What's the derivation in a moving magnet & conductor problem?

[tade]

In the Wikipedia page of the moving magnet and conductor problem, it asserts "This results in: E' = v x B", but does not elaborate why.

What's the full derivation?

 

[BvU]

A derivation of the Lorentz force would be somewhat circular: the expression for the LF is derived from observations. That's the way it is ...

If you find this view unsatisfactory, I can understand. But digging deeper doesn't really change this situation, I think. Would be interested in a theoretician's opinion.
@Orodruin , @vela, @Nugatory ?

 

[Orodruin]

What is being referred to is the Maxwell-Faraday equation
$$
\nabla \times \vec E' = - \frac{\partial \vec B'}{\partial t}
$$
leading to the given relation. The behaviour of $\vec B'$ is given in the equation before, but that equation seems incomplete. Approximately (for small velocities),
$$
\vec B'(\vec r', t) = \vec B(\vec r'+\vec v t).
$$
This leads to
$$
\partial_t \vec B' = \partial_t \vec B(\vec r+ \vec v t) = (\vec v \cdot \nabla)\vec B.
$$
For the left-hand side with $\vec E' = \vec v \times \vec B$ you would have
$$
\nabla \times \vec E' = \nabla \times (\vec v \times \vec B) = \vec v (\nabla \cdot \vec B) - (\vec v \cdot \nabla) \vec B = - (\vec v \cdot \nabla) \vec B,
$$
because the magnetic field is divergence free. Thus $\vec E' = \vec v \times \vec B$ solves the Maxwell-Faraday equation.

The "better" way I think is to use the Lorentz transformation properties of the electromagnetic field.

 

[tade]

What is being referred to is the Maxwell-Faraday equation
$$
\nabla \times \vec E' = - \frac{\partial \vec B'}{\partial t}
$$
leading to the given relation. The behaviour of $\vec B'$ is given in the equation before, but that equation seems incomplete. Approximately (for small velocities),
$$
\vec B'(\vec r', t) = \vec B(\vec r'+\vec v t).
$$
This leads to
$$
\partial_t \vec B' = \partial_t \vec B(\vec r+ \vec v t) = (\vec v \cdot \nabla)\vec B.
$$
For the left-hand side with $\vec E' = \vec v \times \vec B$ you would have
$$
\nabla \times \vec E' = \nabla \times (\vec v \times \vec B) = \vec v (\nabla \cdot \vec B) - (\vec v \cdot \nabla) \vec B = - (\vec v \cdot \nabla) \vec B,
$$
because the magnetic field is divergence free. Thus $\vec E' = \vec v \times \vec B$ solves the Maxwell-Faraday equation.

The "better" way I think is to use the Lorentz transformation properties of the electromagnetic field.

Thanks, though sorry, could you explain the meaning of this notation: $$(\vec v \cdot \nabla)\vec B$$

And also the intermediate steps required to go from: $$\partial_t \vec B(\vec r+ \vec v t)$$ to: $$(\vec v \cdot \nabla)\vec B.$$

Hope its not too much of a hassle with a lot of TeX

 

[tade]

A derivation of the Lorentz force would be somewhat circular: the expression for the LF is derived from observations. That's the way it is ...

If you find this view unsatisfactory, I can understand. But digging deeper doesn't really change this situation, I think.

I would like the derivation so I can understand how the two formulas give the same numerical results.

 

[BvU]

Lorentz force is $\vec F_L = q (\vec v\times \vec B)$ and with $\vec F = q \vec E$ you are back at the 'result' expression.

 

[tade]

Lorentz force is $\vec F_L = q (\vec v\times \vec B)$ and with $\vec F = q \vec E$ you are back at the 'result' expression.

As Oroduin mentioned, it is: $$\partial_t \vec B(\vec r+ \vec v t)=(\vec v \cdot \nabla)\vec B$$ with some intermediate steps, and I'd like to figure out the details

 

[vanhees71]

I'm a bit worried about the fact that the OP is mislead by using utterly wrong transformation properties of the em. field. Here Lorentz transformations rather than Galilei transformations have to be applied. BTW it's the very problem Einstein used in his famous 1905 paper on special relativity as a motivation for the reformulation of the space-time model!

 

[tade]

I'm a bit worried about the fact that the OP is mislead by using utterly wrong transformation properties of the em. field. Here Lorentz transformations rather than Galilei transformations have to be applied. BTW it's the very problem Einstein used in his famous 1905 paper on special relativity as a motivation for the reformulation of the space-time model!

oh no, which part is wrong? and what are the steps of the correct derivation? thanks

 

[Orodruin]

oh no, which part is wrong?

