
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!

Please help me understand this equation

  1. Apr 20, 2016 #1
    $$\bar{v}=\nabla \times \psi \hat{k}$$
    The problem is much bigger, i know how a rotor or curl is calculated in cylindrical coordinates, but i'm just asking to see what would be the "determinant" rule for this specific curl.
    2. Relevant equations
    $$\psi$$ is in cylindrical coordinates (r,theta,z) and depends only on r and z, not theta meaning $$\psi(r,z)$$

    3. The attempt at a solution
    I'm asked to write the velocity field in therms of the current function psi, and i know i have to do it with said equation above, i believe the determinant rule used (given this is a demonstration and i already know the answer) was
    (i tried doing a matrix and i couldn't, but the bottom line of the determinant looks like it was 0 & psi & 0, so the result should yield) $$\bar{v}=(\frac{-1}{r} \frac{\partial \psi}{\partial z}, 0, \frac{1}{r}\frac{\partial \psi}{\partial r})$$
  2. jcsd
  3. Apr 20, 2016 #2
    I think i solved it by myself, but if this is wrong I'd like some inputs.
    All the theta derivatives are zero because psi doesn't depend on theta (1)
    The middle therm is zero because the flux is axisymmetric, therefore the derivatives on each side are the same, cancelling each other (2)
  4. Apr 21, 2016 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Did you mean ##\nabla \times (\psi \hat\theta)## instead of ##\nabla \times (\psi \hat k)##?

    You can tell your answer isn't dimensionally correct. If ##r## and ##z## have units of length, the derivative should only only introduce one power of length in the denominator. Your result has two.
  5. May 10, 2016 #4
    I'm answering this absolutely late but i solved this doubt the next day asking my tutor
    The equation, even though given as an "absolute" in the textbook (it implies it can be used everywhere) it depends on the flux, it is
    $$\bar{v}=\nabla \times \phi(*)$$ where the asterisk (*) is the coordinate the current function does not posses, in this case phi is a function of r and z, therefore the asterisk must be a theta.
    In any case i find much more comfortable using
    $$\bar{v}=\nabla \phi \times (*)$$
    I believe it's a similar equation, with another syntax, that relates velocity field with current function
    Thanks for your answer anyways
    PS=I'm struggling a lot in this fluid mechanics c(o)urse

Have something to add?
Draft saved Draft deleted