Vector field equality Curl Proof of Moving Magnet & Conductor Problem

[tade]

Of course, it's not true what you want to prove

The moving magnet and conductor problem hinges upon the notion that if the curls of the fields are equal, the fields are equal.

Einstein thought, they have equal magnitudes but totally different origins, how silly, they are both the exact same phenomena of a relative motion.

So its not correct then?@Delta2

 

[PeterDonis]

So its not correct then?

What are you asking? Are you asking if the observed results of the experiment with the moving magnet and the conductor are different depending on which frame we use to analyze the experiment? The answer to that is obvious: no, they aren't.

If your question is something else, it would be a good idea to restate it, since this thread has been dormant for a while.

 

[tade]

What are you asking? Are you asking if the observed results of the experiment with the moving magnet and the conductor are different depending on which frame we use to analyze the experiment? The answer to that is obvious: no, they aren't.

If your question is something else, it would be a good idea to restate it, since this thread has been dormant for a while.

vanhees71 said, "Of course, it's not true what you want to prove"And these are the relevant posts:

Another possible solution might be $\vec{E}^\prime = \vec{v} \times \vec{B} + \vec{F_1}$
where $\vec{F_1}$ is a curl-free vector field

but weren't they able to show it pre-relativity?

to be able to show, pre-relativity, that not only are the curls equal, the original fields are equal as well

 

[PeterDonis]

vanhees71 said, "Of course, it's not true what you want to prove"

I'm not sure what he was referring to, because I'm not sure anyone in the previous discussion was entirely clear about what the issue was. That's why I asked you to restate the issue.

not only are the curls equal, the original fields are equal as well

If you mean, to show that $\vec{E}^\prime = \vec{v} \times \vec{B}$ (with the prime on the $\vec{E}^\prime$ to indicate that it is the field in a different frame), try looking at the Lorentz transformation equations for the fields.

 

[tade]

I'm not sure what he was referring to, because I'm not sure anyone in the previous discussion was entirely clear about what the issue was. That's why I asked you to restate the issue.

i see, i thought you were asking for his sake

If you mean, to show that $\vec{E}^\prime = \vec{v} \times \vec{B}$ (with the prime on the $\vec{E}^\prime$ to indicate that it is the field in a different frame), try looking at the Lorentz transformation equations for the fields.

but, as mentioned, I'd like to figure out how they proved it pre-relativity

as it was cited by Einstein in his 1905 seminal SR paper

 

[PeterDonis]

I'd like to figure out how they proved it pre-relativity

The fact that Maxwell's Equations are Lorentz invariant was known pre-relativity. Einstein in his 1905 paper was not telling anyone anything they did not already know when he said that Maxwell's Equations were Lorentz invariant. The new element in that paper was figuring out how to make ordinary mechanics, i.e., descriptions of things like measuring rods and clocks, Lorentz invariant, not electrodynamics.

 

[tade]

If you mean, to show that $\vec{E}^\prime = \vec{v} \times \vec{B}$ (with the prime on the $\vec{E}^\prime$ to indicate that it is the field in a different frame), try looking at the Lorentz transformation equations for the fields.

for confirmation, its this one right?

\begin{align}
E'_x &= E_x & \qquad B'_x &= B_x \\
E'_y &= \gamma \left( E_y - v B_z \right) & B'_y &= \gamma \left( B_y + \frac{v}{c^2} E_z \right) \\
E'_z &= \gamma \left( E_z + v B_y \right) & B'_z &= \gamma \left( B_z - \frac{v}{c^2} E_y \right). \\
\end{align}

 

[PeterDonis]

its this one right?

If you mean for a Lorentz boost in the $x$ direction with velocity $v$, yes. The more general transformations are here:

https://en.wikipedia.org/wiki/Class...rmation_of_the_fields_between_inertial_frames

 

[tade]

If you mean for a Lorentz boost in the $x$ direction with velocity $v$, yes. The more general transformations are here:

https://en.wikipedia.org/wiki/Class...rmation_of_the_fields_between_inertial_frames

thanks, were these transformations first derived for Lorentz ether theory?

 

[PeterDonis]

were these transformations first derived for Lorentz ether theory?

