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Field lines

  1. Aug 21, 2006 #1
    I'm not sure under what heading this belongs but I'm going to ask here:

    Show that the field lines y = y(x) of a vector function

    [tex] \mathbf F(x,y) = \mathbf i F_x(x,y)+\mathbf j F_y(x,y)[/tex]

    are solutions of the differential equation

    [tex]\frac {dy} {dx} = \frac {F_y(x,y)} {F_x(x,y)} [/tex]

    Could someone suggest a way to get started on this one?
    Last edited: Aug 21, 2006
  2. jcsd
  3. Aug 21, 2006 #2


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    Yes. Use the fact that for a field line [itex]\vec{r}(t)[/itex], we have the property that [itex]\vec{r}(t)\times \vec{F}(\vec{r}(t))=0[/itex]. What do you get out of this property when the variable of parametrization t is simply the coordinate x?
  4. Aug 22, 2006 #3


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    Oops! My eyes just wandered on the sheet of paper on which I had sketched the solution to your problem and sudenly I realized "Hey, I this is not what I wrote in my post!"

    The correct condition is that the derivative of r(t) is parallel to the field line:

    [tex]\frac{d\vec{r}}{dt}(t)\times \vec{F}(\vec{r}(t))=0[/tex]
  5. Aug 23, 2006 #4
    Let me stress the geometric meaning of our problem.
    For a function y=y(x) the inclination of the tangent [tex]\alpha_y [/tex]
    to the line representing the function in any point x is its derivative in that point:

    [tex]\alpha_y = \frac {dy} {dx} [/tex]

    For a vector field F(x,y) the inclination of the vector [tex]\alpha_F [/tex] in any point (x,y) is the ratio of its y and x components:

    [tex]\alpha_F = \frac {F_y(x,y)} {F_x(x,y)} [/tex]

    Hence, your equation

    [tex]\frac {dy} {dx} = \frac {F_y(x,y)} {F_x(x,y)} [/tex]

    can be seen as equating the two tangents:

    [tex]\alpha_y = \alpha_F [/tex].

    In other words, that equation states that the inclination of the line y(x) and the vector F(x,y) in the same point are equal. That's the definition of a field line.
  6. Aug 25, 2006 #5
    Thanks. Quasar you had me scratching my head there for a while, but it makes sense now.
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