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Lorentz force in two frames

  1. Jun 17, 2005 #1
    Consider a magnetic field with a charge at rest in inertial frame S. From this reference point the particle is not moving wrt the magnetic field and will not experience a Lorentz force. Now consider a frame S' moving wrt S at a constant speed. An observer in this frame would see the charge moving and hence it would be experiencing a force given by the Lorentz force equation. If you attached an accelerometer to the particle would it indicate that the charge had accelerated on not?
     
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  3. Jun 17, 2005 #2

    dextercioby

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    Well,what does the last phrase mean in the context of Newtonian physics ?Does a magnetic Lorentz force accelerate the charge particle and if so,how?

    There's a trick here.In the Lorentz boosted frame,there will be a nonzero electric field which would appear to accelerate the electric charge,even though the magnetic one won't linearly accelerate it.

    Daniel.
     
  4. Jun 17, 2005 #3
    The Lorentz force will cause the charge to move in a circle. The electric field is a result of the accelerating charge but how can the field act on the charge that is creating it?

    The point is though that according to the observer in the rest frame the charge is at rest and therefore would expect the accelerometer to register no acceleration at all. The person in the boosted frame would say the charge moved in a circle and therefore was accelerated.
     
  5. Jun 17, 2005 #4

    dextercioby

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    The field is exernal.If you meant the radiation reaction force (Abraham-Lorentz force),that's something different.

    Nope,in the boosted frame,the situation is a bit more complicated.Write Newton's law in tensor form in the boosted frame and u'll see what i mean.

    Daniel.
     
  6. Jun 17, 2005 #5
    I am more interested in a conceptual explanation. There appears to be a paradox here. The principle of relativity says that all inertial reference frames equally valid in terms of the laws of physics being the same in all. Here it appears that we can say who is moving and who is not so there seems to be a problem.

    As far as the tensor stuff goes are you referirefereinge electromagnetic field tensor from special relativity?
     
  7. Jun 17, 2005 #6

    robphy

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    Does "moving" imply "accelerating"?

    Could this charge be "moving" with constant velocity?
    If so, then the net-force on the charge would be zero.
     
  8. Jun 17, 2005 #7

    pervect

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    When you transform the magnetic field to the moving frame, you will find that you now have both an electric and magnetic field acting on the particle.

    If you calculate the magnitude of the force exerted by the transformed field on the particle, you should find that the total force is still zero - the force exerted by the magnetic field is exactly cancelled out by the force exerted by the electric field.
     
  9. Jun 17, 2005 #8
    A charge moving in a magnetic field in any direction that is not parallel to the magnetic field will experience a force given by:
    [itex]
    F=qvBsin\theta
    [/itex]
    where [itex]\theta[/itex] is the angle between the direction of the velocity and the magnetic field.
     
  10. Jun 18, 2005 #9
    does this mean that the charge does not actually move in a circle? In high school physics we are taught that a moving charge in a magnetic field experiences a force perpendicular to the velocity and hence moves in a circle. If the net force is zero the charge would not move in a circle? So what is the path of the charge?
     
  11. Jun 18, 2005 #10

    robphy

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    The trajectory is a straight line in space.... an object at rest in an inertial frame S has the trajectory of a straight line (with constant velocity) in another inertial frame S'. [Lorentz Transformations are linear transformations... straight worldlines are mapped to straight worldlines.]

    While it is true that "a moving charge in a magnetic field experiences a force perpendicular to the velocity" (part of the Lorentz Force Law), it is the net-force that determines the acceleration of the object (Newton II). As mentioned above, there will be in S', in addition to a magnetic force, an electric force (also part of the Lorentz Force Law and the transformation of the EM fields). In order for the trajectory to be a constant-velocity one, these two forces must balance.
     
  12. Jun 19, 2005 #11
    So it doesn't move in a circle? This goes against everything we are taught in high school physics.
     
  13. Jun 19, 2005 #12

    ZapperZ

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    Not ALL moving charges in a magnetic field do that. If the charge is moving only along the SAME direction as the B-field, it experiences no Lorentz force. Zero. The v x B term is zero.

    Zz.
     
  14. Jun 19, 2005 #13
    This has been stated above.
     
  15. Jun 19, 2005 #14
    If all you have is a charged particle and an external magnetic field, the particle will move in a circle (or along a helix, which looks like a circle when viewed along its axis) if it has a non-zero velocity in a direction other than along a magnetic field line. However, if you now add some other force to the particle it can move along other paths.

    Consider a wire in a magnetic field - the electrons in the wire will want to move in a circle as much as possible, but will also be constrained to move only along the path of the wire by other forces; they don't all jump off of the wire in order to follow a circular path.

