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Does the action always exist?

  1. Jul 2, 2007 #1
    That was discussed before but the thread died inconclusively.

    Anybody knows of necessary or sufficient conditions the differential equations of motion of a given system have to satisfy to be derivable by variations of some action?
     
  2. jcsd
  3. Jul 4, 2007 #2
    I am not sure whether this is a sufficient condition, but it is necessary that the net force acting on the system must be conservative.
     
  4. Jul 4, 2007 #3
    My question is more mathematical than physical. Someone writes some differential equations for some quantities that describe some imaginary system. I need the mathematical conditions these equations have to satisfy to be derivable from action principle.
     
  5. Jul 5, 2007 #4
    Non-conservative forces such as 'friction' can be incorporated imposing several constraints to the system in this case you have a Lagrangian

    [tex] \mathcal L =T(\dot x) -V(x)+\lambda U(\dot x , x ,t) [/tex]

    from this you can construct the Hamiltonian, of course the H-J equation with Constraints goes in a similar way to QM with constraints by Dirac so i guess you should keep a look at.

    http://en.wikipedia.org/wiki/Second_class_constraints
     
  6. Jul 5, 2007 #5

    olgranpappy

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    here's another obvious counterexample: electromagnetism.
     
  7. Jul 5, 2007 #6
    Non-conservative forces such as 'friction' can be incorporated imposing several constraints to the system in this case you have a Lagrangian

    [tex] \mathcal L =T(\dot x) -V(x)+\lambda U(\dot x , x ,t) [/tex]

    from this you can construct the Hamiltonian, of course the H-J equation with Constraints goes in a similar way to QM with constraints by Dirac so i guess you should keep a look at.

    http://en.wikipedia.org/wiki/Second_class_constraints
     
  8. Jul 6, 2007 #7
    When you're dealing with electromagnetism, the net force is obviously conservative. Here's why:
    The definition of a conservative vector field in 3 dimensions is that the curl of the field must be zero, or equivalently, that the field be the gradient of some scalar field. The electric field is conservative because it is the gradient of electric potential. Therefore, the electric force is also conservative since it is proportional to the electric field. An alternate definition of conservative force is that the work around a closed path must be zero. This definition is obviously satisfied by the magnetic force because the work done by the magnetic force along any path, whether open or closed, is 0.
    It is for this reason that the net force for electromagnetism is conservative.
     
  9. Jul 6, 2007 #8

    olgranpappy

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    wrong. you are talking about electrostatics.
     
  10. Jul 6, 2007 #9
    You're right, I was only talking about the case in which electric and magnetic fields are constant. However, if you consider an arbitrary closed path, then the work done along that path by the changing electric field always cancels exactly the work done along the path by the changing magnetic force. Therefore, the net electromagnetic force is conservative even when the fields are changing.
     
  11. Jul 6, 2007 #10

    olgranpappy

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    still wrong

    :rolleyes:
    ...didn't you just say [quite correctly] that the magnetic force does no work? Anyways, you should stop trying to show that the electromagentic force is conservative because it simply can not be derived from a potential... hence it is not conservative. hence, as you said, I am right and you are still wrong. Let me help you out a bit; in electromagnetism the force is given by:
    [tex]
    \vec F = q \left(\vec E + \vec v \times \vec B\right)
    [/tex]
    where q is charge and B is mag field and most importantly
    [tex]
    \vec E = -\nabla \phi - \frac{\partial}{\partial t}\vec A\;.
    [/tex]
    In the above equation you see the 2nd term there with the vector potential [tex]\vec A[/tex] that is not zero in general... That term keeps the force from being conservative because there is no way to write it as the gradient of a scalar... because if there were then [tex]\nabla\times E[/tex] would be zero, which it is not.
     
  12. Jul 6, 2007 #11
    The work done by the force associated with a static magnetic field is 0. However, the work done by the force associated with a changing magnetic field is not necessarily 0. If you consider a closed path, this nonzero work cancels with the nonzero work produced by the nonconservative electric force. So the net electromagnetic force is still conservative.
    Here's a simple example of a changing magnetic field doing nonzero work:
    Consider two parallel current-carrying wires. There is an attractive magnetic force between them. For this reason, the will move closer together, and the magnetic field produced by each wire will do nonzero work on the other wire. The reason that work can be done is that as the wires move, the magnetic field produced by each wire constantly changes.
     