The transformation rules are being given in the low-velocity approximation, not in its full relativistically invariant glory.

and what are the steps of the correct derivation?

The "better" way I think is to use the Lorentz transformation properties of the electromagnetic field.

 

[tade]

The transformation rules are being given in the low-velocity approximation, not in its full relativistically invariant glory.

I see, I guess that's not "utterly wrong" though.Back to my previous question, as the low-velocity approximation is fine for me, I'd like to know the notation's meaning and the intermediate steps, thanks

Thanks, though sorry, could you explain the meaning of this notation: $$(\vec v \cdot \nabla)\vec B$$

And also the intermediate steps required to go from: $$\partial_t \vec B(\vec r+ \vec v t)$$ to: $$(\vec v \cdot \nabla)\vec B.$$

Hope its not too much of a hassle with a lot of TeX

 

[tade]

I'm a bit worried about the fact that the OP is mislead by using utterly wrong transformation properties of the em. field. Here Lorentz transformations rather than Galilei transformations have to be applied. BTW it's the very problem Einstein used in his famous 1905 paper on special relativity as a motivation for the reformulation of the space-time model!

oh no, which part is wrong? and what are the steps of the correct derivation? thanks

@vanhees71 I'm good with the low-velocity approximations, keeping it simple :)

 

[vanhees71]

It's utterly wrong not only in a physical sense but also in a didactical. Classical electrodynamics is among the most difficult subjects in the undergraduate curriculum. It's not necessary to make it even more complicated by using oldfashioned concepts which have been solved more than 110 years ago by the development of special relativity, finalized by Minkowski in 1908.

 

[Orodruin]

It's utterly wrong not only in a physical sense but also in a didactical. Classical electrodynamics is among the most difficult subjects in the undergraduate curriculum. It's not necessary to make it even more complicated by using oldfashioned concepts which have been solved more than 110 years ago by the development of special relativity, finalized by Minkowski in 1908.

I do not think it needs to be didactically wrong as long as one is clear about being in the low velocity limit. To the contrary, keeping only the leading order terms in a small parameter expansion is an important tool in phenomenology.

 

[vanhees71]

There is not one low-velocity limit in the sense of Galilei invariant electromagnetics but (at least) two, i.e., the electro-quasi-static and the magneto-quasi-static approximation. Neither is complete. Already for the most important application of these approximations in engineering, i.e., AC circuit theory, you need both. I'm not sure, but isn't precisely this "moving-magnet problem" the paradigmatic example for the fact that neither works, and that's why Einstein put it in the introductory paragraphs of his 1905 paper on the subject?

Another example is the homopolar generator, which is describable only with the correct relativistic version of the constituent equations a la Minkowski (i.e., taking into account the Hall effect in Ohm's Law, which is the correct relativistic form of it) although no large velocities are involved, see

https://th.physik.uni-frankfurt.de/~hees/pf-faq/homopolar.pdf

 

[BvU]

@vanhees71 seems to forget that there are millions and millions for whom Ampere law, Faraday law etc are well beyond their zenith in abstraction. Yet they make their living with everyday applications of the Maxwell equations that they design and realize. A curriculum as he proposes is indeed ideal for a very, very select group.

 

[tade]

@BvU @vanhees71 @Orodruin thanks guys, though sorry, I’m trying to figure out #4

 

[vanhees71]

I didn't get where the problem is. The operator is self-explaining by notation. Maybe it helps to write it down in the concrete Ricci calculus for Carstesian coordinates $x^j$:
$$(\vec{v} \cdot \vec{\nabla}) B^j=v^k \partial_k B^j.$$

 

[Orodruin]

Also, the only thing necessary to reach that expression is the chain rule for derivatives.

 

[tade]

I didn't get where the problem is. The operator is self-explaining by notation. Maybe it helps to write it down in the concrete Ricci calculus for Carstesian coordinates $x^j$:
$$(\vec{v} \cdot \vec{\nabla}) B^j=v^k \partial_k B^j.$$

Also, the only thing necessary to reach that expression is the chain rule for derivatives.

Thanks, though sorry, could you explain the meaning of this notation: $$(\vec v \cdot \nabla)\vec B$$

And also the intermediate steps required to go from: $$\partial_t \vec B(\vec r+ \vec v t)$$ to: $$(\vec v \cdot \nabla)\vec B.$$

Hope its not too much of a hassle with a lot of TeX

sorry, I’m not clear about what the operator $$(\vec v \cdot \nabla)$$ means

 

[Orodruin]

It is a directional derivative in the direction of $\vec v$. It is written explicitly how it is expressed in post #18. I am sorry, but I do not see how it can be any clearer than that ...