    Now, when you have a purely magnetic field in one inertial frame and you perform a Lorentz transformation to another inertial frame, you will see both magnetic and electric fields. The magnetic field in the second frame will still try to move the particle in a circle, but the electric field will also create a force on the particle. It is the net force of these two added together that causes the particle to move along some path other than a circle. For the situation you are describing, this path will be a straight line (and at constant velocity - no acceleration measured on the accelerometer).
     
  16. Jun 19, 2005 #15

    robphy

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    The following might help clear things up:

    A more accurate statement of "what the magnetic field does to a charged particle" is this:
    a magnetic field tends to "turn" (and not change the speed of) a moving charged particle...
    since [tex]\vec v \times \vec B[/tex] is always perpendicular to [tex]\vec v[/tex]. (Of course, if [tex]\vec v \times \vec B[/tex] is zero, then the magnetic force does not deflect the particle.)

    When that magnetic field happens to be uniform in space (and time) and
    when there happen to be no other forces are acting,
    i.e., "a charged particle only under the influence of a uniform magnetic field",
    then that charged particle travels in a circle (assuming [tex]\vec v \times \vec B[/tex] is nonzero).
    (Obviously, this result is a very special situation.)

    It's great that you learned that result in high-school.
    Now, it's time to learn WHY that result is true... what laws of physics are being used (Lorentz Force, Newton II) and what special assumptions were made (static uniform magnetic field and no other fields or forces)?

    The problem you posed considered the case of a charged particle at rest in a magnetic field and how that situation is viewed from another inertial reference frame. Asking about what happens in this other frame of reference invokes another law (or princple) of physics: the principle of relativity. As has been mentioned numerous times above, there is an electric field [as well as a different magnetic field] seen by the moving observer. To that moving observer, the situation is NOT that of "a charged particle only under the influence of a uniform magnetic field".... so that observer should not expect that charge to travel in a circle.

    In fact, as I hinted above, the particle travels with constant velocity in a straight line in S'. Specifically, if the particle is at rest in frame S, and the observer S' moves with constant velocity [tex]\vec U[/tex] with respect to S, then, from frame S', the particle must move with constant velocity [tex]-\vec U[/tex] with respect to S'. For this to happen, S' must see a zero net-force on the charged particle.
     
    Last edited: Jun 19, 2005
  17. Jun 20, 2005 #16

    Meir Achuz

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    So much confusion. Pervec gave the correct, simple answer.
    A simpler one is that a LT cannot produce acceleration if there was no acceleration
    in the original frame. That is why SR was conceived, why the LT was made a linear transformation, and what determines the transformation of E and B.
     
  18. Jun 20, 2005 #17

    robphy

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    Yes, pervect did give the correct, simple answer.

    IMO, part of the "confusion" is that the original poster has the misconception that the particle must have a circular orbit. In spite of being given the correct answers, this misconception needed to be addressed.

    My second post (#10) gave this "simpler" answer.
     
  19. Jun 20, 2005 #18

    pervect

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    I guess I must have missed metrictensor's follow up question in #9, so I wasn't aware he was still confused.

    Hopefully the other very clear and detailed responses from other posters have set him on the right path.

    The only thing I would add is that

    http://scienceworld.wolfram.com/physics/ElectromagneticFieldTensor.html

    is a good source which describe how electric and magnetic fields transform in relativity. This writes down the transformation laws component by component for Ex, Ey, Ez, Bx, By, Bz. The only thing it doesn't explain properly is that the tranformation laws written down in this manner are for a Lorentz boost (velocity change) which occurs in the 'z' direction.
     
  20. Jun 20, 2005 #19
    Yes. Notice, however, that the particle at rest in S and the charge in S' would cancel out and leave room to make the charged object yield the a force on the particle if we were to measure such a thing. If tghe force was not just right then there would be a residual charge to attract or repel.

    I recommend reworking this notion many times over while changing the currents and charges.

    Pete
     
  21. Jun 21, 2005 #20
    This is very helpful. I never thought there was a paradox in SR. I just didn't know how to figure it out. I did the transformations but I get that the magnetic and electric forces in the Lorentz boosted frame are acting in the same direction instead of canceling. BTW, there is a very good book on SR, Introduction Specal Relativity by W.S.C. Williams. Good mix of physical and mathematical understanding.

    PS I think I corrected the mistake. If you look at the problem from the point of view of the observer in S' the charge is moving in the negative Z direction to the transformed E-field is in fact in the negative y-direction.

    this is a truly facinating result and gets to the heart of SR. I had read that electricity and magnetism (and velocity dependent forces) were Einstein's initial inspiration for the theory. Such a great theory.

    I am starting a new thread on the importance of SR and QM...
     
    Last edited: Jun 21, 2005
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