Harmless Lorentz transformation question

[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:

[tex] \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.

 

[cepheid]

Originally Posted by Goldstein, pg 277
...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...

Right, I understand that.That is a statement of what Einstein called the Principle of Relativity. However, there is something about the buildup to SR that I have never understood. People other than Einstein were aware of the problem that the principle of relativity had become at odds with the law of the velocity propagation of light in vacuo. However, their favoured resolution to the problem completely flew in the face of the principle of relativity! Why is it that nobody else had a problem right from the start with the notion of an absolute rest frame, and differences in the perceived speed of light perpendicular and // to Earth's orbit? If the predictions of the Michaelson-Morely experiment were bourne out, would people have been willing to throw out the principle of relativity completely? This despite the fact that it was something so logically intuitive that nobody had even thought to question it since the time of Galileo and Newton! Everybody seemed willing to ignore this glaring problem and support a prediction that preserved the Galilean transformation, under which Maxwell's equations are not invariant.

 

[JesseM]

Right, I understand that.That is a statement of what Einstein called the Principle of Relativity. However, there is something about the buildup to SR that I have never understood. People other than Einstein were aware of the problem that the principle of relativity had become at odds with the law of the velocity propagation of light in vacuo. However, their favoured resolution to the problem completely flew in the face of the principle of relativity! Why is it that nobody else had a problem right from the start with the notion of an absolute rest frame, and differences in the perceived speed of light perpendicular and // to Earth's orbit? If the predictions of the Michaelson-Morely experiment were bourne out, would people have been willing to throw out the principle of relativity completely? This despite the fact that it was something so logically intuitive that nobody had even thought to question it since the time of Galileo and Newton! Everybody seemed willing to ignore this glaring problem and support a prediction that preserved the Galilean transformation, under which Maxwell's equations are not invariant.

No one is bothered by the fact that soundwaves in air don't look the same in every reference frame. As long as electromagnetic waves are vibrations in a physical medium, the ether, then the fact that the laws of electromagnetism don't look the same in every reference frame doesn't mean there are any fundamental laws of nature that disobey the principle of relativity, since the laws governing soundwaves in ether wouldn't be seen as any more fundamental than the laws governing soundwaves in air, it could be seen as just a matter of historical accident that the ether has the particular rest frame it does (you'd be under no obligation to believe the ether is at rest in absolute space, for example).

I don't know if 19th-century scientists thought of it this way though--maybe they just didn't consider the principle of relativity as fundamental as we do (after all, Newton himself believed in absolute space, since that seemed to be the only way to explain why acceleration is absolute).

 

[dextercioby]

If the predictions of the Michaelson-Morely experiment were bourne out, would people have been willing to throw out the principle of relativity completely? This despite the fact that it was something so logically intuitive that nobody had even thought to question it since the time of Galileo and Newton!

Why would they ??Would the Principle of Relativity ("all laws of mechanics are invariant under the Galieli group") be proven wrong,if the Michaelson-Morley experiment would have had another turn out...?

Daniel.

 

[pervect]

Why is it that nobody else had a problem right from the start with the notion of an absolute rest frame, and differences in the perceived speed of light perpendicular and // to Earth's orbit? If the predictions of the Michaelson-Morely experiment were bourne out, would people have been willing to throw out the principle of relativity completely?

While I'm not sure how accurate my insight into the mindset of those times is, my impression was that the usual practice was to compare ether waves with sound waves. So while I don't think the abandonment of the principle of relativity would be complete, I also don't think it would have the same emphasis as it does today.

However, I think a significant number of people even at the time were rather disturbed by the rather strange collection of properties that were being ascribed to the ether - that it was simultaneously present even in empty space, offering no resistance to the passage of uncharged bodies, but yet so rigid that it would transmit vibrations at the speed of light.

So people did what I think was a very sensible thing- they looked for some actual evidence that it (the ether) existed.

