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Why magnetic potential on a moving charge=q v.A ?

  1. Mar 31, 2006 #1
    Hi,
    Can ny one direct me to a reference that explains in detail that the magnetic potential on a moving charge = q v.A or at least why is the generalized force on a charge in a velocity dependent potential(U) in terms of the generalized coordinates: Fj = - dU/dqj+d/dt (dU/dvj) ? vj is the velocity in direction j , some d's are actually partial differentiations.

    Thanks.
     
  2. jcsd
  3. Mar 31, 2006 #2

    Meir Achuz

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    Your second equation is Lagrange's equation. V=qv.A is the potential energy needed in Lagrange's equation to get the correct magnetic force.
    This is treatred in graduate EM texts, like Jackson "Classical Electrodynamics" or Franklin "Classical Electromagnetism". Look up "Lagrangian" in thr index.
     
  4. Mar 31, 2006 #3
    Using this potential in Lagrange's equation will give us the correct Lorentz force ,provided that the generalized force is given by the expression outlined in my message. ( This is the condition to use any potential in the Lagrangian- see Goldstein , 3rd ed. page 22) So I was asking for a proof for this expression of the force as an alternative way.
    The Lagrangian in Jackson utilizes this form of potential directly and doesn't prove it. ( section 12.1 - 3rd ed.)
    Thanks
     
  5. Apr 1, 2006 #4

    Meir Achuz

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    I am still not sure what you want. Your second equation is just the usual form of Lagrange's equation. Do you want a new derivation of that?
    The result V=-qv.A follows (in NR CEM) from writing the Lorentz force equation, and then writing E and B in terms of phi and A. The full derivation is just Eq. (15.21) in Franklin.
    The relativistic derivation is neater. The interaction Lagrangian is uniquely given as the only posible scalar linear in the potentials and the velocity. This is Eq. (15.25). You are right. Jackson assumes things without always saying so.
     
  6. Apr 1, 2006 #5
    Unfortunately, I don't have access to Franklin's book. Does it derive the potential q v.A only from Biot-Savart, Lorentz Force and vector identities of the magnetic field? The derivation of Lagarange's equation doesn't prove the the force takes the form - dU/dqj+d/dt (dU/dvj), rather it says that this is the condition for the lagrangian to be written in the form L=T-U and to follow the famous eqaution - dL/dqj+d/dt (dL/dvj)=0. My question again is why the velocity dependent potential is related to the force by Fj= - dU/dqj+d/dt (dU/dvj) while velocity indpenedent potential is related to its force merely by F=grad(U). ?
     
  7. Apr 1, 2006 #6
    The only proofs I have seen rely on the Lagrangian principle. E.g.

    The Lagrangian

    [tex]L=T-V=\frac{1}{2}mu^2 + q\vec{u} \cdot \vec{A} [/tex]

    Subsituted in the Euler Lagrange equations

    [tex]\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=\frac{\partial L}{\partial x}[/tex]

    Using generalised coordinates x, y, z with generalized velocities [tex]\dot{x}, \dot{y}, \dot{z}[/tex] and [tex]u^2=\dot{x}^2+\dot{y}^2+\dot{z}^2[/tex]

    yields

    [tex]\frac{d}{dt}(m\dot{x}+qA_x)=q\frac{\partial}{\partial x}(\dot{x}A_x+\dot{y}A_y+\dot{z}A_z)[/tex]
    [tex] \ddot{x}+q \frac{\partial A_x}{\partial t} =q(\dot{x}\frac{\partial A_x}{\partial x}+\dot{y}\frac{\partial A_y}{\partial x}+\dot{z}\frac{\partial A_z}{\partial x})[/tex]

    or after rearranging

    [tex]m \ddot{x}=-q(\frac{\partial \phi}{\partial x}+\frac{\partial A_x}{\partial t})+q[\dot{y}(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})+\dot{z}(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z})]=qE_x+q(\dot{y}B_z-\dot{z}B_y)[/tex]

    Which is exactly the x-component of the usual Lorentz force. So the potential term [tex]V=-q\vec{u} \cdot \vec{A} [/tex] can be motivated by the fact that it yields the right equation of motion.




