# PRoblem with differential form of Maxwell's third equation?

1. May 11, 2006

### leright

Why does it seem as if the standard differential form of Maxwell's third equation (Faraday's Law) for time varying fields not take into account motional EMF. The differential form simply says that the curl of E is equal to minus the time rate of change of B field. However, there could be a curl of E even in cases where B is constant with a time varying loop in a static B field, so the differential form fails in motional emf, correct? Why isn't the differential motional emf term added to the right hand side of Maxwell's third equation (in differential form)? If it were added in, Maxwell's third equation would work in all cases.

The integral form of Maxwell's third equation works fine in all cases, as long as the derivative is kept OUTSIDE of the flux integral. When it is brought in then there is trouble, and clearly two different answers will be obtained for the emf when comparing the derivative being inside of outside the integral (when the derivative is only applied to the B-field, and not ds). I feel as if this is a flaw in the logic in the derivation of the differential form of the third equation.

This is not independent research. I am just looking for answers to a question.

Thanks.

2. May 11, 2006

### leright

Any opinions?

He said that the differential motional emf term that I added was just an approximation at low velocities. Also, when Maxwell formulated the third equation he assumed the moving charge to be the reference frame. However, it seems that it wouldn't be too difficult to add in a term that describes the motional emf at a reference frame other than the bar.

The motional EMF term I added was integral(u x B)*dl. Stokes theorem can be applied to obtain integral(curl(u x B))*dS Therefore, the term curl(u x B) can be added to the right hand side of the differential form of the third equation.

Last edited: May 11, 2006
3. May 11, 2006

### vanesch

Staff Emeritus
No, the differential form doesn't fail, because BY DEFINITION, you have to calculate the curl (or the circular integral in the integral formula) in a time-fixed loop, that is, a curve which is time-independent wrt the coordinate frame in which you are expressing the E and B fields.

Why this apparent paradox ? The reason is that the E and B fields do not transform as vectors when you switch from a coordinate frame to another, moving frame (in Lorentz speak, a boost).
E and B aren't really vectors in fact (they only behave as vectors under translations and rotations of coordinates, but not under boosts). They are really elements of a 2-tensor.
So if you want to consider a "moving curve" in frame A, then you should first go to the frame B that is static wrt the curve, and then transform your E and B field from frame A into frame B. And now comes the miracle:

A pure B field in frame A transforms into a B field PLUS an E field in frame B. The extra E field is exactly what corresponds to the EMF.
Indeed, the B-field will be "moving" but that is just "changing" in the B frame, and hence there will be a changing B field and thus a curl of the E-field in the frame B.

4. May 11, 2006

### leright

I know that the curl is only calculated in a time fixed frame and I realize that if the derivative of B field in a time fixed frame is 0 then the B field does not induce an E field, and if the derivative of B field in a time fixed frame is NOT 0 then the B field DOES induce an E field. My issue is that an e-field CAN be induced by a constant B-field in the event of the motion of free charge (non-free space conditions) in the field.

I think that the commonly used differential form of Maxwell's equation assumes free space conditions where there is no moving free charge.

I also understand that the commonly used differential form of Maxwell's equation assumes that if there IS moving free charge the reference frame should be taken to be the free charge itself, and this is why Maxwell's equation in differential form "works", but it simply does not convey all possible imformation.

5. May 11, 2006

### leright

message has been deleted.

Last edited: May 12, 2006
6. May 11, 2006

### Hurkyl

Staff Emeritus

EMF is being induced -- not an E-field.

In the case of the time-varying magnetic field, the reason EMF gets induced is because the time-varying magnetic field induces an electric field, and that electric field starts pushing your (stationary) charges around.

In the case of motional emf, what happens is that the magnetic field itself starts pushing your (moving) charges around. It has absolutely nothing to do with the electric field.

In fact, my textbook specifically says that the integral form:

$$\oint \vec{E} \cdot d\vec{l} = - \frac{d \Phi_B}{dt}$$

is "valid only if the path around which we integrate is stationary."

