Harmless Lorentz transformation question

[dextercioby]

That is only at the surface,i won't ask u what trully happens to the photons...
What about the acceleration...?

Daniel.

 

[JesseM]

Jesse, see my challenge to Hurkyl. Also, if Maxwell's equations are unconditionally true, why don't they predict superconductivity?

Regards,

Guru

No one thinks Maxwell's laws are unconditionally true, but they were among the most accurate known laws in 1905. A hypothesis of relativity is that any new fundamental laws that come along in the future will also have the property of Lorentz-invariance, and this is indeed true; the laws of quantum electrodynamics (along with all other quantum field theories) are also Lorentz-invariant, for example.

 

[JesseM]

Hurkyl, using the definition of an inertial reference frame, can you derive the conclusion that the speed of light is c in all such frames?

What definition of an inertial frame? If an inertial frame is explicitly defined as one in which the laws of physics work the same way as in other inertial frames, then of course since light moves at c in slower-than-light inertial frames, any frame satisfying this defintion must also be a slower-than-light frame.

In other words, you have to use Maxwellian electrodynamics to prove the following:

If F is an inertial reference frame and V denotes the speed of an "electromagnetic wave" or "photon" in frame F then V = \frac{1}{\epsilon_0 \mu_0}.

I would enjoy seeing such a proof, I anticipate that it will make some reference, either explicitly or implicitly to Faraday's law of induction.

Are you just asking for a proof that Maxwell's laws prove than an electromagnetic wave (not a photon, they don't exist in classical electromagnetism) must always travel at \frac{1}{\epsilon_0 \mu_0}? If so, there's a proof at the bottom of this page, and another on this page.

Also, one more thing here is JCSD's quote:

A reference frame in which the two postulates of special relativity hold true. Any frame traveling at constant velcoity rleative to an inertial frmae is also an inertial frame.

Hurkyl... if JCSD is correct, then since any frame attached to a photon is moving at a constant speed in an inertial reference frame, it follows that the rest frame of a photon is an inertial reference frame.

He defined an inertial frame as one where the two postulates of relativity hold true (one of which is the postulate that all the fundamental laws of physics obey the same equations in all inertial frames), and this would not be true of any coordinate system you could define where the photon is at rest. He misspoke when he said any frame moving at constant velocity relative to another is also an inertial frame...this would be true in Newtonian physics, but in relativity it would only be true for coordinate systems that have a velocity v<c.

 

[Hurkyl]

Hurkyl, using the definition of an inertial reference frame, can you derive the conclusion that the speed of light is c in all such frames?

No. As you point out, I would invoke the principle of relativity. In particular, that Maxwell's equations are valid in any IRF.

I would enjoy seeing such a proof

It's rather straightforward, at least when using the differential form. You simply start with \partial^2 \vec{E} / \partial t^2 and make the appropriate substitutions using Maxwell's equations. (And the same for B).

Also, I would like to caution you in advance, classical EM does not predict superconductivity, and don't reason on the converse.

It doesn't predict photons either.

Hurkyl... if JCSD is correct, then since any frame attached to a photon is moving at a constant speed in an inertial reference frame, it follows that the rest frame of a photon is an inertial reference frame.

As I mentioned, a coordinate chart in which a photon is at rest can't be a reference frame because it has a degenerate time axis. (There is no unit vector in the time direction)

Since photons can change direction, they can be accelerated.

First off, photons don't even have 4-velocity vectors. Secondly, they can't be accelerated in SR.

Well for one, a mirror will reflect light that hits it.

The photon that comes back isn't the same photon that strikes the mirror.

 

[JesseM]

As I mentioned, a coordinate chart in which a photon is at rest can't be a reference frame because it has a degenerate time axis. (There is no unit vector in the time direction)

That'd be true if you tried to use the Lorentz transform to get such a coordinate system, but it would be possible to find a non-degenerate coordinate system in which every point is moving at constant velocity relative to Lorentzian frames and in which the photon is at rest, just by applying a Galilei transformation to a coordinate system constructed according to the relativistic rules...but obviously the laws of physics would not work the same way in this coordinate system, and neither postulate of relativity would be satisfied, so if these conditions are part of the definition of "inertial frame", it wouldn't qualify as an inertial reference frame.

