Relativity fails with Magnetic Force

Discussion in 'Physics & Math' started by martillo, May 24, 2009.

  1. martillo Registered Senior Member

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    Last edited: May 24, 2009
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  3. andbna Registered Senior Member

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    [SNIP]


    Edit 3: So, waking up a bit... I have completely redone my post.

    Remember that other force that effects charges? The electrostatic? Yah; you need to take that into account.

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    Let's simplify the problem: 2 electrons moving parallel; or at rest, depending on the frame.

    In the rest frame, we would expect to use Coulomb's law to find that they repulse each other.
    In the moving frame, we would expect to see the Lorentz force pull them closer to each other... but we would also see the effects of time dilation.

    So, just how strong, exactly, is the Lorentz force, when compared to the electrostatic force, at classical speeds? What about relativistic speeds? The speed of light?

    The answer in words: Really fricken weak at classical speeds; so weak, infarct, that time dilation becomes a factor if it is taken into consideration.
    At relativistic speeds: not so weak... but it still doesn't overpower the electrostatic force... and time dilation is now easily seen as a factor.
    At the speed of light? The electrostatic and Lorentz forces are equal! Who would have thought?

    What does this mean? It means that when the website treated the particles as moving at classical speeds; and thus treated Time dilation as negligible, it should have also treated the Lorentz force as negligible, when compared to the massive electrostatic repulsion of the electrons.
    There-in lies the err.

    Of course, what would we have if not for some calculations?

    To simplify, we shall consider the electron stream as simply consisting of an individual electron (A) and the particle, electron B.

    So, I shall do 3 calculations: 1 the electrostatic repulsion, as per Coulomb's law; valid in all reference frames.

    2. The Lorentz force at a 'classical' reference frame, of 1m/s.

    3. The Lorentz force when v=c.

    1:
    F= (9*10^9)*q1*q2/(*r*r)

    Where 9*10^9 is coulomb's constant (rounded to 1 digit.)
    r is the distance between the electrons, and Q1 and Q2 are equal; the elementary charge.

    So picking a reasonable r, say 1 m/s (easier to calculate) we have:

    F = (9*10^9)*q^2
    F = (9*10^9)*(1.6*10^-19)^2
    F = 2.3*10^-28

    A fairly small force... but the electrons are pretty far away; and are also pretty tiny (smaller forces still accelerate them a good bit.)

    Anyway, we are interested in comparing that force, to the Lorentz force:

    2.
    F = q*v*B
    Magnetic field due to a moving point charge, B = (u/4*pi)*q*v/(r*r) (again, removing the cross product because we have our directions defined, with B being perpendicular to v)

    Now, v=v and q=q since our charges are each 1 electron, and moving at the velocity. u is the permeability of free space.

    Thus: F = (u/4*pi)*q*q*v*v/(r*r)
    Velocity is 1m/s, r is 1m again, so:

    F = (4*10^-7/(4*pi))*q^2
    F = (4*10^-7/(4*pi))*(1.6*10^-19)^2

    F= 8.1*10^-46

    WAY smaller than the Lorentz force.
    Adjusting the units to give a better sense of scale:
    thats:
    2.3*10^0 compared to 8.1*10^-18
    Or, a car's length compared to something smaller than a proton!!!
    In other words: at classical speeds; the Lorentz force is extremely negligible.

    3.
    Rather than plug in numbers, I will mathematicaly show that the Lorentz force equals the electrostatic force. (Just computing the rough numbers will yield the two to be on the same order of magnitude, but let's have perfect accuracy.)

    My assumption:
    F1=F2, when v=c.

    We shall see if that is true

    First, let's put Coulomb's constant in terms of the permeability of free space:
    F1= (9*10^9)*q1*q2/(*r*r)
    F1= (u*c^2 / (4*pi))*q1*q2/(*r*r)
    Notice that c^2 ? Some alarm bells should be going off now

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    F2 = (u/4*pi)*q*q*v*v/(r*r)

    for both, v=c, r=1. (and all q's are equal)

    F1= (u*c^2 / (4*pi))*q1*q2/(*r*r)
    F1= (u*c^2 / (4*pi))*q^2
    rearrange that a bit and we have:
    F1= (u/(4*pi))*q^2*c^2

    F2 = (u/4*pi)*q*q*v*v/(r*r)
    F2 = (u/4*pi)*q^2*c^2
    Notice any similarities?

    F1=(u/4*pi)*q^2*c^2 = F2=(u/4*pi)*q^2*c^2

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    Now, we could factor the Lorentz factor, into the Lorentz force... but I think by now it's pretty clear that relativity is not inconsistent.
    And on that note, special relativity was pretty much a result of electromagnetics: the whole Aether shebang came from the problems that the permeability/permittivity of free space posed. Those two constants governed the speed of light, and hence, if there was either an absolute reference frame (the aether,) or that all observers should measure the speed of light to be the same (as well as the rest of electromagnetics.) in all reference frames

    Guess which one was correct?