Ultimately, science is not about how long an idea has been around, but how well it works. If the ether theory had made testable predictions that were borne out by experiment, I think the theory would have stuck around. But it didn't.

 

[reilly]

Physicsguru -- Read the literature. Physics has had a century to study and dissect relativity and E&M. Your objections have certainly been raised many times. How could you dispute the Lorentz invariance of Maxwell's Eq.? That invariance is central to much of modern physics, which, after all, is at least modestly successful. The debate was over certainly by the 1930s, and Einstein and Maxwell came through unscathed. Lasers, radar and masers, communication with space probes, radio and tv broadcasting, the Compton effect, photoelectric phenomena, atomic spectra and the classical theory of radiation all require the standard approach to realtivity and electrodynamics.

If you are serious about your objections then
1. Make sure you fully understand the electrodynamics and relativity that most of us treasure and love.(If you do not, few if any will pay attention to you)
2. Base your objections on experimental evidence. After all, physics is an empirical science.

There's no way a photon can be at rest in an inertial frame. If you can demonstrate that this is not true, buy your ticket to Stockholm.

Regards,
Reilly Atkinson

 

[Physicsguru]

Physicsguru -- Read the literature. Physics has had a century to study and dissect relativity and E&M. Your objections have certainly been raised many times. How could you dispute the Lorentz invariance of Maxwell's Eq.? That invariance is central to much of modern physics, which, after all, is at least modestly successful. The debate was over certainly by the 1930s, and Einstein and Maxwell came through unscathed. Lasers, radar and masers, communication with space probes, radio and tv broadcasting, the Compton effect, photoelectric phenomena, atomic spectra and the classical theory of radiation all require the standard approach to realtivity and electrodynamics.

If you are serious about your objections then
1. Make sure you fully understand the electrodynamics and relativity that most of us treasure and love.(If you do not, few if any will pay attention to you)
2. Base your objections on experimental evidence. After all, physics is an empirical science.

There's no way a photon can be at rest in an inertial frame. If you can demonstrate that this is not true, buy your ticket to Stockholm.

Regards,
Reilly Atkinson

Suppose that the charge density $$\rho$$ which aappears in Coulomb's law satisfies the following postulate:

Postulate:

$$\nabla ^2 \rho = \frac{1}{c^2} \rho_t_t$$

From the postulate above, we have:

$$\nabla ^2 \rho = \nabla \bullet \nabla \rho = \frac{1}{c^2} \frac{\partial}{\partial t} \frac{\partial \rho}{\partial t}$$

Recall the continuity equation which holds if electric charge is conserved:

$$\nabla \bullet \vec J = - \frac{\partial \rho}{\partial t}$$

Therefore, if the postulate is true & electric charge is conserved then:

$$\nabla ^2 \rho = \nabla \bullet \nabla \rho = - \frac{1}{c^2} \frac{\partial}{\partial t} \nabla \bullet \vec J$$

From which it follows that:

$$\nabla \bullet \nabla \rho = \nabla \bullet - \frac{1}{c^2} \frac{\partial \vec J}{\partial t}$$

From which it follows that:

$$\nabla \rho = -\frac{1}{c^2} \frac{\partial \vec J }{\partial t}$$

Now, consider the formula for the electric field at (x,y,z) due to the presence of some non-zero charge density $$\rho$$ which satisfies the postulate above. We have:

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

Now, take the curl of both sides of the expression above to obtain:


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

Recall the following mathematical theorem:

Theorem:

$$\nabla X (f\vec A) = \nabla f \times \vec A + f\nabla \times \vect A$$

For any scalar function f, and any vector function A.

