- #1

feynman1

- 435

- 29

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter feynman1
- Start date

- #1

feynman1

- 435

- 29

- #2

Dale

Mentor

- 33,684

- 11,260

Don’t forget to describe the electric field and electric force. If you are going to be changing reference frames you need both.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.

I am not sure what you mean here. Can you write this explanation?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.

- #3

feynman1

- 435

- 29

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.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?

- #4

Dale

Mentor

- 33,684

- 11,260

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.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.

- #5

- 22,038

- 12,952

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.

- #6

feynman1

- 435

- 29

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.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.

- #7

Dale

Mentor

- 33,684

- 11,260

Why not? The magnet is moving. What does Faraday’s law say?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.

- #8

feynman1

- 435

- 29

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.Why not? The magnet is moving. What does Faraday’s law say?

- #9

feynman1

- 435

- 29

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?

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.

- #10

Dale

Mentor

- 33,684

- 11,260

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.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.

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

- 435

- 29

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.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.

- #12

Dale

Mentor

- 33,684

- 11,260

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.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.

- #13

feynman1

- 435

- 29

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.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.

- #14

Dale

Mentor

- 33,684

- 11,260

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.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.

Last edited:

- #15

- 22,038

- 12,952

Sure, Faraday's Law says (in Heaviside-Lorentz units)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.

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

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

- 5,695

- 2,473

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 proofAn 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.

https://en.wikipedia.org/wiki/Faraday's_law_of_induction#Proof

Be sure to click on [show] link to see the details.

- #17

- 22,038

- 12,952

$$\frac{1}{c} \partial_t \vec{B}+\vec{\nabla} \times \vec{E}=0.$$

- #18

- 5,695

- 2,473

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

$$\frac{1}{c} \partial_t \vec{B}+\vec{\nabla} \times \vec{E}=0.$$

$$EMF=\frac{d}{dt}\int_{S(t)}\mathbf{B}\cdot d\mathbf{S}$$

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.

Last edited:

- #19

- 22,038

- 12,952

$$\mathcal{E}=\int_{\partial S} \mathrm{d} \vec{x} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right)=-\frac{1}{c} \mathrm{d}_t \Phi_{\vec{B}}.$$

For moving surfaces you have the full electromotive force, including the electric and the magnetic piece, as it must be.

- #20

Dale

Mentor

- 33,684

- 11,260

All of which leads me to prefer the differential expression

- #21

- 22,038

- 12,952

$$\int_{S} \mathrm{d}^2 \vec{f} \cdot (\vec{\nabla} \times \vec{E})=\int_{\partial S} \mathrm{d} \vec{x} \cdot \vec{E}=-\frac{1}{c} \int_{S} \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}$$

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

$$\mathrm{d}_t \Phi_B=-\int_{\partial S} \mathrm{d} \vec{x} \cdot (c \vec{E} + \vec{v} \times \vec{B}).$$

- #22

feynman1

- 435

- 29

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.In my notation it reads

$$\mathrm{d}_t \Phi_B=\mathrm{d}_t \int_S \mathrm{d}^2 \vec{f} \cdot \vec{B}=\int_S \mathrm{d}^2 \vec{f} \cdot (\partial_t \vec{B} + \vec{v} (\vec{\nabla} \cdot \vec{B})-\int_{\partial S} \mathrm{d} \vec{x} \cdot (\vec{v} \times \vec{B}).$$

- #23

feynman1

- 435

- 29

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?$$\frac{1}{c} \mathrm{d}_t \Phi_{\vec{B}} = -\int_{\partial S} \mathrm{d} \vec{x} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right),$$

where ##\vec{v}(t,\vec{x})## is the velocity of the line elements of the boundary ##\partial S##.

- #24

- 22,038

- 12,952

No, the formula given is correct. For a proof, seeI 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.

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

- 435

- 29

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.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.

But the above technicality I don't care at this point. I really care about post #23.

- #26

- 22,038

- 12,952

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.

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.

Last edited:

- #27

feynman1

- 435

- 29

So nice of you translating all this!I don't have an English version yet. So here it is

- #28

- 5,695

- 2,473

- #29

- 22,038

- 12,952

Thanks! I've corrected the typos.

Share:

- Replies
- 9

- Views
- 667

- Replies
- 9

- Views
- 454

- Replies
- 8

- Views
- 356

- Replies
- 12

- Views
- 1K

- Replies
- 6

- Views
- 604

- Last Post

- Replies
- 9

- Views
- 357

- Replies
- 3

- Views
- 464

- Replies
- 76

- Views
- 2K

- Replies
- 94

- Views
- 4K

- Replies
- 17

- Views
- 739