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I Field strength variation of different types of fields

  1. Jul 7, 2016 #1
    The field strength of gravitational, electric and magnetic fields vary as the inverse square of the distance from the source.
    Is this because all of the above fields are generated by fermions and they behave identically regardless of the nature of the fields ?
    Do the above fermions have rest mass ? :oops:
     
  2. jcsd
  3. Jul 9, 2016 #2

    BiGyElLoWhAt

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    I think it comes from conservation of energy. If you look at the point source of a field, it will generate a field out radially. Try to imagine a charge being created.
    It gets created, and the field propagates out radially, making a spherical wave front with area 4 pi r^2. The energy on that surface should be the same as the energy at a previous surface, or any later surface, since if it were to annihilate, it will have "generated" a finite amount of energy, that energy will continue to propagate in the same direction (barring strong gravitational fields and the like). This means the energy density must decrease.
    If we compare the total energy of some small area, call it ##\Delta A## at some small value of r, we will find it to be rather high. Now if we move to larger r, but keep the same ##\Delta A## we'll find it to be lower. Requiring the total energy through the surface at ##r_1## to be the same as that of ##r_2##, we get that ##E_1 = E_2## and ##\rho_1 *4\pi r_1^2 = \rho_2 *4\pi r_2^2## and ##\frac{\rho_1}{\rho_2} = \frac{r_2^2}{r_1^2}## which is a form of the inverse square law. So ##\rho_1 \propto \frac{1}{r_1^2}## It literally just comes from the fact that a spheres surface area is proportional to r^2.
    Fermions do have rest mass, at least all of the ones I'm aware of. Electrons, protons, etc.

    It's worth noting that the same logic can be applied to any radially propagating conserved quantity.
     
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