Harmless Lorentz transformation question

[Hurkyl]

There is no internal inconsistency in this theory, it just isn't supported by the evidence.

What is that this theory? Of course I've heard of people talking vaguely about it, but I've never seen anyone try and present formulae for it.

 

[Physicsguru]

Because the vast majority of your posts in this thread indicate a poor understanding of the border between the two theories,and hence if the theories as a whole.

Daniel.

P.S.No offense meant.

Is simultaneity absolute or relative, this requires total understanding of the border between the two theories. Prove the answer.

Kind regards,

Guru

 

[JesseM]

What is that this theory? Of course I've heard of people talking vaguely about it, but I've never seen anyone try and present formulae for it.

Presumably the theory is that there is a frame S where Maxwell's laws hold exactly, and that in another frame S' moving at velocity v relative to S, you'd perform a Galilei transform on Maxwell's equations--replace x by x' + vt', replace dx/dt by dx'/dt' + v, and so forth. This would give a new set of equations that would describe how the laws of electromagnetism work in frame S'.

 

[JesseM]

Is it possible that the speed of light is c only relative to the source of that which emits the photons/radiation, and that is the correct physical interpretation of the otherwise mysterious parameter \frac{1}{\sqrt{\epsilon_0 \mu_0}}?

It's logically possible, but it would mean there is no frame where Maxwell's laws work, since it can be proved that in any frame where Maxwell's laws hold, the velocity of any electromagnetic wave is \frac{1}{\sqrt{\epsilon_0 \mu_0}}, regardless of the velocity of the source in that frame. And if you're going to throw out Maxwell's laws, what theory do you want to replace them with? It will have to be one that reproduces all the predictions of Maxwell's laws that have been verified by experiment, yet says something completely different about how electromagnetic waves work--I doubt it's possible to construct such a theory.

 

[quantumdude]

What is that this theory? Of course I've heard of people talking vaguely about it, but I've never seen anyone try and present formulae for it.

It's Maxwell's electrodynamics coupled with Galilean relativity. The formulae are nothing other than Maxwell's equations and the Galilean transformation.

 

[Physicsguru]

It's logically possible, but it would mean there is no frame where Maxwell's laws work, since it can be proved that in any frame where Maxwell's laws hold, the velocity of any electromagnetic wave is \frac{1}{\sqrt{\epsilon_0 \mu_0}}, regardless of the velocity of the source in that frame. And if you're going to throw out Maxwell's laws, what theory do you want to replace them with? It will have to be one that reproduces all the predictions of Maxwell's laws that have been verified by experiment, yet says something completely different about how electromagnetic waves work--I doubt it's possible to construct such a theory.

What part of Maxwell's original theory, prevents the equations from only being true in a frame where the emitter which was previously at rest, has just been accelerated (or frames moving at constant velocity relative to such a frame)?

Kind regards,

Guru

 

[JesseM]

What part of Maxwell's original theory, prevents the equations from only being true in a frame where the emitter which was previously at rest, has just been accelerated (or frames moving at constant velocity relative to such a frame)?

Suppose you have two emitters moving at different velocities. Whatever frame you use, Maxwell's laws should be able to make predictions about both emitters from the perspective of that frame (after all, there's no magnetism unless you have a charge that's moving in your frame), and they will predict that both emitters create electromagnetic waves that move at the same speed when they accelerate.

 

[Hurkyl]

It's Maxwell's electrodynamics coupled with Galilean relativity. The formulae are nothing other than Maxwell's equations and the Galilean transformation.

That gives you a preferred frame where all light travels at c... not a theory where the speed of light is c relative to the motion of the source. (Using Galilean velocity addition)

 

[JesseM]

That gives you a preferred frame where all light travels at c... not a theory where the speed of light is c relative to the motion of the source. (Using Galilean velocity addition)

In the post where I said "There is no internal inconsistency in this theory, it just isn't supported by the evidence", I wasn't referring to a theory where the velocity of electromagnetic waves depended on the velocity of the source--I said "it is certainly logically possible that Maxwell's laws would only hold in a single preferred reference frame (the rest frame of the ether, as physicists used to think of it) and that in other frames they'd have to be modified by a Galilei transform, which would insure that any observer moving at v relative to this preferred frame would see light moving at v+c in one direction and v-c in the other."

 

[Hurkyl]

Whoops, I misread then. My curiosity still stands, though.

