Is Lorentz Force Truly Conservative?

[gulsen]

[tex]\mathbf \nabla \times (\mathbf E + \mathbf v \times \mathbf B)[/tex]
pluggin stuff from Maxwell equations
[tex]= -\frac{\partial B}{\partial t} + \mathbf v (\mathbf \nabla \cdot B) - \mathbf B (\mathbf \nabla \cdot v)[/tex]
Since
[tex]\frac{\partial}{\partial t}(\mathbf \nabla \cdot \mathbf r) = 0[/tex]
it's
[tex]= -\frac{\partial B}{\partial t}[/tex]
which is not zero in general. Or am I doing something wrong??

Can this be the field is also holding energy?

 

[Meir Achuz]

The Lorentz force is not conservative if dB/dt is not zero, as you have just shown. The dB/dt force is how the betatron accelerates particles.
The circuit producing dB/dt puts energy into the system.

 

[masudr]

Indeed. In fact, only conservative force fields can be written in terms of the gradient of a scalar field. Which is why

[tex]\vec{E} = -\vec{\nabla}\phi[/tex]

only applies when [itex]\partial B / \partial t = 0.[/itex]

 

[gulsen]

What?!?
But... wait! Isn't electromagnetism (and all fundamental forces) supposed to be a conservative? What happens to [tex]\partial B / \partial t[/tex]?

 

[samalkhaiat]

What?!?
But... wait! Isn't electromagnetism (and all fundamental forces) supposed to be a conservative?

NO, who told you that?

Lorentz force is not conservative in the sense
[tex]\nabla \times \vec{F} = 0[/tex]
This is because (except for static field) the work done by the EM-force on a charge around a closed path is not zero;
[tex]\oint \vec{F}.\vec{dr} \neq 0[/tex]

This situation is similar to that of frictional forces in mechanics.
Lorentz force is derivable from a velocity-dependent "generalized" potential according to
[tex] E_{i} + (\vec{v} \times \vec{B})_{i} = -\partial_{i}U + \frac{d}{dt}(\frac{\partial U}{\partial v_{i}})[/tex]

with
[tex]U=\phi - \vec{A}. \vec{v}[/tex]

or if you like 4-vectors:

[tex]f^{\mu} = -\partial_{\nu}T^{\nu\mu}[/tex]

where

[tex]f^{\mu} \equiv (J.E,\rho E_{i} + (J \times B)_{i})[/tex]

is the Lorentz force density and [tex]T^{\mu\nu}[/tex] is the symmetric energy-momentum tensor of the EM-field.


regards

sam

 

[gulsen]

NO, who told you that?

Richard Feynman.
I can't remember where exactly, probably in lectures on physics. It may take while for me to find where exactly.

 

[gulsen]

Found it:

14-4 Nonconservative Forces

We have spent a considerable time discussing conservative forces; what about nonconservative forces? We shall take a deeper view of this than is usual, and state that there are no nonconservative forces! As a matter of fact, all the fundamental forces in nature appear to be conservative. This is not a consequence of Newton's laws. In fact, so far as Newton himself knew, the forces could be nonconservative, as friction apparently is. When we say friction apparently is, we are talking a modern view, in which it has been discovered that all the deep forces, the forces between the particles at the most fundamental level, are conservative.

...

EM force is the last force I'd expect to be non-censervative since almost everything we observe is chemical which is by means of EM forces, yet conservative when observed carefully.

From the "work-energy relation in EM" in Griffiths:

Poynting's theorem says, then, that, the work done on the charges by the EM force is equal to the decrese in energy stored in the field, less the energy that flow out through the surface.

OK, if the curl of force comes out zero, we say "this's conservative" right away. But what is the meaning of the term if it doesn't come out zero. Ie, I wonder the meaning of the Faraday term which came out.

Please, can't anyone explain what's going on in this curl?!?

 

[masudr]

It's usually because you are not considering the whole system:

Poynting's theorem tells us how energy is transported by the electromagnetic field: this is done via EM radiation. Really, energy is conserved in EM, but only when one considers the whole universe as the system (which is rarely done). e.g. usually one doesn't include the battery which supplies power nor accounts for energy lost as radiation, all problems of non-conservation will disappear.

 

[gulsen]

So that extra term has something to do with energy flow to/from the field? Poynting's theorem seem to do the job, but it's a bit cumberstone.
Can we show that Faraday term has to do with field?

Sorry for such lame questions, I just haven't taken EMT course yet.

