Results 1 to 10 of 10
Like Tree4Thanks
  • 1 Post By romsek
  • 1 Post By romsek
  • 1 Post By romsek
  • 1 Post By topsquark

Thread: Divergence in cylindrical coordinates

  1. #1
    Member
    Joined
    Mar 2010
    Posts
    221

    Divergence in cylindrical coordinates

    Hello, I would like to know how do we get from (2) to (3):

    $\displaystyle
    (1)\ \ \nabla \cdot B = 0
    $

    $\displaystyle
    (2)\ \ \frac{1}{r}\frac{\partial r B_r}{\partial r}\frac{\partial B_z}{\partial z} = 0
    $

    $\displaystyle
    (3)\ \ rB_r = -\int_{0}^{r} r \frac{\partial B_r}{\partial z} dr
    $

    After (3) I know how to integrate, I just want to know how do we get from (2) to (3), if someone can do it slow with comments and as much steps.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,519
    Thanks
    2863

    Re: Divergence in cylindrical coordinates

    I suspect that line (2) is equivalent to

    $\dfrac 1 r \dfrac{\partial}{\partial r}(rB_r)+\dfrac{\partial B_z}{\partial z} = 0$

    $\dfrac 1 r \left(B_r + r \dfrac{\partial B_r}{\partial r}\right)+ \dfrac{\partial B_z}{\partial z} = 0$

    work that through and see where it goes.
    Last edited by romsek; Apr 27th 2019 at 09:54 AM.
    Thanks from Nforce
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2010
    Posts
    221

    Re: Divergence in cylindrical coordinates

    Oh sorry, there is a mistake. The second equation is missing a plus sign:

    Quote Originally Posted by Nforce View Post
    Hello, I would like to know how do we get from (2) to (3):

    $\displaystyle
    (1)\ \ \nabla \cdot B = 0
    $

    $\displaystyle
    (2)\ \ \frac{1}{r}\frac{\partial r B_r}{\partial r} + \frac{\partial B_z}{\partial z} = 0
    $

    $\displaystyle
    (3)\ \ rB_r = -\int_{0}^{r} r \frac{\partial B_r}{\partial z} dr
    $

    After (3) I know how to integrate, I just want to know how do we get from (2) to (3), if someone can do it slow with comments and as much steps.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,519
    Thanks
    2863

    Re: Divergence in cylindrical coordinates

    This seems pretty straightforward actually

    $\dfrac 1 r \dfrac{\partial}{\partial r}(r B_r) + \dfrac{\partial B_z}{\partial z} = 0$

    $ \dfrac{\partial}{\partial r}(r B_r) = -r\dfrac{\partial B_z}{\partial z} $

    $rB_r = - \displaystyle \int_0^R r\dfrac{\partial B_z}{\partial z}~dr$

    $R$ is some constant that comes from the specific problem. I don't think it's intended to be the variable $r$
    Thanks from Nforce
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2010
    Posts
    221

    Re: Divergence in cylindrical coordinates

    If I understand correctly then here:

    $ \dfrac{\partial}{\partial r}(r B_r)$

    If we integrate:

    $$\int \dfrac{\partial}{\partial r}(r B_r) dr$$

    Then by the fundamental theorem of calculus, if we integrate a derivative we get the integrand?

    Does this apply also to partial derivatives? I always thought that then we have to expose the partial derivative operator. So, in this case, the partial derivative is a normal derivative.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Mar 2010
    Posts
    221

    Re: Divergence in cylindrical coordinates

    Sorry, did I ask something stupid?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,519
    Thanks
    2863

    Re: Divergence in cylindrical coordinates

    Quote Originally Posted by Nforce View Post
    Sorry, did I ask something stupid?
    sorry missed the question.

    You can always integrate w/respect to any variable of a multivariable function (provided that integral exists).
    In this case they are just integrating with respect to the variable $r$
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Mar 2010
    Posts
    221

    Re: Divergence in cylindrical coordinates

    Why don't we write then:

    $$\int \dfrac{\partial}{\partial r}(r B_r) \partial r$$

    ?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,519
    Thanks
    2863

    Re: Divergence in cylindrical coordinates

    I just don't know Jim.... I'm only a country doctor....
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,466
    Thanks
    882
    Awards
    1

    Re: Divergence in cylindrical coordinates

    Quote Originally Posted by Nforce View Post
    Why don't we write then:

    $$\int \dfrac{\partial}{\partial r}(r B_r) \partial r$$

    ?
    My two cents:

    If r can be written as a function of, say, two other variables p and q then
    $\displaystyle dr = \dfrac{\partial r}{\partial p} dp + \dfrac{\partial r}{\partial q} dq$

    We can't do that with $\displaystyle \partial r$.

    I realize that that isn't the case here but it possibly explains the difference in notation.

    -Dan
    Thanks from Nforce
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 16th 2017, 01:18 PM
  2. Divergence Theorem Cylindrical Co-ordinates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 9th 2011, 11:05 AM
  3. Replies: 6
    Last Post: Aug 30th 2010, 03:34 PM
  4. Covert Cartesian coordinates eq. into Cylindrical coordinates
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Aug 6th 2010, 06:56 AM
  5. Cylindrical coordinates..
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Aug 2nd 2008, 08:44 PM

/mathhelpforum @mathhelpforum