Feynman's derivation of Maxwell equations, http://uk.arxiv.org/abs/hep-ph/0106235

Some times ago Dyson published a paper (Feynman's proof of the Maxwell equations, Am. J. Phys. 58 (1990), 209-211) about unusual proof of homogeneous Maxwell equations that Feynman had shown him in 1948. This approach can be generalized to the case of extra spatial dimensions, but only in seven dimensional space. See the paper at the indicated link.

Hi silagadze

The link .. ( non-subscribers have to pay for it.)
http://scitation.aip.org/getabs/servlet/Ge...=cvips&gifs=yes

any free discussion links?

Best wishes,

-C2.

Hello!
For discussion try the following link:
www.lns.cornell.edu/spr/1999-09/msg0018217.html

Hi silagadze,

( www.lns.cornell.edu/spr/1999-09/msg0018217.html )

I'm interested but way out of my depth on this.

From the link http://www.lns.cornell.edu/spr/1999-09/msg0018239.html

> Moataz H. Emam wrote:
> > Yesterday I came across a paper by Freeman Dyson publishing Feynman's
> > proof of Maxwell's equations. He starts by:
> >
> > F=ma
> > and the [x,x] m[x,dx/dt] commutators
> >
> > and ends up with Maxwell's homogeneous equations.

The moderator deleted the bits in between.

Hmm

Sorry I'm no help.

Best wishes,

-C2.

Because you asked, here is my post number 1000:

Ok, this is going to mix classical and quantum concepts, so it's not really a mathematical derivation but more like a back-of-envelope type math that Physicists like Feynman are really good at, with the good math to show that the result is rigorous only coming later.

Let's start with the classical Newton's second law:
(eq1) m ∂v/∂t = F(x,v,t)
and the commutation relations from (non-relativistic) quantum mechanics:
(eq2) [ x_i, x_j ] = 0
(eq3) [ x_i, v_j ] = i(h/2πm) ∂_ij
which imply for any two functions f(x,v,t) and g(x,t) the results below hold:
(eq4) [ x_i, f(x,v,t) ] = i(h/2πm)(∂f/∂v_i)
(eq5) [ v_i, g(x,t) ] = -i(h/2πm)(∂g/∂x_i)
Differentiating eq3 with time we have:
[ ∂x_i/∂t, v_j ] + [ x_i, ∂v_j/∂t ] = 0
or
[ v_i, v_j ] + [ x_i, ∂v_j/∂t ] = 0
or, using eq1
[ v_i, v_j ] + [ x_i, (1/m) F_j(x,v,t) ] = 0
or
[ v_i, v_j ] + (1/m)[ x_i, F_j(x,v,t) ] = 0
or, using eq4
[ v_i, v_j ] + (1/m)i(h/2πm)(∂F_j(x,v,t)/∂v_i) = 0
or
(eq6) ∂F_j(x,v,t)/∂v_i = i(2πm^2/h)[ v_i, v_j ]

On a side note, we use the Jacobi identity
[x_k, [ v_i, v_j ] ] + [v_i, [ v_j, x_k ] ] + [v_j, [ x_k, v_i ] ] = 0
and eq3 to calculate
[x_k, [ v_i, v_j ] ] + [v_i, constant_1 ] + [v_j, constant_2 ] = 0
or
[x_k, [ v_i, v_j ] ] = 0
which implies
∂([ v_i, v_j ])/∂v_k
so the right hand side of eq6 is does not depend on any component of velocity. So then let use define this as field B(x,t) such that
i(2πm^2/h)[ v_i, v_j ] = - ε_ijk B_k(x,t)
or
(eq7) B_i(x,t) = -i(2πm^2/h)ε_ijk[ v_j, v_k ]
where ε_ijk B_k is the usual antisymmetric tensor which is related to the cross-product and the Einstein summation convention where a duplication of indicies requires summation is applied.
Integrating eq6, and accounting for all components we have:
(eq8) F(x,v,t) = < v x B(x,t) > + E(x,t)
where E is the constant of integration and < v x B(x,t) > is a Weyl-ordered quantum-operator-safe expression: 1/2 ( v x B(x,t) - B(x,t) x v )

This looks like the Lorentz force law (we chose B and E for that reason) so lets identify E and B with the electric and magnetic field, respectively. But do they obey Maxwell's equations?

div B should be related to [v_i, B_i] or ε_ijk[v_i, [ v_j, v_k ]] or 1/3 ε_ijk ( [v_i, [ v_j, v_k ]] + [v_j, [ v_k, v_i ]] + [v_k, [ v_i, v_j ]] ) = 0
So no problem there.

