Maxwell Equations

Basic Concepts and Principles

Maxwell's equations are partial differential equations governing electromagnetic phenomena in vacuum. These equations are
General Maxwell's Equations

1) \( div\vec{E} = \frac{\rho}{\epsilon_{0}} \)
2) \( div\vec{B} = 0 \)
3) \( rot\vec{B} = \frac{\partial \vec{B}}{\partial t} \)
4)\( rot\vec{B} = \mu_{0}\vec{j}+\epsilon_{0}\mu_{0}\frac{\partial \vec{E}}{\partial t} \)
In absence of electric charges and currents

1) \( div\vec{E} = 0 \)
2) \( div\vec{B} = 0 \)
3) \( rot\vec{B} = \frac{\partial \vec{B}}{\partial t} \)
4) \( rot\vec{B} = \epsilon_{0}\mu_{0}\frac{\partial \vec{E}}{\partial t} \)
Where E y B are the intensity of electric and magnetic fields respectively, j is the electric current density, ρ is the charge density, and ε0 (vacuum permittivity constant or electric constant), μ0 (vacuum permeability ) the constants

$$ \epsilon_{0}=8,854.10^{-12} m^{-2}.N^{-1}.C^{2}$$ $$ \mu_{0}=1,257.10^{-6} mk.C^{-2}$$

These two constants are related to the speed of light in vacuum, c, exactly

$$ c^{2}=\frac{1}{\epsilon_{0}\mu_{0}}$$
If we take Maxwell's equations in a vacuum, we can obtain the wave's equation as follows: first, we derive for t in last one and substituting in the second:

$$ \epsilon_{0} \mu_{0} \frac{\partial^{2} \vec{E}}{\partial t^{2}} = rot \: rot \vec{E} $$

Using the relation
$$\Delta \vec{F} = \triangledown (div\vec{F}) - rot \: rot \vec{F} \Rightarrow rot \: rot \vec{F} = \triangledown (div\vec{F}) -\Delta \vec{F} $$
So we obtain
$$ \epsilon_{0} \mu_{0} \frac{\partial^{2} \vec{E}}{\partial t^{2}} = \triangledown (div\vec{E}) -\Delta \vec{E} $$
Applying the first of equations (in vacuum) we finally obtain:

$$ \epsilon_{0} \mu_{0} \frac{\partial^{2} \vec{E}}{\partial t^{2}} = c^{2} \Delta \vec{E} $$
This is the the wave's equation. Note that the wave has speed=c. Thus, if for an observer the speed of light is not c then Maxwell's equations could not be hold for him.

Suppose now two inertial observers, called A and B. A moving uniformely to half speed of light in a radial direction to B and assume that B observer turns on a flashlight directed to A.

B notes that the the ray of light away from him at a speed c = 300.000 km/s.

Logically, A should measure the ray of light travels at a speed

$$c - \frac{c}{2} = \frac{c}{2} = 150.000 km/s (aprox.) $$

This would mean that for observer A, Maxwell's equations fails, at least as way we have seen before, because ligth-speed is not c. However, A observer might say that he is no moving at all. Who is moving is B in opposite direction!!.

Moreover, if A were moving to B he could measure the speed of light even higher than c!!!

In an attempt to explain this discrepancy, scientists of the nineteenth century created the "ether light" theory . On This theory speed of light in vacuum would only make sense respect to ether, this, in an analogy with the sound on earth whose speed of 340m / s is only meaningful relative to air.
Several experiments (like Michelson-Morley experiment in 1887) showed that this substance could not exist. Thus, classical mechanics and in particular the kinematics should be reviewed.

The theory of special relativity was born to solve these questions among others, teaching us that there is no privileged (inertial) observer and that for all of them it is true that the speed of light in a vacuum is invariably equal to, approximately 300,000km / s.

We will see the formulas in detail, but anticipating we will say now that time becomes relative to observer who measure it. The faster we move in our spaceship and we approach the speed of light, for an external observer from outside the ship, would see how our clocks move more slowly, that is, time is dilated and, at the same time, the coordinates spacecraft suffer a contraction, external oberver would see our spacecraft "shorter."





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Berndt Barkholz:

2012-12-01 03:53:18
You write (Maxwells equations) : c = 1/ (ε0 μ0) I hope you know this is wrong, it should be : c = 1/ (ε0 μ0)^0.5 or c^2 = 1/ (ε0 μ0)..... best regards

Mathstools Administrator:

2012-12-02 11:51:12
Absolutelly, Berndt. You are right, we change this. Thanks




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