> But it's not at all obvious to me why the exponent a is 1 in nature
i think we can rule out any exponent just by dimensional analysis if you allow powers of q, then K has ambiguous units. Same reason you cant exponentiate unitful quantities
More specifically I think we can rule out even exponents by anti-symmetry of charge. That is q^2n = (-q)^2n which know is ruled out by experiment.
Show by experiment that the force of charge0 against charge1+charge2 equals the force of charge0 against charge1 plus charge0 against charge2. Induce an additive-homomorphic property F(r,q0,q1+q2) = F(r,q0,q1)+F(r,q0,q2). Then, exponent 1 follows.
I figured this out by listing a bunch of mathematical properties. I couldn't see how the author jumps from zero-preserving to multiply-charges, and I still don't know how, but we can call it out of scope lol
r : distance between p and q
q0 : charge 0
q1 : charge 1
F : coulomb force function
charge-commutative: F(r,q0,q1) = F(r,q1,q0)
zero-preserving: 0 = F(r,q0,0)
additive-homomorphic: F(r,q0,q1+q2) = F(r,q0,q1)+F(r,q0,q2)
homogenous-degree-1: F(r,q0,n*q1) = n*F(r,q0,q1)
multiplicative-separability: F(r,q0,q1) = K*R(r)*Q(q0,q1)
multiply-charges: F(r,q0,q1) = K*R(r)*(q0*q1)^a
Given F(r,q0,q1) = K*R(r)*(q0*q1)^a, charge-commutative, zero-preserving, additive-homomorphic.
Induction using additive-homomorphic proves homogenous-degree-1. (For example, F(r,q0,2*q1) = F(r,q0,q1+q1) = 2*F(r,q0,q1))
Equational proof follows from homogenous-degree-1:
K*R(r)*(q0*n*q1)^a = n*K*R(r)*(q0*q1)^a
(q0*n*q1)^a = n*(q0*q1)^a
n^a*(q0*q1)^a = n*(q0*q1)^a
n^a = n
n = 0 or a = 1
n≠0, therefore a=1.
I feel like that should be gettable from rotational symmetry and somehow mentioning L2 norm to get the square. Either that or conservation of E-field flux.
Rotational symmetry or L^2 norm only matter in vector formulations. I assumed a scalar formulation.
Conservation of E-field flux certainly implies additive-homomorphism (addition of charges equals addition of forces). But that seems a bit ahistorical because Maxwell would develop field theory 70 years after Coloumb. Either way, the axiom you choose requires empirical justification, and I think a home hobbyist could more easily demonstrate by experiment that adding charges will add the forces.
Or, if you meant F(r,1,1) = K/r^2, then yeah, conservation of E-field flux could give you an inverse square law. But again, that requires an experiment to justify the axiom.
nh23423fefe · · focus · HN ↗
i think we can rule out any exponent just by dimensional analysis if you allow powers of q, then K has ambiguous units. Same reason you cant exponentiate unitful quantities
More specifically I think we can rule out even exponents by anti-symmetry of charge. That is q^2n = (-q)^2n which know is ruled out by experiment.
amavect · · focus · HN ↗
I figured this out by listing a bunch of mathematical properties. I couldn't see how the author jumps from zero-preserving to multiply-charges, and I still don't know how, but we can call it out of scope lol
nh23423fefe · · focus · HN ↗
F(r,q0,q1) = F(r,q0,q1+0) = F(r,q0,q1)+F(r,q0,0) = F(r,q0,q1) + 0
amavect · · focus · HN ↗
nh23423fefe · · focus · HN ↗
amavect · · focus · HN ↗
Conservation of E-field flux certainly implies additive-homomorphism (addition of charges equals addition of forces). But that seems a bit ahistorical because Maxwell would develop field theory 70 years after Coloumb. Either way, the axiom you choose requires empirical justification, and I think a home hobbyist could more easily demonstrate by experiment that adding charges will add the forces.
Or, if you meant F(r,1,1) = K/r^2, then yeah, conservation of E-field flux could give you an inverse square law. But again, that requires an experiment to justify the axiom.