Both answerers do not remember what every serious mechanics course teaches. The whole program of Lagrangean and Hamiltonian mechanics also works in the presence of magnetic fields; the scalar product of magnetic vector potential and particle velocity times particle charge is the term which has to be added to the Lagrangean to generate the magnetic force. Form H (Hamiltonian) from this Lagrangean in the standard way and you will see that H is constant (i.e. forces are conservative) if the fields are constant in time.

Who does not believe should google, e.g. Hamiltonian Lagrangean magnetic vector potential.

A forcefield is conservative, if its rotor vanishes, in your case that would mean that:

rot(B) = 0.

if you look at Maxwell's equations, it holds that: rot(B) = µ_0*j + µ0*e0*dE/dt

(where µ0 ... magnetic constant, e0 ... vacuum permittivity, dE/dt ... in this case the PARTIAL derivative of your electrical field, j ... your current density).

So, I would say, no, the magnetic force field is in general not conservative.

Magnetic force is not conservative. The force (Lorentz force) on a charged particle q equals qv x B, where v is the velocity and B the magnetic field. The magnetic force is perpendicular to the magnetic field as well as the moving particle's direction and depends on the position of the particle q, as well as on its velocity. By definition then, the magnetic force is not conservative.

Both answerers do not remember what every serious mechanics course teaches. The whole program of Lagrangean and Hamiltonian mechanics also works in the presence of magnetic fields; the scalar product of magnetic vector potential and particle velocity times particle charge is the term which has to be added to the Lagrangean to generate the magnetic force. Form H (Hamiltonian) from this Lagrangean in the standard way and you will see that H is constant (i.e. forces are conservative) if the fields are constant in time.

Who does not believe should google, e.g. Hamiltonian Lagrangean magnetic vector potential.

Dear Ulrich,

a very good remark - I had to think about it for a while :)

H = const. only means that the energy of your system is conserved (which you can achieve in every case, as long as your system is "large" enough (or entirely closed). But the B-field is clearly not conservative as you can see from the non-vanishing rot(B) in Maxwell's equations (which is also taught in a good mechanics course). Furthermore you can see that in a B-field in general the force (i.e. the Lorentz force) is dependent on the velocity, which also indicates a non-conservative field. You can also show the non-conservativeness by integrating a closed loop in a magnetic field which delivers you a non zero-work.

But I admit this is not very easy to see and it give me a hard night thinking about it.

Dear Johannes,

you probably need a second hard night of thinking to get rid of your misconception.

I will have to admit that the misconception was on my side if you can substantiate your claim:

>

The essence of the Lagrangean/Hamiltonian treatment of magnetic fields is to show that there is a special class of velocity dependent forces (those that are perpendicular to the velocity) for which the widespread association of velocity-dependent forces with dissipation of energy is not correct. Your rot-argument applies to force fields, but B is not a force field (it would be a force field for magnetic monopoles but not for electrical charges).

Hi Ulrich, It is an interesting problem, and like Johannes I thought about your answer last night. In a general sense, one would say that gravity, spring force, electric field, magnetic force etc. are conservative. Until you look at it a bit closer, then you find that general relativity is non-conservative (i.e. our gravity model) and the electric field needs a time-independent magnetic field. A moving conductor in a magnetic field has eddy currents induced in it, which is turned into heat (changing its potential energy, and net work done is not zero), and alters its movement, so that the sum of the potential and kinetic energy is not conserved. Moving from one point to another in the magnetic field then implies that the work done between two points is not independent of the taken path. So by definition the magnetic field is non-conservative.

I suppose, that in the case of any of these conservative/non-conservative forces, certain conditions can be found where they do not pass the criteria of falling exactly and always within their 'defined' force types.