I don't know that they were first derived for any particular theory other than electrodynamics itself.

 

[tade]

If you mean, to show that $\vec{E}^\prime = \vec{v} \times \vec{B}$ (with the prime on the $\vec{E}^\prime$ to indicate that it is the field in a different frame), try looking at the Lorentz transformation equations for the fields.

what do you think of using a general solution to the curl equation

$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.$$

 

[vanhees71]

The moving magnet and conductor problem hinges upon the notion that if the curls of the fields are equal, the fields are equal.

Einstein thought, they have equal magnitudes but totally different origins, how silly, they are both the exact same phenomena of a relative motion.

So its not correct then?@Delta2

I obviously still miss to understand the question. Of course Einstein's famous introduction of his 1905 paper about special relativity is correct, and SR solves the apparent problem with the relative motion between the magnet and the coil (conductor) completely. That there was a problem before relativity is because Maxwell's equations are not Galilei but Poincare invariant, and that the Poincare transformation rather than the Galilei transformation is the correct realization of the special principle of relativity is the solution of this problem.

There is no problem with the non-uniqueness of the electromagnetic potentials given a physical situation either, because electromagnetism is a gauge theory. I hope, my long previous posting #30 makes this clear.

 

[vanhees71]

thanks, were these transformations first derived for Lorentz ether theory?

A slightly different predecessor of the Lorentz transformation were first derived from purely mathematical considerations by Woldemar Voigt in 1887, but he (as well as Poincare and Lorentz) missed the physical implications, which were Einstein's ingenious insight of 1905 (finalized by Minkowski in revealing the full mathematical structure of special-relativistic spacetime as the "Minkowski space").

 

[vanhees71]

what do you think of using a general solution to the curl equation

$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.$$

It's not uniquely solvable either. You need all four Maxwell equations together with appropriate boundary and initial-value conditions to get a unique answer for $\vec{E}$ and $\vec{B}$ with given sources $\rho$ and $\vec{j}$. These solutions are given either in terms of the retarded solutions of the wave equations for the potential in Lorenz gauge (or any gauge equivalent solution in any other gauge) or equivalently by the Jefimenko equations:

https://en.wikipedia.org/wiki/Retarded_potential
https://en.wikipedia.org/wiki/Jefimenko's_equations

 

[tade]

I obviously still miss to understand the question. Of course Einstein's famous introduction of his 1905 paper about special relativity is correct, and SR solves the apparent problem with the relative motion between the magnet and the coil (conductor) completely.

I was referring to when you said, "Of course, it's not true what you want to prove".

 

[tade]

It's not uniquely solvable either. You need all four Maxwell equations together with appropriate boundary and initial-value conditions to get a unique answer for $\vec{E}$ and $\vec{B}$ with given sources $\rho$ and $\vec{j}$. These solutions are given either in terms of the retarded solutions of the wave equations for the potential in Lorenz gauge (or any gauge equivalent solution in any other gauge) or equivalently by the Jefimenko equations:

https://en.wikipedia.org/wiki/Retarded_potential
https://en.wikipedia.org/wiki/Jefimenko's_equations

so how would you show that E' = v × B without resorting to the Lorentz transformations of the E and B fields?

 

[vanhees71]

Please, could you ask a clear question? Of course, if you want to change from one inertial reference frame to another you have to use the Lorentz transformation to transform the electromagnetic field components with respect to one of these frames to the other, and using the Lorentz transformation rather than the Galilei transformation resolves all apparent problems with Maxwell's equations with regard to the special principle of relativity.

 

[tade]

Please, could you ask a clear question? Of course, if you want to change from one inertial reference frame to another you have to use the Lorentz transformation to transform the electromagnetic field components with respect to one of these frames to the other, and using the Lorentz transformation rather than the Galilei transformation resolves all apparent problems with Maxwell's equations with regard to the special principle of relativity.

for example, in the derivation in #5, the curls are shown to be equal without resorting to the Lorentz transformation

 

[PeterDonis]

in the derivation in #5, the curls are shown to be equal without resorting to the Lorentz transformation

No, in the derivation in #5 (which, btw, your quote does not provide a link to--it would be really helpful if you could give a link to the actual thread so we could see the context) shows the curls to be equal if it is assumed that $\vec{E}^\prime = \vec{v} \times \vec{B}$. As I pointed out many posts ago. Nobody anywhere that I can see has shown that the curls are equal without assuming that $\vec{E}^\prime = \vec{v} \times \vec{B}$.

