Motional EMF in the frame of a moving wire

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  • #1
feynman1
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When a magnetic field is moving (moving magnet) while a wire isn’t---in the reference frame of the wire, there’s no known magnetic force/effect on the wire before the introduction of motional emf and relativity. If viewed in the reference frame of the magnetic field instead where the wire is moving, there’s no need of introducing motional emf as Lorentz force can be used to explain. So may I confirm if the perspective in the reference frame of the wire is the reason cause for introducing motional emf?
 

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  • #2
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When a magnetic field is moving (moving magnet) while a wire isn’t---in the reference frame of the wire, there’s no known magnetic force/effect on the wire before the introduction of motional emf and relativity.
Don’t forget to describe the electric field and electric force. If you are going to be changing reference frames you need both.

If viewed in the reference frame of the magnetic field instead where the wire is moving, there’s no need of introducing motional emf as Lorentz force can be used to explain.
I am not sure what you mean here. Can you write this explanation?
 
  • #3
feynman1
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Don’t forget to describe the electric field and electric force. If you are going to be changing reference frames you need both.

I am not sure what you mean here. Can you write this explanation?
When the wire is moving, there's a similar Hall effect driven by qvB to deflect charges that ultimately turns into an emf. So there's no need of invoking a changing magnetic flux that induces an emf. That a changing magnetic flux induces an emf is considered a new concept.
 
  • #4
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When the wire is moving, there's a similar Hall effect driven by qvB to deflect charges that ultimately turns into an emf. So there's no need of invoking a changing magnetic flux that induces an emf. That a changing magnetic flux induces an emf is considered a new concept.
Thank you for clarifying the second part of my question. What about the first? You didn’t describe the E fields or forces in the wire frame.
 
  • #5
vanhees71
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There's one electromagnetic field, and that's it. You have to always work with the complete field, if you change from one frame of reference through the other, and you have to use the Lorentz transformation of spacetime-fourvectors as well as the electromagnetic field components, which are the components of a 2nd rank antisymmetric tensor in Minkowski spacetime.

In the reference frame, where the magnet is moving you have both electric and magnetic components. The electric field acts on the electrons in the wire even when they are still at rest and thus a current can start to flow and then of course you need the complete Lorentz force including electric and magnetic field components.

In the reference frame where the magnet is at rest and the wire is moving, you have a pure magnetostatic field and the Lorentz force acts with the magnetic field acts on the electrons within the wire.

This is precisely the problem Einstein started his famous paper on special relativity in 1905. With SR all the apparent problems were finally completely understood, and no asymmetries in the description of the situation in either the moving-magnet or magnet-at-rest frame and a wire (or wire loop) occur anymore, as Einstein promised in the introduction to this paper.
 
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  • #6
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Thank you for clarifying the second part of my question. What about the first? You didn’t describe the E fields or forces in the wire frame.
In the frame of the wire, there’s no known electric field/force to explain an observed induced current if there were no concept like motional emf.
 
  • #7
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In the frame of the wire, there’s no known electric field/force to explain an observed induced current if there were no concept like motional emf.
Why not? The magnet is moving. What does Faraday’s law say?
 
  • #8
feynman1
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Why not? The magnet is moving. What does Faraday’s law say?
So I think there’s nothing but Faraday’s law that can explain the induced current in the frame of the wire, but I consider Faraday’s law to be new. In the frame of the magnetic field, the wire is moving so qvB alone can explain the induced current without invoking Faraday’s law.
 
  • #9
feynman1
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There's one electromagnetic field, and that's it. You have to always work with the complete field, if you change from one frame of reference through the other, and you have to use the Lorentz transformation of spacetime-fourvectors as well as the electromagnetic field components, which are the components of a 2nd rank antisymmetric tensor in Minkowski spacetime.

In the reference frame, where the magnet is moving you have both electric and magnetic components. The electric field acts on the electrons in the wire even when they are still at rest and thus a current can start to flow and then of course you need the complete Lorentz force including electric and magnetic field components.

In the reference frame where the magnet is at rest and the wire is moving, you have a pure magnetostatic field and the Lorentz force acts with the magnetic field acts on the electrons within the wire.

This is precisely the problem Einstein started his famous paper on special relativity in 1905. With SR all the apparent problems were finally completely understood, and no asymmetries in the description of the situation in either the moving-magnet or magnet-at-rest frame and a wire (or wire loop) occur anymore, as Einstein promised in the introduction to this paper.
Yes what you say is sth way after Faraday’s law. This thread is trying to understand the logic when Faraday’s law came into being. I think, in the frame of the magnetic field, qvB can explain the induced current; in the frame of the wire, nothing can explain the induced current before Faraday. Do you agree?
 
