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Force on a particle , length contraction

  1. Aug 19, 2013 #1
    Okay because my previous post of similar kind was I guess rather too full of thought experiments and ideas that nobody actually responded to it except for one user then I will try to explain the same thing here only much simpler.
    A few situations.

    so there is this site http://physics.weber.edu/schroeder/mrr/MRRtalk.html
    Now scrolling a little bit down you can see the picture with the charged plates and charges.
    Instead of accelerating the plates like told in the site what would happen if

    A) Two parallel plates , electrostatic field between them , the plates themselves insulated and a flow of charges like ions between them with electrodes at both ends to drive the current.The question is do those flowing charges flowing perpendicular to the E field experience length contraction , also do charges flowing perpendicular to an E field create a magnetic field which then acts as a "back" emf or counter electromotive force and pushes against the flow of charges in this situation which created it in the first place , so does this back EMF then works to make the length contraction even stronger?

    B) if say our point charge is a proton , now if it would experience length contraction that would mean that it's field infront of the particle in the direction of motion would get weaker right?
    does that mean that if two protons would be on a collision course being length contracted they would require less energy to smash together as the e field of them would be weaker in the parts of impact ?
     
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  3. Aug 19, 2013 #2

    Simon Bridge

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    When considering relativity, you must include the frame of the observer.
    i.e. fast objects do not, themselves, "experience length contraction" - when you go fast you don't notice yourself getting shorter in the direction of your motion. You notice everyone else getting shorter.

    You can use special relativity to compare constant relative velocity frames. accelerations use general relativity (though there are ways to do the latter with the former).

    The mean separation of charges in the frame of the plates is less than the separation in the frame of the charges - that what you mean?

    Magnetism depends on the reference frame.
    In the frame of the plates, the flow of charges produces a magnetic field.
    A constant current does not produce a changing magnetic field, therefore there is no back-emf.

    In the frame of the proton, the proton is not length contracted.

    For a single particle, I think you need to specify a model.

    Are you thinking that the electrostatic equipotentials will turn into ellipsoids - squashed in the direction of relative motion?

    The energy needed to smash two protons together in a particle accelerator does not depend on the speed of the protons - just the normal at-rest Coulomb repulsion. But if they are already going quite fast, you do need less energy. But that is not a relativistic effect, just normal energy conservation: you need less energy to accelerate something to 5m/s from 2m/s than from rest.
     
  4. Aug 20, 2013 #3
    okay so if the charged particles go through between the plates of electrostatic field then that means that from the reference point of the plates the field is just about as it used to be on the plates and the particle flow creates a magnetic field , from the particle point of reference they are just as they used to be but for them the electric field from the plates now seems contracted and hence pushing against their flow in the direction in which they are flowing right? If this is right the what about the magnetic field from the flowing charge reference point because from the plates point there is a induced magnetic field due to charged particles cutting E field lines along their way but what do the particles that flow see themselves?

    Or as you said about the separation of charges I guess hence the particles are moving not the plates it should then be that from the plate reference point their charge is as usual but from the particle that are moving point the charges on the plates seem more compressed together right?
     
  5. Aug 20, 2013 #4

    Simon Bridge

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    In the reference frame of the particles, the plates are length contracted. That's right.
    The particles are stationary in that reference frame, the plates are moving.
    Observers stationary wrt the particles do not see a magnetic field from the particles - this is expected since the particles are not moving.

    The charge density in the beam will be less than that in the reference frame of the plates.

    The charges in the plates that give rise to the electric field, however, are moving - so there will be an associated magnetic field.

    Unfortunately you can't do relativity by rule of thumb.
    A consistent EM transformation is handled by the Faraday tensor.
     
  6. Aug 21, 2013 #5
    Okay so from different perspective it looks like either the plates are contracted or the particles.

    Now this may not be relativity anymore but one more thing interests me ,
    when those particles flow perpendicular to the plates and to the E field they do experience repulsion since their moving perpendicular to the E field and that repulsion also is the reason why they create a magnetic field , as if those particles would flow parallel to the E field there wouldn't be magnetic field ? I hope I'm gettin this straight.
     
  7. Aug 21, 2013 #6

    Simon Bridge

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    It is unhelpful to use "looks like" in relativity - it tend to trick people into thinking it is some kind of illusion - as if there is something "really" happening and there is someone who knows what that is.

    You could set it up a different way and have have a row of charges stuck in an insulator, and some charged plates zip by. Alice, stationary wrt the charges, measures the linear charge density as s and the the length of the plates as x. Bob, stationary wrt the plates measures the linear charge density as d and the plate length as y. When Alice and Bob compare notes, they discover that their measurements differ, and that y/x = d/s = γ, but they cannot tell who is seeing the real picture.

    Where did you get this idea from?

    A current makes a magnetic field all by itself, without passing through an electric field.

    Charges want to move along the electric field lines like masses want to move along gravitational field lines. In a wire, there is an electric field along the wire. The current goes along the wire. There is still a magnetic field. It sounds to me like you are trying to use a qualitative description of two different situations as if they worked together. I think I've warned you about this before.

    http://ocw.mit.edu/courses/nuclear-...tic-interactions-fall-2005/readings/chap2.pdf
     
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