Astronuc said:
Moving charges induce a magnetic field, regardless of speed.
yeah, but, for a fixed charge, the strength of the magnetic field
does depend on the speed.
Maxwell's equations can be derived from SR, to which dextercioby alluded.
i can't do tensors, so my understanding of the math used in SR is limited to sophmore/junior introductory modern physics. but this thought experiment (which is
not general, only an illustrative example) persuaded me that magnetic fields (from a classical physics POV) come from the sole electrostatic field with SR taken into account:
The simplest hypothetical experiment would be two identical parallel infinite lines of charge (with charge per unit length of \lambda ) and some non-zero mass per unit length of \rho separated by some distance R. If the lineal mass density is small enough that gravitational forces can be neglected in comparison to the electrostatic forces, the static non-relativistic repulsive (outward) acceleration (at the instance of time that the lines of charge are separated by distance R) for each infinite parallel line of charge would be:
a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho}
If the lines of charge are moving together past the observer at some velocity, v, the non-relativistic electrostatic force would appear to be unchanged and this
would be the acceleration that an observer traveling along with the lines of charge would observe.
Now, if special relativity is considered, the clock of the observer moving along with the lines of charge would be ticking at a relative
rate (ticks per unit time or 1/time) of \sqrt{1 - v^2/c^2} from the point-of-view of the stationary observer because of time dilation. Since acceleration is proportional to (1/time)
2, the at-rest observer would observe an acceleration scaled by the square of that rate, or by {1 - v^2/c^2}, compared to what the moving observer sees. Then the observed outward acceleration of the two infinite lines as viewed by the stationary observer would be:
a = \left(1 - v^2 / c^2 \right) \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho}
or
a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} - \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} = \frac{ F_e - F_m }{\rho}
The first term in the numerator, F_e, is the electrostatic force (per unit length) outward and is reduced by the second term, F_m, which with a little manipulation, can be shown to be the classical magnetic force between two lines of charge (or conductors).
The electric current, i_0, in each conductor is
i_0 = v \lambda
and \frac{1}{\epsilon_0 c^2} is the magnetic permeability
\mu_0 = \frac{1}{\epsilon_0 c^2}
because c^2 = \frac{1}{ \mu_0 \epsilon_0 }
so you get for the 2
nd force term:
F_m = \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} = \frac{\mu_0}{4 \pi} \frac{2 i_0^2}{R}
which is precisely what the classical E&M textbooks say is the magnetic force (per unit length) between two parallel conductors, separated by R, with identical current i_0.
so you can look at it two ways: one is purely classical where the magnetic force exists and has a separate origin from the electrostatic force and effects of relativity are not considered. the other is where there is
only the electrostatic force but the effects if special relativity are considered. both results appear the same to the "stationary" observer.