  • #10
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So I think there’s nothing but Faraday’s law that can explain the induced current in the frame of the wire, but I consider Faraday’s law to be new. In the frame of the magnetic field, the wire is moving so qvB alone can explain the induced current without invoking Faraday’s law.
It doesn’t work that way. You have to start with a complete solution to all of Maxwell’s equations in one reference frame. Then you can transform it to another frame to get a complete solution to Maxwell’s equations in that frame.

You cannot take part of a solution and transform it to get anything meaningful. Faraday’s law is an essential part of Maxwell’s equations, you cannot pick and choose parts to leave out and expect to get something that can be meaningfully transformed.
 
  • #11
feynman1
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It doesn’t work that way. You have to start with a complete solution to all of Maxwell’s equations in one reference frame. Then you can transform it to another frame to get a complete solution to Maxwell’s equations in that frame. You cannot take part of a solution and transform it to get anything meaningful. Faraday’s law is an essential part of Maxwell’s equations, you cannot pick and choose parts to leave out and expect to get something that can be meaningfully transformed.
I’ve been speaking under the context as if Maxwell’s eqs weren’t established yet. This thread is to ponder on the advent of Faraday’s law.
 
  • #12
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I’ve been speaking under the context as if Maxwell’s eqs weren’t established yet. This thread is to ponder on the advent of Faraday’s law.
You should have mentioned that in the OP, but I don’t think that changes my response. Without Faraday’s law you cannot meaningfully transform the fields.
 
  • #13
feynman1
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You should have mentioned that in the OP, but I don’t think that changes my response. Without Faraday’s law you cannot meaningfully transform the fields.
An alternative way of asking my question. The change in magnetic flux has 2 parts, 1 the change in B, the other the change in area. The latter can be explained by qvB. The former has to resort to a new law: Faraday.
 
  • #14
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An alternative way of asking my question. The change in magnetic flux has 2 parts, 1 the change in B, the other the change in area. The latter can be explained by qvB. The former has to resort to a new law: Faraday.
This doesn’t make any sense to me. Flux is just a quantity. I am not sure how any part of it is explained by one formula or another. Formulas explain relationships between quantities, not the individual quantities themselves. I don’t see what it means in this context for flux to be explained by either Faraday’s law or the Lorentz force law.
 
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  • #15
vanhees71
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So I think there’s nothing but Faraday’s law that can explain the induced current in the frame of the wire, but I consider Faraday’s law to be new. In the frame of the magnetic field, the wire is moving so qvB alone can explain the induced current without invoking Faraday’s law.
Sure, Faraday's Law says (in Heaviside-Lorentz units)
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.$$
The complete integral law for moving surfaces ##S## and their boundary ##\partial S## uses Stokes's integral theorem and Gauss's Law for the magnetic field, ##\vec{\nabla} \cdot \vec{B}=0##, leading to
$$\frac{1}{c} \mathrm{d}_t \Phi_{\vec{B}} = -\int_{\partial S} \mathrm{d} \vec{x} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right),$$
where ##\vec{v}(t,\vec{x})## is the velocity of the line elements of the boundary ##\partial S##.
 
  • #16
Delta2
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An alternative way of asking my question. The change in magnetic flux has 2 parts, 1 the change in B, the other the change in area. The latter can be explained by qvB. The former has to resort to a new law: Faraday.
Faraday's law of induction (not to be confused with Maxwell-Faraday equation) explains the emf due to both parts of the change in magnetic flux. That is Faraday's law of induction includes the case where the magnetic flux is changing due to the change of area enclosed or due to the magnetic field changing. Look at this proof
https://en.wikipedia.org/wiki/Faraday's_law_of_induction#Proof
Be sure to click on [show] link to see the details.
 
  • #17
vanhees71
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What is Faraday's law of induction vs. Maxwell-Faraday equation? The fundamental form are the local Maxwell equations, and there's one Faraday's law, which reads
$$\frac{1}{c} \partial_t \vec{B}+\vec{\nabla} \times \vec{E}=0.$$
 
  • #18
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What is Faraday's law of induction vs. Maxwell-Faraday equation? The fundamental form are the local Maxwell equations, and there's one Faraday's law, which reads
$$\frac{1}{c} \partial_t \vec{B}+\vec{\nabla} \times \vec{E}=0.$$
That's the Maxwell-Faraday equation in differential form. In its integral form the surface and the boundary of the surface are constant in time. In Faraday's law of induction the boundary of the surface (and the surface) can vary in time and the time derivative is in front of the surface integral, that is Faraday's law of induction is
$$EMF=\frac{d}{dt}\int_{S(t)}\mathbf{B}\cdot d\mathbf{S}$$
while the Maxwell-Faraday equation in integral form is
$$\int_{\partial S} \mathbf{E}\cdot d\mathbf{l}=\int_S\frac{\partial \mathbf{B}}{\partial t}\cdot d\mathbf{S}$$
They reduce to the same equation if the boundary of surface S does not depend on time, but they are not the same thing if the boundary of the surface S changes with time.
 