7. May 11, 2006

### leright

In order for there to be an induced EMF (capacity to do work on charges) there MUST be an corresponding e-field. In order for there to be an induced EMF that can drive current around a closed path there MUST also be a "non-conservative" e-field produced, and the only way a non conservative e-field can be produced is from the energy stored in a b-field. The b-fields themselves do not do work. The energy in the b-fields INDUCE e-fields that can do work, and work over a closed path at that (conservation of energy is not violated since the energy allowing the e-field to do work over the closed path is coming from the b-field)

And in motional EMF, the magnetic field is not doing the work. The magnetic field is storing the ENERGY to produce an e-field that does the work. The force on a moving charge or current element is due to an induced e-field which comes from the energy in the b-field.

This is all getting more and more clear to me, and the more I think about it, the more things make sense. Before, the more I thought about it and the deeper I went, the more confused I got. I've reached the point where the deeper I get and the more I think about it, the more clearly things become.

Just because something has energy in some form does not mean that it itself can do work, even though energy is defined as the capacity to do work. The energy has to be converted to other forms of energy that can do the work. For instance, b-fields themselves cannot do work because the force is ALWAYS perpendicular to the direction of motion, and the b-field itself cannot increase the kinetic energy of a charge. However, the energy in the b-field induces an e-field which does the work for it. So, even if the energy itself cannot do work that doesn't imply that it isn't really energy. It can be converted to other forms of energy which can the work. So b-fields aren't DOING work, but they are facilitating work by inducing e-fields.

Last edited: May 11, 2006
8. May 11, 2006

### Hurkyl

Staff Emeritus
Work it out yourself:

(1) Electric field that is zero everywhere.
(2) Constant magnetic field.
(3) Conducting wire.

When the wire starts to move, you have moving charges in a magnetic field. Thus, those charges start to move along the wire, and you get current, even though the electric field is zero, and the magnetic field is constant.

Your entire line of thought is based on the fact that magnetic forces cannot do work, correct? So all that means is that something else is doing the work. Why do you assume it's the electric field? There's a much simpler explanation that doesn't involve rejecting established physical laws!

I wonder how that wire started moving...

9. May 11, 2006

### leright

ok, the b-field still is not doing WORK on the charges in the rod. When the rod moves in the constant magnetic field motional emf occurs. Motional emf is caused by an e-field that pushes the charges and that e-field is induced by the magnetic field. I am not rejecting any physical laws.

It is an electric field that is induced perpendicular to the magnetic field that causes the emf. The force is perpendicular to the magnetic field so it is not doing work...the induced e-field is doing the work, but the energy to produce the e-field is coming from the b-field.

Why is the force due to a magnetic field called Fsubm and when you divide Fsubm by Q it's called Esubm? Perhaps because it is an e-field doing the work. Also, take a look at the lorentz transformations (relationships between E and B fields with relativistic considerations) and you'll notice a UxB term when getting the e-field from b-field. The UxB is the force due to the B-field.

http://farside.ph.utexas.edu/teaching/em/lectures/node123.html - look at the determination of the perpendicular e-field due to the b-field.

Last edited: May 11, 2006
10. May 12, 2006

### Galileo

A moving (conducting) loop in a constant magnetic field (like a generator) will induce an EMF, but not because of Faraday's law. It's simply because in the loop you now have moving charges in a magnetic field which are pushed along by the Lorentz force and this causes the current. The magnetic force is not doing work ofcourse, but the person pulling the loop is.

11. May 12, 2006

### arunbg

I believe Galileo is right, the phenomenon is similar to Hall's effect which is attributed to Lorentz's forces.

However what I don't understand is that in many cases such as a rod rotating in a uniform magnetic field the induced emf across the ends of the rod can also be calculated by the eqn for motional "induced" emf
E = Blv (it is done so in my text) even though there is no change in net flux.I calculated the answer independently using lorentz forces and got the exact same answer which seemed the right approach to me.

Does this imply some kind of relation or is this simply a freak coincidence?

12. May 12, 2006

### leright

It is because of Faraday's law. Faraday's law says that the induced EMF is equal to the time rate of change in the flux through a conducting loop. This is what Faraday's law says. You can produce a change in the magnetic flux by changing the area of the loop, even in a constant B field.

The induced EMF can be because of a number of physical phenomena, but all of the physical phenomena is contained in Faraday's law.