 

[jcsd]

This is false. Since photons can change direction, they can be accelerated. That which accelerates has a center of inertial mass. Therefore, a photon has a center of mass. Since you can write formulas which are true in a reference frame in which the center of mass is at rest, you can write laws of physics which are true in reference frames in which a photon is at rest. So light, i.e. photons have frames.

Kind regards,

Guru

Unfortunately that is wtrong for a lot of reasons:

1) photons don't accelrate but irrelvant

2) what is the inertial mass of a photon? You actually need a 3x3 matrix to represent somehting akin to inertial mass in special rleativty.

3) photons don't have rest mass and there is no frame in which they are at rest.

 

[jcsd]

That'd be true if you tried to use the Lorentz transform to get such a coordinate system, but it would be possible to find a non-degenerate coordinate system in which every point is moving at constant velocity relative to Lorentzian frames and in which the photon is at rest, just by applying a Galilei transformation to a coordinate system constructed according to the relativistic rules...but obviously the laws of physics would not work the same way in this coordinate system, and neither postulate of relativity would be satisfied, so if these conditions are part of the definition of "inertial frame", it wouldn't qualify as an inertial reference frame.

The problem is a Galilean transformation is only the limit of a Lorentz transformation in special relativity, so it's not valid to use it to transform from one frame to the other. Hurkyl's reasiong is correct as it shows not only can you not define an inertial frame for light, but also that you can't even define a non-inertial frame for light.

 

[Hurkyl]

That'd be true if you tried to use the Lorentz transform to get such a coordinate system, but it would be possible to find a non-degenerate coordinate system in which every point is moving at constant velocity relative to Lorentzian frames and in which the photon is at rest, just by applying a Galilei transformation to a coordinate system constructed according to the relativistic rules

Not true: any vector lying along the time axis will have magnitude zero.

(In particular, note that the inner product is no longer given by a diagonal bilinear form)

 

[JesseM]

Not true: any vector lying along the time axis will have magnitude zero.

Well, it depends how you define the magnitude of vectors in the new coordinate system, if you define it in such a way as to insure it will agree with the magnitude of the same vector as seen in a valid inertial coordinate system, then of course this will be true. Is this all that the phrase "degenerate time axis" means? Any two events which have distinct coordinates in an inertial system will still have distinct coordinates in this non-inertial coordinate system, even if the two events lie along the worldline of a light ray.

(In particular, note that the inner product is no longer given by a diagonal bilinear form)

I don't know what the term "diagonal bilinear form" means...but like I said above, if you want to insure that the magnitude of a vector in the new coordinate system is the same as the magnitude of the same vector in an inertial coordinate system, you can't assume that the maginitude of two vectors (t0, x0, y0, z0) and (t1, x1, y1, z1) is given by -t0*t1 + x0*x1 + y0*y1 + z0*z1.

 

[JesseM]

The problem is a Galilean transformation is only the limit of a Lorentz transformation in special relativity, so it's not valid to use it to transform from one frame to the other.

It's "not valid" in the sense that the resulting coordinate system won't be an inertial one, but it will be an example of a non-inertial coordinate system which is moving at constant velocity relative to all inertial coordinate systems, which is why I think your earlier statement "Any frame traveling at constant velcoity rleative to an inertial frmae is also an inertial frame" was incorrect (unless by 'frame' you specifically meant 'inertial frame').

 

[jcsd]

Well, it depends how you define the magnitude of vectors in the new coordinate system, if you define it in such a way as to insure it will agree with the magnitude of the same vector as seen in a valid inertial coordinate system, then of course this will be true. But how do you define "degenerate"? After all, it is also true that certain vectors in inertial coordinate systems have magnitude zero, namely those that lie along the path of a light ray. I don't know what the term "diagonal bilinear form" means...but like I said above, if you want to insure that the magnitude of a vector in the new coordinate system is the same as the magnitude of the same vector in an inertial coordinate system, you can't assume that the maginitude of two vectors (t0, x0, y0, z0) and (t1, x1, y1, z1) is given by -t0*t1 + x0*x1 + y0*y1 + z0*z1. But likewise, you couldn't necessarily assume this in an accelerating coordinate system...would you say that accelerating coordinate systems are degenerate?

