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Find the potential fucntion, phi

  1. Jun 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider the vector field given by

    [itex]F(x, y, z) = yz \hat{i} + xz \hat{j} + (xy + 3z^{2})\hat{k}[/itex]

    a. Calculate ∇xF and show that F is a conservative field. Done, result = <0,0,0> which implies the vector field is conservative.

    b. The way we were taught this is to set

    [itex]
    ∂\phi /∂x = yz, \\
    ∂\phi /∂y = xz, \\
    ∂\phi /∂z = xy + 3z^{2}
    [/itex]

    Then find the integrals of all 3 equations to get,

    [itex]
    \phi = xyz + C_{x},\\
    \phi = xyz + C_{y},\\
    \phi = xyz + z^{3} + C{z}
    [/itex]

    Finally, look for similar features and construct [itex]\phi[/itex] like so

    [itex]\phi = xyz + x^{3} + C[/itex]

    Is that correct?
     
    Last edited: Jun 13, 2013
  2. jcsd
  3. Jun 13, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Is that really how you were taught? It will work but that "look for similar features" looks ambiguous to me.
    In fact, in this particular problem, it is impossible to find such a [itex]\phi[/itex]. It simply cannot exist.
    Recall that (for "nice" functions, having continuous second derivatives as this does) the order of differentiation in mixed derivatives is irrelevant. That is [tex]\frac{\partial}{\partial x}\left(\frac{\partial\phi}{\partial z}\right)= \frac{\partial}{\partial z}\left(\frac{\partial\phi}{\partial x}\right)[/tex]
    But if [tex]\frac{\partial \phi}{\partial x}= yz[/tex] then [tex]\frac{\partial}{\partial z}\left(\frac{\partial\phi}{\partial x}\right)= y[/tex] while if [tex]\frac{\partial \phi}{\partial z}= xy+ 3x^2[/tex] then
    [tex]\frac{\partial}{\partial x}\left(\frac{\partial\phi}{\partial z}\right)= y+ 6x[/tex].

    Those are NOT the same so this is impossible. If that "[itex]3x^2[/itex]" in the derivative with respect to z were "[itex]3z^2[/itex]" then it would be possible.

    Here's how I would do it.

    From [itex]\dfrac{\partial \phi}{\partial x}= yz[/itex], yes we have [itex]\phi(x,y,z)= xyz+ u(y, z)[/itex] where u(y,z) (your "[itex]C_x[/itex]") is a function of y and z only (constant with respect to x). Differentiating that with respect to y we have
    [tex]\dfrac{\partial\phi}{\partial y}= xz+ \dfrac{\partial u}{\partial y}[/tex]

    But we are told that [itex]\dfrac{\partial \phi}{\partial y}= xz[/itex] so we must have [itex]xz+ \dfrac{\partial u}{\partial y}= xz[/itex]. The "xz" terms cancel leaving [itex]\dfrac{\partial u}{\partial y}= 0[/itex] so that u does NOT, in fact, depend on y and we can write [itex]\phi(x, y, z)= xyz+ u(z)[/itex]. That is, we now know that u can only depend on z. Differentiating that with respect to z we get [itex]\frac{\partial\phi}{\partial z}= xy+ \frac{\partial u}{\partial z}[/itex].

    But we are told that [itex]\dfrac{\partial \phi}{\partial z}= xy+ 3z^2[/itex]. So we must have [itex]xy+ \frac{du}{dz}= xy+ 3z^2[/itex]. The "xy" terms now cancel leaving [itex]\frac{du}{dz}= 3z^2[/itex]. (I can now write "ordinary" derivatives because I know that u depends on z only.) That gives [itex]u(z)= z^3[/itex] so that [itex]\phi(x,y,z)= xyz+ u= xyz+ z^3[/itex].

    If that "[itex]xy+ 3x^2[/itex]" were, in fact, correct, here is what would happen. We would repeat everything up to the last paragraph which would now be:
    " But we are told that [itex]\dfrac{\partial \phi}{\partial z}= xy+ 3x^2[/itex]. So we must have [itex]xy+ \dfrac{du}{dz}= xy+ 3z^2[/itex]. The "xy" terms now cancel leaving [itex]\dfrac{du}{dz}= 3x^2[/itex]. That is impossible- if u is a function of z only, its derivative cannot be a function of x. There is no [itex]\phi(x,y,z)[/itex] having those partial derivatives.
     
  4. Jun 13, 2013 #3
    Thank you, good sir. That makes sense and seems slightly more intuitive than the method we were taught. This is a great explanation and it should hopefully help me for my examination tomorrow =/
     
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