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Inverse square laws?

  1. Jan 26, 2012 #1
    Why are Newtons law of universal gravitation,

    [itex]F=G\frac{m_{1}m_{2}}{r^{2}}[/itex]

    and Coulombs law,

    [itex] F = K_{e}\frac{q_{1}q_{2}}{r^{2}}[/itex]

    inverse square laws? I understand why they are inverse because the force decreases with distance but why is the distance, r, squared?

    Thanks
    AL
     
  2. jcsd
  3. Jan 26, 2012 #2
    In gravitation and electricity the force can be represented by lines radiating from a point in 3 dimensions. If you can accept that the strength of the force is something to do with the number of lines passing through a square metre then as distance increases the number of lines per square metre decreases and the force weakens. The surface through which all the lines pass is a sphere (if a point source) and the surface area is proportional to r^2 therefore the number of lines per square metre decreases as 1/r^2.
    This simplified explanation assumes that the lines are not absorbed or stopped in any way (true for gravity but less true for electrical lines)
    hope this helps
     
  4. Jan 26, 2012 #3

    vanhees71

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    This is a quite deep question about the physical laws. The answer is the mathematical structure of space-time. The laws you are looking at are all in the approximation of Newtonian mechanics or the special theory of relativity. In both space-time models, there always exists (by assumption!) at least one frame of reference, where the special principle of relativity, i.e., the principle of inertia is valid, i.e., a force-free particle always moves in straight lines or stays at rest. In addition, for any inertial observer, i.e., an observer who is at rest relatively to such an inertial frame, space is described by a Euclidean space.
    Particularly space is symmetric under arbitrary rotations around any point (isotropy of space, i.e., no direction in space is special). It is also homogeneous, i.e., it's invariant under translations: There's no special place in space. This means that on a fundamental level all laws of nature must be described by equations that are consistent with homogeneity and isotropy of space.

    Another very successful concept of physics is the description of forces by local fiel equations, and if you work out the very equations that obey the above mentioned symmetry principles you come to quite simple general forms of such equations. One such equation is Laplace's or Poisson's equation,

    [tex]\Delta \Phi(\vec{x})=-\rho(\vec{x}).[/tex]

    It's fundamental solution reads

    [tex]\Phi(\vec{x})=\int \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.[/tex]

    Particularly, if the source term on the right-hand side is taken as a pointlike unstructured object at the origin of the coordinate system, you get

    [tex]\Phi(\vec{x})=\frac{Q}{4 \pi |\vec{x}|}.[/tex]

    The corresponding vector field, describing the forces is given by the gradient, and this leads to radial forces which decrease with the square of the distance,

    [tex]\vec{F}=-\vec{\nabla} \Phi=\frac{Q}{4 \pi} \frac{\vec{x}}{|\vec{x}|^3}.[/tex]
     
  5. Jan 26, 2012 #4
    Thanks technician and vanhees71.

    I think I understand it now, maybe not the mathematical reason as vanhees presented.

    Would it be fair to say that in laymans terms as the distance increases the force is acting over a larger area so therefore its weaker because there is less force acting upon a given area?
     
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