13. May 12, 2006

### arunbg

There is no time rate of change of flux through a fixed conducting loop moving in a uniform magnetic field.

14. May 12, 2006

### Hurkyl

Staff Emeritus
My book doesn't use that notation, so I don't know what you mean by F_m and E_m.

But we're not doing a change of reference frame, so the Lorentz transformations are irrelevant. :tongue:

The Lorentz transformations say that the EMF induced by motion through a pure magnetic field in one frame has to have the same effect as the EMF induced by the electric field in some other reference frame where there is no motion.

But that does not mean that an electric field spontaneously appears in the original reference frame!

No -- there is no work being done in the production motional EMF. The energy to create the current comes from the kinetic energy of the wire. Something caused the conducting wire to start moving -- the magnetic field acts to change that motion into a current. If there was no friction, then given enough time and space, the magnetic field will convert all of the wire's kinetic energy into current, and the wire will stop moving.

You're rejecting the third of Maxwell's laws. And, it seems, the mathematical theorem that the integral form is true if and only if the differential form is true.

15. May 12, 2006

### leright

no no no....my whole point in this thread is that the differential form of the third equation FAILS to take into consideration motional EMF (and yes, motional EMF induces an E field with a curl), which is where there is a change in reference frame. That is the correction to the equation I am making, and for this you would need Lorentz transformations.

Last edited: May 12, 2006
16. May 12, 2006

### Staff: Mentor

If the loop rotates (as in a simple generator), the flux through the loop does change. For a fixed-area loop in a uniform magnetic field:

$$\Phi = AB \sin \theta$$

$$\frac{d \Phi}{dt} = AB \frac {d}{dt} (\sin \theta) = AB \cos \theta \frac {d\theta}{dt}$$

17. May 12, 2006

### leright

ok, well then take a time varying conducting loop!

18. May 12, 2006

### Hurkyl

Staff Emeritus
And my answer is that your point is based upon flawed assumptions.

The third equation does not fail to take into consideration motional EMF -- motional EMF does not act as a source of the electric field, and thus (correctly) does not appear in the equation.. :tongue:

This is true both in the differential and integral forms. The equation:

$$\oint_C \vec{E} \cdot d\vec{l} = - \frac{d \Phi_B}{dt}$$

is only valid for a fixed contour C. This equation cannot tell you anything about motional EMF.

If you do not understand this, then allow me to demonstrate some of the many problems that you will encounter, both mathematical and physical.

Suppose that

$$\oint_S (\nabla \times \vec{E}) \cdot d\vec{s} = - \frac{d \Phi_B}{dt}$$

is true for a time-varying surface S. Then we can consider two different time-varying surfaces, $S_1$ and $S_2$. Suppose that at t = 0 that $S_1 = S_2$. Then, the above equation tells us that the time derivative of the flux through $S_1$ is equal to the time derivative of the flux through $S_2$ -- but that can't be right, because those surfaces can be varying in vastly different ways!

Some physical problems: you want a velocity to appear in Maxwell's third equation. Well, velocity of what?

Let's suppose you're right -- some electric field does spontaneously appear due to the velocity (of something), and that's what causes motional EMF.

Well, what if our moving conducting wire passes right next to a stationary conducting wire. Well, since the moving conducting wire (somehow) causes there to be an electric field, then that should induce an EMF in the other wire in the same direction. Does that happen? No! In fact, once we account for all effects, don't we see the exact opposite?

If we had two conducting wires passing by each other moving at equal speeds, your argument would say that electric fields are induced... but since they're in opposite directions they cancel, and we see no motional EMF at all. But that's not what happens: we will see a current in both wires.

For the sake of completion, I will admit that you are right (but as far as I can tell, for the wrong reasons): motional EMF does induce an E-field with curl, but for reasons that can be entirely explained via Maxwells equations, differential or otherwise.

Motional EMF is, at the lowest level, nothing more than the acceleration of moving charged particles.
Accelerating charged particles (generally) result in nonstationary currents.
Because current acts as a source of the magnetic field, a nonstationary current gives rise to a time-varying magnetic field.
A time-varying magnetic field acts as a source of the electric field.
And in this way, motional EMF does induce an electric field.

