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Can a solid sphere only be considered a point for inverse-square forces?

  1. Aug 3, 2011 #1
    I am curious as to whether the force of solid sphere can be considered to be originating entirely at its center when the force is not of [itex]\frac{1}{r^2}[/itex] nature.

    It is said that the field inside a uniform spherical shell is zero for any [itex]\frac{1}{r^2}[/itex] type force and not for any others. It would seem likely that the other such conclusions would not hold for any force that was not proportional to the inverse square of the distance.

    If so, are there any similar conclusions or symmetries that are independent of the type of decay of the force?
     
  2. jcsd
  3. Aug 3, 2011 #2
    If you have a force that is NOT proportional to the inverse of r^2 then your dealing with some sort of crazy scenario where energy or space is not symmetrical.

    So you can't treat the sphere as a point without taking into consideration the modifications you have made to the system to make it non-symmetrical.
     
  4. Aug 3, 2011 #3
    To my knowledge there existed a fair amount of forces that are not inverse-square in nature.

    When electric dipoles are involved, for instance, the decay becomes one of [itex]\frac{1}{r^3}[/itex]. Also, the macroscopic functions for intermolecular forces such as those due to dipole-dipole and Van der Waals attractions are usually proportional to [itex]\frac{1}{r^8}[/itex].

    Would the proof for considering the force from solid sphere to be from its center (from outside the sphere) fall apart without the inverse-square force decay?
     
  5. Aug 3, 2011 #4
    Actually as long as it proportional to r of any order than it should not fall apart. Because if each part of the sphere that you move further away you also move another part the same distance closer when treating the sphere as a point at the center.
     
  6. Aug 3, 2011 #5

    uart

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    But do those things have spherical symmetry?
     
  7. Aug 3, 2011 #6
    You raise a very good point. The dipole-dipole force falls off as [itex]\frac{1}{r^3}[/itex] only along the plane equidistant between the two poles. Doh.

    As far as the LDF intermolecular forces go, the attraction is entirely independent of direction and therefore spherically symmetrical.
     
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