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Questions regarding Lorentz force

  1. Sep 18, 2013 #1
    Why doesn't a stationary charge experience a magnetic force? And why does a moving charge produce a magnetic field? From a previous thread on the topic I understand it is from the relativistic interpretation of its electrostatic field. So then why shouldnt it affect a stationary charge?
     
    Last edited: Sep 18, 2013
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  3. Sep 18, 2013 #2

    BruceW

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    The relativistic interpretation can explain why a moving charge produces a magnetic field. But "why doesn't a stationary charge experience a magnetic force?" is a different question really. This seems to me like you are wondering why the magnetic field and electric field are different in how they interact with matter. There are no magnetic monopoles, this is really the only reason for the difference between how the electric and magnetic fields act.

    p.s. there is also intrinsic spin, so a stationary charge can experience a magnetic force, but that is quantum physics, so I'm guessing it's not really relevant to your question.

    edit: about the 'no magnetic monopoles' thing: we say we have a stationary charge, right. But if there were magnetic monopoles, then we would have to be specific in saying whether our stationary charge was electric or magnetic. So a stationary magnetic charge would produce a magnetic field, but no electric field.
     
    Last edited: Sep 18, 2013
  4. Sep 18, 2013 #3

    jtbell

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    Because we define the magnetic force as the part of the electromagnetic force that depends on the motion of the charge. The part that doesn't depend on the motion of the charge is the electric force.
     
  5. Sep 18, 2013 #4
    The best way to understand the magnetic field as a relativistic E field is with the example of a current-carrying wire. Griffiths explains it best: "A current-carrying wire that is electrically neutral in one inertial system will be charged in another."

    Suppose you put a "stationary" electric test charge near the current carrying wire. The reason why it does not experience a force is because the wire is uncharged in the "stationary" frame. But, as soon as the electric test charge starts moving, the charges in the wire appear Lorentz-contracted to the test charge, and so the wire will appear to have a net charge.
     
    Last edited: Sep 18, 2013
  6. Sep 18, 2013 #5

    vanhees71

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    An elementary way, i.e., without the use of the Hamilton principle of least action and symmetry arguments, to find the correct force law for electric point charges in an electromagnetic field is to recall the covariant form of the equation of motion for a point particle (using Heaviside-Lorentz units with [itex]c=1[/itex] for simplicity)
    [tex]\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=K^{\mu},[/tex]
    where [itex]p^{\mu}[/itex] is the energy-momenum-four vector given by
    [tex]p^{\mu}=m u^{\mu}=m\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.[/tex]
    Further [itex]\tau[/itex] is the proper time of the particle, defined by
    [tex]\mathrm{d} \tau=\sqrt{\mathrm{d} x^{\mu} \mathrm{d} x_{\mu}}.[/tex]
    From this it immediately follows that
    [tex]p_{\mu} p^{\mu}=m^2=\text{const},[/tex]
    which implies
    [tex]p_{\mu} \frac{\mathrm{d} p^{\mu}}{\mathrm{d} m}=p_{\mu} K^{\mu}=0. \qquad (1)[/tex]
    Next we must recall that the force of a particle at rest is proportional to its charge and given by [tex]\vec{F}=q \vec{E},[/tex]
    where [itex]\vec{E}[/itex] are the electric components of the electromagnetic field.

    In relativistically covariant notation, the electromagnetic field is given by the antisymmetric tensor
    [tex]F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu},[/tex]
    where [itex]A^{\mu}[/itex] is the four-potential of the electromagnetic field. Thus, to find the relativistic generalization of the static case, we note that [itex]F_{\mu \nu}[/itex] consists of the electric and magnetic field components and that we can form a vector from it, which is (a) proportional to the field strengths and (b) to the electric charge of the particle and (c) fulfills the on-shell constraint (1), we come to the ansatz
    [tex]K^{\mu}=q {F^{\mu}}_{\nu} u^{\mu}.[/tex]
    This also has the right dimensions, because [itex]u^{\mu}[/itex] is a velocity (i.e., dimensionless in our natural units).