 

[samalkhaiat]

Found it:


EM force is the last force I'd expect to be non-censervative since almost everything we observe is chemical which is by means of EM forces, yet conservative when observed carefully.

"all fundamental forces. appear to be conservative" said Feynman

Well, if "by conservative", Feynman meant [tex]\mathbf \nabla \mathbf F = 0[/tex] ,then he is wrong, because
1) the very concept "conservative F" loses its meaning when one talks about the weak and strong "fundamental" forces.We simplely don't have F.
2) it is easy to show that Lorentz force is not conservative.
I think Feynman was talking about "conservation laws" in fundamental interactions, i.e, energy conservation. in this case, his statement is correct.

OK, if the curl of force comes out zero, we say "this's conservative" right away. But what is the meaning of the term if it doesn't come out zero. Ie, I wonder the meaning of the Faraday term which came out.

OK, Let us start from the begining:

If a force field is such that the work done around a closed path is zero,i.e,

[tex]\oint \mathbf F.d \mathbf r = 0[/tex]

then the force is said to be conservative
By Stokes' theorem;
[tex]\oint \vec{F}.d\vec{r} = \int \vec{\nabla} \times \vec{F} . d\vec{S}[/tex]
the above condition for conservative forces, is equivalent to

[tex]\vec{\nabla}\times \vec{F}= 0[/tex]

Now, for charged particle in EM-field, we have;

[tex]\mathbf \nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}[/tex]
[tex]\mathbf \nabla \times (\mathbf v \times \mathbf B) = -(\mathbf v .\mathbf \nabla ) \mathbf B[/tex]

Thus

[tex]\mathbf \nabla (\mathbf E + \mathbf v \times \mathbf B) = -\frac{d\mathbf B}{dt}\neq 0[/tex]

This,now, can be used to prove (the experimental fact) that the work done by Lorentz force upon the charge around a closed circuit is equal to the rate of change in the magnetic flux through the area enclosed by the circuit;

[tex]\oint (\vec{E} + \vec{v}\times \vec{B}) . d\vec{r} = -\frac{d}{dt}\int \vec{B}.d\vec{S}[/tex]

Now, if you put [tex]\mathbf B = \mathbf \nabla \times \mathbf A[/tex] in the RHS and use Stokes' theorem, you get

[tex]\oint (\vec{E}+\vec{v} \times \vec{B}+\frac{d\vec{A}}{dt}).d\vec{r} = 0[/tex]

Notice that the integrand in this equation is a conservative "force-like" object.But it is not gauge invariant, so it can not be the EM force.

Mathematically, it is always possible to construct conservative-object from non-conservative force.

I have the "courage" to call the above equation; D'Alembert's principle in electrodynamics. I can use it (as a postulate) to derive the equation of motion of a charged particle in the EM field:

From "D'Alembert's principle"

[tex]\oint \frac{d}{dt} (m\vec{v} + \vec{A}). d\vec{r} = 0[/tex]

one can find a scalar function [tex]U[/tex] such that,

[tex]m\frac{d}{dt}\vec{v} = -\frac{\partial}{\partial t} \vec{A} - (\vec{v}.\vec{\nabla}) \vec{A} - \vec{\nabla}U[/tex]

Now, if you use the identity

[tex]\vec{\nabla}(\vec{v}.\vec{A}) - \vec{v} \times (\vec{\nabla} \times \vec{A}) = (\vec{v}.\vec{\nabla}) \vec{A}[/tex]

and "define"

[tex]U + \vec{A}.\vec{v} \equiv \phi[/tex]

[tex]\vec{\nabla}\times \vec{A} \equiv \vec{B}[/tex]

and

[tex]-\frac{\partial}{\partial t} \vec{A} -\vec{\nabla} \phi \equiv \vec{E}[/tex]

,you find

[tex]m\frac{d}{dt}\vec{v} = \vec{E} + \vec{v} \times \vec{B}[/tex]

regards

sam

 

[Sojourner01]

Forces are conservative when you 'fix' whatever object is supposedly creating them. If you imagine an infinite, constant electric field independent of the charges that produce it - that is, in a vacuum - it's conservative, because no matter what you do in the field, there's no way to change it and thus it'll be exactly the same when you go back in the opposite direction.

In practice, any object that 'feels' an electric field also creates one of its own. This field affects the particles that generate the field you're considering, so the gross field itself is altered by your motion. It's no longer conservative.