  13. Jul 6, 2007 #12

    olgranpappy

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    no. that's just wrong.
     
  14. Jul 7, 2007 #13
    Why is it wrong? My reasoning seems to be perfectly valid.
     
  15. Jul 7, 2007 #14
    Consider a source of the emf causing the current in the conductors. Just for fun, let's say it's an ideal DC generator being driven from by some perfectly constant mechanical driver. Let the mechanical driver be coupled to the generator by an infinitely precise torque meter. Imagine we somehow keep the wires perfectly stationary in the presence of the attractive force, do you think that we should read some constant value on the torque meter (again, assume ideal conditions so we don't have to get crazy over irrelevent details). What would happen to the meter reading if the conductors were suddenly allowed to respond to the attraction?

    Next, let's assume we use a high frequency generator for the source emf. Now we have some highly varying [electro]magnetic fields. How would the torque meter respond if we repeated the above experiment?
     
    Last edited: Jul 7, 2007
  16. Jul 7, 2007 #15
    I'm not sure what kind of situation you're imagining. I'm talking about two straight, infinitely long, parallel wires with currents flowing in the same direction.
     
  17. Jul 8, 2007 #16
    Please forgive me for being unclear. I'll take a different approach. It is my understanding that as the wires move toward each other (conductors' currents in same direction), each will be moving in an area of increasing flux density that surrounds the other wire. Then, each would experience an increasing change in flux which would induce an increasing emf in the direction that would oppose the wires' motion (Lenz's Law). Therefore, I don't see any connection between the changing magnetic field and any work done in moving the wires. The magnetic field deflects the moving charges' direction of motion in the conductor but it is the prime mover of the charges (the source of the potential driving the charges in the conductors) that ends up doing the work in moving the wires. The induced emf from the changing flux due to the wires motion ends up opposing that motion. As a result, I believe that even for changing magnetic fields, their influence will still always be at right angles to the motion of any particular charge and, therefore, will not do any work throughout the conductor. If I'm incorrect in this, I hope someone will show me.
     
    Last edited: Jul 8, 2007
  18. Jul 8, 2007 #17
    Unfortunately, I'm still having trouble with your explanation. Faraday's Law (in integral form) states that the magnitude of the induced emf ALONG A CLOSED PATH is equal to the rate of change of magnetic flux through that CLOSED PATH. When you say that each wire "would experience an increasing change in flux," what CLOSED PATH are you considering?
     
  19. Jul 8, 2007 #18
  20. Jul 9, 2007 #19
    I have a couple questions maybe someone (anyone) could help me with. I'm kinda confused about the possibility of induction in this infinite parallel straight conductor scenario. Since we are talking conductors where there's no loop by which to perform the normal closed path integrals related to induction, as the wires move due to the Lorrentz force, each conductor is still moving through a region of changing B field density. So here's my questions:

    1. This question is strictly about the Lorrentz force. Assume the wires are being attracted (conductor currents in the same direction). Does each conductor experience an increasing Lorrentz force as it moves closer to the other wire and thus their motion is accelerated? It seems to me that it would since at a smaller distance the other wire's B field magnitude is stronger than it is at a greater distance.

    2. Even though we have no actual conductor loop in this hypothetical scenario, would there still be induction in each conductor as it moves through the other conductor's magnetic field - especially since that field strength varies as a function of the distance from its source? Or, owing to the geometry of the two wire's spatial relationship, there can be no difference in B field magnitude anywhere inside the conductor (except across its radial plane) at any instant in time, so no induction can take place?

    3. If induction is possible, would the induced emf be in the direction that would tend to oppose the wire's motion? i.e. buck the other wire's current?
     
  21. Jul 9, 2007 #20
    I could be mistaken, but this is my reasoning:
    F is proportional to Bd, where B is the magnetic force and d is the distance between the wires. However, B is proportional to 1/d. Therefore, F should be constant as the wires move closer together, however counterintuitive this may seem. And it is highly counterintuitive; how can two tiny wires a thousand miles away from each other exert the same force on one another as if they were two inches from each other?
     
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