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Question about the curl with a specific type of field

  1. Feb 11, 2014 #1
    Hello,

    I'm a bit stuck with a case in which the curl gives a vector that does not transform under rotation.

    As an example, let's have a field with only [itex]\hat{x}[/itex] direction (but this does not mean that it's a scalar field!). The field has this expression:

    F(x,y,z)= A*exp(-[itex]\sqrt{y^{2}+z^{2}}[/itex] [itex]\hat{x}[/itex]

    Obviously, you have to take the positive value of the square root.

    If you evaluate ∇x F in the x axis you will get that the curl is:

    ∇x F = A (-[itex]\hat{y}[/itex]+[itex]\hat{z}[/itex]).

    Now, it's straightforward to see that since F depends really on the radius ([itex]y^{2}+z^{2}[/itex]), if you rotate the YZ axis, the curl will not rotate but continue been the same.

    I have heard that the curl is a pseudovector but I didn't expected to find that in some cases it does not rotates. Have I done something wrong?


    Best regards,
    Sergio
     
  2. jcsd
  3. Feb 11, 2014 #2

    pasmith

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    Pseudovectors behave as vectors under rotations. It's under reflections that they behave differently.
     
  4. Feb 12, 2014 #3
    Yes pasmith, it's because of this that I am stuck
     
  5. Feb 15, 2014 #4

    sal

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    If I found the curl right, then on the x axis your curl equation is formally 0/0. In short, you've got what looks like a major discontinuity at the x axis. Anyplace off the axis, when z is zero, so's the z component of the curl; ditto for y. Yet on the axis, depending on how you take the limit, it seems like the y and z components are both one ... or both zero. So the derivatives you're looking at are probably nonsense when you're on axis.

    Think about it for a minute, though. You've got rotational symmetry in the vector field about the x axis, so the curl must have rotational symmetry too; so, how can the curl on the axis point in any direction except parallel the x axis? And it's obviously not pointing parallel to the x axis, so it must actually be zero there. If that's not the answer you got then check your work again and look out for division by zero.
     
  6. Feb 19, 2014 #5
    Hi Sal,

    It's true that in the x axis the curl is formally 0/0 but this limit can be solved to one regarding that if you derivate in the z axis, for example, you have 2z/[itex]\sqrt{y^2+z^2}[/itex] and when moving dz, the y coordinate is absolutely 0, so the limit will be 1.

    Anyway, the rotational symmetry on the x axis will appear with other functions that do not have this 0/0 limit, for example:

    F(x,y,z)=exp(z+y)[itex]\hat{x}[/itex].

    This function, although it discriminates the [itex]\hat{y}[/itex] and [itex]\hat{z}[/itex] directions over other directions in the ZY plane, it does not have the 0/0 limit but it still have rotational symmetry.


    In fact I got this question when doing a quite weird thing: try to see if a radial function:

    g = [itex]\hat{r}[/itex] / [itex]r^{2}[/itex] x ( "x" is a cross product)

    Could be the Green function for the curl operator ∇x.

    Of course the curl of the radial function cancels at everywhere, but assuming that the function is equal to zero in the very zero point (because here [itex]\hat{r}[/itex] is undefined), and assuming that you can threat the derivatives in a discrete way (taking dx, dy or dz -> 0)... under this assumptions you get that the curl of the radial function is not zero on the origin, but then it appears this rotational symmetry :(.


    Best regards,
    Sergio
     
  7. Feb 19, 2014 #6

    sal

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    Just one very quick response to one part of your message:

    That's not exactly correct. The limit depends on the direction from which you approach the x axis. If you move dy instead, you find the limit is 0, not 1. Ergo it's not well defined, and you cannot conclude that the value for the curl on the x axis should be [itex]A (\hat{y}+\hat{z})[/itex]. You can't just decide to ignore a bunch of directions because they don't give you the limit you want.

    As to:

    The sum 'z+y' isn't rotationally symmetric about the x axis. The value at (y,z)=(1,0) is 1; the value at (0,1) is 1; but the value at [itex]\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)[/itex] is [itex]\frac{2}{\sqrt{2}}[/itex].

    In any case the point isn't that rotational symmetry (coupled with lack of dependence on the x coordinate) leads to curl of 0/0, it's that it leads to a curl which lies along the x axis.
     
  8. Feb 22, 2014 #7
    Hi sal,

    First of all, sorry for taking so long to answer. Then, I would like to comment some of your statements.

    Since this work is related with the study of Green functions I treat the derivatives in some discrete way (with the differentials tending to zero). So I can assume that when moving from the X axis in Z direction, I can ignore the Y coordinate.


