboy errors (ofcourse the fact that I'm still at school is not the point!:biggrin: ). I miss read your notation. Thank you anyway for checking my result. I think I should be able to calculate the divergence and the curl of the vector field [tex]f = xyi + yzj + x^2zyk [/tex] would be:

    [tex] \nabla \cdot f = y + z + yx^2 [/tex]
    [tex] \nabla \times f = -yi - 2xzj - xk [/tex]

    Once again, thanks for all the help Patrick (hope you don't mind me calling you that), but I must ask once more (I promise this will be the last time), have I made anymore foolish errors?
     
  19. May 22, 2006 #18

    nrqed

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    You are welcome (and of course call me Patrick :smile: )

    I agree with your divergence but not your curl. Your z component of the curl is ok. The y component seems to be missing a factor of y. As for the x component I think you forgot the partial derivative of V_z with respect to y so it should be (x^2 z - y).
    (it looks as if you simply forgot the y in the z component of your vector field a few times)


    Regards

    Patrick
     
  20. May 22, 2006 #19

    nrqed

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    By the way, the trick to remember the expression for the curl of a vector field [itex] {\vec V} [/itex] is to write a 3 by three matrix with on the first row the unit vectors i, j and k, on the second row the partial derivatives [itex] \partial_x , \partial_y [/itex] and [itex] \partial_z [/itex] and on the thrid row the components of your vector field, V_x, V_y and V_z. Now take the determinant of that 3 by 3 matrix and you will see that you get the curl of the vector field.
     
  21. May 24, 2006 #20
    Sorry for not responding soon, but I've been very busy (I've got very important exams coming up, and have to do well).

    The use of the 3x3 matrix determinant makes this much easier. I have had a go at some more practise questions and seem to be getting the hang of this. You must be bored of me saying this, but thanks again for all the help, much appreciated, cheers Pat
     
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