 

[samalkhaiat]

Forces are conservative when you 'fix' whatever object is supposedly creating them. If you imagine an infinite, constant electric field independent of the charges that produce it - that is, in a vacuum - it's conservative, because no matter what you do in the field, there's no way to change it and thus it'll be exactly the same when you go back in the opposite direction.

In practice, any object that 'feels' an electric field also creates one of its own. This field affects the particles that generate the field you're considering, so the gross field itself is altered by your motion. It's no longer conservative

Is this, by any chance, what Mr. Spok says in STAR TREK II?
Seriously, you need to know something about conservative forces.
You can read about them in any elementary physics book (mechanics).

sam

 

[Sojourner01]

Mhm, ok. What was wrong about what I said?

 

[Sojourner01]

Anyone? I'm curious now.

 

[StatMechGuy]

The EM field is conservative, but you've only accounted for the energy stored in the particle. If you look at the total energy stored in an EM field, you get an energy conservation equation, that is unless you start pumping the system from the outside, but that's generally true.

This is an interesting point, actually. If you start out with the lagrangian for a uniform magnetic field, you get that the total energy equals the hamiltonian, but there's no dependence on [tex]\vec{A}[/tex] when it all comes out. That's because you haven't written down the "whole hamiltonian" for the entire system, just the hamiltonian for how this charged particle interacts with an external field.

 

[Galileo]

There is no violation of conservation of energy in EM.
Frankly I do not see the meaning of taking the curl of the lorentz force. Taking the curl applies to vector fields (like the E and B fields). The lorentz force describes the force on a charged particle, it's not a field.

Anyhoo, since the curl E = -dB/dt and curl B=uJ+(1/c^2)d(E/dt) they are not conservative in the sense that their curl is zero.
However if you consider the entire system of charged particles and fields you do have conservation of energy. Whatever energy is lost in the fields is gained by the particles and vice versa.

 

[quasar987]

Great post samalkhaiat. Keep spreading the knowledge.

There's just one step I don't understand...

[tex]\oint \frac{d}{dt} (m\vec{v} + \vec{A}). d\vec{r} = 0[/tex]

one can find a scalar function [tex]U[/tex] such that,

[tex]m\frac{d}{dt}\vec{v} = -\frac{\partial}{\partial t} \vec{A} - (\vec{v}.\vec{\nabla}) \vec{A} - \vec{\nabla}U[/tex]

Shouldn't the conclusion be simply "one can find a sclalar function
[itex]U[/itex] such that

[tex]m\frac{d}{dt}\vec{v} = -\frac{\partial}{\partial t} \vec{A} - -\vec{\nabla}U[/tex]"

?? What's the [itex](\vec{v}.\vec{\nabla}) \vec{A}[/itex] doing there? Thx?

 

[quasar987]

There is no violation of conservation of energy in EM.
Frankly I do not see the meaning of taking the curl of the lorentz force. Taking the curl applies to vector fields (like the E and B fields). The lorentz force describes the force on a charged particle, it's not a field.

Couldn't you see it as a force field changing its value everywhere in space as v changes?

Anyhoo, since the curl E = -dB/dt and curl B=uJ+(1/c^2)d(E/dt) they are not conservative in the sense that their curl is zero.
However if you consider the entire system of charged particles and fields you do have conservation of energy. Whatever energy is lost in the fields is gained by the particles and vice versa.

Poynting theorem.

 

[samalkhaiat]

Great post samalkhaiat. Keep spreading the knowledge.

There's just one step I don't understand...




Shouldn't the conclusion be simply "one can find a sclalar function
[itex]U[/itex] such that

[tex]m\frac{d}{dt}\vec{v} = -\frac{\partial}{\partial t} \vec{A} - -\vec{\nabla}U[/tex]"

?? What's the [itex](\vec{v}.\vec{\nabla}) \vec{A}[/itex] doing there? Thx?

Hi,

That term comes from the relationship between the total and partial derivatives:

[tex]\frac{d}{dt} = \frac{\partial}{\partial t} + \frac{dx_{i}}{dt} \frac{\partial}{\partial x_{i}} = \partial_{t} + \vec{v} . \nabla [/tex]

this is what I did:

[tex]m\frac{d}{dt}\vec{v} + \frac{d}{dt}\vec{A} = -\nabla U[/tex]

then I applied the above-mentioned relation to the second term on the left.


regargs

sam

 

[samalkhaiat]

There is no violation of conservation of energy in EM.
Frankly I do not see the meaning of taking the curl of the lorentz force. Taking the curl applies to vector fields (like the E and B fields). The lorentz force describes the force on a charged particle, it's not a field.