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  • #19
vanhees71
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This is correct (up to a sign) only if ##S## and thus ##\partial S## are not moving, but if they are moving from the "Reynolds transport theorem" for surface integrals you get from the local form of Faraday's Law together with Gauss's Law for the magnetic field in local form the correct equation
$$\mathcal{E}=\int_{\partial S} \mathrm{d} \vec{x} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right)=-\frac{1}{c} \mathrm{d}_t \Phi_{\vec{B}}.$$
For moving surfaces you have the full electromotive force, including the electric and the magnetic piece, as it must be.
 
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  • #20
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All of which leads me to prefer the differential expression
 
  • #21
vanhees71
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Indeed. The great confusion is that in many textbooks the authors use Stokes's integral theorem, which leads to the (still correct!) equation
$$\int_{S} \mathrm{d}^2 \vec{f} \cdot (\vec{\nabla} \times \vec{E})=\int_{\partial S} \mathrm{d} \vec{x} \cdot \vec{E}=-\frac{1}{c} \int_{S} \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}$$
and then without much explanation and a thick red caveat sign interchange the time derivative with the integral on the right-hand side. This is allowed if AND ONLY IF the surface ##S## and its boundary are at rest.

Otherwise you have to use the "transport theorem" for surface integrals. I don't know an online proof for it except in the German Wikipedia (it's the only example, where the German Wikipedia is more complete than the English I now of):

https://de.wikipedia.org/wiki/Transportsatz#Transportsatz_für_Flächenintegrale

In my notation it reads
$$\mathrm{d}_t \Phi_B=\mathrm{d}_t \int_S \mathrm{d}^2 \vec{f} \cdot \vec{B}=\int_S \mathrm{d}^2 \vec{f} \cdot (\partial_t \vec{B} + \vec{v} (\vec{\nabla} \cdot \vec{B})-\int_{\partial S} \mathrm{d} \vec{x} \cdot (\vec{v} \times \vec{B}).$$
Now using ##\vec{\nabla} \cdot \vec{B}=0## and the differential form of Faraday's Law you get
$$\mathrm{d}_t \Phi_B = -c \int_S \mathrm{d}^2 \vec{f} \cdot (\vec{\nabla} \times \vec{E}) --\int_{\partial S} \mathrm{d} \vec{x} \cdot (\vec{v} \times \vec{B}).$$
Once more using Stokes's theorem you get the equation quoted above
$$\mathrm{d}_t \Phi_B=-\int_{\partial S} \mathrm{d} \vec{x} \cdot (c \vec{E} + \vec{v} \times \vec{B}).$$
 
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  • #22
feynman1
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In my notation it reads
$$\mathrm{d}_t \Phi_B=\mathrm{d}_t \int_S \mathrm{d}^2 \vec{f} \cdot \vec{B}=\int_S \mathrm{d}^2 \vec{f} \cdot (\partial_t \vec{B} + \vec{v} (\vec{\nabla} \cdot \vec{B})-\int_{\partial S} \mathrm{d} \vec{x} \cdot (\vec{v} \times \vec{B}).$$
I think when applying transport theorem, there's an error with the position of the dot product in $$\vec{v} (\vec{\nabla} \cdot \vec{B})$$. The dot should've been between v and nabla.
 
  • #23
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$$\frac{1}{c} \mathrm{d}_t \Phi_{\vec{B}} = -\int_{\partial S} \mathrm{d} \vec{x} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right),$$
where ##\vec{v}(t,\vec{x})## is the velocity of the line elements of the boundary ##\partial S##.
Good. What the OP was asking is that v*B term wasn't a new thing thanks to Faraday's law, but E*dx term was observed in experiment and invented in theory after Faraday's law. Would you agree?
 
  • #24
vanhees71
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I think when applying transport theorem, there's an error with the position of the dot product in $$\vec{v} (\vec{\nabla} \cdot \vec{B})$$. The dot should've been between v and nabla.
No, the formula given is correct. For a proof, see

https://itp.uni-frankfurt.de/~hees/publ/mameth-l3.pdf

Sect. 3.16.2. Though it's in German, I guess the formula density should be high enough that it's understandable.
 
  • #25
feynman1
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No, the formula given is correct. For a proof, see

https://itp.uni-frankfurt.de/~hees/publ/mameth-l3.pdf

Sect. 3.16.2. Though it's in German, I guess the formula density should be high enough that it's understandable.
That's nice of you. In order to tell what I missed, I'd better check the English version. Any? For from what I see, simply take the derivative of B(x,t) w.r.t. time yields a 2nd term I proposed, in line with transport theorem in fluids.
But the above technicality I don't care at this point. I really care about post #23.
 