Hello Ulrich,

>

but this is just Ampere's law (However I should have written "a non-zero term" instead of "work" for more clarity).

you can find a more mathematical treatment here:

http://www-astro.physics.ox.ac.uk/~sr/lectures/multiples/Lecture7reallynew.pdf

@Johannes:

You finally got the point: it delivers zero work. This is what counts for conservativity, not the value of the integral of B along some line (which is nonzero when the line is a loop which encloses a current - I heard about that :-).

@Ludwig:

If it is an electric charge that is moving (the case to which my Hamiltonian/Lagrangean statement refers) then this work is zero and thus independent of the path taken.

So the answer is simple,

1.The curl of the vector field must be zero everywhere and the vector field irrotational for the magnetic force to be conservative,

2. However, the curl of the magnetic force is not zero everywhere as there is always current density or a time varying electric field present at the point experiencing the magnetic force, and

3. in the case of an electric charge as per Ulrich (i.e. moving electron) the electric charge still produces its own time dependent electric field,

Thus, as per the definition of a conservative force, the magnetic force is non-conservative.

In summary, a conservative magnetic force is mathematically possible, but in reality the magnetic force is non-conservative.

@Ludwig

These are my remarks to your points 1-3:

1.,2. There is no vector field for the magnetic force on an electric charge (since this force does not only depend on position and time but also on the velocity of the charge). So you can't speak about the curl of this field. The tricky part that probably excaped your attention is that the magnetic fields B and H a r e force fields for magnetic forces on magnetic poles (which always come in pairs, which should not matter here). But they don't do work on a moving electric charge. So their non-vanishing curl does not make the system 'one electric point charge moving in a time-independent external electromagnetic field' a non-conservative one. What this exactly means that the total energy of the charge is a constant of motion. It can be written as

E = gamma*m*c² + q*Phi,

where gamma is the relativistic factor sqrt(1 - (v/c)²), m is the rest-mass of the particle, q its electric charge, and Phi the electrostatic potential. That the magnetic field does not shine up in the energy expression does not mean that the motion is not influenced by it. The last section of the Wikipedia link gives more on it.

3. Of course, an 'external electromagnetic field' is an idealization. It ignores the more subtle phenomena of radiation damping and self-interaction such as actions of the field generated by a charge on that charge, that you mention. This idealization of 'external electromagnetic fields' has wide technical applications. All electron microscopes (which may employ electrostatic or electromagnetic focussing) are designed with this model in mind. Of course, there is more than one electron in a working electron microscope. But electromagnetic particle-particle interaction, which is a subtle point for the theory, can here easily taken into account approximately.

http://en.wikipedia.org/wiki/Hamiltonian_mechanics

Hi Ulrich, thanks, yes interesting. From the Hamiltonian one can directly calculate the Lorentz force law. So it all adds up, I agree the Hamiltonian works, and the Lorentz force law works. It also shows then that the magnetic field can do no work on a single, isolated charge as the force is perpendicular to the motion of the charge, and that the electric field of the charged particle does the work, and not the magnetic field.However, the problem still remains that the 'particle' (and now I choose the particle), can interact with the magnetic field, say it is a flat disk made up of steel. First of all, we end up with the steel being magnetized by just introducing it into the field (realignment of atoms), then we have eddy currents, heat etc. Zero work? Are we not assigning the conservative nature of the magnetic force under certain, specified conditions?

Dear Ulrich,

I don't quite understand your point - the fact that int(B dl) over a closed loop delivers a finite result is one of the definitions of a non-conservative field - see here, for example:

http://en.wikipedia.org/wiki/Conservative_force#Mathematical_description

here are some more examples of explanations why the B-field is not conservative (I guess lectures from Oxford University or Rutgers can be considered reliable).

http://www-astro.physics.ox.ac.uk/~sr/lectures/multiples/Lecture7reallynew.pdf

http://www.physics.rutgers.edu/ugrad/227/L13%20Magnetic%20field%20Magnetic%20force%20on%20charges.pdf

and here is even an experiment (a very nice one, I think) which shows the non-conservativenes of the B-field with quite a simple circuit - it from the MIT:

http://web.mit.edu/8.02/www/Spring02/lectures/lecsup3-15.pdf

Hi Ludwig,

the interaction of magnetic fields with macroscopic bodies, such as lumbs of steel is a completely different matter. Of course there is magnetization, electric polarization, eddy currents (I once got a patent for a eddy current-based device) which all are accompanied with energy dissipation (so certainly not zero work). This kind of 'non-conservativness', however, has nothing to do with the question whether the curl of the magnetic field vanishes or not.