Anyhoo, since the curl E = -dB/dt and curl B=uJ+(1/c^2)d(E/dt) they are not conservative in the sense that their curl is zero.
However if you consider the entire system of charged particles and fields you do have conservation of energy. Whatever energy is lost in the fields is gained by the particles and vice versa

Hi,

If you did not understand the material in my post #10, then I would suggest you pick up a book on mecanics and look for the definition of conservative force.
By the way, who said that the total energy is not conserved in non-conservative systems (Systems with frictional or Lorentz force)? There is something called the Work-energy theorm which I also stated in my post #10.

regards

sam

 

[Galileo]

Hi,

If you did not understand the material in my post #10, then I would suggest you pick up a book on mecanics and look for the definition of conservative force.

I know very well the definition of a conservative force field, thank you. But don't immediately relate the mathematical definition of a conservative force field with the physical principle of conservation of energy.
First: B doesn't fit the in the kind of force fields when considering conservative fields, since the FORCE in never in the direction of B.
Second: The fields are time varying, and there are charged particles which also carry energy. Only when you consider the energy in the entire kaboodle will you have the physical conservation of energy.

By the way, who said that the total energy is not conserved in non-conservative systems (Systems with frictional or Lorentz force)? There is something called the Work-energy theorm which I also stated in my post #10.

I didn't, so where do our posts contradict? (apart from my statement that taking the curl of the lorentz force is not meaningful in this case, since it doesn't not a physical vector field. There is no field that changes everywhere because the particle changes velocity).

And non-conservative systems only exist if you choose to neglect parts which interact with your physical system.

 

[Sojourner01]

Aha. I understood what I meant, just gave a bad example which now I think about it, only works when you make some assumptions. What I was getting at was that an electric field (say) is conservative, but when you include the matter generating the field in the problem, energy will be transferred into that matter and thus the work done on an object in the field will no longer be path-independent.

As for how this relates to the Lorentz force, is it right that it is by definition the combined electric and magnetic field contributions to a body in the vicinity of an electrode? That is, is it required that if you're considering a Lorentz force you assume it's being generated by a 'real' system?

If so, I believe the force is nonconservative both practically and mathematically. Mathematically, it may be path-dependent based on the contribution of the magnetic force - I haven't worked it out.

Practically though, regardless of the form of the magnetic force, any motion in the lorentz system is still path-dependent due to the nature of the system generating it. That is, you won't get back the work you put in and vice versa, because you're dealing with a lossy system.

 

[masudr]

Indeed, the best formalism with which to examine this, by far, is the Lagrangian/Hamiltonian types. If the Lagrangian has global spacetime translation symmetries, then you've got energy-momentum conservation.

This is so useful that it becomes an easy way of seeing if a system is independent from the rest of the universe or not: if its Lagrangian has these global symmetris, then yes it is independent.

 

[samalkhaiat]

I know very well the definition of a conservative force field, thank you.

Do you? OK, answer the followings;

Given a force, any force
[tex]\vec{F}=m d\vec{v}/dt[/tex]

1) Is it "meaningful" to calculate the work done by F around a closed path?
[my answer; Yes, if we need to do meaningful physics]

2) What do you call F when
[tex]\oint \vec{F}.d\vec{r} \neq 0[/tex]
[my answer; non-conservative force]

3) Is the statement in (2) equivalent to
[tex]\nabla \times \vec{F} \neq 0[/tex]
[my answer;Yes, this follows from Stocks' theorem]

4) Is it "meaningful to take curl of F?
[my answer; what a garbage, of course it is. we already said this in (1) and (1) is equivalent to (3). Plus, the curl is a mathematical operator, so we can apply it to any vector]

5) Is F a vector field?
[my answer; In classical mechanics, any function of (r,t) defines a field, So yes F(r,t) is a vector field]

6) Is F a "physical" field?
[my answer; first, this is a meaningless question because "physical" field is not well-defined concept in classical mechanics, i.e. it would be meaningless to answer (6) by Yes or No. second, such question has no relevance to the issues in (1) to (5)]

7) Does Lorentz force satisfy our only condition on F? that is
[tex]\vec{F}=m\frac{d}{dt}\vec{v}[/tex]
[my answer; Yes, Lorentz force is no different from any force in mechanics]

Now, if your answer to any of the above questions is different from mine, then you have problem with understanding mechanics. However, if your answers are same as mine, then, Wooops, this means that taking the curl of Lorentz force is "meaningful"
Look, calculating the work done by Lorentz force around a closed path is a very impotrant business and it involves taking the curl.

