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Why all forces are subjected 1/r^2?

  1. Feb 11, 2004 #1
    Is it not strange thing that all forces are subjected 1/r^2 but not 1/r?
    Why their relationships are nonlinear? It must have the some logical explanation or imaginary picture .
     
  2. jcsd
  3. Feb 11, 2004 #2

    russ_watters

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    Last edited: Feb 11, 2004
  4. Feb 11, 2004 #3
    But 3-d already assumes a linear dependence, square dependence and cubic dependence.
     
  5. Feb 11, 2004 #4

    Chi Meson

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    FIrst of all, all forces are NOT inverse square proportional to distance. THe strong nuclear force actually increases with distance (up to a point).

    THe E-M force's inverse-square proportionality is easily visualized with teh concept of "flux": Imagine a particle that has a force field around it. Imagnine this force field as lines that radiate from the particle (This is the model of electric field lines, for example). The strength of the force on another particle depends onthe density of the field lines from the first particle.

    As you move away from the first particle, the same total number of lines will exist but they will spread further and further apart. As distance increases, the density of the lines gets smaller and smaller. Density is "total number of lines" divided by "the surface area of the sphere with a radius equal to the distance."

    Area of a sphere is proportional to the square of the radius, density (and thus strength of the force field) is proportional to the inverse square of the distance.
     
  6. Feb 11, 2004 #5

    turin

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    1/r is nonlinear, too.
     
  7. Feb 11, 2004 #6
    May be a Gravity force, for example, must be equal to
    G (4pi) M1*M2/(4pi)r^2
    I.e. a gravity force must be inversely to area of sphere with radius r. Then G is not correct and should be multiplied on 4pi.
     
  8. Feb 11, 2004 #7

    russ_watters

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    Re: Re: Why all forces are subjected 1/r^2?

    D'oh - can't believe I missed that.
    I don't understand what you mean: the diagram I linked shows a square relationship between area and distance in 3d space. That geometric relationship is likely the reason we see square (or inverse square) relationships so often in physical laws.
     
  9. Feb 11, 2004 #8
    Forces are subject to whatever potential energy function defines them, via a gradient:

    [itex]\vec{F} = -\vec{\nabla} U[/itex]

    Central potentials (1/r) yield inverse-square forces.

    But yes, it actually has to do with the dimensionality of the spaces as well. In fact, in modern theories of large extra dimensions, the Newtonian potential/force is expected to deviate from inverse/inverse-square at submillimeter scales.

    moderator edit: fixed TeX
     
    Last edited by a moderator: Feb 11, 2004
  10. Feb 11, 2004 #9

    Chi Meson

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    Similar to the way that the original "Coulomb constant", k, turned out to be a variation of the permeability of free space: k = 1/(4 pi epsilon)
     
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