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Central force understanding

  1. May 19, 2007 #1
    General form of a central force is F(r)=F(r) (r^)

    [Note that This form of central force satisfies L=rxp=0 as well]

    But the isotropic or centro-symmetric form is

    F(r)=F(r) (r^)

    I found in a book that the second form of a central force is conservative.OK,this can be proved easily.What about the first expression?It is NOT centro-symmetric...depends on the position vector r it is acting on.

    Why is it NOT conservative always?

    Actually,I am not sure whether the same curl operation will do...Please check it...I am getting stuck in the differentiation of the r vector wihin the bracket while taking the curl.I feel confusion if the curl in two cases can be done in exactly similar way.
     
  2. jcsd
  3. May 24, 2007 #2

    siddharth

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    I don't completely understand your question. Do you need to show that the force of the form

    [tex] F(\vec{r}) \hat{e}_r [/tex]

    is not conservative while

    [tex] F(||\vec{r}||) \hat{e}_r[/tex]

    is conservative. Is this what you're trying to do?
     
    Last edited: May 24, 2007
  4. May 24, 2007 #3

    Meir Achuz

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    [tex]\nabla\times[{\vec r}f({\vec r})]
    =-{\vec r}\times\nabla f({\vec r})[/tex].
     
    Last edited: May 24, 2007
  5. May 24, 2007 #4

    jambaugh

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    Recall that conservative forces yield path independent work functions.
    This defines a relative potential difference between any two points as the unambiguous work required to move a test particle between the points.

    Thus (unless you let force also be a function of velocity as in EM) the force must be the gradient of a scalar potential (this by --essentially-- the fundamental theorem of calculus). This dictates that for a spherically symmetric force the potential be a function only of the radial coordinate r (not the vector). Thence:

    [tex] \mathbf{F}(\mathbf{r}) = \nabla V(r) = V'(r)\nabla r = V'(r)\hat{\mathbf{r}}[/tex]

    Intuitively consider that if the force has a directional dependence in magnitude even though the direction is always radial then you could do more work lifting in one direction than in another. Put a "ferris wheel" so that objects going up on one side of the wheel experience less radial force than objects going down on the other side. The wheel will spin faster and faster getting energy from "nowhere". I.e. you have built a perpetual motion machine and are violating conservation of energy.

    Electromagnetism gets around this problem because the Lorentz force is always perpendicular to the velocity of the test particle hence no actual work is done by the (magnetic) force. It only redirects the momentum without changing its magnitude.

    Regards,
    James Baugh
     
  6. May 24, 2007 #5
    I am sorry as I was unable to keep the touch with you...

    siddharth:Yes,you are right...I was talking of that.

    Meir Achuz:That is nice...gradient of f(r) is not going to be || to r...In another forum,I found this approach.

    jambaugh:A fragrance of physics.Very nice...
     
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