But don't immediately relate the mathematical definition of a conservative force field with the physical principle of conservation of energy.

Energy conservation IS NOT the issue of this thread even though it is related (The energy loss/gain is determined by the value of the integral
[tex]\int \vec{F}.d\vec{r}[/tex]
so the total energy is conserved only when we take the value of this integral into consideration)

I hope it is now clear to you that the mathematical definition of conservative force answers three "equivalent" questions:

1) is the force derivable from velocity-independent potential?

[tex]\vec{F} = - \vec{\nabla} U(r)[/tex]

2) what is the value of

[tex]\oint \vec{F}.d\vec{r}[/tex]

3) what is the value of

[tex]\nabla \times \vec{F}[/tex]

B doesn't fit the in the kind of force fields when considering conservative fields

I do not understand what this means.

since the FORCE in never in the direction of B.

Who cares about the direction. The direction of force has no relevance in taking the curl.

Second: The fields are time varying, and there are charged particles which also carry energy. Only when you consider the energy in the entire kaboodle will you have the physical conservation of energy.

Yeh, so what is the connection to conservative FORCE?

I didn't, so where do our posts contradict?

Didn't you start post #16 by saying "there is no violation of conservation of energy in EM" Well we all know this, You did not need to bring about energy conservation, because it is NOT the issue in here.

(apart from my statement that taking the curl of the lorentz force is not meaningful in this case, since it doesn't not a physical vector field. There is no field that changes everywhere because the particle changes velocity).

You also said (post #16) "..curl applies to vector field like the E and B field"

What so special about E and B, have you heard about the conservative and non-conservative VELOCITY fields in Hydrodaynamics. Let me tell you this again; you can apply the curl to any vector full stop.

And non-conservative systems only exist if you choose to neglect parts which interact with your physical system

Definition: a system is called non-conservative, if it is acted upon by at least one force that is not derivable from a velocity-independent potential.

So we call the system with frictional force, non-conservative, not because we "choose to neglect" friction, but because of the fact that the force of friction is not derivable from velocity-independent potential (non-conservative force)

You always seem to mix the meaning of the terms conservative and non-conservative (force,system), with energy conservation. Energy is always conserved in all physical systems.

I believe I have fully answered all relevant (and sometime irrelevant) questions raised in this thread. All answers can be found in post #10.

regards

sam

 

[samalkhaiat]

If so, I believe the force is nonconservative both practically and mathematically. Mathematically, it may be path-dependent based on the contribution of the magnetic force - I haven't worked it out.

I did work this out for you in post #10.

sam

 

[Galileo]

Sam, the whole reason a vector field with zero curl is called conservative is exactly because those are the kind of fields for which you can define a potential energy function leading to conservation of energy. The connection between conservative fields and conservation of energy is only natural, which is why I (and many others is the thread) think this is what the thread IS about.

Anyway, I think you may have missed the essence behind my remarks, so I hope to clarify that here.

Given a force, any force
[tex]\vec{F}=m d\vec{v}/dt[/tex]

1) Is it "meaningful" to calculate the work done by F around a closed path?

No, it's not. Because [tex]\vec{F}=m d\vec{v}/dt[/tex] is Newton's second law. It tells you how the force on a particle relates to its acceleration. It is not a vector field (not a function of position), just like the lorentz force law is not a vector field.
The [tex]\vec F[/tex] here is different from the [tex]\vec F[/tex] in, say Coulomb's law: [tex]\vec F(\vec r)=Kq_1q_2/r^2 \hat r[/tex], in the sense that this IS a vector field. And calculating the curl, or the work done using this field IS meaningful.

4) Is it "meaningful to take curl of F?
[my answer; what a garbage, of course it is. we already said this in (1) and (1) is equivalent to (3). Plus, the curl is a mathematical operator, so we can apply it to any vector]

Correction: we can apply it to any vector FIELD.

5) Is F a vector field?
[my answer; In classical mechanics, any function of (r,t) defines a field, So yes F(r,t) is a vector field]

Exactly my point. Is the F in the lorentz force law a function of position? Me thinks not.