 

[tade]

It is a directional derivative in the direction of $\vec v$. It is written explicitly how it is expressed in post #18. I am sorry, but I do not see how it can be any clearer than that ...

I’m not familiar with Ricci calculus sadly

 

[Orodruin]

Then just look at it as a directional derivative
$$
(\vec v \cdot \nabla) f(\vec x) = \lim_{\epsilon\to 0}\left(\frac{f(\vec x + \epsilon \vec v) - f(\vec x)}{\epsilon}\right),
$$
where $f$ is any field (scalar, vector, tensor, etc).

 

[tade]

I’m not familiar with Ricci calculus sadly

@vanhees71

 

[tade]

Then just look at it as a directional derivative
$$
(\vec v \cdot \nabla) f(\vec x) = \lim_{\epsilon\to 0}\left(\frac{f(\vec x + \epsilon \vec v) - f(\vec x)}{\epsilon}\right),
$$
where $f$ is any field (scalar, vector, tensor, etc).

ε is the quantity of time right?

 

[Orodruin]

ε is the quantity of time right?

No. This is just the standard definition of a derivative, it has nothing to do with any physical quantity. As I already said, you will also need the chain rule for derivatives.

 

[vanhees71]

Again, in Cartesian components it reads
$$\vec{v} \cdot \vec{\nabla}) \vec{B} = \vec{e}_j v^k \partial_k B^j.$$

 

[tade]

No. This is just the standard definition of a derivative, it has nothing to do with any physical quantity. As I already said, you will also need the chain rule for derivatives.

As in, in this specific case of the moving magnet, ε represents t?

 

[tade]

Again, in Cartesian components it reads
$$\vec{v} \cdot \vec{\nabla}) \vec{B} = \vec{e}_j v^k \partial_k B^j.$$

Sorry, not familiar with Ricci calculus.

@Orodruin @vanhees71 Is it correct to say that the x-component of $$(\vec{v} \cdot \vec{\nabla}) \vec{B}$$ reads $$v_x (\partial_x B_x)+v_y (\partial_y B_x)+v_z (\partial_z B_x)$$ ?

 

[Delta2]

Sorry, not familiar with Ricci calculus.

@Orodruin @vanhees71 Is it correct to say that the x-component of $$(\vec{v} \cdot \vec{\nabla}) \vec{B}$$ reads $$v_x (\partial_x B_x)+v_y (\partial_y B_x)+v_z (\partial_z B_x)$$ ?

yes I believe that is correct.

 

[Orodruin]

As in, in this specific case of the moving magnet, ε represents t?

No. It is just a number. Again, the relation to $t$ comes from the chain rule.

@Orodruin @vanhees71 Is it correct to say that the x-component of $$(\vec{v} \cdot \vec{\nabla}) \vec{B}$$ reads $$v_x (\partial_x B_x)+v_y (\partial_y B_x)+v_z (\partial_z B_x)$$ ?

Yes.

 

[vanhees71]

Sorry, not familiar with Ricci calculus.

@Orodruin @vanhees71 Is it correct to say that the x-component of $$(\vec{v} \cdot \vec{\nabla}) \vec{B}$$ reads $$v_x (\partial_x B_x)+v_y (\partial_y B_x)+v_z (\partial_z B_x)$$ ?

Yes, that's the x-component of what I wrote from the very beginning. Sorry, I was not aware that you didn't know Ricci calculus. In the usual matrix-vector notation you have
$$(\vec{v} \cdot \vec{\nabla} \vec{B})=\begin{pmatrix}
v_x \partial_x B_x + v_y \partial_y B_x + v_z \partial_z B_x \\
v_x \partial_x B_y+ v_y \partial_y B_y + v_z \partial_z B_y \\
v_x \partial_x B_z + v_y \partial_y B_z + v_z \partial_z B_z
\end{pmatrix}.$$

 

[tade]

$$\nabla \times \vec E' = \nabla \times (\vec v \times \vec B) = \vec v (\nabla \cdot \vec B) - (\vec v \cdot \nabla) \vec B = - (\vec v \cdot \nabla) \vec B,$$
because the magnetic field is divergence free. Thus $\vec E' = \vec v \times \vec B$ solves the Maxwell-Faraday equation.

by the way, the divergence of E should be zero right?

 

[Orodruin]

by the way, the divergence of E should be zero right?

Yes, this follows directly from $\vec B$ being divergence free.

 

[tade]

Yes, this follows directly from $\vec B$ being divergence free.

I was thinking about the divergence of v × B. It's a triple product, so it is equal to -v ⋅ (∇ × B).

If the divergence of E is zero, then so must the divergence of v × B.

But the curl of B is not always zero, as it is proportional to the current density.