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Why do E and G both degrade by 1/r^2?

  1. Nov 7, 2015 #1
    Both the electric and gravitational fields 'die off' at 1/r^2 ... why?

    Is this 1/r^2 value based on a geometric property (area of a sphere) alone? If so, then why don't all the fundamental forces degrade by this same value?

    Thanks
     
  2. jcsd
  3. Nov 7, 2015 #2
    The 1/r^2 is purely due to geometry - if you have any charge or mass distribution other than a sphere or point source, the 1/r^2 goes away and the field is more complicated.
     
  4. Nov 7, 2015 #3

    davenn

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    can you name one that doesn't ?
     
  5. Nov 7, 2015 #4
    Then why doesn't the strong force also degrade by 1/r^2 ?
     
  6. Nov 7, 2015 #5

    davenn

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    that's correct, the strong nuclear force and also the weak nuclear force degrade differently
    Basically, it is all to do with the mediating particles and if they have mass or not
    The strong nuclear force which has the mediator particle as a gluon is only strong at atomic distances.
    as is the weak nuclear and its mediator particles the W and Z bosons

    EM radiation is mediated by the photon and has no mass and as a result the range of an EM field is infinite

    Im not a particle physicist ... I don't want to go any deeper than that :wink:

    Dave
     
  7. Nov 7, 2015 #6
    That helps, so it's not just geometry, or more specifically, the geometric reasoning only applies when the mediating particles are mass-less... Thanks.
     
  8. Nov 8, 2015 #7
    the 1/r^2 thing is basically the way something radiating from a point decreases in a three dimensional space. A line source decreases at 1/r and if we lived in four spatial dimensions instead of three, a point source would decrease as 1/r^3.
     
  9. Nov 8, 2015 #8
    But that's not the complete answer (1/r^2 only applies to 2 of the 4 forces.) See Davenn's response above.
     
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