 

[dextercioby]

The equations of Hertz's G-invariant electrodynamics (vacuum case) are obtained from the ones of Maxwell by the simple G-invariant transformations
\nabla\rightarrow \nabla ' =\nabla

\frac{\partial}{\partial t} \rightarrow \frac{\partial}{\partial t'}=\frac{\partial}{\partial t}+\vec{u} ' \cdot \nabla

The equations of H.Hertz are very valid for small velocities wrt "c".

Daniel.

 

[pervect]

Yes Jesse, I am just asking for a proof that Maxwell's laws prove that an electromagnetic wave must always travel at \frac{1}{\sqrt{\epsilon_0 \mu_0}} in an inertial reference frame.

I fail to see how the argument presented in the link above, leads to the conclusion that, "In any inertial reference frame, the speed of an electromagnetic wave is \frac{1}{\sqrt{\epsilon_0 \mu_0}}.

Regards,

Guru

The argument presented in the link shows that in any coordinate system where Maxwell's equations work, the speed of an electromagnetic wave is 'c'.

Note that this is a property of Maxwell's equations themselves, it has nothing to do with inertial frames.

If Maxwell's equations work, they represent a system where waves propagate at 'c'.

The link

dervies the equation

\nabla^2 E = \frac{1}{c^2} \frac{\partial ^2 E}{\partial t^2}

where c = \mbox{1/\sqrt{\epsilon_0 \mu_0}}

from Maxwell's equations

It's easiest to see that this represents a wave traveling at 'c' by considering a case with only one spatial dimension. The results are very similar for three dimensions.

http://planetmath.org/encyclopedia/WaveEquation.html

goes through the solution process for the 1d equation. One simply lets
a = x-ct and b=x+ct, then one arrives at D'Alembert's solution

u(x,t) = C1 f(x-ct) + C2 f(x+ct)

It's easy to see this is a solution by direct substitution, the uniquness theorems for PDE's guarantee that any solution which satisfies the equations can be put into this form.

You may need to consult a textbook on differential equations to find a detailed proof of the uniqueness theorem.

 

[Physicsguru]

Go back and look at the London derivation of the magnetic flux (what you wrote doesn't count). Classical electrodynamics is fully used to show this.

It is the single-valuedness requirement (i.e. pure boundary conditions) that is causing the quantized flux. This is not unusual because Maxwell equations only give you the recipe to solve for SPECIFIC situation! Without such equations, it is impossible to show such quantization!

I prefer to formulate the laws of physics based upon experiments, rather then let them rest upon the theory of functions of a complex variable. In other words, I wouldn't go looking for an ad hoc mathematical reason why magnetic flux is quantized, I would simply say I inferred this fact from our experiments. There is nothing logically/epistemologically objectionable to such an approach.

Let f denote a real valued scalar function. Here is the gradient theorem:

\int^b_a \nabla f \bullet d \vec l = f(b) - f(a)

So now, suppose we integrate around a closed path. In that case, f(b)=f(a) so that we have:

\oint \nabla f \bullet d \vec l = 0

Which as you can see, is not +/- 2n \pi

Now, I must admit, I have no clue about this "derivation using classical electrodynamics" which you speak of, so I would really appreciate you showing me how classical electrodynamics is used to reach the conclusion that magnetic flux is quantized... in a cooper pair no less.

Let me preface this by saying that there is nothing "classical" about any derivation of any formula which either uses complex variables or deliberately inserts the parameter 'h'.

Regards,

Guru

 

[Physicsguru]

The argument presented in the link shows that in any coordinate system where Maxwell's equations work, the speed of an electromagnetic wave is 'c'.

Note that this is a property of Maxwell's equations themselves, it has nothing to do with inertial frames.

If Maxwell's equations work, they represent a system where waves propagate at 'c'.

I have some questions, particularly as regards the variables which are differentiated, and in what kind of frame they are differentiated in.

We have x,y,z,t for some field point, and we have (x`,y`,z`,t`) for some moving charged particle.

1) Is it possible to use classical electrodynamics to infer that, "if coordinate system F is an inertial reference frame then Maxwell's equations are true in F?"

2) Is it possible to use classical electrodynamics to infer that, "if Maxwell's equations are true in coordinate system F, then F is an inertial reference frame?"

The link

dervies the equation

\nabla^2 E = \frac{1}{c^2} \frac{\partial ^2 E}{\partial t^2}

where c = \mbox{1/\sqrt{\epsilon_0 \mu_0}}

from Maxwell's equations.

You forgot something.

