Undergrad Field strength variation of different types of fields

Click For Summary
The field strength of gravitational, electric, and magnetic fields decreases with the inverse square of the distance from their source, a phenomenon linked to the propagation of energy from point sources. This decrease in energy density occurs because the energy must remain constant across spherical surfaces, leading to a relationship where energy density is inversely proportional to the square of the radius. The discussion highlights that fermions, which include particles like electrons and protons, possess rest mass and contribute to the generation of these fields. The underlying principle is that as energy propagates radially, the surface area of the sphere increases, resulting in a decrease in energy density. This concept applies to any conserved quantity that propagates outward in a similar manner.
petrushkagoogol
Messages
28
Reaction score
4
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:
 
Physics news on Phys.org
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.
 
  • Like
Likes petrushkagoogol
I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
12
Views
2K