reilly said:
Physicsguru -- Read the literature. Physics has had a century to study and dissect relativity and E&M. Your objections have certainly been raised many times. How could you dispute the Lorentz invariance of Maxwell's Eq.? That invariance is central to much of modern physics, which, after all, is at least modestly successful. The debate was over certainly by the 1930s, and Einstein and Maxwell came through unscathed. Lasers, radar and masers, communication with space probes, radio and tv broadcasting, the Compton effect, photoelectric phenomena, atomic spectra and the classical theory of radiation all require the standard approach to realtivity and electrodynamics.
If you are serious about your objections then
1. Make sure you fully understand the electrodynamics and relativity that most of us treasure and love.(If you do not, few if any will pay attention to you)
2. Base your objections on experimental evidence. After all, physics is an empirical science.
There's no way a photon can be at rest in an inertial frame. If you can demonstrate that this is not true, buy your ticket to Stockholm.
Regards,
Reilly Atkinson
Suppose that the charge density [tex]\rho[/tex] which aappears in Coulomb's law satisfies the following postulate:
Postulate:
[tex]\nabla ^2 \rho = \frac{1}{c^2} \rho_t_t[/tex]
From the postulate above, we have:
[tex]\nabla ^2 \rho = \nabla \bullet \nabla \rho = \frac{1}{c^2} \frac{\partial}{\partial t} \frac{\partial \rho}{\partial t}[/tex]
Recall the continuity equation which holds
if electric charge is conserved:
[tex]\nabla \bullet \vec J = - \frac{\partial \rho}{\partial t}[/tex]
Therefore, if the postulate is true & electric charge is conserved then:
[tex]\nabla ^2 \rho = \nabla \bullet \nabla \rho = - \frac{1}{c^2} \frac{\partial}{\partial t} \nabla \bullet \vec J[/tex]
From which it follows that:
[tex]\nabla \bullet \nabla \rho = \nabla \bullet - \frac{1}{c^2} \frac{\partial \vec J}{\partial t}[/tex]
From which it follows that:
[tex]\nabla \rho = -\frac{1}{c^2} \frac{\partial \vec J }{\partial t}[/tex]
Now, consider the formula for the electric field at (x,y,z) due to the presence of some non-zero charge density [tex]\rho[/tex] which satisfies the postulate above. We have:
[tex]\vec E = \frac{1}{4\pi\epsilon_0} \int \rho d\tau \frac{\hat R}{R^2}[/tex]
Now, take the curl of both sides of the expression above to obtain:
[tex]\nabla \times \vec E = \frac{1}{4\pi\epsilon_0} \int \nabla \times (\rho d\tau \frac{\hat R}{R^2} )[/tex]
Recall the following mathematical theorem:
Theorem:
[tex]\nabla X (f\vec A) = \nabla f \times \vec A + f\nabla \times \vect A[/tex]
For any scalar function f, and any vector function A.
Since the charge density rho is a scalar function, and the volume current density is a vector function, it follows that:
[tex]\nabla \times \vec E = <br />
<br />
\frac{1}{4\pi\epsilon_0} \int \nabla \rho \times \frac{\hat R}{R^2} d\tau <br />
<br />
+<br />
<br />
\frac{1}{4\pi\epsilon_0} \int \rho \nabla \times \nabla (\frac{-1}{R}) d\tau <br />
[/tex]
Since the curl of a gradient is zero, the second term above drops out, and we are left with:
[tex]\nabla \times \vec E = \frac{1}{4\pi\epsilon_0} \int \nabla \rho \times \frac{\hat R}{R^2} d\tau[/tex]
Now, substitute the expression which was derived earlier from the postulate for the gradient of the charge density in the equation above to obtain:
[tex]\nabla \times \vec E = \frac{1}{4\pi\epsilon_0} \int <br />
<br />
<br />
-\frac{1}{c^2} \frac{\partial \vec J }{\partial t}<br />
<br />
\times \frac{\hat R}{R^2} d\tau[/tex]
Suppose that the classical Maxwellian result for the speed of light is correct, so that:
[tex]\frac{1}{c^2} = \epsilon_0 \mu_0[/tex]
We can now rewrite the curl of E as:
[tex]\nabla \times \vec E = - \frac {\partial}{\partial t}<br />
<br />
<br />
\frac{\mu_0}{4\pi} \int \vec J \times \frac{\hat R}{R^2} d\tau[/tex]
Where we have arrived at the Biot-Savart expression for B:
Biot-Savart Law (definition of Magnetic field B)
[tex]\vec B = \frac{\mu_0}{4\pi} \int \vec J \times \frac{\hat R}{R^2} d\tau[/tex]
Thus, we have arrived at one of Maxwell's equations, namely:
[tex]\nabla \times \vec E = - \frac{\partial \vec B}{\partial t}[/tex]
The forumal above is true, provided that all the postulates used in the derivation are true. Namely:
Postulate 1: Charge density obeys a wave equation, namely:
[tex]\nabla ^2 \rho = \frac{1}{c^2} \rho_t_t[/tex]
where the wavespeed is given by c, where:
[tex]c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}[/tex]
Postulate 2: The total electric charge of the universe cannot vary for any reason.
By experiment it is known that the speed of light through matter is less than the speed of light in vaccuum. The two speeds are related through the well known "index of refraction" n. Let v denote the speed of light through some substance with index of refraction n, we have:
[tex]v = \frac{c}{n}[/tex]
So, if instead, the wave equation for a charge density wave is:
[tex]\nabla ^2 \rho = \frac{1}{v^2} \rho_t_t[/tex]
where
[tex]v = \frac{c}{n}[/tex]
Then instead we have:
[tex]\nabla ^2 \rho = \frac{n^2}{c^2} \rho_t_t[/tex]
This would have altered the previous derivation above slightly. Instead, we would have had:
[tex]\nabla \times \vec E = \frac{1}{4\pi\epsilon_0} \int <br />
<br />
<br />
-\frac{n^2}{c^2} \frac{\partial \vec J }{\partial t}<br />
<br />
\times \frac{\hat R}{R^2} d\tau[/tex]
Let c denote the speed of light relative to an emitter, and let the speed of light relative to the emitter satisfy:
[tex]\frac{1}{c^2} = \epsilon_0 \mu_0[/tex]
Therefore, instead we have:
[tex]\nabla \times \vec E = - n^2 \frac{\partial \vec B}{\partial t}[/tex]