\nabla \bullet \vec E = 0
\nabla \bullet \vec B = 0
\nabla \times \vec E = - \frac{\partial \vec B}{\partial t}
\nabla \times \vec B = \epsilon_0 \mu_0 \frac{\partial \vec E}{\partial t}

Recall the following purely mathematical theorem:

Theorem:

\nabla \times \nabla \times V = \nabla (\nabla \bullet \vec V ) - \nabla^2 \vec V

Using the previous theorem, take the curl of both sides of the fourth Maxwell equation above, to obtain:

\nabla \times \nabla \times B = \nabla (\nabla \bullet \vec B ) - \nabla^2 \vec B = \epsilon_0 \mu_0 \frac{\partial }{\partial t} (\nabla \times \vec E)

Noting that the divergence of B is zero, and that the curl of E is negative partial B with respect to t (in the reference frame in question), we have:

- \nabla^2 \vec B = -\epsilon_0 \mu_0 \frac{\partial }{\partial t} \frac{\partial \vec B}{\partial t}

From which it follows that:

\nabla^2 \vec B = \epsilon_0 \mu_0 \frac{\partial^2 \vec B }{\partial t^2}

Which is also true in a coordinate system in which Maxwell's equations are true, assuming they are true in any at all, keep in mind that if Maxwell's equations lead to even one contradiction, Maxwell's equations aren't simultaneously true in even one reference frame (inertial or otherwise).

Regards,

Guru

 

[JesseM]

1) Is it possible to use classical electrodynamics to infer that, "if coordinate system F is an inertial reference frame then Maxwell's equations are true in F?"

2) Is it possible to use classical electrodynamics to infer that, "if Maxwell's equations are true in coordinate system F, then F is an inertial reference frame?"

As I've pointed out several times already, you can't use classical electrodynamics to prove that Maxwell's laws are true in every inertial frame. There is no internal inconsistency in the theory that Maxwell's laws only hold in the rest frame of the ether, and that in other inertial frames they have to be modified by a Galilei transformation; but this theory is not supported by the evidence, and it is a lot less elegant to boot.

 

[pervect]

I have some questions, particularly as regards the variables which are differentiated, and in what kind of frame they are differentiated in.

We have x,y,z,t for some field point, and we have (x`,y`,z`,t`) for some moving charged particle.

1) Is it possible to use classical electrodynamics to infer that, "if coordinate system F is an inertial reference frame then Maxwell's equations are true in F?"

2) Is it possible to use classical electrodynamics to infer that, "if Maxwell's equations are true in coordinate system F, then F is an inertial reference frame?"

If by classical electrodynamics you mean relativistic electrodynamics, it is possible to say that Maxwell's equations are true in any inertial frame. But it's not clear if you mean relativistic electrodynamics when you say "classical electrodynamics". As other posters have pointed out, before relativity, it was thought that there was only one frame in which the speed of light was constant. The Michelson Moreley experiment and other later experiments disproved this set of theories.

You forgot something.

[snip derivation]

From which it follows that:

\nabla^2 \vec B = \epsilon_0 \mu_0 \frac{\partial^2 \vec B }{\partial t^2}

Which is also true in a coordinate system in which Maxwell's equations are true, assuming they are true in any at all, keep in mind that if Maxwell's equations lead to even one contradiction, Maxwell's equations aren't simultaneously true in even one reference frame (inertial or otherwise).

Regards,

Guru

You've just shown that in a vacuum, B obeys the wave equation, just as E does. This is just further confirmation of the fact that the speed of light as given by Maxwell's equations is a constant, c, in a vacuum.

 

[Physicsguru]

As I've pointed out several times already, you can't use classical electrodynamics to prove that Maxwell's laws are true in every inertial frame. There is no internal inconsistency in the theory that Maxwell's laws only hold in the rest frame of the ether, and that in other inertial frames they have to be modified by a Galilei transformation; but this theory is not supported by the evidence, and it is a lot less elegant to boot.

I am not looking to theorize that Maxwell's equations are true only in one preferred frame of reference, one at rest with respect to the 'ether'.

 

[JesseM]

I am not looking to theorize that Maxwell's equations are true only in one preferred frame of reference, one at rest with respect to the 'ether'.

Well, if you agree that Maxwell's equations are true in every inertial frame, the only possible way for this to be true is if you use the Lorentz transform to translate between the coordinates of different inertial frames.

 

[Physicsguru]

If by classical electrodynamics you mean relativistic electrodynamics, it is possible to say that Maxwell's equations are true in any inertial frame.

