Re: why lorentz transformation?


David McAnally wrote:
> "Sue..." <suzysewnshow@xxxxxxxxxxxx> writes:
>
> >David McAnally wrote:
> >> "Sue..." <suzysewnshow@xxxxxxxxxxxx> writes:
> >>
> >> >David McAnally wrote:
> >> >> "Sue..." <suzysewnshow@xxxxxxxxxxxx> writes:
> >> >>
> >> >> >francisco wrote:
> >> >> >> galileo's principle of relativity states that the laws of mechanics should
> >> >> >> be the same for all inertial observers. and indeed, newtonian mechanics is
> >> >> >> unchanged under galilean transformations. the problem is that maxwellian
> >> >> >> electrodynamics is not the same in every inertial frame under that
> >> >> >> transformation. so what to do? find a set of transformations under which
> >> >> >> both mechanics and electrodynamics are the same for all inertial frames.
> >> >> >> this leads to the lorentz transformation.
> >> >>
> >> >> >Since Maxwell's equations don't predict radiation either
> >> >>
> >> >> As the softrat points out, Maxwell's Equations do predict the existence of
> >> >> electromagnetic radiation. What line of reasoning could possibly have led
> >> >> you to believe that they don't? Seeing that you have made this
> >> >> demonstrably false claim about Maxwell's Equations, how could anybody
> >> >> trust your word on anything else?
> >>
> >> >Science is not the business of knowing who to trust.
> >> >The line of reasoning is in the two well considered
> >> >papers offered
> >>
> >> At no point in either of those papers was there any statement to the
> >> effect that Maxwell's Equations do not predict the existence of radiation.
> >> The first of the papers that you listed was connected to gauges for the
> >> electromagnetic potential, and the fact that the physical properties of
> >> the electromagnetic field are not affected by the choice of gauge.
> >>
> >> >and you have offered nothing
> >> >*demonstrable*.
> >>
> >> I shouldn't need to demonstrate the fact that Maxwell's Equations predict
> >> the existence of radiation. The fact is very well documented, and has
> >> been known since the nineteenth century. If, in the twenty-first century,
> >> this fact, which has been known for well over 100 years, escapes you, then
> >> that says more about your incompetence than it says about Maxwell's
> >> Equations.
> >>
> >> I will give a small part of the derivation, although, with your level of
> >> competence, I have no doubt that it will be a waste of time to explain it
> >> to you.
> >>
> >> In vacuo, and in the absence of sources, Maxwell's Equations reduce to
> >>
> >> @E_x/@x + @E_y/@y + @E_z/@z = 0,
> >>
> >> @B_z/@y - @B_y/@z = (1/c^2) @E_x/@t,
> >>
> >> @B_x/@z - @B_z/@x = (1/c^2) @E_y/@t,
> >>
> >> @B_y/@x - @B_x/@y = (1/c^2) @E_z/@t,
> >>
> >> @B_x/@x + @B_y/@y + @B_z/@z = 0,
> >>
> >> @E_z/@y - @E_y/@z = - @B_x/@t,
> >>
> >> @E_x/@z - @E_z/@x = - @B_y/@t,
> >>
> >> @E_y/@x - @E_x/@y = - @B_z/@t,
> >>
> >> where E_x is the x-component of the electric field, B_x is the x-component
> >> of the magnetic field, etc, c is a constant with the dimensions of speed
> >> (with known value 299 792 458 m s^{-1}), and for any field f, @f/@x
> >> denotes the derivative of f with respect to x, etc.
> >>
> >> Upon differentiation of the first equation above with respect to x, the
> >> second with respect to r, the seventh with respect to z, and the eighth
> >> with respect to y, we get
> >>
> >> @^2 E_x/@x^2 + @^2 E_y/(@x @y) + @^2 E_z/(@x @z) = 0,
> >>
> >> @^2 B_z/(@t @y) - @^2 B_y/(@t @z) = (1/c^2) @^2 E_x/@t^2,
> >>
> >> @^2 E_x/@z^2 - @^2 E_z/(@z @x) = - @^2 B_y/(@z @t),
> >>
> >> @^2 E_y/(@y @x) - @^2 E_x/@y^2 = - @^2 B_z/(@y @t).
> >>
> >> By an appropriate linear combination of these four statements, it follows
> >> that
> >>
> >> @^2 E_x/@x^2 + @^2 E_x/@y^2 + @^2 E_x/@z^2 = (1/c^2) @^2 E_x/@t^2.
> >>
> >> This is just the statement that E_x satisfies the wave equation.
> >> Similarly, E_y,E_z, B_x, B_y and B_z satisfy the wave equation. This
> >> immediately implies the existence of electromagnetic radiation which
> >> moves at a speed of c.
> >>
> >> Alternatively, in vacuo and in the absence of sources,
> >>
> >> div E = 0,
> >>
> >> curl B = (1/c^2) @E/@t,
> >>
> >> div B = 0,
> >>
> >> curl E = - @B/@t.
> >>
> >> It follows that
> >>
> >> curl curl B = (1/c^2) @(curl E)/@t = - (1/c^2) @^2 B/@t^2,
> >>
> >> and
> >>
> >> curl curl B = grad div B - del^2 B = - del^2 B,
> >>
> >> where del^2 is the Laplacian operator, i.e.
> >>
> >> del^2 f = @^2 f/@x^2 + @^2 f/@y^2 + @^2 f/@z^2.
> >>
> >> It follows that del^2 B = (1/c^2) @^2 B/@t^2. Similarly,
> >> del^2 E = (1/c^2) @^2 E/@t^2. So both E and B satisfy the
> >> wave equation, and so the existence of electromagnetic
> >> radiation automatically follows from these considerations.
> >>
> >> -----
>
> >I don't see any 1/r^2 term in your work.
>
> That is about a SOLUTION of the equations. Can't you even tell the
> difference between equations and their solutions, or is this distinction
> too hard for your minuscule intellect?
>
> Look up the Lienard-Weichert potential sometime,

<< R is vector from the retarded position of the electron (i.e., the
position at which is was "seen" from position r at time t), >>

http://scienceworld.wolfram.com/physics/Lienard-WiechertPotential.html
<and also the
> electromagnetic field associated with it. The Lienard-Weichert potential
> obeys the Lorenz gauge condition. The Coulomb formula for the electric
> field of a stationary charge is derivable from the Lienard-Weichert
> potential, and the Lienard-Weichert potential is derivable from Maxwell's
> Equations.

Can we see that ?

Sue...

The radiation from an accelerating charge (in which the
> strength of the electromagnetic field drops as 1/r) is also derivable from
> the Lienard-Weichert potential.
>
> >Try again:
> >http://arxiv.org/abs/physics/0204034
>
> That paper is completely irrelevant to any point that you think that you
> are making. That paper is about different potentials corresponding to
> solutions of Maxwell's Equations, and about the different gauges. The
> paper also notes the fact that all physical quantities are independent of
> the gauge, a fact which is well-known. The idiotic thing here is that
> you have already been told this fact. Perhaps you were too obtuse to
> understand what you were being told at the time.
>
> ------

.

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