    Certainly x+y is not rotationally symmetric but the point is that this is another example where you find a coordinate-system depending curl (i.e. does not change properly under rotation of the YZ plane)

    From the 'z+y' example, we see that it's not necessary rotational symmetry to get this effect.

    Best regards,
    Sergio
     
  9. Feb 22, 2014 #8

    sal

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    For the last time: No, you can't. Regardless of your purpose in looking at this, it's a three dimensional system and you can't just ignore one dimension. The limit is well defined only if it's the same from all angles of approach, and it's not. So, it is not well defined, and in fact if you go back to the definition of curl and try to sketch the arrows and derive it from first principles it's clearly zero on the axis in this case. (No, I won't include a drawing.)

    In general, when the magnitude plot of a unidirectional vector field "looks like" a parabola, the curl at the origin in the direction from which you're viewing it is going to be zero, because, as you pass from one side of the origin to the other, the magnitude doesn't change. The origin is a minimum (or maximum). And in this case, due to the rotational symmetry, the magnitude of the field does look that way when viewed from either the Y axis or the Z axis. Ergo, the curl in both Y and Z directions must be zero, and if you're not getting zero for the answer, you're doing it wrong.

    Wrong conclusion. It transforms just fine. But neither the original field, nor the curl, is rotationally symmetric in this case, so rotating the coordinate system around the X axis makes a difference. When a vactor field is rotationally symmetric, and independent of the X coordinate, then rotating about the X axis does not make a difference and the curl must lie along the X axis.

    Look, the math is consistent, and it works. If you're not getting it to work, and you're getting inconsistent results, then you are making a mistake, and your goal should be to find your error, not to show other people how the math is broken. It's not.
     
  10. Feb 22, 2014 #9
    The properties of physical systems change when you change the number of dimensions so I suppose I cannot ignore any dimension. I will meditate more about this.

    I will have to review my stuff to see why I don't get a zero. Probably because in the absolute zero I assume the function is zero for being undefined. I will check if this assumption is correct.




    So if the original field is not rotationally symmetric the curl will not change properly under rotations.
     
  11. Mar 1, 2014 #10
    Hello again,

    Let it be in this case

    ∇xF= ∂F/∂y [itex]\hat{z}[/itex] - ∂F/∂z [itex]\hat{y}[/itex]

    If we use the classical definition of the derivative:

    [itex]\frac{dF}{dy}[/itex]=lim(dy->0)[itex]\frac{1}{dy}[/itex]*[ F(x,dy,0)-F(x,0,0) ]

    Since F(x,dy,0) = exp(-dy) = 1 - dy .

    The absolute derivate respect y gives -1.

    On the other hand, if we make:

    [itex]\frac{dF}{dy}[/itex]=lim(dy->0)[itex]\frac{1}{dy}[/itex]*(F(x,0,0)-F(x,dy,0)

    The result is not -1 but 1.

    Of course, there is the same situation for [itex]\frac{dF}{dz}[/itex].

    So there is really a discontinuity on the curl and therefore it cannot be calculated. On the other hand, after moving a small Δy, the derivate respect z vanishes and therefore the curl is in the [itex]\hat{z}[/itex] direction.
     
  12. Mar 2, 2014 #11

    vanhees71

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    What you have given is the vector field (it's a field, not imply a vector),
    [tex]\vec{F}(\vec{x})=A \exp(-\sqrt{y^2+z^2}) \vec{e}_x.[/tex]
    It's written in a specific Cartesian coordinate system and behaves under rotations as a vector field
    [tex]\vec{F}'(\vec{x}')=\vec{F}(\vec{x}).[/tex]

    The curl can be simply evaluated by taking the appropriate derivatives (with the exception along the [itex]x[/itex] axis, where the field is nonanalytic due to the non-analyticity of the square root at 0):
    [tex]\vec{\nabla} \times \vec{F}(\vec{x})=(\partial_y F_z-\partial_z F_y) \vec{e}_x + (\partial_z F_x-\partial_x F_z) \vec{e}_y + (\partial_x F_y - \partial_y F_x) \vec{e}_z.[/tex]
    In your case you thus have
    [tex]\vec{\nabla} \times \vec{F}=-A \frac{z}{\sqrt{y^2+z^2}} \exp(-\sqrt{y^2+z^2}) \vec{e}_y + A \frac{y}{\sqrt{y^2+z^2}} \exp(-\sqrt{y^2+z^2}) \vec{e}_z =\frac{A}{\sqrt{y^2+z^2}} \exp(-\sqrt{y^2+z^2}) (y \vec{e}_z-z \vec{e}_y).[/tex]
    This is, of course also a vector field under rotations (it's an axial vector field under parity, supposed [itex]\vec{F}[/itex] is a polar vector field under parity).
     
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