Your resumee of the state of affairs misses a point which I consider the most relevant one in our context. Not only that the Hamiltonian formalism works (for point charges in an electro-magnetic field) in the sense that we can find a mathematical expression (the Hamiltonian) from which we can get the equations of motion by differentiation (this is what you acknowledge), but also that the Hamiltonian considered as a physical quantity (which is something different from a mathematical expression: an expression has sub-expressions a quantity not) and for this quantity the theory says that it is constant in time. This is what the word 'conservative' refers to: Energy is 'conserved', i.e. constant, not dissipated, ..... The usual way physics texts express this result of Hamilton's theory is:

d H / d t = partial_d H / partial_d t

where in a stationary environment the partial derivative on the right-hand side vanishes by definition. Then by virtue of this theorem the Hamiltonian is a conserved quantity (also called constant of motion). For your steel disk in a magnetic field there is no conserved quantity irrespective of the curl of the field. So the system is not conservative. This shows that the quality 'conservative' can more naturally associated with a well-defined physical system than with an abstractum like 'the magnetic force'.

I hope this frees your spirit for a spectacular progress in your exciting observational project. Good Luck.

Johannes,

I hate to say the same thing trice. Whether the magnetic field B is conservative in the sense of general vector field theory (to which you refer) has no influence on whether a charged particle moving in B is a conservative system. Why this is so, should be clear from any of my previous contributions (or, preferably, from your own physical understanding).

Since the original question was always about the magnetic force and not about the magnetic field, all observations concerning the line integral of B are irrelevant to the original question.

Dear Ulrich,

yes, I see now where the misunderstanding was - I was only referring to the B-field while you were also talking about charged particles and the Lorentz force.

Now everything makes quite sense - sorry it took me a while :)

best regards

Hi Ulrich, excellent, then we agree! Your statement "This shows that the quality 'conservative' can more naturally associated with a well-defined physical system than with an abstractum like 'the magnetic force'." makes me feel free indeed, I was starting to feel a bit stupid at a stage!

I am delayed in this discussion, but I think U. Mutze has already claryfied the question. I woud like to add only this: from the general physics point of view the magnetic force is conservative as it is perpendicular to the velocity (its work is zero). From the Classical Mechanics point of view, the magnetic force depending from the veocity is not defined as usually as minus gradient (according to coordinate derivatives) of a scalar potential, but it can be defined as minus gradient acccording to coodinate derivatives of a scalar potential (q.v.A) plus a time derivative of the gradient according to the velocity components of this potential. (Pease see the Lagragean definition in the electromantic fied and electrical charges presence). Although the magnetic force is not curl-free of the classical form, it can be considered consevative because it can be derived in some way from a potential.

I found the whole debate very interesting. With reference to the comment of Bejo, I would like to submit the following remark: from analytical mechanics (as exposed, e.g., in Goldstein, "Classical Mechanics" and Lanczos, "The Variational Principles of Mechanics") one is led to assume that the Lorentz force on a single charge q(E+v \times B/c) is monogenic, i.e., it is derivable from a velocity-dependent potential U, which is a function of the particle velocity and of both scalar and vector potentials. Then the Lagrangian is L= T-U, while the conserved quantity is not T+U but the Hamiltonian of the system. Accordingly, one might state that the Lorentz force is "conservative" in this more general sense. Though this view is mathematically consistent, the physical nature of velocity-dependent potentials seems elusive.