
Feynman’s proof of the Maxwell equations - luisb
http://fermatslibrary.com/s/feynmans-proof-of-the-maxwell-equations
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effie
I think this paper has a very misleading title, as the result it presents is
not as remarkable as to be actually proof of what is known as Maxwell's
equations (they were formulated by Heaviside btw.) Basically what the paper
does is that it defines operators for ρ, j, E based on the hindsight knowledge
of Maxwell's equations and non-relativistic equation of motion for a particle;
then it shows that it is possible to have both sets as Heisenberg equations.
That's definitely _not_ something a physicist would call deriving Maxwell's
equations. Most of the thing claimed to be derived in the title is actually
defined/assumed, the result is merely that the Heisenberg formalism, the
commutation relations and non-relativistic equation of motion do not seem
incompatible. The speed of light in the result is there purely because Dyson
knows the result he needs to get.

~~~
evanpw
I disagree. I would call this a proof but not a derivation. In other words,
each step follows logically from the previous ones, even if there's no way
that you would discover the final equations this way if you didn't already
know or guess them.

In general, this is part of what makes reading mathematical papers so
difficult; the steps of the proofs are almost never written down in the order
they were discovered. All of the false starts and a lot of the intuition is
discarded to yield a shorter but sometimes totally mysterious path from the
assumptions to the conclusion.

~~~
eru
The same applies to the more theoretical aspects of programming. (Because
Computer Science is a subset of math..)

See [http://blog.sigfpe.com/2006/08/you-could-have-invented-
monad...](http://blog.sigfpe.com/2006/08/you-could-have-invented-monads-
and.html) for someone trying to show the way, and not just the result.

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cschmidt
See also Feynman’s Derivation of the Schrödinger Equation discussed a couple
of weeks ago...

[https://news.ycombinator.com/item?id=10974803](https://news.ycombinator.com/item?id=10974803)

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noobermin
To be honest, the best part isn't the proof, it's the editor's note from
Dyson.

~~~
ciot1CDM
(Not a physicist) So what's the result of applying Feynman's axioms to the
analysis of whatever gedanken experiment that lead Einstein to special
relativity? Does it further enlighten whatever paradox existed that required a
resolution via special relativity?

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betolink
I was lucky enough of having a good math teacher that understood and taught
Lorentz and Einstein equations, he used to say "Once that you get it, you will
laugh at them", here I am several years later still waiting for some of those
laughs.

~~~
lightlyused
I think that was the joke.

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theophrastus
Here is the pdf (downloadable form) of this fascinating paper:
[http://signallake.com/innovation/DysonMaxwell041989.pdf](http://signallake.com/innovation/DysonMaxwell041989.pdf)

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Steuard
Aside: I'm not signed up for the site (not interested in tying that login to
other accounts), but I wanted to comment that luisb's explanation of Eq. (13)
seems to miss the point a little. That equation doesn't need to be _derived_
from Eq. (4). Instead, Eq. (13) is just a statement that any antisymmetric 3x3
matrix can always be recharacterized as a (pseudo)vector "times" the totally
antisymmetric Levi-Civita symbol epsilon_{ijk}. As the text says, Eq. (13) is
the _definition_ of H. Only after we've defined what H is can we then use Eq.
(4) to provide a definition of E.

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amelius
It would be nice if somebody could post a recommended bibliography required to
understand this proof, assuming graduate level physics background.

Edit: Sorry I meant undergraduate.

~~~
aroberge
The proof starts from two sets of equations:

1\. Newton's second law (first year undergraduate)

2\. Commutation relations in QM (2nd or 3rd year undergraduate)

To make the desired connection, one must have seen Maxwell's Equations (which
some first year undergraduate physics textbooks include).

There is no need to assume graduate level physics; someone having done a B.Sc.
in Physics should know more than enough to follow this derivation.

~~~
pc86
Yeah my Political Science degree is not much help at the moment.

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diebir
There are some interesting comments below the article. The article claims to
derive Maxwell equations based on Newtonian mechanics (Galilean
transformations), which sounded a little strange. I remember pretty well that
one had to consider Lorentz transformations to obtain Maxwell equations. Sure
enough, the comments indicate that this is required and the proof does not
quite succeed.

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effie
OK I have a problem with this step: _[x '_1,H_1]+[x'_2,H_2]+[x'_3,H_3] is
equivalent to ∂_1H_1 + ∂_2H_2 + ∂_3H_3 = 0._ In the whole paper until the eq.
18 Dyson talks about operators being functions of particle coordinate
operators x_k and velocities x'_k. There is no mention of space coordinates.
Suddenly he claims that some operator is equivalent to divergence of a field H
that was never mentioned before. It is not clear how does he calculate what
∂_1H_1 is. I do not see how this makes any sense.

~~~
eusebio
He is not claiming that the commutator is equivalent to the divergence, what
he is saying is that [xl',Hl] = 0 is equivalent to div H = 0, which is
different. If you click on the annotation on the left you will see that
[xl',Hl] = 0 means that Hl is not a function of xl and so ∂_lH_l will be zero
and consequently div H = Sum ∂_lH_l is also zero.

~~~
effie
I know, I forgot to check the formula before clicking the submit button. I
meant it as you explain. The problem is that this commutator equation implies
nothing about partial derivatives of H with respect to spatial coordinates
that EM fields are function of, because Dyson's x_l are not spatial
coordinates (which are always real numbers), but they are non-relativistic
Heisenberg operators of coordinates of considered particle (infinite
matrices).

The difference is as important as the difference between coordinate of a space
point and coordinate of a particle at some definite time. This confusion was
probably caused by the unfortunate choice of notation, where particle
coordinates were denoted as x_l, notation better spent on the general spatial
coordinates. If we use, say, r_l to denote Heisenberg operators for the
particle coordinates, it is clear that (not being a function of operators r_l)
implies nothing about (not being a function of position x_l).

Let me simplify this argument. What Dyson is saying seems no different from
saying that if you have a field v_l(x) giving velocity of water at general
point x, this field is obviously not a function of coordinates r_l of a test
particle and it follows that sum of derivatives ∂_lv_l is zero. Behold, we
arrive at the conclusion that the flow must be incompressible, no need to
assume anything from experience. All water must be incompressible! The
absurdity of this conclusion is pretty obvious, and the reason it was obtained
is clear: v_l is independent of r_l, but it depends on x_l, so nothing about
partial derivatives ∂/∂x_l can be easily inferred.