The magnitude of vectors is invariant, it's only decided by the metric tensor at that point. The metric tensor is a diagonal bilinear form.

Onm way of looking at it is that in every coordinate system we have four numbers to define each event, but the problem is that in our hypotehical refrenbce frma eof light one of those numbers is always zero, so we have three numbers to define each point in four dimensional space which cannot be done in a useful way.

The time basis vector in the rest frmae of an object is the unit vector tangent to it's worldline, in the case of light their is no unit vector tangent to it's wolrdline as any vector tangent to the worldline of light is null. No time basis vector, no coordinate system, no frame.

 

[jcsd]

It's "not valid" in the sense that the resulting coordinate system won't be an inertial one, but it will be an example of a non-inertial coordinate system which is moving at constant velocity relative to all inertial coordinate systems, which is why I think your earlier statement "Any frame traveling at constant velcoity rleative to an inertial frmae is also an inertial frame" was incorrect (unless by 'frame' you specifically meant 'inertial frame').

No it isn't, infact as a Galilean transformation presevres time we haven't done anything particularly interesting by performing a Galilean transformation, we're just using a different spatial coordinate system whose spatial origin (the points (t,0,0,0) vary with t). This is certainly not the refrence frame of light.

 

[JesseM]

No it isn't, infact as a Galilean transformation presevres time we haven't done anything particularly interesting by performing a Galilean transformation, we're just using a different spatial coordinate system whose spatial origin (the points (t,0,0,0) vary with t). This is certainly not the refrence frame of light.

I don't know what you mean by the phrase "the reference frame of light", I am certainly not claiming this coordinate system has any physical significance whatsoever, I'm just saying it's a non-inertial coordinate system which is moving at a constant velocity relative to all inertial frames, and in which an electromagnetic wave could be at rest.

 

[JesseM]

Onm way of looking at it is that in every coordinate system we have four numbers to define each event, but the problem is that in our hypotehical refrenbce frma eof light one of those numbers is always zero, so we have three numbers to define each point in four dimensional space which cannot be done in a useful way.

The time basis vector in the rest frmae of an object is the unit vector tangent to it's worldline, in the case of light their is no unit vector tangent to it's wolrdline as any vector tangent to the worldline of light is null. No time basis vector, no coordinate system, no frame.

Hmm, I hadn't thought about trying to express coordinates in terms of multiples of basis vectors...but why isn't it just as valid to express coordinates in terms of a coordinate transformation from some inertial system? Why can't you just pick an inertial system with coordinates x,y,z,t and then say:

x'=x - ct
y'=y
z'=z
t'=t

For any two points with distinct x,y,z,t coordinates, they'll be mapped to two points in this system with distinct x',y',z',t' coordinates.

 

[jcsd]

If anyhting it's a very odd inertial coordinate system. You can't go around labelling just anything speed d(x' + ct)/dt = c is still the velocity of an object with constant spatial coordinates in this coordinate system

 

[pervect]

One quick comment - there is such a thing in GR as a "null tetrad". This is something that I've been meaning to learn more about, it's closely related to Newmann-Penrose formalism and spinors. Instead of spacelike and timelike vectors, one has some (or perhaps even all) of the basis vectors being null vectors, which have a Lorentz interval of zero.

There's some discussion of this at

http://membres.lycos.fr/pvarni/dirac/node10.html

I've seen other examples where one mixes together null basis vectors with non-null vectors (i.e. ones that are space-like or time-like).

 

[jcsd]

Hmm, I hadn't thought about trying to express coordinates in terms of multiples of basis vectors...but why isn't it just as valid to express coordinates in terms of a coordinate transformation from some inertial system? Why can't you just pick an inertial system with coordinates x,y,z,t and then say:

x'=x - ct
y'=y
z'=z
t'=t

For any two points with distinct x,y,z,t coordinates, they'll be mapped to two points in this system with distinct x',y',z',t' coordinates.

