This post will be related to the discussion on uncertainty in simultaneous measurement of electro-magnetic fields,as sketched out in Gottfried p.10-11.

Personally,I did not find earlier the notion of uncertainty in the simultaneous measurement of EM fields up to arbitrary accuracy.The author claims the simultaneous measurement of E-M fields up to arbitrary accuracy will tell us the position and momentum of the charge emanating those fields simultaneously with arbitrary accuracy.

Now,from classical electrodynamics,the most general electric and magnetic fields are given by Griffths equation (11.62) & (11.63).

$\vec{E}\ =\frac{\ q}{4\pi\epsilon_0}\frac{\ R}{(\vec{\ R}\cdot\vec{\ u})^3}\ [(\ c^2\ -\ v^2)\vec{u}\ +\vec{R}\times(\vec{u}\times\vec{a}\)]$

where $\vec{R}\ =(\vec{r}\ -\vec{r"})\$

$\vec{u}\ =(\ c\hat{R}\ -\vec{v})\$

And $\vec{B}\ =\frac{\hat{R}\times\vec{E}}{\ c}\$

They are complicated,as expected.But what does the present author mean by saying that "position and momentum" can be simultaneously measured? I hope it is difficult to invert the formulae above,if not impossible.

The answer is not that straightforward.After a second thought,I found that given $\vec{E}\$ and $\vec{B}\$,there might be a way to solve for the momentum and velocity.But,we must refer to relativistic results.

In relativity,the law of motion is expressed in terms of three force and three velocities etc. as

$\vec{F}\ =\frac{\ m}{\sqrt(\ 1\ -\frac{\ v^2}{\ c^2}) }[\vec{a}\ +\frac{\vec{v}(\vec{v}\cdot\vec{a})}{\ c^2\ -\ v^2}]\$

[Griffiths,eqn-12.73;this is not difficult to prove]

Since the electric and magnetic fields are known to us,the force on the charged particle or currents may also assumed to be known(I must check this point).Then,the above becomes a differential equation in the first time derivative of $\vec{v}\$.It is not possible to say that this equaton is solvable.However,under certain condition on the net force,velocity can be found as a function of time;and hence,the position.One such example is worked out in Griffiths 12.10,though the context is entirely different.

That the speed can be found as a function of time is also clear from the result of problem 12.
36.It uses the formula written above and says

$\vec{a}\ =(\frac{\ q}{\ m})\sqrt(\ 1\ -\frac{\ v^2}{\ c^2}) [\vec{E}\ +(\vec{v}\times\vec{B})\ -\frac{\vec{v}(\vec{v}\cdot\vec{E}}{\ c^2}]\$

Again,the form of the equation is not so illuminating to solve.But under certain condition (I do not anticipate what the conditions could be at present),$\vec{a}\ =\frac{\ d\vec{v}}{\ dt}\$ can be solved for $\vec{v}\$.
This would give the momentum and hence,position...

Again,the uncertainty product for electromagnetic fields as written there is somewhat strange.The uncertainty is said to evolve from the interference of one test body in one space-time volume $\Omega_1\$ with another test body (how does it come into picture?) in space-time volume $\Omega_2\$ (the space-time regions can connected by light signal).This is unlike the uncertainty principle we are familiar with...position and momentum cannot be measured simultaneously with arbitrary accuracy.There we need not worry about the presence of any other body anywhere.

However,we will be in a position to justify the claim later when we read the relevant chapter.