Sojourner01 said:
Quite; I don't really understand how the classical magnetic field is supposed to come about, but I've been told that it's to do with a relativistic transform of the 'electric' field - so they're actually one and the same. Presumably if viewing an electrostatic field relativistically yields a new field, viewing a gravitational field will do the same. This had never occurred to me but is fascinating.
this is from Wikipedia (or used to be):
According to special relativity, electric and magnetic forces are part of a single physical phenomenon, electromagnetism; an electric force perceived by one observer will be perceived by another observer in a different frame of reference as a mixture of electric and magnetic forces. A magnetic force can be considered as simply the relativistic part of an electric force when the latter is seen by a moving observer.
A thought experiment one can do to show this is with two identical infinite and parallel lines of charge having no motion relative to each other but moving together relative to an observer. Another observer is moving alongside the two lines of charge (at the same velocity) and observes only electrostatic repulsive force and acceleration. The first or "stationary" observer seeing the two lines (and second observer) moving past with some known velocity also observes that the "moving" observer's clock is ticking more slowly (due to time dilation) and thus observes the repulsive acceleration of the lines more slowly than that which the "moving" observer sees. The reduction of repulsive acceleration can be thought of as an attractive force, in a classical physics context, that reduces the electrostatic repulsive force and also that is increasing with increasing velocity. This pseudo-force is precisely the same as the electromagnetic force in a classical context.
here's the quantitative version (from the talk page):
The classical electromagnetic effect is perfectly consistent with the lone electrostatic effect but with special relativity taken into consideration.
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_e}{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 that
would be the acceleration an observer traveling along with the lines of charge would observe.
Now, if special relativity is considered, the in-motion observer's clock 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 \.
I'll eat my last sentence, then
