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Finding a field line of a vector field

  1. Aug 5, 2015 #1
    1. The problem statement, all variables and given/known data

    Find the field line of

    [tex] \vec{E}(\vec{r}) = \frac{m}{4 \pi r^3} (2 \cos\theta, \sin\theta, 0)[/tex]

    through the point (a, b, c)

    (Spherical coordinates).

    m is a constant

    I know the answer, but I don't see what I do wrong.

    3. The attempt at a solution

    [tex] \frac{d\vec{r}}{d \tau} = C \cdot \vec{E}(\vec{r}(\tau)) [/tex]

    [tex] \frac{dr}{d \tau} = C \cdot \frac{m}{4 \pi r^3} 2 \cos\theta[/tex]
    [tex] \frac{d\theta}{d \tau} = C \cdot \frac{m}{4 \pi r^3} \sin\theta[/tex]
    [tex] \frac{d\phi}{d \tau} = 0[/tex]

    By setting [itex] C = \frac{4 \pi}{m}[/itex] I get

    [tex] \frac{dr}{d \tau} = \frac{ 2 \cos\theta}{r^3}[/tex]
    [tex] \frac{d\theta}{d \tau} = \frac{\sin\theta}{r^3} [/tex]
    [tex] \frac{d\phi}{d \tau} = 0[/tex]

    To get rid of [itex]r^3[/itex] I divide [itex]\frac{d\theta}{d \tau}[/itex] by [itex]\frac{dr}{d \tau}[/itex] (must not be zero and so on) and get

    [tex]2 \tan^{-1}\theta \frac{d\theta}{d \tau} = \frac{dr}{d \tau} [/tex]

    Then I multiply both sides with [itex]d\tau[/itex] (which is a somewhat mysterious operation to me).

    After integration I obtain [itex]2 \log (\sin\theta) = r + const[/itex]. I could of course determine the const and so on, but this isn't the answer anyway.

    Where's the error and why?

    Scale factors? But if so, why?
  2. jcsd
  3. Aug 5, 2015 #2


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    ##\text{tan}^{-1} \theta## is the inverse tan function. That is, it is the angle that has tan equal to ##\theta##.

    I think you want ##\text{tan} \theta \frac{dr}{d\tau} = 2 \frac{d\theta}{d\tau}## meaning ##\frac{dr}{d\theta} = \frac{1}{2} \text{tan} \theta##
  4. Aug 5, 2015 #3


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    EDIT: See my post below.
    Last edited: Aug 5, 2015
  5. Aug 5, 2015 #4


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    After staring at this for a bit, I think the whole problem statement is wrong. Could you please provide the original problem statement as it was given with no modifications?

    I say this because you should have started with something like ##\vec E(x, y, z, t) = E_x \space \hat i + E_y \space \hat j + E_z \space \hat k##.

    I'm guessing this would reduce to ##\vec E = E_x \space \hat i + E_y \space \hat j## due to the zero component.

    Then you would simply solve ##\frac{dy}{dx} = \frac{E_y}{E_x}##.
  6. Aug 5, 2015 #5
    Ok, the notation might not be the best, [itex]\tan^{-1} \theta [/itex] above is meant to be [itex]1 / \tan\theta[/itex]

    The problem statement is unmodified, with the exception of the coordinates of interest that I generalized. It's given in spherical coordinates (or practically in polar as you said)

    I get the right answer if I use the scale factor $$r$$ (used in your expression):

    $$\frac{dr}{d \theta} = r \frac{E_r}{E_{\theta}}$$

    So I think I made my error here

    $$\frac{d\vec{r}}{d \tau} = C \cdot \vec{E}(\vec{r}(\tau))$$

    which seems to hold for cartesian coordinates only.

    To generalize I seem to require something like this, but I can't convince myself of it.

    $$\frac{d\vec{r}}{d \tau} = C \cdot \frac{1}{h_i} \vec{E}(\vec{r}(\tau))$$

    $$h_{\theta} = | \frac{ \partial{\vec{r}} }{ \partial{\theta} }| = r$$
    Last edited: Aug 5, 2015
  7. Aug 5, 2015 #6
    Or is the field expressed in spherical coordinates?
  8. Aug 5, 2015 #7
    Yes my bad if it's unclear, everything is in spherical coordinates.

    theta would be the angle measured from the z axis

    Ok, in a general case a change in cartesian coordinates can be described like this with other (normalized, orthogonal) base vectors [itex]\vec{e_i}[/itex]

    $$ d\vec{r} = \sum_{i=1}^{3} h_i \vec{e_i} du_i $$

    So I guess

    $$\frac{d\vec{u}}{d \tau} = C \cdot \frac{1}{h_i} \vec{E}(\vec{u}(\tau))$$

    could make sense.
    Last edited: Aug 5, 2015
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