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Is Magnetic force conservative?

  1. Nov 7, 2015 #1
    The work done by magnetic force, [itex]F=qv\times B[/itex], is zero. Can we infer that the magnetic force is conservative? If so, how can we show that [itex]\nabla \times F=0 [/itex]?
     
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  3. Nov 7, 2015 #2

    Vanadium 50

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    Is this homework?

    Where have you started?
     
  4. Nov 7, 2015 #3
    No. This in not homework. The magnetic force is perpendicular to the velocity of the particle so the work down by the force is zero always. Therefore the work doesn't depend on the path between the two points, say A and B. I want to know whether the magnetic force is conservative or not.
     
  5. Nov 7, 2015 #4
    Conservative forces conserve energy. That's where the name comes from. You already know that this is the case for Lorentz force.
     
  6. Nov 7, 2015 #5
    Thanks, how to show [itex]\nabla \times F=0[/itex] with [itex]F=qv\times B[/itex]?
     
  7. Nov 7, 2015 #6
    You can show that for force fields only.
     
  8. Nov 7, 2015 #7
    I did that using the Formula [itex]\nabla \times (v\times B) =(\nabla . B)v-(\nabla . v)B+(B. \nabla)v-(v. \nabla)B[/itex] and I can only eliminate the first term ([itex] \nabla .B=0[/itex]). How to show that the other three terms are zero or cancel each other?
     
  9. Nov 7, 2015 #8
    As pointed out earlier (DrStupid), the magnetic force is not a force field, since it depends on the velocity. There is good discussion about this in wikipedia: https://en.wikipedia.org/wiki/Conservative_force, and also in the book on Classical Mechanics by John R Taylor.
    To say that a conservative force is one that conserves energy is not quite correct, since the magnetic force does zero work, and therefore certainly conserves energy. So does the normal force exerted by a surface in contact with an object. Neither of these forces is conservative, that is, they cannot be derived from a potential energy. The safest way to define a conservative field seems to be as is done by Taylor, by stating two conditions:
    1. The force depends only on the position of the particle, and not on any other variable, such as velocity, or time.
    2. The work done by the force as the particle moves from point 1 to point 2, for any two points, is the same for all paths connecting 1 and 2.
     
  10. Nov 8, 2015 #9

    vanhees71

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    By definition a conservative force is a force that's expressible (at least locally) as the gradient of a scalar field. Since on the other hand all forces (interactions) on a fundamental level are relativistic that's a very rare case. E.g., in electrodynamics it's the case for electrostatics, where the electric field and thus also the electric force on a test charge is conservative.

    In contrast to this the conservation of energy (and inseparable from it also momentum) is much more general. It's implicit already in the very foundations of the special relativistic spacetime model. Thus, as long as you can neglect gravitation (which you can very often, because it's very weak compared to the other fundamental interactions) you have energy-momentum and angular-momentum conservation for any closed system.
     
  11. Nov 8, 2015 #10
    One hears this frequently, but it is a misconception. The point is that the fundamental quantity in analytical mechanics isn't the potential energy function but the work function. As long as the work function ##U(q_1,q_2,\cdots, q_n,\dot{q}_1, \dot{q}_2, \dot{q}_n)## is independent of time, the ##\textit{generalised}## forces are conservative and may be calculated from this single scalar function ##U##, taking the form ##F_i = \frac{\partial U}{\partial q_i} - \frac{d}{dt}\frac{\partial U}{\partial \dot{q}_i}##. For time independent systems (##U## independent of t as well as the kinetic energy function, i.e. the constraint equations do not contain t explicitly) the total energy is conserved and can still be expressed as the sum of the kinetic energy and the ##\textit{potential}## energy ##V##, provided however that ##V## is defined as a Legendre transformation of the work function: $$V=\sum_i \frac{\partial U}{\partial \dot{q}_i}\dot{q}_i - U.$$
    This was actually part of the intent of the problem I posted in the "interesting E-M problems" thread that was initiated by another member just over a week ago!
     
  12. Nov 8, 2015 #11

    vanhees71

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    No, these forces are not conservative but energy is conserved, if ##U## is not explicitly time dependent. Again, a force is called conservative iff it's the gradient of a scalar potential, independent of the (generalized) velocities (or canonical momenta in the Hamiltonian formulation).
     
