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Magnetism: which reference frame?

  1. Mar 30, 2007 #1
    A friend and I have been discussing what we think is an error in our textbook (either that or a misunderstanding on our part). The book gives an example in which two protons are initially traveling parallel to one another in the same direction with equal velocity.

    We think that since the protons are at rest with respect to one another, initially they will feel no magnetic force--only electrostatic repulsion. Then, once the protons begin to move away from one another, one proton will generate a magnetic field because of its velocity with respect to the other--but neither proton will feel a magnetic force because the component of the magnetic fields are zero on the axis along which the protons have velocity.

    However, the book says that the protons DO feel a magnetic force because they have a velocity (this velocity is with respect to some seemingly arbitrary coordinate system). How can this be true? If it were, different observers in different inertial reference frames would see different magnetic fields acting on these protons; for example, a person traveling with the protons would see no magnetic field due to their initial velocity and would predict that the protons will only be influenced by electrostatic forces, whereas a person moving in the opposite direction of the protons would predict that, because the protons have a velocity, they will exert magnetic forces in opposition to the electrostatic--this leads to different proton trajectories depending on which reference frame you're in--clearly not possible!
     
  2. jcsd
  3. Mar 30, 2007 #2
    The basic thing with magnetism is that it's a coordinate dependent interpretation of a real physical effect. In one frame the acceleration may be due to magnetic forces; in another frame the forces may be electrostatic, but the acceleration/result will be the same.
     
  4. Mar 30, 2007 #3

    mjsd

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    Homework Helper

    these statements are very unclear... firstly, you said in the proton rest frame, nothing is moving with respect to each other.. then later the protons begin to move away from each other....In which frame are you talking about here?

    remember Special Relativity and electro-magnetism go hand in hand.
     
  5. Mar 30, 2007 #4
    When we talk about magnetism, do we really take movement of a charge particle depending on the observer or do we take it form the point of absolute spacetime as our reference point? If it were the latter then it would make sense what you book is trying to present.
     
  6. Mar 30, 2007 #5
  7. Mar 30, 2007 #6
    In the reference frame moving with the protons, sorry if that was unclear.

    I think now I know why there appeared to be a discrepancy; maybe someone can confirm it. I think I was neglecting the fact that the moving observer would experience relativistic effects because of his velocity. Bear with me, I haven't actually studied relativity yet, but I think I sort of have a way to demonstrate this.

    The stationary observer sees the protons traveling initially parallel to one another in the +x direction. According to the stationary observer, the protons will generate a magnetic field of magnitude
    [tex]
    B = \frac{\mu_0 q v}{4 \pi r^2}
    [/tex]
    at the position of the other proton, where r is the distance between the protons.
    The electrostatic force and magnetic force on each proton act in opposing directions; therefore, the average acceleration of one proton will be
    [tex]
    a = (F_E - F_B)/m
    [/tex]
    [tex]
    = ( \frac{q^2}{4 \pi \epsilon_0 r^2} - \frac{\mu_0 q v B}{4 \pi r^2} ) / m
    [/tex]
    [tex]
    = (\frac{q^2}{4 \pi \epsilon_0 r^2} - \frac{\mu_0 q^2 v^2}{4 \pi r^2}) / m
    [/tex]
    [tex]
    = \frac{q^2}{m 4 \pi r^2} \bigg( 1/\epsilon_0 - \mu_0 v^2 \bigg)
    [/tex]
    [tex]
    = \frac{q^2}{m 4 \pi \epsilon_0 r^2} \bigg( (1 - \mu_0 \epsilon_0 v^2) / \epsilon_0 \bigg)
    [/tex]
    [tex]
    = \frac{q^2}{m 4 \pi \epsilon_0 r^2} \bigg( 1 - v^2/c^2 \bigg)
    [/tex]
    The distance the protons move apart (this is in the y direction) is
    [tex]
    \Delta y = \frac{1}{2}a t^2 = \frac{q^2}{8 \pi \epsilon_0 r^2 m} \bigg( 1 - v^2/c^2 \bigg) t^2
    [/tex]
    Then if we know that the protons must move the same distance apart in both reference frames (but perhaps in different times, as time differs between the reference frames-- _moving is time or distance as observed by the moving observer, _stationary is that as observed by the stationary observer)
    [tex]
    \frac{\Delta y_{moving} }{\Delta y_{stationary} } = 1 = 1 \bigg( \frac{1 - 0}{1 - v^2/c^2} \bigg) \frac{t_{moving}^2}{t_{stationary}^2}
    [/tex]
    and
    [tex]
    t_{moving} = t_{stationary} \sqrt{1 - v^2/c^2}
    [/tex]
    which is the equation for determining time dilation.

    Here is a question: are time dilation and length contraction actually two different ways to represent the same thing? In other words, you might measure a scaling of time between reference frames if you assume that distance remains constant between frames, whereas you might measure a length contraction if you assume time remains constant between frames?

    Forgive my ignorance; I enjoy working things like this out on my own!
     
    Last edited: Mar 31, 2007
  8. Mar 30, 2007 #7
    Answer is no: time dilation and length contraction must both occur, simultaneously, in order to avoid contradictions. However, magnetism and electr..icity are two different ways to represent the same thing.
     
    Last edited: Mar 31, 2007
  9. Mar 31, 2007 #8

    mjsd

    User Avatar
    Homework Helper

    relativistic effects come in when velocity is very high otherwise newtonian mechanics is good. ok back to electro-magnetism. The reason I say SR and EM go hand in hand not because the "velocity" comes into your situation, but rather the choice of reference frames. relativity is simply a statement that the coordinate systems that you choose is unimportant. what is important is that once you have picked a coordinate chart you must stay with it until the end of a particular calculations, you shouldn't change to something else mid-calculations unless you carefully transform everything you have worked out so far in the new system before continuing. (ie. don't mix results from different coordinate system). I have got a feeling that your current confusion came from "mixing" the perspective of different frames.

    by the way electric or magnetic fields are vectors and they have a direction (you probably know this). ok, back to your situation. you have a moving proton generating a B-field (lab frame), and there is another proton moving
    so it will feel a magnetic force as long as
    [tex]\vec F = \frac{\mu_0 q q'}{4\pi r^2} \vec v \times (\vec v' \times
    \hat{\vec r}) \neq 0[/tex]

    anyway, E-field effect would be much stronger than B-field effect unless the charges are moving very rapidly (ie. close to speed of light), of course however, in situation when E-field effect is cancelled, B-field can come in on its own.

    final note: magnetic force can be regarded as generated by a transformation of reference frame applied to an electric force.
     
  10. Mar 31, 2007 #9
    Yes, I think that was exactly the case...I was trying to apply the same time/length across reference frames when these would actually differ as a result of special relativity.

    Thank you for your explanations!
     
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