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Conservative force field

  1. Jul 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that the force field [itex]\vec{F}=f(r)\vec{r}[/itex] is conservative. [itex]f(r)[/itex] is a scalar field. [itex]r=|\vec{r}|[/itex]


    2. Relevant equations
    [itex]curl(\vec{F})=0[/itex]


    3. The attempt at a solution
    I tried calculating the cross product, in cartesian coordinates, but how do i treat f(r) when doing the partial derivatives?
    Would it be better to use spherical coordinates?
     
  2. jcsd
  3. Jul 9, 2014 #2

    Orodruin

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    Yes.
    *edit* Well, to elaborate - while you have to introduce some scale factors in order to do this, the only partial derivative you will need to care about is ##\partial_r##. At the same time, your field only has a component in the radial direction ...
     
  4. Jul 9, 2014 #3
    Does that mean in spherical coordinates my field looks like this: [itex]\left(f(r),0,0\right)[/itex] ?
     
  5. Jul 9, 2014 #4

    Orodruin

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    If the first component is the one associated with the radial unit vector: almost.

    Remember that ##\vec r = r \vec e_r##. This implies that ##f(r) \vec r = f(r) r \vec e_r##. However, this will matter very little for the computation of the curl.
     
  6. Jul 9, 2014 #5

    ehild

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    Treat f(r) as f(√(x2+y2+z2)).

    Recall that ##\nabla \times (f(r)\vec r)= \left(\nabla f(r) \right)\times \vec r + f(r) \left(\nabla\times\vec r\right)##

    ehild
     
  7. Jul 9, 2014 #6

    Orodruin

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    Of course it can be done in Cartesian coordinates. However, it is much simpler in spherical coordinates assuming that the general expression for the curl in curvilinear coordinates is known and the scale factors are known or easily calculable as is the case here. There is no need to resort to using the expression for r in terms of Cartesian coordinates.
     
  8. Jul 9, 2014 #7

    ehild

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    You do not need to use any system of coordinates, only the properties of the nabla operator and those of cross product are needed.
    The gradient can be handled as differentiation with respect to vector ##\vec r##. ##r=\sqrt{\vec r ^2}##, so ##\nabla r = \frac{d\sqrt{\vec r ^2}}{d\vec r}= \frac{\vec r}{r}##, ##\nabla f(r) = \frac{df}{dr }\frac {\vec r}{ r}## And the cross product of a vector by itself is zero.

    ehild
     
  9. Jul 9, 2014 #8
    Got it now thanks, since theta and phi are zero and since f(r) only depends on r, all the partial derivatives will be zero. Saw it after I looked up the curl in spherical coordinates. Thanks for the help :)
    Oh and please correct me if what I just said is wrong.
     
  10. Jul 9, 2014 #9
    Thanks for all the replies :)
     
  11. Jul 9, 2014 #10

    Orodruin

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    Well, your first line seemed to indicate you wanted the OP to do it in Cartesian coordinates.

    The coordinate free approach is based on
    $$
    \nabla f(r) = \frac{df}{dr} \nabla r
    $$
    and thus the knowledge that ##\nabla r = \vec e_r## (or more generally ##h_s \nabla s = \vec e_s##, where ##s## is a coordinate). I am not sure that we can assume that the OP has seen this.
     
  12. Jul 9, 2014 #11

    Orodruin

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    Only thing: ##\theta## and ##\phi## as coordinates are not zero. It would be more correct to say that the components of the vector field in the ##\theta## and ##\phi## directions are zero. Perhaps this is what you meant, but just to make sure.
     
  13. Jul 9, 2014 #12

    ehild

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    But it is easy in Cartesian coordinates knowing that r=√(x2+y2+z2).

    ehild
     
  14. Jul 9, 2014 #13

    Orodruin

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    But it is easier in spherical coordinates knowing that ##h_r = 1##. :)

    Regardless it is a simple matter of computation in either case.
     
  15. Jul 10, 2014 #14

    ehild

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    One usually remembers how to get gradient, divergence, curl in Cartesian coordinates. To use these in spherical polar coordinates, you need to find the formulas which are utterly complicated.

    It is useful to find out and remember the gradient of r, the divergence and curl of ##\vec r##.

    ehild
     
  16. Jul 10, 2014 #15

    Orodruin

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    "Utterly complicated"? I think that is pushing it a bit. All that is needed to remember is the general forms (which are not utterly complicated) and the scale factors. In this particular case the relevant scale factor is even one ... I will agree that it is useful to know ##\nabla r,\ \nabla\cdot\vec r##, and ##\nabla\times\vec r##. For some reason I just don't like differentiating square roots.
     
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