Now I've had more time to think on it, what I guess you have there is a skew coordinate system (I.e. there will be off-diagonal entries in the metric in thta coordinate system). The reason I said it is inertial is thta it's time axis is the worldline of an inertial observer.

The thing with special relativity is that it is not general covraiant theory so arbitary coordinate transformations should be avoided as it means re-formulating the laws of physics. Though we can use other coordinate systems if we wish we should genrally base our our coordimante systems around Lorentz coordinates. Even non-inertial coordiante systems are generally going to be based around Lornetz coordinates by taking the worldline of our observer as as the time axis and then building the system using theat observers momentarily comoving inertial frames. In some ways the results can be unsatisfctory for non-inertial observers as non-inertial coordinate systems constructed in such a way are not generally well-behaved, BUT the observations of a non-inertial observer are not going to be 'well-behaved' so in this sense the coordinate system just refelcets the observations of the observer.

In special relativity the time basis vector is always the unit vector tangent to the observers worldline, so this is enough to exclude light from having a frame of reference.

I suppose i really didn't think about bases with null vectors, probably becasue of the unforunate terminology as the term null vector is often used to describe the additive identity (the zero vector) in a vector space and a set of n vectors containing the zero vector cannot span an n-dimensional space, wheras I guess null vectors can span Minkowski space.

 

[Physicsguru]

How about another question:Why don't the very praised Maxwell equations predict the Bohm-Aharonov effect...??

Daniel.

P.S.The answers are identical to both questions:mine & yours.

Short answer: Classical electrodynamics doesn't predict quantization of magnetic flux... a quantum effect exhibited by superconductors.

\Phi = magnetic flux = \oint \vec B \bullet d\vec a = n \Phi_0 = n(\frac{h}{2e})

Where \frac{h}{2e} is the magnetic flux quantum, n a positive integer, B the magnetic field.

Long answer:

Suppose that in some inertial reference frame, a charged particle is moving through a region of space at speed v in the frame, and that in the region of space local to the particle, there is a nonzero electric field OR a nonzero magnetic field, due to some non-local charge configuration nearby. According to classical EM, the particle will be accelerated in this inertial reference frame, and the total force acting upon this particle to accelerate it in this inertial frame, is the Lorentz force, which is given by:

\vec F = q[\vec E + \vec v \times \vec B]

Let us assume the classical relation above is a true statement, as regards the acceleration of the charged particle in the IRF, and the real force which gives it its acceleration in the IRF.

If the statement above is correct, then a charged particle which is moving through a region of space which is totally devoid of electric and magnetic fields (and other fields as well), will not experience an external force, and therefore, will not be accelerated. Hence, the charged particle will obey Galileo's Law of inertia (Newton's first law) in the inertial reference frame. That is, the particle will continue moving in a straight line at a constant speed.

Suppose now, that we design an experiment, where we have a charged particle pass near a solenoid. Using classical EM, the magnetic field exterior to the solenoid should be zero, and as long as the current in the solenoid is constant, there will also be no induced electric field either.

So then, we can control whether or not the solenoid is on or off in this experiment, but in either case, once the solenoid is on, or once the solenoid is off, there is no external B or E field due to the presence of the solenoid in our experimental setup, hence turning it on or off should not change the trajectory of charged particles which are passing near it.

Consider now a double slit experiment, in which electrons are impinging on the slits, and just behind the tiny wall separating the two slits, we have a tinier solenoid hidden there, which we can turn on, or off.

If the Lorentz force formula is correct, turning the solenoid on or off, should produce no effect on the trajectory of the electrons as they pass through the slits.

The experimental result is a diffraction pattern on the far wall, hence turning the solenoid on or off, should produce no noticable change in the observed diffraction pattern.

This is not what happens, instead the diffraction pattern is altered, the maxima and minima are switched, and the reason for this non-classical observation is called the Aharonov-Bohm effect.