Since the charge density rho is a scalar function, and the volume current density is a vector function, it follows that:

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

Since the curl of a gradient is zero, the second term above drops out, and we are left with:

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

Now, substitute the expression which was derived earlier from the postulate for the gradient of the charge density in the equation above to obtain:

$$\nabla \times \vec E = \frac{1}{4\pi\epsilon_0} \int -\frac{1}{c^2} \frac{\partial \vec J }{\partial t} \times \frac{\hat R}{R^2} d\tau$$

Suppose that the classical Maxwellian result for the speed of light is correct, so that:

$$\frac{1}{c^2} = \epsilon_0 \mu_0$$

We can now rewrite the curl of E as:

$$\nabla \times \vec E = - \frac {\partial}{\partial t} \frac{\mu_0}{4\pi} \int \vec J \times \frac{\hat R}{R^2} d\tau$$

Where we have arrived at the Biot-Savart expression for B:

Biot-Savart Law (definition of Magnetic field B)

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

Thus, we have arrived at one of Maxwell's equations, namely:

$$\nabla \times \vec E = - \frac{\partial \vec B}{\partial t}$$

The forumal above is true, provided that all the postulates used in the derivation are true. Namely:

Postulate 1: Charge density obeys a wave equation, namely:

$$\nabla ^2 \rho = \frac{1}{c^2} \rho_t_t$$

where the wavespeed is given by c, where:

$$c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}$$

Postulate 2: The total electric charge of the universe cannot vary for any reason.

By experiment it is known that the speed of light through matter is less than the speed of light in vaccuum. The two speeds are related through the well known "index of refraction" n. Let v denote the speed of light through some substance with index of refraction n, we have:

$$v = \frac{c}{n}$$

So, if instead, the wave equation for a charge density wave is:

$$\nabla ^2 \rho = \frac{1}{v^2} \rho_t_t$$

where

$$v = \frac{c}{n}$$

Then instead we have:

$$\nabla ^2 \rho = \frac{n^2}{c^2} \rho_t_t$$

This would have altered the previous derivation above slightly. Instead, we would have had:

$$\nabla \times \vec E = \frac{1}{4\pi\epsilon_0} \int -\frac{n^2}{c^2} \frac{\partial \vec J }{\partial t} \times \frac{\hat R}{R^2} d\tau$$

Let c denote the speed of light relative to an emitter, and let the speed of light relative to the emitter satisfy:

$$\frac{1}{c^2} = \epsilon_0 \mu_0$$

Therefore, instead we have:

$$\nabla \times \vec E = - n^2 \frac{\partial \vec B}{\partial t}$$

 

[dextercioby]

1.Your last two expressions coincide.So edit your post.EDIT:You did.
2.How did u get that formula giving $\vec{E}$ as a function of $\rho$
3.What are your calculations trying to prove...?

Daniel.

 

[Physicsguru]

1.Your last two expressions coincide.So edit your post.EDIT:You did.
2.How did u get that formula giving $\vec{E}$ as a function of $\rho$
3.What are your calculations trying to prove...?

Daniel.

Suppose there is a blob of matter at rest in some inertial reference frame, which has a nonzero electric charge, and pretend its inertia is infinite. Let all other matter in the universe be electrically neutral, except for one electron which happens to find itself in the electric field of the blob. The idea is that the electron, which otherwise would coast at a constant speed in a straight line experiences a force, and is accelerated, and the formula relating the inertia of the electron, to its change in speed, and the impressed force upon it is:

$$\vec F = \frac{d(m\vec v)}{dt}$$

And also that this force is related to the electric charge q of the electron, and the value of the electric field of the blob local to the electron, as:

$$\vec F = q \vec E$$

Couloumb's law for the electric force between two charged objects:

Let one object have electric charge Q1, and let the other object have charge Q2. By experiment, the electric force between them is:

$$\vec F = KQ1Q2 \frac{\hat R}{R^2}$$

Postulate: The total electric charge of anybody is quantized. A unit of electric charge is given by:

$$e = -1.6021773 x 10^{-19} C$$

So if there are N electric charges on one object, and n electric charges on another object, we have:

Q1 = ne
Q2 = Ne

Let the object with electric charge Q2 be the object with an infinite inertia, and let the other object have n=1, and call the other object an electron.