You always seem to mix the meaning of the terms conservative and non-conservative (force,system), with energy conservation. Energy is always conserved in all physical systems.

It's only natural to relate the two, as others (and most likely the OP) in the thread have done, since the very definition of a conservative force field is inspired by the physical principle of conservation of energy.

 

[quasar987]

What Galileo said is consistent with the definition of "consrvative force" given by Symon in his book "Mechanics". He says that

A force which function of position alone, and whole curl vanishes, is said to be conservative, because it leads to the theorem of conservation of energy.

On the other hand, he laters adds, if a force is function of position and also explicitely of time, and if it so happens that the curl of such a force vanishes at all points for a given time, the force is not to be called conservative because energy is not conserved.

(Because the potential will be calculated at the fixed time t_0 for whic the curl vanishes, but it takes time for a particle to go from point A to point B, and the force changes during that time, such that the total work done on the particle during its trip does not necessairily match the difference in potential calculated at the fixed time t_0.)

 

[samalkhaiat]

Sam, the whole reason a vector field with zero curl is called conservative is exactly because those are the kind of fields for which you can define a potential energy function leading to conservation of energy. The connection between conservative fields and conservation of energy is only natural, which is why I (and many others is the thread) think this is what the thread IS about.

I hope, you have noticed that I only gave the precise "definition" of conservative (non-conservative) force. I chose not to say anything about the "reason" for calling (naming) them conservative(non-conservative), because the names are;
1) trivial. Personally I prefer using the name curl-less (curl-full) force instead of coservative (non-conservative) force.
2) misleading. Thanks to the ambiguous statements of some textbooks, students often get confused about the status of energy conservation in non-conservative systems.
Energy conservation of any system follows from (what I call) the golden rule of mechanics:
" the energy is given by the first integral of the equation of motion"
If one integrates Newtons 2nd law once, one finds

[tex]E = \frac{1}{2}mv^2 - \int \vec{F}.d\vec{r} = c[/tex]
where c is constant.

F can be purely conservative, i.e.,

[tex]\vec{F} = - \nabla V(r)[/tex]

(this leads to the familiar energy conservation T+V=const.)

or purely non-conservative;

[tex]\vec{F'} \neq \nabla U(r)[/tex]

But, realistically, F is the sum of both;

[tex]\vec{F} = -\nabla V(r) + \vec{F'}[/tex]

This leads to energy conservation of the form;

[tex]E = T + V - \int \vec{F'}d\vec{r}[/tex]
(for frictional force the integral in the above equation represents the amount of heat generated during the motion)
Notice that we can not, completely, eliminate friction. So, in real life, physical systems are non-conservative in the above sense eventhough the energy is conserved.
Therefore, a system is conservative (forces are derivable from a scalars), if and only if

[tex] |\int \vec{F'}.d\vec{r}| \ll |T + V|[/tex]

No, it's not. Because [tex]\vec{F}=m d\vec{v}/dt[/tex] is Newton's second law. It tells you how the force on a particle relates to its acceleration. It is not a vector field (not a function of position), just like the lorentz force law is not a vector field.
The [tex]\vec F[/tex] here is different from the [tex]\vec F[/tex] in, say Coulomb's law: [tex]\vec F(\vec r)=Kq_1q_2/r^2 \hat r[/tex], in the sense that this IS a vector field. And calculating the curl, or the work done using this field IS meaningful.

I do not see any logic in here. Newtons 2nd law is the definition of force (any force). It does not tell you that F is not a function of (r,t). If it does, mechanics would end at school's level.
ALL FORCES must satisfy Newtons law this includes your coulombs force which is nothing but a special case of Lorentz force.
You must have learned in kindergarten that calculating the work done by any force is a very important business. and for Lorentz force calculating the work(taking the curl) has an industrial applications.

Correction: we can apply it to any vector FIELD.

Thank you for this valuable piece of information

Exactly my point. Is the F in the lorentz force law a function of position? Me thinks not.

Me knows you are wrong. take the following EM-fields (solutions of Maxwell's equations);

[tex]\vec{E} = \vec{E_{0}} e^{i(k.r - \omega t)}[/tex]
[tex]\vec{B} = \vec{B_{0}} e^{i(k.r - \omega t)}[/tex]

put them in Lorentz force, you get;

[tex]\vec{F} = e^{i(k.r - \omega t)}(\vec{E_{0}} + \vec{v} \times \vec{B_{0}})[/tex]

What is this then? a Hamburger or function of r.



regards

sam