  • #26
vanhees71
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I don't have an English version yet. So here it is

Let the time-dependent surface be described by the parametrization
$$\vec{x}=\vec{\xi}(t,q_1,q_2).$$
The surface element is
$$\mathrm{d}^2 \vec{f} = \mathrm{d}^2 q \partial_{q_1} \xi \times \partial_{q_2} \xi.$$
For an arbitrary vector field ##\vec{W}=\vec{W}(t,\vec{x})##
$$\Phi_W(t)=\int_F \mathrm{d}^2 \vec{f} \cdot \vec{W}=\int_Q \mathrm{d}^2 q (\partial_{q_1}\vec{\xi} \times \partial_{q_2} \vec{\xi}) \cdot \vec{W}(t,\vec{\xi}).$$
Then
$$\dot{\Phi}_W=\int_Q \mathrm{d}^2 q [(\partial_{q_1} \dot{\vec{\xi}} \times \partial_{q_2} \vec{\xi} +\partial_{q_1}\vec{\xi} \times \partial_{q_2} \dot{\vec{\xi}})\cdot \vec{W} + (\partial_{q_1}\vec{\xi} \times \partial_{q_2} \vec{\xi}) \cdot (\partial_t \vec{W} + (\dot{\vec{\xi}} \cdot \nabla)\vec{W}))].$$
Now we interpret ##\vec{v}=\dot{\vec{\xi}}## as a function of ##\vec{x}##. Then we have
$$\partial_{q_j} \dot{\vec{\xi}} = (\partial_{q_j} \vec{\xi} \cdot \vec{\nabla}) \vec{v}.$$
Plugging this in above after some longer calculation in the Ricci calculus (see my manuscript, for this part you really don't understand the German text), gives
$$\mathrm{d}^2 q \vec{W} \cdot (\partial_{q_1} \dot{\vec{\xi}} \times \partial_{q_2} \vec{\xi} +\partial_{q_1}\vec{\xi} \times \partial_{q_2} \dot{\vec{\xi}}) = \mathrm{d}^2 \vec{f} \cdot [\vec{W} (\vec{\nabla} \cdot \vec{v}) - (\vec{W} \cdot \vec{\nabla}) \vec{v}).$$
So finally we have [EDIT: corrected in view of #28]
$$\dot{\Phi}_W=\int_F \mathrm{d}^2 \vec{f} \cdot [\partial_t \vec{W} + (\vec{v} \cdot \vec{\nabla}) \vec{W} + \vec{W} \vec{\nabla} \cdot \vec{v} - (\vec{W} \cdot \vec{\nabla}) \vec{v})].$$
We can simplify this a bit by using
$$\vec{\nabla} \times (\vec{W} \times \vec{v}) = (\vec{v} \cdot \vec{\nabla} \vec{W} - \vec{v} (\vec{\nabla} \cdot \vec{W}) + \vec{W} (\vec{\nabla} \cdot \vec{v})-(\vec{W} \cdot \vec{\nabla})\vec{v},$$
which is also shown most clearly with a longer calculation in the Ricci calculus. Thus you can write [EDIT: Corrected in view of #28]
$$\dot{\Phi}_W=\int_F \mathrm{d}^2 \vec{f} \cdot [\partial_t \vec{W} + \vec{v} (\vec{\nabla} \cdot \vec{W}) + \vec{\nabla} \times (\vec{W} \times \vec{v})].$$
Finally we apply Stokes's theorem to the curl term:
$$\dot{\Phi}_W=\int_F \mathrm{d}^2 \vec{f} \cdot [\partial_t \vec{W}+\vec{v}(\vec{\nabla} \cdot \vec{W})] + \int_{\partial F} \mathrm{d} \vec{x} \cdot (\vec{W} \times \vec{v}).$$
Setting ##\vec{W}=\vec{B}## and using
$$\vec{\nabla} \cdot \vec{B}=0, \quad \partial_t \vec{B}=-c \vec{\nabla} \times \vec{E}$$
leads to the general Faraday Law in integral form
$$\frac{1}{c} \dot{\Phi}_B=-\int_{\partial F} \mathrm{d} \vec{x} \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right)=-\mathcal{E},$$
where ##\mathcal{E}## is the complete electromotive force including both contributions from the electric and magnetic field.
 
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  • #27
feynman1
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I don't have an English version yet. So here it is
So nice of you translating all this!
 
  • #28
Delta2
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Thanks a ton @vanhees71 for the nice derivation. Just to correct some small typos, I believe in the formula you write after "So finally we have" the third term should be $$\vec{W}(\vec{\nabla}\cdot\vec{v})$$. And also about 5 lines down after "Thus you can write" again the third term should be $$\vec{\nabla}\times(\vec{W}\times\vec{v})$$
 
  • #29
vanhees71
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Thanks! I've corrected the typos.
 

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