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A Charge moving with a constant linear velocity...

  1. Aug 15, 2017 #1
    Charge moving in constant linear velocity does not produce magnetic field...
    If not, please provide an explanation...
     
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  3. Aug 15, 2017 #2

    Orodruin

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    Yes it does.
    Your question is unclear. You labeled this thread "A". Which part of Maxwell's equations do you not understand?
     
  4. Aug 15, 2017 #3

    vanhees71

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    The most simple way to get the electromagnetic field (which of course has non-zero electric and magnetic components) is to use Lorentz invariance. Just solve Maxwell's equations for the charge at rest (leading of course to a Coulomb field) and then do a Lorentz boost. You'll get both electric and magnetic field components! Of course, it's easier to first use the four-potential and then take the derivatives to get the field-strength tensor.
     
  5. Aug 17, 2017 #4
    OK, so if I place a magnetometer probe next to the running belt fo Van De Graff generator I should detect magnetic field, or should I? Also if constant linear velocity of a charge generate magnetic field then it means that absolute motion can be measured and the concept of aether is reality... Still one needs experimental confirmation but I cannot find any...
     
  6. Aug 17, 2017 #5

    vanhees71

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    Of course not! There's no aether. Contrary to the Newtonian spacetime model special relativity (i.e., the Minkowskian spacetime model) precisely makes the description of electromagnetism possible without assuming a preferred reference frame (aka. "restframe of the aether" from the ancient days before Einstein).

    Of course if there's a moving charge in a reference frame, the observer at rest wrt. this frame will observe both electric and magnetic field components. In a frame, where the charge stays at rest, he/she will observe only electric field components. The electromagnetic field is of course invariant, i.e., frame independent (it's an antisymmetric tensor field, the Faraday tensor).
     
  7. Aug 24, 2017 #6
    Does that mean that if charge has magnetic field around it, it (the magnetic field) indicates an absolute motion?
     
  8. Aug 24, 2017 #7
    Just like to point that an abstract quantities like kinetic energy can change wrt to an observer but magnetic field is a physical quantity and does not change due to relative motion of an observer thus correct logic deduction suggests that constant linear velocity of charge particle does not produce magnetic field...
    Also absence of magnetic field around charged particle would indicate an absolute stop motion... Feather more all objects with excess charge located on the Earth
    would generated their own magnetic field due to the orbital motion of around 500 m/s...
     
  9. Aug 24, 2017 #8

    Ibix

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    No. Whether a charge has a magnetic field or not is a frame-dependent question. If it has a magnetic field then it is in motion in that frame. If it does not have a magnetic field it is not in motion in that frame.

    It always has an electromagnetic field. In some frames this may have a magnetic component or be purely electrostatic.
     
  10. Aug 24, 2017 #9

    Ibix

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    This is incorrect, as I wrote above.
     
  11. Aug 24, 2017 #10
    Laws of physics must hold across all frames of motion and cannot be frame dependent...
     
  12. Aug 24, 2017 #11
    It would mean that charged particle in motion produces magnetic field which intensity is observer dependent thus completely hiding magnetic field for moving observer and showing magnetic field for non moving observer... That would be totally illogical...
     
  13. Aug 24, 2017 #12

    Ibix

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    True. Whether or not a charge possesses a magnetic component to its electromagnetic field is not a law of physics. The relevant laws of physics are Maxwell's equations, which have the same form in all frames.
    If you think that then you need to revise your notion of what constitutes "logical". It's trivial to show that it happens by carrying out a Lorentz transform on the electromagnetic field strength tensor.
     
  14. Aug 24, 2017 #13
    Logic also suggests that mathematical models have no effect on physics of reality thus incorrect models will not be supported by experimental evidence...
    In short if you can point to real experiment where charge moving in constant linear velocity produces magnetic field then that will prove that your Lorentz transform correctly represents mathematical model of the phenomena in question...
     
  15. Aug 24, 2017 #14

    Ibix

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    A straight wire carrying a constant current.
     
  16. Aug 24, 2017 #15

    vanhees71

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    Indeed the electromagnetic field,
    $$\boldsymbol{F}=F^{\mu \nu} \boldsymbol{e}_{\mu} \otimes \boldsymbol{e}_{\nu}$$
    is a tensor field and thus invariant under Lorentz transformations. There's no way to infer an absolute coordinate frame, i.e., there's no aether in special relativity. That's how special relativity has been discovered by Lorentz, Fitzgerald, Poincare, and finally Einstein.
     
  17. Aug 24, 2017 #16
    OK, we have the model (still parameter descriptions missing) and now all we need is the experimental evidence... Or is there any at all? Also the above model should be solved for magnetic field not a force...
     
  18. Aug 24, 2017 #17

    Ibix

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  19. Aug 24, 2017 #18
    There is not a single electron moving in constant linear velocity in a wire carrying constant current...
     
  20. Aug 24, 2017 #19

    vanhees71

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    Now let's finally do the calculation, which hasn't occured in this thread. The most simple way is to just do a Lorentz boost of the four-potential.

    So let ##x^{\mu}## inertial coordinates, where the charge is at rest sitting in the origin of the spatial frame. Then, in (1+3) notation we have simply a Coulomb field, which is described in Lorenz gauge by the potentials
    $$A^0(x)=\frac{q}{4 \pi |\vec{x}|}, \quad \vec{A}(x)=0. \qquad (1)$$
    The four-potential in Lorenz gauge transforms as a four-vector, i.e.,
    $$A^{\prime \mu}(x')={\Lambda^{\mu}}_{\rho} A^{\rho}(x).$$
    To make our life easier, instead of doing the combersome transformation we rather write everything in a covariant way. The situation is described in a covariant way by the charge ##q## of the particle (which is a scalar) and the particles' constant velocity, which is described covariantly by the four-velocity ##(u^{\mu})=(\gamma,\gamma \vec{v})##, where ##\gamma=1/\sqrt{1-\vec{v}^2}## (I set ##c=1## for convenience).

    Thus, the building blocks for our field, given in the particles's restframe, where ##(u^{\mu})=(1,0,0,0)##, are only ##x^{\mu}## and ##u^{\mu}##. It's clear that ##A^{\mu} \propto u^{\mu}## and finally we need ##\vec{x}^2##. It's easy to project out the time component from the four vector ##x^{\mu}##:
    $$(x^{\mu}-u^{\mu} u \cdot x)=(0,\vec{x})$$
    and thus
    $$-\vec{x}^2=[x-u(u \cdot x)]^2=x^2-(u \cdot x)^2.$$
    So finally we have
    $$A^{\mu}=\frac{q}{\sqrt{(u \cdot x)^2-x^2}} u^{\mu}.$$
    In the reference frame where the particle moves with three-velocity ##v## we have
    $$u^{\prime \mu}=\gamma (1,\vec{v})$$
    and
    $$(u \cdot x)^2-x^2=\gamma^2 (t'-\vec{v} \cdot \vec{x})^2-t^{\prime 2}+\vec{x}^{\prime 2}.$$
    With this you get the field components in the primed frame by taking the usual differential operators,
    $$\vec{E}'=-\dot{\vec{A}}'-\vec{\nabla} A^{\prime 0}, \quad \vec{B}'=\vec{\nabla}' \times \vec{A}'.$$
     
  21. Aug 24, 2017 #20

    vanhees71

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    Hm, that's a good question. I'm not aware of an experiment which has ever measured the electromagnetic field of a single charged body moving with constant velocity. Of course the measurable macroscopic fields used in everyday electrical engineering all follow Maxwell's equations, and thus we can be pretty sure that everything works also for this special case.
     
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