Harmless Lorentz transformation question

[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.

 

[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.

 

[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 = <br /> <br /> \frac{1}{4\pi\epsilon_0} \int \nabla \rho \times \frac{\hat R}{R^2} d\tau <br /> <br /> +<br /> <br /> \frac{1}{4\pi\epsilon_0} \int \rho \nabla \times \nabla (\frac{-1}{R}) d\tau <br /> <br />

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 <br /> <br /> <br /> -\frac{1}{c^2} \frac{\partial \vec J }{\partial t}<br /> <br /> \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}<br /> <br /> <br /> \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 <br /> <br /> <br /> -\frac{n^2}{c^2} \frac{\partial \vec J }{\partial t}<br /> <br /> \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