There is enough information above to solve for the electric field created by the object with electric charge Q2, and we find that:

$$\vec E = KQ2 \frac{\hat R}{R^2}$$

The formula above is obtained if

$$Q2 = \int \rho d\tau$$

Giving us the classical result that:

$$\vec E = K \int \rho d\tau \frac{\hat R}{R^2}$$

And letting $$K= \frac{1}{4\pi\epsilon_0}$$
we finally have the formula you asked about.

P.S. And yes I know, classically, electric charge is not quantized.

Regards,

Guru

 

[dextercioby]

And how's that stuff related to the previous discussion ?And that $\vec{E}$ in what reference frame is it considered...?

Daniel.

 

[Physicsguru]

And how's that stuff related to the previous discussion ?And that $\vec{E}$ in what reference frame is it considered...?

Daniel.

If you read carefully, the speed of the test charge (a single electron in the example here) is being defined in the rest frame of the blob.

P.S.

The blob has an infinite mass, and therefore cannot be accelerated in this frame, regardless of whether or not it emits particles.

The electrons in the blob are the things that are responsible for a nonzero electric field local to the test charge, which is assumed far away from the center of mass of the blob, so that this force is a long range force.

If the electrons in the blob are at rest in this frame, then the electric field local to the electron way far away is irrotational, in otherwords, E would be an electrostatic field.

However, if the postulate about charge density waves is correct, then the blob could be supporting a charge density wave.

Regards,

Guru

 

[controlfreak]

My question is simple.

If, after time amount of time $$\Delta t$$ has elapsed according to some clock which is permanently at rest in S, the photon has coordinates (L,0,0) in frame S, what is my x` coordinate in frame S`?

For the sake of definiteness, suppose that exactly one second has ticked according to a clock at rest in frame S. Therefore, the location of the photon in frame S is given by (299792458 meters, 0,0).

Let (M,0,0)` denote the location of the origin of inertial reference frame S, in reference frame S`, at the instant that the clock at rest in frame S strikes one. Solve for M.

Guru

I read all arguments, but then finally what is the answer to the first question raised by Guru?

What is M? and
(I am adding this) What is the speed of the photon in the frame S'?!

How I feel the question can be answered (possible ways) are:

a) Give exact values for M and Vs'
b) Give probability distribution of possible values for M and Vs'
c) Assert and prove that the Conceptualization of a reference frame attached to the photon, is meaningless and thus refrain from answering the question. (Here whether lorentz transformation holds good in that reference frame is besides the question as we do have some thing (photon) which moves at 'c' the lorentz constant and in our minds we can think of it as a reference frame. What I feel one needs to prove here is that thought is meaningless and doesn't make sense).

I didnt go thru the entire thing, but I suppose I haven't missed the answer to the initial question.

-cf

 

[controlfreak]

If, after time amount of time $$\Delta t$$ has elapsed according to some clock which is permanently at rest in S, the photon has coordinates (L,0,0) in frame S, what is my x` coordinate in frame S`?
Guru

I think here the key is not the x' coordinate, but the t' coordinate. The light will move with the speed 'c' in the S' reference frame also, as the light speed is constant. But the time coordiante will stand still at 0. As no time will be elaspsed inspite of the fact the photon travels so fast, the photon will still be at the origin of S' frame as measured from S' frame.

The lorentz transformation becomes indefinite when when v/c goes to 1 if we do mere substitution,but I feel if we are able to take limits of the lorentz transformation as v/c goes to 1, through rules similar to L'Hospitals, it would be possible to find the t' and x' coordinates of the photon as limits and my guess is it will be equal to 0 and 0 resply always and in a way doesn't depend on t and x.

Also the location of the origin of S' reference frame measured in S reference frame will be moving at the speed of ct.

I haven't dwelled on any quantum reasoning here and I am sticking to classical physics.