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 Thanks for getting back to me. 1)I wasn't trying to ask about absolute position here, so perhaps I should try and make myself clearer. In electrostatics, I can locate a charge at a point x and it experiences a force that depends upon the separation of that point from another point x'; if I think about x' as being somehow fixed, then the force a charge experiences is a function of position x, so it makes sense to write F^[electrostatic](x). Now in magnetostatics, both the source of the magnetic force and the object that experience it are extended objects. The only sense in which it makes sense to talk about a loop being located at a point is if I take it to be sufficiently small that I can neglect its length. Now this sort of approximation is straightforward if I have something like a small charged ball and I approximate it by a charge at the centre. But the force law (1.11) instructs me to integrate over both loops, taking orientation into account. I can't simply take a random point on each loop to be "the location" of the loop, because if I change these points then the resulting tangent vectors to the loop will have totally different orientations, and hence their cross products will give totally different results. So I naturally want to think about two current elements at x and x', compute a two-body force between them, and integrate over both loops (parametrised as functions of x and x'). But doing that integration should totally eliminate any functional dependence on the positions of points on either curve. So what is the meaning of writing F(x)? Question 2) is related to 1). Eq. 1.15 uses the "magnetostatic force F(x)" to define the magnetic field B(x). If I integrate over the loops in F(x), and wash out the functional dependence on x in the process, how do I define B(x)? It seems to me like the way to proceed is to define a force F(x) that acts on a current-carrying element located at a point, but that isn't what Thidé does in eq 1.11 because he integrates over both loops, so I'm not sure if I'm missing something. Thanks.