Let's rewrite eq7 as
B(x,t) = -i(2πm^2/h)( v x v )
and calculate the total time derivative:
Left hand side:
dB/dt = ∂B/∂t + < v dot grad B >
where
d B_i/ dt = ∂B_i/∂t + 1/2 (v_j ∂B_i/∂x_j + ∂B_i/∂x_j v_j )
and the Right hand side:
-i(2πm^2/h)( ∂v/∂t x v + v x ∂v/∂t )
= -i(2πm/h)( F x v + v x F )
= -i(2πm/h)( E x v + v x E + < v x B > x v + v x < v x B > )
And by eq5, we have
-i(2πm/h)( E x v + v x E ) = - curl E
Evaluating the remaining terms we have for the ith component of < v x B > x v + v x < v x B >
(< v x B > x v + v x < v x B >)_i
= ε_ijk[ v_j, ( < v x B > )_k ]
= (1/2) ε_ijk ε_kmn [v_j, v_m B_n + B_n v_m ]
= (1/2) [v_j, v_i B_j + B_j v_i - v_j B_i - B_i v_j ]
= (1/2) ( [v_j, v_i] B_j + B_j [v_j, v_i] ) - 1/2 ( v_j[v_j, B_i] + [v_j, B_i]v_j )
But [v_j, v_i] B_j + B_j [v_j, v_i] goes like ε_ijk(B_k B_j + B_j B_k ) = 0
and so - 1/2 ( v_j[v_j, B_i] + [v_j, B_i]v_j ) = i(h/2πm)(1/2)(v_j ∂B_i/∂x_j + ∂B_i/∂x_j v_j ) = i(h/2πm) < v dot grad B >
Going back to our Left and Right hand sides and setting them equal to each other, we have
∂B/∂t + < v dot grad B > = - curl E + < v dot grad B >
or
∂B/∂t = - curl E
or
∂B/∂t + curl E = 0
which is the Faraday law.
div B = 0 and ∂B/∂t + curl E = 0
are the source-free equations (the Bianchi set) of the Maxwell equations, which according to Dyson: "a thing which baffles everybody including Feynman, because it ought not be possible"

[1] F. J. Dyson, "Feynman's proof of the Maxwell equations," Am. J. Phys. 58, 209-211 (1990).
[2] F. J. Dyson, "Feynman at Cornell," Phys. Today 42 (2), 32-38 (1989).

This post has been edited by rpenner on Nov 30 2006, 08:50 PM

Another way of deriving them using slightly more powerful but elegant methods is found in this PDF I uploaded 2 or 3 weeks ago to prove to Zephir you can derive the existence of Maxwell's Equations in a quantum field theory easily.

Equation (6) shows the F_mu.nu tensor along with something involving f^abc. In QED you're working with abelian gauge generators so f^abc = 0 and you have the usual Maxwell tensor, giving you electromagnetic from the simple fact you want a transformation invariant theory

That particular Lagrangian is for scalar QCD, not fermionic QED but the requirement for a D_mu is identical.

Keeping f^abc gives you gluon behaviour.

Yes, thank you, but I was trying to work with the limitation of this forum

I didn't mean it as a replacement for your post, more just "the math which came later" thing. As usual, you've more patience and explaination in your posts than I would have done

The method I did I doubt Feynman did initially. It relys on stuff which probably didn't come till a few years later. As you said, your post shows the true flare that Feynman had because he could see so many steps ahead and come out with the end result from quite a long method. One of those people Zephir doesn't think exists, a person with amazing ability in both maths and physical intuition