Your question, is "why doesn't classical electrodynamics predict the effect?"

The answer is, because classical EM has the force which acts upon the charged particles as being caused solely by the electric or magnetic fields, when in an experiment in which the AB effect appears, since B=0 exterior to the solenoid whether the solenoid is on or off, it is the magnetic vector potential local to the particle which changes (and therefore is responsible for the force which gives rise to the AB effect), hence the Lorentz formula is wrong (not the whole story, because the potential gave rise to a force). In the next post, I will derive the formula for the magnetic vector potential.

Regards,

Guru

 

[dextercioby]

It's far easier the other way around,sorry to dissapoint you.

Maxwell equations are much more easily (and correctly when it comes to quantum behavior,like the B-A effect) expressible & integrable in terms of the potentials...

Daniel.

 

[dextercioby]

You didn't read my post.To me,what u did (namely extracting the \vec{A} once knowing the \vec{B}) means nothing.
Sorry.But that's the way i see it.

Daniel.

P.S.And yes,u still have some 6 equations to correct wrt the latex format.

 

[Physicsguru]

Derivation of magnetic vector potential A

Recall the answer:

\vec A = \frac{\mu_0}{4\pi} \int \frac{\vec J}{R} d\tau

We can derive the formula for the magnetic vector potential from the Biot-Savart law, provided we know the following purely mathematical theorem:

Theorem:

\frac{\hat R}{R^2} = \nabla (- \frac{1}{R} )

Derivation of Magnetic vector potential from Biot-Savart Law

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

Where J is the volume current density, and d tau is a differential volume element dx dy dz, and R is a vector from a charged particle in that which has some non-zero J, to an arbitrary field point (x,y,z).

Recall the following theorem of the vector differential calculus:

Theorem:
For any scalar function f, and any vector function J:

\nabla \times (f \vec J) = \nabla f \times \vec J + f \nabla \times \vec J


From the previous theorem, it follows that:

\vec J \times \nabla f = f \nabla \times \vec J - \nabla \times f \vec J

Making use of the first theorem, and letting f = -1/R, we can rewrite the Biot-Savart law as:

\vec B = \frac{\mu_0}{4\pi} \int \vec J \times \nabla f d\tau

And now using the second theorem we have:

\vec B = \frac{\mu_0}{4\pi} \int f \nabla \times \vec J - \nabla \times f \vec J d\tau

Since J is not a function of the field points x,y,z, it follows that del X J is zero, and so we obtain the following equation:

\vec B = - \frac{\mu_0}{4\pi} \int \nabla \times (f \vec J) d\tau

And since f=-1/R, we have the following:


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

Pulling the del symbol all the way out to the front we have:

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


We can now solve for the magnetic vector potential A, using the following:

\vec B = \nabla \times \vec A

The result is the classical answer for A.

Regards,

Guru

 

[Physicsguru]

What Are you just asking for a proof that Maxwell's laws prove than an electromagnetic wave (not a photon, they don't exist in classical electromagnetism) must always travel at \frac{1}{\epsilon_0 \mu_0}? If so, there's a proof at the bottom of this page

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

 

[ZapperZ]

There are some really serious error here:

Short answer: Classical electrodynamics doesn't predict quantization of magnetic flux... a quantum effect exhibited by superconductors.

\Phi = magnetic flux = \oint \vec B \bullet d\vec a = n \Phi_0 = n(\frac{h}{2e})

Where \frac{h}{2e} is the magnetic flux quantum, n a positive integer, B the magnetic field.

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!

Suppose now, that we design an experiment, where we have a charged particle pass near a solenoid. Using classical EM, the magnetic field exterior to the solenoid should be zero, and as long as the current in the solenoid is constant, there will also be no induced electric field either.

So then, we can control whether or not the solenoid is on or off in this experiment, but in either case, once the solenoid is on, or once the solenoid is off, there is no external B or E field due to the presence of the solenoid in our experimental setup, hence turning it on or off should not change the trajectory of charged particles which are passing near it.