But this is an electric field caused by motional EMF. The electric field has absolutely nothing to do with causing the motional EMF. Motional EMF is caused entirely by the magnetic field: moving charges in a magnetic field are deflected in a direction perpendicular to their motion.

(Of course, the electric field could indirectly cause motional EMF... say, by having the time variations of the electric field induce a magnetic field)

Last edited: May 12, 2006
19. May 13, 2006

### leright

OK, I see your point regarding point charges moving (non stationary current) which in turn produces a time variant magnetic field, which in turn produces an E field when in the presence of a B field. You may be correct on this statement, but I feel this only holds for moving CHARGES. Let us now consider a loop of conducting wire (but no current) that is rotated in free space with absolutely no B field present. This wire has no net charge OR current so as it moves it is not producing any b-field or changing b field. We know that no E field or EMF will be induced (or do we?) Let us assume for a moment that it doesn't produce a B field. It would be nice to have some experimental results handy claiming this is true. Someone should rotate an electrically neutral loop of wire somehow in a controlled environment (no B field, no E field, and in a vacuum) to and see if a B field is produced.

Now, let us move this same conducting loop of wire into a time invariant B field. We can rotate it in this B field or do whatever we want and produce a changing flux (not due to time variance of the B field, but time variance of the surface area of the loop....as we change the angle of the loop in the B field we are changing the magnetic flux in the B field even if the scalar area of the loop stays constant, since the angle between the surface vector of the loop and the time invariant B field changes we are changing the magnetic flux). We know from experiment that this changing FLUX induces a curl in E which produces EMF (or that this induces an EMF which in turn produces a curl in E, depending on how you look at it). However, there is no current or moving charges or anything, so that originally static field is still static. In this case we are changing ONLY the magnetic flux through the loop and not the B field. However, we know that there is an EMF induced by experiment, and if there is an EMF there is an E-field, PERIOD! NOWHERE in the differential form of Maxwell's third equation (and a changing B field doesn't pop out in any of the other equations in differential form, so your previous agument fails in this case) is this scenario accounted for, but the integral form covers it. The error (or assumption, perhaps) Maxwell made (or did Heaviside do this, I can't remember) when going from the integral form to the differential form was when he brought a partial derivative into the integral of B over a surface and didn't account for a changing area over time (ds). I don't know how many electrodynamics problems you've worked through, but there are COUNTLESS problems where the diffential form fails to yield the same result as the integral form. You can attempt some problems by computing the curl of the E field using the differential form and then integrating the curl over a surface to get the EMF and you DO NOT get the same answer as the integral form. This is a FACT, and is clear proof of the failure of the differential form of Maxwell's equation. Maxwell assumed that no motional EMF was taking place when he went from differential form to integral form. In fact, he assumed a lot of things. The equations are not wrong if this assumption is made clear whenever the equations are written down, but most textbooks FAIL to make this clear. In fact there are tons of websites out there with modified differential forms of maxwell's equations taking things like motional EMF, material space, etc into account (and even quantum effects if you want to get into QED). Trust me, the differential form of Maxwell's equation (3rd and 4th, really) has plenty of errors (or assumption that were made, so I don't insult Maxwell. :tongue: )

Your comment about about how the EMF causes the electric field, but an electric field does not cause the EMF I feel is incorrect, simply because of the way EMF is defined. I don't feel one necessarily causes the other, but I simply feel like they COMPLEMENT each other and occur simultaneously (an E field doesn't appear by itself in a time invariant frame and the EMF due to that E field appears in the next frame). If you have an EMF it is because of energy in a magnetic field allowing a non conservative E field to be produced. Remember, EMF is the closed path integral of a an E field. And I think it is pointless to argue over this (if the e-field produces emf, or if the emf produces the e-field). Under maxwell's equations, the condition of the field is taken in a time invariant frame so they are both really occuring simultaneously. This isn't what I am arguing about, really.

Last edited: May 13, 2006
20. May 13, 2006

### leright

http://www.wolfram-stanek.de/maxwell_equations.htm

here's a great website with all of the extensions on Maxwell's equations. Notice the ones on the top are the equations in question, and they fail in the case of moving objects, according to the website.

Last edited: May 13, 2006