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I Partial derivative of potential energy and work

  1. Apr 3, 2016 #1
    For a conservative force [itex]\vec{F}=-\vec{\nabla} U \implies dW=-\vec{\nabla}U \cdot d\vec{s}[/itex]

    Where [itex]d\vec{s}[/itex] is the infinitesimal vector displacement.

    Does the following hold?

    [itex]-\frac{\partial U}{\partial \vec{s}}=-\vec{\nabla} U \cdot d\vec{s}=d W[/itex], i.e. the infinitesimal work is minus the directional derivative of [itex]U[/itex] in the direction of [itex]\vec{s}[/itex].
     
  2. jcsd
  3. Apr 3, 2016 #2
    No this doesn't make sense, you've written what the spatial derivative of the potential is, its ##\vec{F}## not dW.
     
  4. Apr 3, 2016 #3
    Your equation is wrong. It should read ##dU=\vec{\nabla} U \cdot \vec{ds}=-dW##
    The interpretation is the infinitesimal work in the direction of ##\vec{ds}## is equal to minus the directional derivative of U in the direction of ##\vec{ds}##.
     
  5. Apr 3, 2016 #4
    @jamie.j1989 Thanks for the answer, i just applied the rule that uses nabla for directional derivatives, so how to correctly write the directional derivative?


    @Chestermiller Thanks a lot for the reply, where is the exactly the mistake? Is it the fact of not having specified that the infinitesimal work is in the direction of ##\vec{ds}## ? Moreover how is the directional derivative of ##U## written explicitly, if ##\frac{\partial U}{\partial \vec{s}}## is wrong?
     
  6. Apr 3, 2016 #5
    The directional derivative of ##U## is ##\vec{\nabla}U = \hat{i}\frac{\partial U}{\partial x}+\hat{j}\frac{\partial U}{\partial y}+\hat{k}\frac{\partial U}{\partial z}= -\left(\hat{i}F_x +\hat{j}F_y+\hat{k}F_z\right)##
     
  7. Apr 3, 2016 #6

    jtbell

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    Staff: Mentor

    No, ##\vec \nabla U## is the gradient of U (a vector). The directional derivative of U along a direction specified by a unit vector ##\hat n## is ##\vec \nabla U \cdot \hat n## (a scalar). If you go a distance ##ds## in the direction of ##\hat n##, then the change in U is ##dU = (\vec \nabla U \cdot \hat n) ds = \vec \nabla U \cdot (\hat n ds) = \vec \nabla U \cdot \vec{ds}##.
     
  8. Apr 3, 2016 #7
    Thanks a lot for the reply, where is the exactly the mistake? Is it the fact of not having specified that the infinitesimal work is in the direction of ##\vec{ds}## ?
    No. It's that you set dU/ds equal to a differential, which, of course, it is not.

    Moreover how is the directional derivative of ##U## written explicitly, if ##\frac{\partial U}{\partial \vec{s}}## is wrong?
    If ##\vec{s}## is position vector from an arbitrary origin to a position in space, and ##\vec{s}+\vec{ds}## is a position vector from the origin to an adjacent position in space, then the change in potential U between ##\vec{s}## to ##\vec{s}+\vec{ds}## (in the direction of ##\vec{ds}## is given by:
    $$dU=\vec{\nabla} U\cdot \vec{ds}$$The derivative of U with respect to distance in the direction of ##\vec{ds}## is obtained by dividing by the magnitude of the differential postion vector ##\vec{ds}##:
    $$\frac{dU}{|\vec{ds}|}=\vec{\nabla} U\cdot \left(\frac{\vec{ds}}{|\vec{ds}|}\right)=\vec{\nabla} U\cdot \vec{n}_s$$
    where ##\vec{n}_s## is a unit vector in the direction of ##\vec{ds}##.
     
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