  13. Nov 8, 2015 #12
    They are conservative as long as one generalises the common ##F_i=\frac{\partial U}{\partial q_i}## (which is appropriate for velocity independent work functions) to the more general case of velocity dependent functions ##F_i = \frac{\partial U}{\partial q_i} - \frac{d}{dt}\frac{\partial U}{\partial \dot{q}_i}##. If the work function is time-independent, we obtain a class of forces called "conservative" because they satisfy the conservation of energy, as the name suggests and was pointed out by Dr.Stupid.
     
  14. Nov 8, 2015 #13
    Is the point not that a gauge potential field doesn't give rise to a non conservative system, it simply curves the trajectory along which flow (determined by the canonical momentum) occurs? A conservative system is a system whose total energy is conserved over time and if one does not generalise the force expression as described above, then yes one seemingly has the situation of a non conservative force even in the case where the total energy is conserved..But that's only because one hasn't generalised the force expression accordingly.
     
  15. Nov 8, 2015 #14

    vanhees71

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    I can only repeat that the common use of language in physics is that a force is called conservative, if ##\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x})##. It's not a good idea to introduce another convention in these forums.
     
  16. Nov 8, 2015 #15
    Fair enough, if that is the way one defines it.. Apologies for potentially contributing to any form of confusion.
     
  17. Nov 8, 2015 #16
    Do you have a reference for this definition?
     
  18. Nov 8, 2015 #17

    vanhees71

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    https://en.wikipedia.org/wiki/Conservative_force

    although that's not a very good article since in relativistic physics almost all forces are velocity (or better momentum) dependent. The reason is that for a classical particle the four-velocity,
    $$u^{\mu}=\frac{1}{c} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},$$
    fulfills necessarily the constraint
    $$u_{\mu} u^{\mu}=1 \; \Rightarrow \; \frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau} u_{\mu}=0.$$
    A relativistically covariant equation of motion,
    $$m \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}=m c \frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau}=K^{\mu},$$
    necessarily implies that the Minkowski force four-vector fulfills the constraint
    $$u_{\mu} K^{\mu}=0,$$
    and that usually implies that ##K^{\mu}## not only depends on ##x^{\mu}## but also on the four-velocity ##u^{\mu}##.

    The magnetic force on a test charge is thus rather typical than special. The equation of motion of a particle in an electromagnetic field is given by
    $$m c \frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau}=q F^{\mu \nu} u_{\nu}=K^{\mu},$$
    where ##F^{\mu \nu}=-F^{\nu \mu}## is the electromagnetic field. Obviously ##K^{\mu}## fulfills the constraint condition.
     
  19. Nov 8, 2015 #18
    I do not find a clear definition in this reference. The second paragraph says that "A conservative force is dependent only on the position of the object." but it is not clear if this is part of the definition or if the author just had force fields in mind. I also checked the German version of this article and there is no such limitation. Both articles have in common that the work done by a conservative force is independent of the taken path. Thats the definition as I know it and it also applies to the Lorentz force.

    It seems Wikipedia is no proper reference for this particular question. Where can we find the official definition of conservative forces? Does such a definition exist at all or is it just a matter of personal preferences?
     
  20. Nov 9, 2015 #19
    Not that this is an official definition or anything, but the end of the first chapter of Lanczos book discusses this: https://books.google.fr/books?id=cmPDAgAAQBAJ&pg=PT21&lpg=PT21&dq=lanczos+variational+principles+chapter+1&source=bl&ots=QBhzOVPOtd&sig=lKVm9oEL3BfDhK94jMBf8xCrzQc&hl=en&sa=X&ved=0CCMQ6AEwAGoVChMI9orY4N6CyQIVgT0UCh2L4gBY#v=onepage&q=lanczos variational principles chapter 1&f=false
    Specifically sections 7 and 8 of the the first chapter. It is my understanding that conservative forces would arise in systems which he would call scleronomic (a term introduced by Boltzmann apparently). It does seem that it may all be just a matter of preference.
    EDIT: Check the summary of section 7.
     
    Last edited: Nov 9, 2015
  21. Nov 9, 2015 #20

    vanhees71

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    That's always right ;-).

    Take any textbook on classical mechanics, where you'll find the definition of conservative forces quite in the beginning. It's not restricted to the motion in external force fields. Also interactions can be conservative. Usually you have two-body interactions (like Newtonian gravity). Then
    $$\vec{F}_{12}=-\vec{\nabla}_1 V(\vec{x}_1-\vec{x}_2), \quad \vec{F}_{21}=-\vec{\nabla_2} V(\vec{x}_1-\vec{x}_2)=-\vec{F}_{12}.$$
    Of course, that's restricted to non-relativistic physics, because it implies action at a distance and the third law for the point particles alone.
     
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