    Notice that the canonical momentum

    [tex]p_i=\frac{\partial L}{\partial \dot{q_i}}[/tex]

    is [tex]\vec{p}=m\vec{v}+q\vec{A}[/tex] which is conserved (in contrast to just mv) even when magnetic phenomena are involved. The kinetic energy

    [tex]\frac{1}{2}mv^2=\frac{1}{2m}(\vec{p}-q\vec{a})^2[/tex]

    Involves a term [tex]q\vec{u} \cdot \vec{A} [/tex] which indicates that the energy is kinetic in nature.
     
  8. Apr 1, 2006 #7

    Physics Monkey

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    Abu, you have just answered your own question! Suppose you can derive your equations of motion from a Lagrangian. You can always separate L into T - U, where T is the kinetic energy and U is everything else. You can than define, if you like, the generalized force to be [tex] F_j = - \frac{\partial U}{\partial q_j} + \frac{d}{dt}\frac{\partial U}{\partial \dot{q}_j} [/tex]. Now what happens when U is independent of [tex] \dot{q}_j [/tex]? Well, the second term is zero, so you just get [tex] F_j = - \frac{\partial U}{\partial q_j}[/tex].

    The justification for the Lagrangian [tex] L = \frac{1}{2} m v^2 + q \vec{v}\cdot\vec{A} - q \phi [/tex] is a separate issue that I think others have addressed quite nicely.
     
    Last edited: Apr 1, 2006
  9. Apr 1, 2006 #8
    Willem. What I am saying is that we cannot use this as proof since we assume implicitly before using any potential in the Lagrangian that it satisfies the relation [tex] F_j = - \frac{\partial U}{\partial q_j} + \frac{d}{dt}\frac{\partial U}{\partial \dot{q}_j} [/tex]. This is clear in deriving the Lagrange's equations from Newton's decond law F=ma ( see Goldstein, section 1.4 ). So it's not strange that we got the correct expression for Lorentz force by applying Lagrang's equation, it could have been obtained directly from [tex] F_j = - \frac{\partial U}{\partial q_j} + \frac{d}{dt}\frac{\partial U}{\partial \dot{q}_j} [/tex] This is a subtle point, I hope you got it. You may say that Lagrange's equation can be also obtained in a different way from functional analysis by minimizing the action. But this method depends on assuming that the Lagrangian L=T-U without providing a basis for this choice. Also, though outside the topic, you can't know what T is without Newton's laws.
    So my point again is that if we proved that the force of a general potential = [tex] F_j = - \frac{\partial U}{\partial q_j} + \frac{d}{dt}\frac{\partial U}{\partial \dot{q}_j} [/tex]. then the problem is over.
     
    Last edited: Apr 1, 2006
  10. Apr 1, 2006 #9
    There are two kinds of magnetic potentials. One is the magnetic vector potential which always exists and then there is the magnetic scalar potential which exists only for static EM fields. You're thinking of the magnetic vector potential for which no energy is assinged to when multiplied by charge. The vetor potential works because it fits the formula for a generalized potential and hence it is a solution of the Maxwell's equations. Actually there are other forces which fit the generalized velocity dependant potential U such as the coriolis force. It would be instructive to look into that two.

    Pete
     
  11. Apr 2, 2006 #10

    Meir Achuz

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    This wouldn't latex correctly so I leave in in latex form:

    m\frac{d{\bf v}}{dt}=
    q[-\nabla\phi-\partial_t{\bf A}+{\bf v\times(\nabla\times A)}]\\
    = q[-\nabla\phi-\partial_t{\bf A}-(\bf v\cdot\nabla)A
    +\nabla({\bf v\cdot A})]\\
    =\nabla[-q\phi+q{\bf v\cdot A}]-q\frac{d{\bf A}}{dt}
     
    Last edited: Apr 2, 2006
  12. Apr 2, 2006 #11
    Meir, you expanded [tex]{\bf v\times(\nabla\times A)}[/tex] into two terms only and assumed that [tex](\bf A\cdot\nabla)v[/tex] =0 why?
    Now assuming that the expansion is correct then what you have proved actually is that [tex]m\frac{d(P+qA)}{dt}=\nabla[-q\phi+q{\bf v\cdot A}][/tex] . How can this tell that [tex][q\phi-q{\bf v\cdot A}][/tex] is the energy of the charge?

    Thanks.
     
    Last edited: Apr 3, 2006
  13. Apr 3, 2006 #12

    Meir Achuz

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    1. v and r are independent variables here. v(r) only would come out after you solved the equations of motion.
    2. The potential energy V is defined by F=-\grad V.
     
