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Finding Beltrami field in Cartesian coordinates

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Working in Cartesian coordinates (x,y,z) and given that the function g is independent of x, find the functions f and g such that: v=coszi+f(x,y,z)j+g(y,z)k is a Beltrami field.

    2. Relevant equations
    From wolfram alpha a Beltrami field is defined as v x (curl v)=0

    3. The attempt at a solution
    So jumping in, I compute the curl of v and the cross product of the curl of v and v:
    Curl_V.png
    But then the above nasty equation results and surely this would give rise to some equally nasty PDEs?
     
  2. jcsd
  3. Feb 19, 2015 #2

    haruspex

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    So, that last equation gives three equations, since each component must be zero. Look at the first one, the x component. Only one term (f.fx) is a function of x. So what do you get if you differentiate wrt x?
     
  4. Feb 19, 2015 #3
    Thanks for the tip haruspex but I'm a little confused as to what you mean. Differentiating the first component with respect to x will remove the function g(y,z) but still leave a nasty PDE.

    I've found another piece of information which I think allows me to solve the problem and that is that a Beltrami field can also be denoted by curl v = kv where k>0 is just a constant. The second component then simplifies nicely where I've deduced k=1.

    If I'm right I think f(x,y,z)=+-sin(z) and g(y,z)=0. This satisfies the equation. I think I initially doubted such answers as I believed that the functions f(x,y,z) and g(y,z) should contain at least y and z variables!
     
  5. Feb 19, 2015 #4

    haruspex

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    I didn't think it was nasty. Using subscript notation, I got ##{f_x}^2+f_{xx}=0##. Dividing through by fx, this integrates to produce ##f^2=(Ax+B)\hat f(y,z)##.
    That sounds a very useful fact. Continuing my first principles approach, I also found g=0, but my f is a little more general: ##\sqrt{C-\cos^2z}##. Maybe I missed some way of showing C=1.
     
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