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Curl of unit vector r / r^2

  1. Feb 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Not sure if this belongs in homework or general discussion - I found this in reading

    In studying the divergence and curl of the magnetic field (B), I found a statement that I need some help with.

    In the derivation of the divergence of B using the Biot-Savart, I have:


    [tex] \nabla B = \frac{u_0}{4\pi}\int\nabla (dot) (J (cross)\frac{r_u}{r^2}\dt [/tex]

    After applying a product rule
    [tex] \nabla (dot) (J (cross)\frac{r_u}{r^2} = \frac{r_u}{r^2} (dot) \nabla (cross) J – J (dot) (\nabla (cross) \frac{r_h}{r^2}) [/tex]

    It is stated that

    [tex] \nabla x \frac{r_u}{r^2}) = 0 [/tex]

    I do not see this.


    2. Relevant equations


    I do know that [tex] \nabla (dot) \frac{1}{r^2} = 0[/tex] everywhere except at the origin (introduction of Dirac Delta).

    I do not see that del cross (unit vector r / r^2) is zero.

    Can you help explain?

    Thanks
    Sparky_
     
  2. jcsd
  3. Feb 25, 2013 #2

    kreil

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    Gold Member

    You can calculate the curl to see why it is zero.

    For F composed of [itex]\left[ F_x,F_y,F_z \right] [/itex], the curl is defined as the determinant of,

    [tex]\left| \begin{matrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{matrix} \right| [/tex]

    Which results in

    [tex]
    \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \hat{i}+\left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) \hat{j}+\left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\hat{k}
    [/tex]

    Now just calculate it for [itex]F(x,y,z) = \hat{r}/r^2[/itex]
     
    Last edited: Feb 25, 2013
  4. Feb 25, 2013 #3

    TSny

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    Homework Helper
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    Following kreil, if you want to work in Cartesian coordinates, note that ##\large{\frac{\hat{r}}{r^2}=\frac{\vec{r}}{r^3}= \frac{x\hat{i}+y\hat{j}+z\hat{k}}{r^3}= \frac{x\hat{i}+y\hat{j}+z\hat{k}}{(x^2+y^2+z^2)^{3/2}}}##

    Or you can work in spherical coordinates and use the expression for the curl in spherical coordinates. See equation 89 here . Although the expression for the curl in spherical coordinates looks complicated, for your problem it is very easy to evaluate.
     
  5. Feb 26, 2013 #4
    I have calculated the curl of [tex]\frac{\hat{r}}{r^2}[/tex] using the Cartesian coordinates.

    [tex] \begin{matrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{matrix} [/tex]

    For x:

    [tex]\hat{x}(\frac{\partial}{\partial y}(\frac{z}{(x^2+y^2+z^2)^{3/2}} – \frac{\partial}{\partial z}(\frac{y}{(x^2+y^2+z^2)^{3/2}}[/tex]

    [tex]= \frac{\partial}{\partial y} (z(x^2+y^2+z^2)^{-(3/2)} - \frac{\partial}{\partial z}(y(x^2+y^2+z^2)^{-(3/2})[/tex]
    [tex] = \frac{-3z}{2}(x^2+y^2+z^2)^{(-5/2)}(2y) +\frac{3y}{2}(x^2+y^2+z^2)^{(-5/2)}(2z)[/tex]
    = 0
    And likewise for the other 2.

    I would now like to duplicate (for fun) the spherical coordinate path.

    However I do not know how to get started on spherical coordinates version (the intimidating equation in the link). What do I put in for my two different angles? I agree things should drop out and get simpler. I just don’t see how to get started.

    I see angle variables and sin(phi) in the denominator (I can't set that to zero) and so forth.

    I am pretty sure with a function in just “r” much of the spherical terms do not exist and things get simple.

    can you guide some more on how to do the spherical coordinate version?