  14. Apr 4, 2006 #13
    Meir, Now you are using Newton's laws and vector calculus, without refering to Lagrangians at all, So why did you assume that v is independent of r ? Lagrangian mechanics is equivalent to newton's laws and whatever could be proven from the first can be proven from the second. So, let's stick to this method( Newotn's law) to see what we will get.

    Again assuming that the expansion is correct, what does your last relation tell?
    It either tells that If we know (somehow) that the right hand side is the gradient of the energy, then the left side is the change of the total momentum of the system, field momentum and mechanical momentum. This the only coclusion.

    OR, If we know that the left side is the change of total momentum (somehow), then the right side is the gradient of energy.

    Now, what is this "somehow" in your case??
    Thanks.
     
    Last edited: Apr 4, 2006
  15. Apr 4, 2006 #14

    Meir Achuz

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    I can't understand any of your objections. Try someone else.
     
  16. Apr 4, 2006 #15

    arildno

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    Hardly.
    Given F, we have that IF there exists a function U so that the identity between the derivatives of U and F holds THEN Lagrange's principle is strong enough to derive the correct equations of motion.
    It is at the outset by no means certain that such a U exists.

    (By the way, Lagrangian mechanics has never, ever shown itself "strong" enough to recapture the dissipative Navier-Stokes equations.)
     
  17. Apr 6, 2006 #16
    What confused me is that I thought q*Phi - q v.A is the energy that the charge will acquire from the field and I was asking for an evidence for that, but in fact the magnetic field doesn't change the energy of the charge at all since its force is always normal to the velocity. The Hamiltonian (total energy) is still 1/2 m v^2 + q.Phi , The q*Phi- q v.A is not the real potential energy, but it's the quantity (called generalized potential - we could have called it anything else) that has to be used in the Lagrangian since it takes the form L=T-U and U is related to the force by Fj = - dU/dqj+d/dt (dU/dvj) whatever U is. It's not necessarily that U is a real energy as is the case here.

    Now what does qA really represent? Does it represent an extra momentum the charge acquires from the field? It doesn't seem so.
     
  18. Apr 6, 2006 #17

    Meir Achuz

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    After E and B are unified in Max's eqs, A is no longer just a magnetic potential since it also relates to E. Go back to physics 101.
    The PE (U) is defined so that F=-grad U. qv.A is part of the PE.
    The Ham is a formal object in which v does not have a direct meaning. H should be written in terms of p, and then qp.A/m still appears.
     
  19. Apr 6, 2006 #18
    This is the component of the force that cause a change in the kinetic energy of the particel. There may exist another component that doesn't change the kinetic energy as the magnetic component in Lorentz force. To add both components in a single equation you will get the total force [tex] F_j = - \frac{\partial U}{\partial q_j} + \frac{d}{dt}\frac{\partial U}{\partial \dot{q}_j} [/tex]. Here U is not the energy, but it's merely the quantity that makes the Lagrangian takes the form L=T-U.
    The potential energy is still [tex] q \phi [/tex] (See Goldstein, p.342)
    So the hamiltonian which is the total energy is the kinetic energy plus the potential energy [tex] =\frac{1}{2} m v^2 +q \phi [/tex]
     
  20. Apr 6, 2006 #19
    This is the component of the force that cause a change in the kinetic energy of the particle. There may exist another component that doesn't change the kinetic energy as the magnetic component in Lorentz force. To add both components in a single equation you will get the total force [tex] F_j = - \frac{\partial U}{\partial q_j} + \frac{d}{dt}\frac{\partial U}{\partial \dot{q}_j} [/tex]. Here U is not the energy, but it's merely the quantity that makes the Lagrangian takes the form L=T-U.
    The potential energy is still [tex] q \phi [/tex] (See Goldstein, p.342)
    So the hamiltonian which is the total energy is the kinetic energy plus the potential energy [tex] =\frac{1}{2} m v^2 +q \phi [/tex]
     
  21. Apr 6, 2006 #20

    Meir Achuz

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    You are confusing phormalism and fysics.
    In the Hamiltonian formalism, v never appears.
    In the Lagragian formalism F (only dp/dt) never appears.
    The basic physical definition of PE is that F=-grad PE.
    Things can get tricky and ambiguous with velocity dependent potential energy.
     
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