I do not mean relativistic electrodynamics. Here is what I do mean by classical electrodynamics:

We start off with electrostatics. In a frame where a particle with electric charge q1 is permantly at rest (pretend q1 has an infinite inertia), the particle is the source of an electrostatic field in this frame which is expressed mathematically as:

\vec E = \frac{q1}{4\pi\epsilon_0} \frac{\hat R}{R^2}

If another electrically charged particle of electric charge q2 finds itself immersed in the electric field due to q1, this particle will be subjected to an external force F, which obeys:

\vec F = \frac{d\vec P}{dt} = q2 \vec E = \frac{q1q2}{4\pi\epsilon_0} \frac{\hat R}{R^2}

Where P is the particle's momentum, which is given by Mv, where v is its velocity in the frame, and M is the inertial mass of q2.

For a large body containing an integral number of electric charges, the total electric field of the object is governed by:

\vec E = \frac{1}{4\pi\epsilon_0} \int \rho d\tau \frac{\hat R}{R^2}

Where d tau is a differential volume element, and rho is the charge density, measured in coulombs per cubic meter. In electrostatics rho is not a function of time, in electrodynamics rho is a function of time.

Case 1: The charge density is not a function of time, only a function of position. That is:

\rho = \rho (x^\prime,y^\prime,z^\prime)

Using vector calculus we can show that:

\nabla \bullet \vec E = \frac{\rho}{\epsilon_0}

and

\nabla \times \vec E = 0


The second of these two equations is not true if rho can vary in time, as is the case with electric current.

Case 2: The charge density is a function of time t.

\rho = \rho (x^\prime,y^\prime,z^\prime,t)

Let us compute the curl of the electric field.

\nabla \times \vec E = \nabla \times \frac{1}{4\pi\epsilon_0} \int \rho d\tau \frac{\hat R}{R^2}











If many electric charges are moving in a frame, they will contribute to a net volume current density J, which will generate a so called magnetic field B, which satisfies the Biot-Savart law below:

\vec B = \frac{\mu_0}{4\pi} \int \vec J \times \frac{\hat R}{R^2} d\tau

If a charged particle with speed v enters a magnetic field B, it will experience an external force given by:

\vec F = q(\vec v \times \vec B)

 

[Hurkyl]

Don't forget that the current density is supposed to remain constant over time as well. (Thus, magnetostatics)

 

[Physicsguru]

Don't forget that the current density is supposed to remain constant over time as well. (Thus, magnetostatics)

Hey :) i didnt know anyone was out there. And thank you :)

 

[dextercioby]

For the record,CED (or classical electrodynamics) is the theory understood as the Einstein-Minkowski L-invariant electrodynamics and not a subtheory of it.

What's your view on QED (quantum electrodynamics)...?

Daniel.

 

[pervect]

I do not mean relativistic electrodynamics. Here is what I do mean by classical electrodynamics:

We start off with electrostatics. In a frame where a particle with electric charge q1 is permantly at rest (pretend q1 has an infinite inertia), the particle is the source of an electrostatic field in this frame which is expressed mathematically as:

\vec E = \frac{q1}{4\pi\epsilon_0} \frac{\hat R}{R^2}

If another electrically charged particle of electric charge q2 finds itself immersed in the electric field due to q1, this particle will be subjected to an external force F, which obeys:

\vec F = \frac{d\vec P}{dt} = q2 \vec E = \frac{q1q2}{4\pi\epsilon_0} \frac{\hat R}{R^2}

Where P is the particle's momentum, which is given by Mv, where v is its velocity in the frame, and M is the inertial mass of q2.

Unfortunately, it's inconsistent to believe that momentum is given by p=mv, and to also believe that Maxwell's equations apply in all inertial frames.

While you only wrote down a subset of Maxwell's equations, I think it's fair to guess that you believe in all of them.

Maxwell's equations are invariant under the Lorentz transformation, while Newtonian physics with p=mv is invariant under the Gallilean transformation.

One cannot "mix and match" the two. Actual physical law is either invariant under the Lorentz transformation, or under the Gallilean transformation. It's simply not possible for it to be invariant under both sets of transformations.

It turns out as the result of experiment that actual physical law is invariant under the Lorentz transformation, and is not invariant under the Gallilean transformation. At low velocities, though, the two sorts of transformations are very close, the difference becomes apparent only at high velocities.

 

[Hurkyl]

One cannot "mix and match" the two. Actual physical law is either invariant under the Lorentz transformation, or under the Gallilean transformation. It's simply not possible for it to be invariant under both sets of transformations.