This is incorrect. Simply by invoking Lenz's is sufficient to show that a moving charge passing near such a solenoid is sufficient to induce all kinds of fields. This is precisely the self-energy interaction that is present in both classical and quantum fields! The fact that charge particles can induce such fields in coils of wire is the vary method we detect the presence of them! And such induced fields from these coils can, in turn, affects the original charged particles - that's why these are called self-interactions!

Zz.

 

[JesseM]

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}}.

The links I gave were only intended to prove that in any frame where Maxwell's laws hold, electromagnetic waves will travel at \frac{1}{\sqrt{\epsilon_0 \mu_0}}. There is no way to "prove" that Maxwell's laws hold in every inertial reference frame from first principles, it is a postulate of relativity that the laws of physics work the same way in all inertial reference frames, so if Maxwell's laws hold in one frame the postulate says they should hold in all. All the evidence has favored the idea that this postulate is correct. For example, the pre-relativistic theory was that Maxwell's laws would only hold in one preferred reference frame (the rest frame of the ether), and in other frames they'd have to be modified by a Galilei transform. This leads to the prediction that if the Earth is moving relative to the ether rest frame, we should be able to see that light moves at different velocities in different directions, but the experiments to try to detect such an effect all failed.

 

[Physicsguru]

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}} .

... so if Maxwell's laws hold in one frame the postulate says they should hold in all.

This is not the kind of answer I was hoping for Jesse.

I was looking for something more along the following lines:

Postulate 1: Electric charge is quantized. The fundamental unit of electric charge q, is given by q=1.60217733 x 10^{-19} Coulombs.



If v is the speed of an electron in inertial reference frame F and the electron is moving through a region of space where not (B=0) and all other fields are zero there
then the electron will experience a force given by \vec F = q ( \vec V \times \vec B ).

If v is the speed of an electron in inertial reference frame F and the electron is moving through a region of space where not (E=0) and all other fields are zero there then the electron will experience a force given by \vec F = q \vec E.

If v is the speed of an electron in an inertial reference frame F and the electron is moving through a region of space where not (E=0) and not (B=0) and all other fields are zero locally then the electron will experience a force given by \vec F = q ( \vec E + \vec V \times \vec B ).

(insert argument here)

Therefore, if v is the speed of an electron in an inertial reference frame F, and not (dv/dt = 0) then electromagnetic waves will radiate spherically outwards from the initial location of the electron's center of mass, and their speed in a frame in which that location is at rest will be given by \frac{1}{\sqrt{\epsilon_0 \mu_0}} AND in any other reference frame in which that location is moving in a straight line at a constant speed, the speed of the "electromagnetic waves" will also be \frac{1}{\sqrt{\epsilon_0 \mu_0}}.

Regards,

Guru

 

[dextercioby]

I would advise Physicsguru to read in parallel and extract the correct conclusions from 2 rock-solid sources on Hertz G-invariant electrodynamics and Einstein-Minkowski L-invariant electrodynamics.For simplicity,in vacuum.


Daniel.

 

[JesseM]

This is not the kind of answer I was hoping for Jesse.

Well, the type of argument you were apparently hoping for is impossible--you can't prove from Maxwell's equations alone that the speed of light is the same in every reference frame, because 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. There is no internal inconsistency in this theory, it just isn't supported by the evidence.

 

[Physicsguru]

I would advise Physicsguru to read in parallel and extract the correct conclusions from 2 rock-solid sources on Hertz G-invariant electrodynamics and Einstein-Minkowski L-invariant electrodynamics.For simplicity,in vacuum.


Daniel.

Why?

Regards,

Guru

 

[Physicsguru]

Well, the type of argument you were apparently hoping for is impossible--you can't prove from Maxwell's equations alone that the speed of light is the same in every reference frame, because 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. There is no internal inconsistency in this theory, it just isn't supported by the evidence.

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}}?

In this case, there are multiple reference frames moving relative to one another, in which the speed of some photons is c, and the speed of other photons isn't c, and the Galilean transformations are the correct coordinate transformations from one inertial reference frame to another?

Regards,

Guru

 

[dextercioby]

Why?

Regards,

Guru

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.