The quote you provide in #5 gives no indication of why one would want to assume that $\vec{E}^\prime = \vec{v} \times \vec{B}$, but I would expect that to come from the context of the thread, which, as noted just now, you have not provided a link to. But one obvious way to motivate such an assumption would be to look at the Lorentz transformation equations and use the low-velocity approximation, in which $\gamma \approx 1$.

 

[tade]

No, in the derivation in #5 shows the curls to be equal if it is assumed that $\vec{E}^\prime = \vec{v} \times \vec{B}$. As I pointed out many posts ago. Nobody anywhere that I can see has shown that the curls are equal without assuming that $\vec{E}^\prime = \vec{v} \times \vec{B}$.

The quote you provide in #5 gives no indication of why one would want to assume that $\vec{E}^\prime = \vec{v} \times \vec{B}$

I believe its the other way round, first using the Maxwell-Faraday equation to show that the curls are equal, and then assuming that the fields are equal, aka assuming that $\vec{E}^\prime = \vec{v} \times \vec{B}$

 

[PeterDonis]

I believe its the other way round

The quote you gave in post #5 makes it clear that the assumption $\vec{E}^\prime = \vec{v} \times \vec{B}$ is used to show that the curls are equal, not the other way around.

Can you give a link to the previous thread? As I noted, your quote in post #5 does not include a link.

 

[tade]

ok searching for it

The quote you gave in post #5 makes it clear that the assumption $\vec{E}^\prime = \vec{v} \times \vec{B}$ is used to show that the curls are equal, not the other way around.

it doesn't appear to start with the assumption that $\vec{E}^\prime = \vec{v} \times \vec{B}$
that would be just two (or one) steps.

instead, it starts with the Maxwell-Faraday equation.
then, expressing the rate of change of the B-field in terms of the uniform velocity field, and finally re-arranging that to give the curl of $\vec{v} \times \vec{B}$

 

[PeterDonis]

it doesn't appear to start with the assumption

The equality of the curls is only stated in one place: right after the assumption that $\vec{E}^\prime = \vec{v} \times \vec{B}$. And it is explicitly stated that the equality of the curls is obtained from that assumption: the quote says "with $\vec{E}^\prime = \vec{v} \times \vec{B}$ you would have [equality of the curls]".

Either give a link to the previous thread or this thread will be closed, since you seem unable to state clearly what it is you are concerned about or even to correctly read things you have quoted yourself.

 

[tade]

The equality of the curls is only stated in one place: right after the assumption that $\vec{E}^\prime = \vec{v} \times \vec{B}$. And it is explicitly stated that the equality of the curls is obtained from that assumption: the quote says "with $\vec{E}^\prime = \vec{v} \times \vec{B}$ you would have [equality of the curls]".

Either give a link to the previous thread or this thread will be closed, since you seem unable to state clearly what it is you are concerned about or even to correctly read things you have quoted yourself.

But we don't need that assumption to show the equality of curls.

Just need to follow these steps, all featured in #5:

instead, it starts with the Maxwell-Faraday equation.
then, expressing the rate of change of the B-field in terms of the uniform velocity field, and finally re-arranging that to give the curl of $\vec{v} \times \vec{B}$

the last step which features the curl of E shows the re-arranging.
And we don't need to assume that $\vec{E}^\prime = \vec{v} \times \vec{B}$ in order to do that re-arranging.

 

[PeterDonis]

the last step which features the curl of E shows the re-arranging.

Yes, after the assumption $\vec{E}^\prime = \vec{v} \times \vec{B}$ has been introduced to derive the equality of curls.

we don't need to assume that $\vec{E}^\prime = \vec{v} \times \vec{B}$ in order to do that re-arranging.

You do to get the equality of curls that then gets rearranged. See above.

Since you continue to make incorrect statements about something you quoted yourself, and are either unable or unwilling to clearly state your issue or to provide a link to the previous thread you are quoting from so we can see the context, this thread is closed.