That's not quite true. You can take an approach similar to GR... they will be invariant under all diffeomorphisms...


The key point, I guess, is that in Special (resp. Gallilean) Relativity, the metric is fixed by a Lorentz (Gallilean) transformation.

 

[pervect]

That's not quite true. You can take an approach similar to GR... they will be invariant under all diffeomorphisms...


The key point, I guess, is that in Special (resp. Gallilean) Relativity, the metric is fixed by a Lorentz (Gallilean) transformation.

I don't see how this is possible. Consider velocity addition, for instance. Under the Lorentz transformation, two velocities add via the relativistic formula, so if B is moving at a velocity v1 relative to A, and C is moving at a velocity v2 relative to B, then the total velocity is given by the relativistic formula v= v1+v2 / (1+v1*v2/c^2).

With the Galilean transformation, the total velocity is v=v1+v2.

The two expressions aren't equal, so if one matches experiment, the other cannot. Only one can give the right answer.

If one believes that Maxwell's equations work in all inertial frames, one has to believe that the speed of light is 'c' in all inertial frames. This isn't possible if velocities add according to the rules of the Galilean transformation. It is possible if velocities add together via the relativistic formula.

 

[Hurkyl]

It doesn't make sense to say the velocity addition formula holds in a frame, since it speaks about multiple frames.


Things like Maxwell's equations would need to be put into a diffeomorphism invariant form, but that can be done. (And would exactly reduce to the usual form when the metric is diagonal)

 

[pervect]

It doesn't make sense to say the velocity addition formula holds in a frame, since it speaks about multiple frames.


Things like Maxwell's equations would need to be put into a diffeomorphism invariant form, but that can be done. (And would exactly reduce to the usual form when the metric is diagonal)

I probably should have been phrasing my remarks in terms of covariance rather than invariance.

But I still don't think it's possilbe to be self-consistent and to combine Newtonian physics (specifically p=mv) with a belief that Maxwell's equations work in all inertial frames, it's just a matter of how to phrase the objection.

Reviewing Goldstein's "Classical Mechanics" for how to best express what I was trying to say, I find the following longish quote.

A long series of investigations, especially the famous experiments of Michelson and Morley, have indicated that the velocity of light is always the same in all directions and is independent of the relative uniform motions of the observer, the transmitting medium, and the source. Since the propagation of light in a vacuum with the speed c is a consequence of Maxwell's equations, it must be concluded that the Galilean transformation ddoes not preserve the form of Maxwell's equations. Now, it is a postulate of physics, implicit since the time of Galielo and Newton, that all phenomena of physics should appear the same in all systems moving uniformly relatively to each other. Measurements made entirely within a given system must be incapable of distinghishing that system from all others moving uniformly with respect to it. This postulate of equivalence requires that physical laws must be phrased in an identical manner for all uniformly moving systems, i.e. be covariant when subjected to a Galilean transformation. The paradox confronting physics at the turn of the twentieth century was that experimentally both Newton's laws and Maxell's equation seemed to satiisfy the equivalence postualte but that theoretically, i.e. according to the Galilean transoformation, Maxwell's equations did not. Einstein, affirming explicitly the postulate of equivalence, concluded that it is the form of Maxwell's equations that must be kept invariant and therefore the Galilean transformation could not be correct. A new relationship between uniformly moving systems, the Lorentz transformation, must be found that preserves the speed of light in all uniformly moving systems. Einstein showed that such a transformation requires revision of the usual concepts of time and simultaneity.

[add]
There's no compatibility issue with Newtonian physics (p=mv) and assuming that Maxwell's equations work in some inertial frame, but in order to believe that Maxwell's equations work in ALL inertial frames, one explicitly assumes the equivalence postulate (one not only assumes that the laws of physics are the same in all frames, one has explicitly written those laws down).

Unfortunately, Maxwell's equations simply ARE NOT covariant under the Galilean transformation.

And, I suppose I should add, p=mv is covariant under the Galilean transformation, but NOT under the Loerntz transfomation.

So the Frankenstein notion of "sewing together" a part of Newtonian physics (p=mv), and a part of relativistic physics (believing that Maxwell's equations work in all inertial frames) just won't work.

 

[jdavel]

pervect,

In post #72 you say,

"The argument presented in the link shows that in any coordinate system where Maxwell's equations work, the speed of an electromagnetic wave is 'c'.
"

But mathematically the equation for the electromagnetic wave is the same as the equation for any classical wave (sound through the air, vibrations on a stretched string, etc.). So if the argument in the link shows that the speed of EM waves is c in any coordinate system where M's equations work, then doesn't it also show that the speed of sound has to be constant in any coordinate system where the wave equation for sound works?

