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Originally posted by 1100f
"Maxwell's equations do not in themselves predict a specific value for the constant (or variable) c which appears in them."
Maxwell's equation do predict that there exist em waves which propagate at the speed of light c.
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That's right. Using the boldface
∂ for the "del" operator, Maxwell's equations
in vacuo are:
∂x
E+(1/c)∂
E/∂t.....(1)
∂x
B-(1/c)∂
E/∂t.....(2)
∂<sup>.</sup>
E=0.....(3)
∂<sup>.</sup>
B=0.....(4)
Making use of the vector identity:
∂x
∂x
A=
∂(
∂<sup>.</sup>
A)-
∂<sup>2</sup>
A,
We can take the curl of equation (1) to obtain:
∂x
∂x
E+(1/c)(∂/∂t)
∂x
B=0
∂(
∂<sup>.</sup>E)-
∂<sup>2</sup>
E+(1/c)(∂/∂t)
∂xB=0
The part in
blue vanishes by virtue of equation (3), and the part in
red can be rewritten as -(1/c)∂
E/∂t, by virtue of equation (2).
This gives us:
∂<sup>2</sup>
E-(1/c<sup>2</sup>)∂<sup>2</sup>
E/∂t<sup>2</sup>=0,
which is a wave equation. Taking the curl of equation (2) and following a similar path will show you that
B satisfies the exact same wave equation.
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Maxwell's equation give the value of this speed and it is indeed c.
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Right again. The components of the plane wave solutions of the wave equation are of the form:
A<sub>i</sub>(
x,t)=A<sub>i0</sub>sin(
k<sup>.</sup>
x-
wt+
f)
where
w/|
k|=c. Since the solutions have constant phase, we can derive the speed of the waves to be c.
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Next in the paper it is said that :"On the basis of this model, a Galilean invariant form of Maxwell's equations is obtained.".
Where are these Galilean invariant form of Maxwell's equations?
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The answer is that they do not exist. If electrodynamics is to be reformulated so that it is Galilean invariant, then the resulting equations will not be Maxwell's equations.
Here's what the reference from Jackson has to say about it. First, assume Galilean relativity. For a moving frame S' and a stationary frame S, we have:
x'=
x-
vt
t'=t
Let the wave equation hold in frame S. What does it look like in S'? We can derive that as follows:
∂/∂x=(∂x'/∂x)∂/∂x'=∂/∂x'
∂/∂y=(∂y'/∂y)∂/∂y'=∂/∂y'
∂/∂z=(∂z'/∂z)∂/∂z'=∂/∂z'
∂/∂t=(∂x'/∂/t)(∂/∂x')+(∂y'/∂t)(∂/∂y')+(∂z'/∂t)(∂/∂z')+(∂t'/∂t)(∂/∂t')
∂/∂t=
v<sup>.</sup>
∂'-(1/c)∂/∂t'
Squaring each operator and writing the equation in the coordinates of S' yields:
(
∂'<sup>2</sup>-(1/c<sup>2</sup>)(∂<sup>2</sup>/∂t'<sup>2</sup>-(2/c<sup>2</sup>)
v<sup>.</sup>
∂'(∂/∂t')-(1/c<sup>2</sup>)(
v<sup>.</sup>
∂)<sup>2</sup>)A<sub>i</sub>=0
where A<sub>i</sub> is any component of either the
E or
B field.
Notice that the above equation is
not a wave equation. That means that, if Galilean relativity is correct, then radio waves emitted from towers should become non-waves when you are driving in your car. If Galilean relativity is correct, then you should not be able to listen to the radio in your car.
The Lorentz transformation, on the other hand,
does preserve the form of the EM wave equation.
This is what none of the preachers of the Anti Relativity Religion understand. Einstein did
not pull length contraction and time dilation out of thin air. They are logically derived consequences of the requirement that the EM wave equation and the speed of light be the same in every frame. The original paper was not even called, "Intro to Special Relativity", it was called, "On the Electrodynamics of Moving Bodies".
Like it or not folks, relativity is correct. If any of you wants to convince thinking persons otherwise, then you will have to argue on these terms, because these are the terms in which relativity was formulated.
edit: fixed various bracket errors.