    Thanks.
    Sparky_
     
  6. Feb 26, 2013 #5

    kreil

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    To convert Cartesian Coordinates to Spherical Coordinates, use:

    [itex] x = r \cos \theta \sin \phi[/itex]
    [itex]y = r \sin \theta \sin \phi[/itex]
    [itex]z = r \cos \phi [/itex]

    The expression may look scary, but alas, in this case we have:

    [tex] F = (F_r,F_{\theta},F_{\phi}) = \left(\frac{\hat{r}}{r^2},0,0 \right)[/tex]

    So that scary expression immediately becomes:

    [tex] \frac{1}{r \sin \theta} \frac{\partial F_r}{\partial \theta} \hat{\phi} - \frac{1}{r} \frac{\partial F_r}{\partial \phi} \hat{\theta}[/tex]

    and these derivatives are clearly zero as well, so the curl is zero.
     
  7. Feb 26, 2013 #6
    Can you help me understand how to deal with the sin(theta) in the denominator?

    Obviously theta is not 0. And I assume it is 90 so that that term goes to 1?

    But if I am thinking about that particular angle correctly it is the angle from the vertical axis to r so it can be any angle?

    Can you help clear this up?

    Thanks again
    Sparky_
     
  8. Feb 26, 2013 #7

    jtbell

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    Staff: Mentor

    First look at the partial derivatives in that expression. Does your ##F_r## depend on either ##\theta## or ##\phi##? So what do those two derivatives work out to? :wink:
     
  9. Feb 26, 2013 #8

    kreil

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    Gold Member

    The goal is not to plug in specific values for these variables. If the function F depended on theta or phi at all it would be in such a way that these terms out in front help to simplify the expression (usually).

    I guess a more suggestive way of saying this is: what is the equation of a sphere in spherical coordinates, and why does it look that way?

    In Cartesian coordinates, a sphere centered at (0,0,0) with radius 2 is described by [itex]x^2+y^2+z^2=4[/itex]

    Using the conversions I gave in my last post, this same sphere is described in spherical coordinates by

    [itex]r^2 \cos^2 \theta \sin^2 \phi + r^2 \sin^2 \theta \sin^2 \phi + r^2 \cos^2 \phi = 4[/itex]

    [itex]r^2 \left[ \sin^2 \phi \left( \cos^2 \theta + \sin^2 \theta \right) + \cos^2 \phi \right]= 4[/itex]

    [itex] r^2 = 4[/itex]


    In essence, this tells us that in order to describe this object (a sphere) all we need to do is restrict the radius and let theta and phi vary from their smallest value to their largest value.

    If we were to plug in specific values for theta or phi, we would suddenly be examining a single point that lies on the sphere and not the object as a whole. The curl is concerned with the object as a whole, and not with specific points.
     
  10. Feb 27, 2013 #9
    Ok,

    So essentially I have
    [tex]\frac{1}{r \sin \theta} \frac{\partial F_r}{\partial \theta} \hat{\phi} - \frac{1}{r} \frac{\partial F_r}{\partial \phi} \hat{\theta}[/tex]

    [tex] \frac{1}{r \sin \theta} (0) \hat{\phi} - \frac{1}{r} (0) \hat{\theta}[/tex]

    and "ignore" because I am multiplying by 0 the [tex] = \frac{1}{\sin \theta}[/tex]

    ??

    Thanks again for walking me through this.
    -Sparky_
     
  11. Feb 27, 2013 #10

    kreil

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    What is being ignored? It is a basic tenet of mathematics that anything multiplied by zero is zero.

    The 1/sin(x) is well behaved everywhere except when sin(x) = 0, which is where it blows up- but again, we are multiplying by zero so this tames the divergence.
     
  12. Feb 27, 2013 #11
    Sorry – poor choice of the word “ignore”

    I got hung up on needing to do something with the 1/sin().

    I was concerned that I was not being mathematically correct.

    I did see that the partial derivatives resulted in zero and you can see from my earlier post that I wanted somehow to set theta equal to 90 degrees or some such so as the possibility of a zero in the denominator goes away.

    Anyway, I have learned a good bit through this discussion and I now see how the text I was reading regarding the Biot Savart law and the statement it made about the curl of (r_hat/r^2) is zero can be made.

    Thanks
    Sparky_
     
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