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Potential of the vector field

  1. Aug 10, 2017 #1
    1. The problem statement, all variables and given/known data

    I have a curve $$\Psi(t) = \hat h_\alpha$$ where the coordinates are $$\alpha=0, \beta=t$$ and $$\gamma=t$$ in the system. Additionaly

    $$x=\sqrt2 ^\alpha \cdot(sin\beta-cos\beta)\cdot \frac{1}{cosh\gamma}$$

    $$y=\sqrt2 ^\alpha \cdot(cos\beta+sin\beta)\cdot \frac{1}{cosh\gamma}$$

    $$z=\sqrt2 ^{\alpha+1} \cdot tanh\gamma$$

    2. Relevant equations

    $$x^2 + y^2 +z^2 =r^2$$

    3. The attempt at a solution

    My job was to derived all normalized frame vectors of this system I did it. Later calculate the arc length $$s$$ and verify that the curve is lying on the sphere.

    My problem is:
    1) How can I define the limits for the integral to calculate the arc length? Is the condition for the curve which is lying on the sphere correct?

    $$x^2 + y^2 +z^2 =r^2$$

    2) Last issue.
    I need to assume the vector field

    $$ G(\alpha,\beta,\gamma)={cosh\gamma\over \sqrt{2}^{\alpha+1}cos\beta} \hat h_\beta+{sinh\gamma\over \sqrt{2}^{\alpha+1}cos\beta} \hat h_\gamma $$

    to be a conservative and determine the corresponding potential $$\phi$$.

    How can I determine this potential.
    Usually I always had 3 factors $$f_x, f_y, f_z$$ here $$f_\alpha, f_\beta, f_\gamma$$ where I was calculating integral of $$f_x$$. Later I was calculating derivatives for $$y$$ and $$z$$ from result of integral of $$f_x$$. I don't know how should I manage it with only two factors. In the beginning I just assumed that $$\int h_\alpha = 0 + C_1(\beta,\gamma)$$ but in the end I'm not getting correct result which should be $$\Phi=ln|cosh(\gamma) \cdot tan(\frac{\Pi}{4}+ \frac{\beta}{2})|+const.$$

    What should I do?
  2. jcsd
  3. Aug 15, 2017 at 7:00 PM #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
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