stevenb said:
However, I'm an engineer, not a mathematician, so I'll just wait until a smart guy shows up to give us the answer.
Well, that smart guy never did show up, so I decided to take a couple of hours and pretend to be one. I think I came up with an illustrative derivation of Biot Savart from Coulomb's Law. It doesn't seem to be fully rigorous because I used a low speed (non relativistic) limit at one point to simplify the derivation. I started off thinking to use a Lorentz transformation of the EM field tensor as a way to convert Coulomb's law to Biot Savart's law. This triggered some old memories where I saw people use the Lorentz length contraction to demonstrate how magnetic forces can be viewed as manifestations of electric forces. So, that's the basic idea, and it turns out to be relatively straightforward, much to my own surprise. What I show here is an outline of the derivation with mention of the key concepts.
Start by considering a long wire with with current I and a test charge Q moving parallel to it with velocity v. The idea is to consider the magnetic force on the test charge due to the current element that is perpendicular to the radial vector, or in other words, the current element closest to the test charge. This will result in a formula for Biot Savart which does not include the cross product (sine of the angle dependence), however, the cross product rule will be intuitively obvious from the geometry and can be inserted in later. I do this just to keep the derivation brief and more understandable.
First, note that the magnetic effects of currents depend on net charge flow, not on the number of charges, type of charge carrier, nor drift velocity, independently. Thus, we can simplify the derivation by assuming positive charge carriers with a suitable linear charge density q chosen to match the assumption that the charges also move with velocity v matched to the test charge. Charge neutrality will require that the charge density of moving positive charges are canceled by an equal, but stationary relative to the conductor, charge density of negative charges.
Now, note that there is no magnetic force on the test charge in a reference frame that moves with the test charge. That is F=QVxB is zero in that reference frame. This requires that the force in that reference frame is due to electric forces only. The electric forces will result, despite the charge neutrality, because of the Lorentz contraction from motion. Even at nonrelativistic speeds this is true. Effectively the average separation between positive charges will increase due to length contraction of the moving conductor (relative to the moving frame). Also the average separation of negative charges will decrease because they are also moving relative to the moving reference frame. Hence, we can write the following net effect of Coulombs law to calculate the electric force. This is the net effect from both positive and negative charges.
{{F}\over{Q}}=E= {{q\; dl}\over{4\pi\epsilon_0 r^2 \sqrt{1- {{v^2}\over{c^2}}} }} - {{q\; dl \sqrt{1- {{v^2}\over{c^2}}} }\over{4\pi\epsilon_0 r^2}}
To simplify, use a first order expansion of the square root using the binomial expansion. This implies a non-relativistic limit for speeds much less than the speed of light.
{{F}\over{Q}}=E\approx {{q\; dl (1+{{v^2}\over{2c^2}} ) }\over{4\pi\epsilon_0 r^2 }} - {{q\; dl (1-{{v^2}\over{2c^2}} ) }\over{4\pi\epsilon_0 r^2}}
This simplifies to
{{F}\over{Q}}=E\approx {{q\;v^2 \;dl }\over{4\pi\epsilon_0 r^2 c^2 }} = {{I v \; dl}\over{4\pi\epsilon_0 r^2 c^2 }}
Noting that the speed of light can be related to permeability and permittivity and reinterpreting electric force as magnetic force QvB in the stationary frame, results in the following.
dB={{\mu_0 I \; dl}\over{4\pi r^2 }}
Now the cross product rule can be restored by noting that the length contraction only occurs in the direction of motion and not perpendicular to that. Hence for a parallel orientation, the electric field is equal from both positive and negative charges, so forces are zero. It's not hard to show that :
dB={{\mu_0 I \sin\theta \; dl}\over{4\pi r^2 }}
EDIT: thinking furthur on this, I think I have a mistake on the above. I think my calculation of electric force is off by a factor of two by not applying length contraction correctly. Then I need to equate this to magnetic force in a reference frame that has no electric force. I'll try this and update the derivation.
EDIT: OK, so here is my correction to the above.
In a reference frame moving with the test charge, the positive charges are at rest and the negative charges appear to move at velocity -v. Hence, the negative charges appear to be separated by a shorter distance due to length contraction. This results in net charge imbalance and stronger field due to negative charges, in a small current element. Applying coulombs law results in the following.
{{dF}\over{Q}}=dE= {{-q\; dl}\over{4\pi\epsilon_0 r^2 \sqrt{1- {{v^2}\over{c^2}}} }} + {{q\; dl }\over{4\pi\epsilon_0 r^2}}
Using the non-relativistic limit to first order results in the following.
{{dF}\over{Q}}=dE\approx {{-q\; dl (1+{{v^2}\over{2c^2}} ) }\over{4\pi\epsilon_0 r^2 }} + {{q\; dl }\over{4\pi\epsilon_0 r^2}}
and simplifying results in the following (which differs by a factor of 2 from before).
{{dF}\over{Q}}=dE\approx {{-q\;v^2 \;dl }\over{8\pi\epsilon_0 r^2 c^2 }} = {{-I v \; dl}\over{8\pi\epsilon_0 r^2 c^2 }}
Now consider a frame of reference moving at half of the speed of the test particle. In this frame the charges in the wire are moving in opposite directions with equal speed v/2. Hence, the Lorentz contraction is the same for each and there is no net electric charge to generate electric field. This means that all force can be interpreted as due to magnetic field. Since the test charge is also moving as speed v/2 in this frame, the magnetic force is dF/Q=vB/2. Equating this to the above relation results in the following, if it is noting that the speed of light can be related to permeability and permittivity.
dB={{\mu_0 I \; dl}\over{4\pi r^2 }}
Considering non perpendicular current elements will result in a cross product relation since length contraction does not occur in a direction perpendicular to motion. The end result is the same as before.
dB={{\mu_0 I \sin\theta \; dl}\over{4\pi r^2 }}