I think the answer is yes. But of course the speed of sound is not constant in all coordinate systems, only the one where the medium (air) is at rest. And prior to the MM experiment everyone (including Maxwell) believed that the same was true for EM waves. That is, they traveled at c with respect to the ether. And if you transform M's equations according to Galilleo/Newton relativity, you find that EM wave speed is just c-v or c+v where v is the speed of the coordinate system wrt the ether, and c and v are parallel.

In short, there is nothing in Maxwell's equations which shows that light speed is frame independent. You need the null result of the MM experiment to show that.

 

[anti_crank]

pervect,

In post #72 you say,

"The argument presented in the link shows that in any coordinate system where Maxwell's equations work, the speed of an electromagnetic wave is 'c'.
"

But mathematically the equation for the electromagnetic wave is the same as the equation for any classical wave (sound through the air, vibrations on a stretched string, etc.). So if the argument in the link shows that the speed of EM waves is c in any coordinate system where M's equations work, then doesn't it also show that the speed of sound has to be constant in any coordinate system where the wave equation for sound works?

I think the answer is yes. But of course the speed of sound is not constant in all coordinate systems, only the one where the medium (air) is at rest. And prior to the MM experiment everyone (including Maxwell) believed that the same was true for EM waves. That is, they traveled at c with respect to the ether. And if you transform M's equations according to Galilleo/Newton relativity, you find that EM wave speed is just c-v or c+v where v is the speed of the coordinate system wrt the ether, and c and v are parallel.

In short, there is nothing in Maxwell's equations which shows that light speed is frame independent. You need the null result of the MM experiment to show that.

This is not correct. While it is true that a classical wave need only satisfy a simple wave equation, which is invariant under a Galileian transformation, electromagnetic waves are more than simply classical waves. It is true that the separated wave equations for the E and B fields are also invariant under Galileian transformations. This, however, is insufficient, for the fields in the EM wave must satisfy the full set of Maxwell equations - there are solutions to the separated wave equations that cannot represent EM waves. To illustrate this, additional constraints for plane waves that follow only from the full set of Maxwell equations are as follows: the E and B fields are in phase, perpendicular to each other and the direction of motion, and their amplitudes are related. When this is included, one finds that the Galileian transformations do not work for EM waves.

There are also problems with defining p=mv; namely that it is not conserved in certain instances where the EM fields carry momentum. This is known as 'hidden momentum' in EM; a Google search should find some info. This having been said, I must apologize for a very hurried post as I must rush to an appointment.

 

[pervect]

pervect,

In post #72 you say,

"The argument presented in the link shows that in any coordinate system where Maxwell's equations work, the speed of an electromagnetic wave is 'c'.
"

But mathematically the equation for the electromagnetic wave is the same as the equation for any classical wave (sound through the air, vibrations on a stretched string, etc.). So if the argument in the link shows that the speed of EM waves is c in any coordinate system where M's equations work, then doesn't it also show that the speed of sound has to be constant in any coordinate system where the wave equation for sound works?

I think the answer is yes.

I agree.

But of course the speed of sound is not constant in all coordinate systems, only the one where the medium (air) is at rest.

I agree again.

And prior to the MM experiment everyone (including Maxwell) believed that the same was true for EM waves. That is, they traveled at c with respect to the ether. And if you transform M's equations according to Galilleo/Newton relativity, you find that EM wave speed is just c-v or c+v where v is the speed of the coordinate system wrt the ether, and c and v are parallel.

In short, there is nothing in Maxwell's equations which shows that light speed is frame independent. You need the null result of the MM experiment to show that.

Yes, exactly. But if you will note the poster that I was responding to was already convinced (for whatever reason) that Maxwell's equations do work in all frames.

I am not looking to theorize that Maxwell's equations are true only in one preferred frame of reference, one at rest with respect to the 'ether'.

I am simply pointing out two logical consequence of the assumption that Maxwell's equations do work in all frames

1) The speed of light is constant in a vacuum in all inertial frames. This follows from the mathematical solution of the wave equations.

2) The Galilean coordiante transformation cannot be valid, because it implies (among other things) that velocities add according to the rule v=v1+v2. This is logically inconsistent with the notion that the speed of light is 'c' in all inertial frames, because the Galilean transformation would demand that the speed of light in a frame moving with velocity v would be v+c.