    That we have also chosen the correct sign can be seen by looking at the spatial components and write everything in terms of the three-dimensional notation, i.e., in terms of
    [tex]\vec{E}=-\vec{\nabla} A^0 - \partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{B}.[/tex]
    In index notation we have
    [tex]E^j=\partial^j A^0-\partial^0 A^j=F^{j0}, \quad B^{j}=-\epsilon^{jkl} \partial^{k} A^{l}=-\frac{1}{2} \epsilon^{jkl} F^{kl}.[/tex]
    From the latter equation we find (by inversion of the 3D Hodge duality):
    [tex]F^{jk} = -\epsilon^{jkl} B^l.[/tex]
    This implies that
    [tex]K^{j}=q (F^{j0} u^0 - F^{jk} u^k)=q (u^0 E^j + \epsilon^{jkl}u^k B^l).[/tex]
    In vector notation this reads
    [tex]\vec{K}=q (u^0 \vec{E}+\vec{u} \times \vec{B}).[/tex]
    The spatial part of the equation of motion thus reads
    [tex]\frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau}=q (u^0 \vec{E}+\vec{u} \times \vec{B}).[/tex]
    This becomes more famliar by using the time derivative instead of the proper-time derivative:
    [tex]\frac{\mathrm{d} \vec{p}}{\mathrm{d} t} = \frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau} \frac{\mathrm{d} \tau}{\mathrm{d} t}=\frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau} \frac{1}{u^0}=q(\vec{E}+\vec{v} \times \vec{B}),[/tex]
    where we have used
    [tex]\vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau} \frac{1}{u^0}=\frac{\vec{u}}{u^0}.[/tex]
    This shows that the Lorentz force law is the natural extension of the electromagnetic force of a particle at rest in an electromagnetic field, which is [itex]\vec{F}=q \vec{E}[/itex], which is correct for a particle at rest (by definition of the electric field as force per unit charge).
     
  7. Sep 18, 2013 #6

    BruceW

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    I've also seen the "on-shell constraint (1)" being called a "pure 4-force".
     
  8. Sep 18, 2013 #7

    The electric field around each free electron in a current carrying wire is Lorentz-contracted. Let's call those fields elementary fields.

    For a static charge near the wire most elementary fields are contracted in such way that the charge experiences a decreased electrostatic force from them.

    But the nearest elementary fields are contracted in such way that an increased electrostatic field is felt by the charge.

    The total electrostatic field experienced by the charge is unchanged.




    Now let's take the point of view of a static charge that is observing a moving charge.

    Static charge says: "Because that other charge is moving, it has a magnetic field, which has no effect on me, because I have no magnetic field, because I'm not moving. So there isn't any magnetic force between me and the other charge."

    From this we can see that it's not necessarily always correct to call the effects caused by Lorentz-contraction of electric field a magnetic effect.
     
    Last edited: Sep 18, 2013
  9. Sep 19, 2013 #8
    Great to get so many replies. I'll try to come to each one one by one. First,

    Okay so can all the phenomena such as why a current carrying conductor in an EM field experiences a force, EM induction, etc. be explained using this ultimately? From what I understand now if a test charge moves parallel to the flow of current in its frame it will experience an electrostatic charge due to Lorentz contractions which is the magnetic force in the lab frame(Is this right?) And lastly, why doesn't aren't charges in the current carrying conductor Lorentz contracted for a stationary charge?
     
  10. Sep 19, 2013 #9
    The charges are not contracted. Its the distance between them that is contracted and that affects the chage density. In the lab frame the distance's are not contracted because the wire is at rest. Which must mean that in the moving electrons own reference frame their distance must be a little bigger so that after Lorentz contraction it exactly matches the distance between the stationary positive charges. leading to a neutral wire at the rest reference frame of the wire as expected.
     
  11. Sep 19, 2013 #10
    Yes, thanks dauto, I misspoke there.

    Pretty much. In relativistic electrodynamics, the [itex]\mathbf{E}[/itex] field and [itex]\mathbf{B}[/itex] field are all the result of the same underlying thing. All magnetic effects can be viewed as electrical effects as viewed in a different frame of reference. However, just note that relativstic E&M is not any more correct than Maxwell's equations, since Maxwell's equations are Lorentz invariant (meaning they have relativity built-in): it's just a little more elegant (IMO).
     
    Last edited: Sep 19, 2013
  12. Sep 21, 2013 #11
    Ok thanks.
     
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