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Conservative force

  1. Jan 2, 2015 #1
    they say that a conservative force can be associated to a potential .. Why is that ? and what does it mean that a force has a potential ???
     
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  3. Jan 2, 2015 #2

    ShayanJ

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    At first, I should say they here we're talking about a force field i.e. a machine that gives you a vector for each point of space.
    As you may know, the work that such a force does on a mass when its moving from point A to point B through a trajectory [itex] \gamma(t) [/itex], is [itex] W=\ \ _\gamma\int_{A}^{B} \vec F \cdot \vec{dr} [/itex] (Where [itex] \vec{dr} [/itex] is an infinitesimal vector which is always parallel to the curve [itex] \gamma(t) [/itex].) Note that I put [itex] \gamma [/itex] under the integral sign because in general the work that the force does on the particle depends not only on the end points of the trajectory but also on the trajectory itself.
    Now there happens to be some force fields for which the work is independent of the trajectory taken by the particle and only depends on the end points. This is stated mathematically by [itex] \oint \vec F \cdot \vec {dr}=0 [/itex] for all closed curves. Now from this equation, its possible to prove that [itex] \vec \nabla \times \vec F=0 [/itex].
    Then, from Helmholtz decomposition, we know that for any reasonable vector field [itex] \vec F [/itex], there exist a scalar field [itex] \Phi [/itex] and a vector field [itex] \vec A [/itex] such that [itex] \vec F=-\vec \nabla \Phi+\vec \nabla \times \vec A [/itex]. Now if I get the curl of this, I'll have [itex]\vec \nabla \times \vec F=-\vec \nabla \times \vec \nabla \Phi+\vec \nabla \times \vec \nabla \times \vec A [/itex]. But I know that [itex] \vec \nabla \times \vec F=0 [/itex] so I should have [itex] -\vec \nabla \times \vec \nabla \Phi+\vec \nabla \times \vec \nabla \times \vec A=0 [/itex].
    From vector calculus, I know that [itex] \vec \nabla \times \vec \nabla \Phi=0 [/itex] identically so only [itex] \vec \nabla \times \vec \nabla \times \vec A=0 [/itex] remains for which [itex] \vec A=0 [/itex] is a reasonable solution. But this means I have [itex] \vec F=-\vec \nabla \Phi[/itex]. Now the scalar field [itex] \Phi [/itex] is called a scalar potential for [itex] \vec F [/itex].
     
  4. Jan 2, 2015 #3
    Sry , but what does it mean that a force has a potential ?
     
  5. Jan 2, 2015 #4

    ShayanJ

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    When we say the force field [itex] \vec F=F_x \hat x+F_y \hat y+F_z\hat z [/itex] has a potential, we mean that we can find a function like [itex] \Phi [/itex] such that [itex] F_x=-\frac{\partial \Phi}{\partial x} [/itex],[itex] F_y=-\frac{\partial \Phi}{\partial y} [/itex] and [itex] F_z=-\frac{\partial \Phi}{\partial z} [/itex]. It sometimes makes things easier because we can work with a scalar instead of a vector. Also it allows us to associate energy with forces so that we can apply conservation of energy.
     
  6. Jan 2, 2015 #5
    Thanks :)
     
  7. Jan 2, 2015 #6

    Stephen Tashi

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    We don't say that a "force" has a potential. We say that a "force field" has a potential. For example, at a point above the surface of the earth there is a "gravitational field", which is a "force field", but unless there is a mass at that point being pulled by gravity there is no "force of gravity" at that point. The concept of a field is a mathematical abstraction.

    A force field is specified by assigning a vector to every point in space. The magnitude of the vector does not represent a force (e.g. newtons). It has units of force per unit-of-something (e.g. newtons per something). As Shyan said, it is often simpler to deal with scalar valued functions than vector valued functions. If you are familiar with multivariable calculus, you know that a scalar valued function can have a gradient, which is a vector valued function. A "potential" function for a force field is a scalar valued function of whose gradient (with respect to position variables) is equal to the force field. It's simpler to solve some problems using the potential function than using the force field itself. Some force fields do not have potential functions.
     
  8. Jan 2, 2015 #7
    thank you :)
     
  9. Jan 4, 2015 #8

    ShayanJ

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    Now that the OP is satisfied with the answer and thanks to @vanhees71, this thread is reminded to me, I want to have a discussion on pedagogy here. I feel that there is a weak point in the argument presented in post #2. Because in the end, I simply accepted [itex] \vec A=0 [/itex] as a solution to [itex] \vec\nabla \times \vec \nabla \times \vec A = 0 [/itex]. But one may ask why can't we use [itex] \vec A=\frac 1 2 a (z^2-y^2) \hat x [/itex]? Because its a solution of [itex] \vec\nabla \times \vec \nabla \times \vec A = 0 [/itex]. But the problem is, it doesn't satisfy [itex] \vec \nabla \times \vec A=0 [/itex] and so it changes [itex] \vec F [/itex]!
    In classical electrodynamics, the answer is provided by gauge invariance but what about Newtonian gravitation?
     
  10. Jan 4, 2015 #9

    vanhees71

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    Of course, the Helmholtz decomposition is not unique. To make it unique you need some boundary conditions at infinity. The point is of course, that in the case that the force field satisfies ##\vec{\nabla} \times \vec{F}=0## in a simply-connected region of space, there exists a scalar potential ##\Phi## such that
    $$\vec{F}=-\vec{\nabla} \Phi.$$
    For Newtonian gravity that's always the case. For a particle of mass ##M## a mass density ##\rho## implies a gravitational force,
    $$\vec{F}=M \vec{g}$$
    such that
    $$\vec{\nabla \times \vec{g}=0, \quad \vec{\nabla} \cdot \vec{g}=-4 \pi \gamma \rho.$$
    This implies that there's a gravitational potential $\phi$ such that
    $$\vec{g}=-\vec{\nabla} \phi$$
    and thus
    $$\Delta \phi=+4 \pi \gamma \rho,$$
    from which you get
    $$\phi(\vec{x})=-\gamma \int_{\mathbb{R}} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
    For a spherically symmetric mass distribution around the origin of total mass ##m## this gives for the region outside of the material Newton's famous law
    $$\phi(\vec{x})=-\frac{\gamma m}{|\vec{x}|}$$
    and
    $$\vec{g}=-\vec{\nabla} \phi(\vec{x})=-m \gamma \frac{\vec{x}}{|\vec{x}|^3}.$$
     
  11. Jan 4, 2015 #10

    ShayanJ

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    But you actually didn't answer to the raised issue. My problem is with this part:
    What you said actually only implies [itex] \vec g=-\vec \nabla \phi+\vec B [/itex] for any vector field [itex] \vec B [/itex] that satisfies [itex]\vec \nabla \times \vec{B}=0 \ \ and \ \ \vec{\nabla} \cdot \vec{B}=0[/itex]. Of course [itex] \vec B=0 [/itex] is a solution but its not the only solution. Can it be proved that having the boundary condition [itex] \displaystyle \lim_{r\to\infty} \vec g=0 [/itex], gives us the unique solution [itex] \vec B=0 [/itex]?
     
  12. Jan 4, 2015 #11

    vanhees71

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    One can prove that the Helmholtz decomposition is unique up to a vectorial constant, if the vector field and its 1st derivatives vanishes at infinity. I don't know, where to find the formal proof of this. I'd have a look in textbooks on mathematical physics. If needed, I can try to find a reference.
     
  13. Jan 4, 2015 #12

    ShayanJ

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    No, its not needed so don't bother. Its just mathematical details. I just wanted to get sure I'm standing on rock here!
    Thanks
     
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