    -Andrew

    And it looks like I was beaten too the answer

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    Last edited: May 25, 2009
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  5. przyk squishy Valued Senior Member

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    martillo,

    As far as your "theoretical inconsistency" is concerned, the velocity appearing in the Lorentz force equation:
    \(\bar{F} \;=\; q \, \bigl[ \bar{E} \,+\, \bar{v} \, \times \, \bar{B} \bigr]\)​
    is simply the charge's (frame-dependent) velocity. This isn't a problem because \(\bar{F}\), \(\bar{E}\), and \(\bar{B}\) are also frame-dependent quantities. Specifically, the way the electric and magnetic fields transform between reference frames is well known (the "electromagnetic field tensor" transforms like any other relativistic tensor) and actually have practical applications: for example, the easiest way of calculating the electric and magnetic fields around a charge in rectilinear motion is probably to Lorentz-transform the electric Coulomb field around the charge in its rest frame.

    The justification for this is pretty trivial: the Lorentz force equation contains the operational definitions of the electric and magnetic fields (the electric and magnetic fields are defined by their action on test charges). Theoretically there isn't actually enough physics in the Lorentz force alone for it to contradict anything: it's only when you add in Maxwell's equations (which tell you what \(\bar{E}\) and \(\bar{B}\) charges and currents will generate) that you have a closed theory that could potentially contradict STR (which of course it doesn't: historically, electromagnetism was found to be Lorentz covariant ten or twenty years before anyone had ever heard of Einstein).

    To specifically address your electron beam problem, this statement:
    is true and exactly what you should expect. To calculate the forces, let's say that the electron beam is also at rest in case 1 (to simplify things) and has a linear charge density \(\lambda\) (so the current is \(I = \lambda v\)). Then the electric field around the beam is \(E \,=\, \frac{1}{2 \pi r \varepsilon_{0}} \lambda\), there's no magnetic field, and the (repulsive) force is:

    \(F \;=\; \frac{q}{2 \pi r} \, \frac{\lambda}{\epsilon_{0}\)​

    In case 2, the magnetic field is given by \(B^{\prime} \,=\, \frac{\mu_{0}}{2 \pi r} I^{\prime} = \frac{\mu_{0}}{2 \pi r} v \lambda^{\prime}\), which will attract the electron. The force is:

    \( \begin{align} F^{\prime} \;&=\; q \, \bigl[ E^{\prime} \,-\, v \, B^{\prime} \bigr] \\ \\ \\ \\ \;&=\; \frac{q}{2 \pi r} \, \biggl[ \frac{\lambda^{\prime}}{\varepsilon_{0}} \,-\, \mu_{0} \, v^{2} \lambda^{\prime} \biggr] \\ \\ \\ \\ \;&=\; \frac{q}{2 \pi r \varepsilon_{0}} \, \bigl[ 1 \,-\, \varepsilon_{0} \mu_{0} \, v^{2} \bigr] \, \lambda^{\prime} \end{align} \)​

    Due to length contraction, the electrons in the electron beam are closer together and the charge density is higher: \(\lambda^{\prime} = \gamma \lambda\). Also, plug in \(\varepsilon_{0} \mu_{0} = \frac{1}{c^{2}}\) and you find:
    \( \begin{align} F^{\prime} \;&=\; \frac{q}{2 \pi r \varepsilon_{0}} \, \Bigl[ 1 \,-\, \frac{v^{2}}{c^{2}} \Bigr] \, \gamma \lambda \\ \\ \\ \\ \;&=\; \frac{q}{2 \pi r \varepsilon_{0}} \, \frac{\lambda}{\gamma} \end{align} \)​

    So the relation between the forces in cases 1 and 2 is \(F^{\prime} = \frac{1}{\gamma} F\). In case 2, the electron will start to accelerate more slowly away from the wire, which is exactly what you'd expect on the basis of relativistic time dilation.
     
    Last edited: May 25, 2009
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  7. rpenner Fully Wired Staff Member

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  8. tsmid Registered Senior Member

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    Well, it is exactly the problem here, although it has nothing to do with Relativity but is due to the quantities v and B in F=q*vxB not being uniquely defined in Maxwell's equations. If you take the usual (frame dependent) definition of the current j (which determines B over the Biot-Savart law) as being given by the product of charge density times velocity, then this is evidently inconsistent as the resultant magnetic field would become dependent on the state of motion of the observer which would make the Lorentz-force a non-linear (i.e. quadratic) function of the velocity, in disagreement with experiments.
    So the current j determining B must be defined frame-independently, and this can only be achieved by taking the relative velocity of different types of charge carriers (i.e. electrons and ions). So it is only the relative motion of electrons and ions in a wire (i.e. the motion in their mutual fields) that produces the magnetic field. Only this defines B frame independently (and thus unambiguously). And the velocity v in F=q*vxB would then have to be the rest frame of the wire (as given by its center of mass).

    If the motion would lead to a higher charge density throughout the whole wire, then this would violate charge conservation (see my page Magnetic Fields and Lorentz Force for more).

    Thomas
     
    Last edited: May 25, 2009
  9. martillo Registered Senior Member

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    przyk:
    But then you would have to admit that just changing the frame of observation (the observer) you would affect the movement of the electron!

    May be I should rewrite the problem in other way...
     
    Last edited: May 25, 2009
  10. martillo Registered Senior Member

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    tsmid:
    The problem if you do that is that in the problem the relative velocity between the electron and the beam is the same and there would be no Magnetic force on the electron in any case ( the isolated elctron always have velocity = 0 relative to the wire) but this would imply that a wire with current would never produce a Magnetic Force on an electron!

    The problem is solved if it is stated that the velocity to be used is a velocity relative to a frame at rest in the Universe.
    If so even if you change the frame you must calculate and use that velocity only wich is unique. When working in other frames you must make the necessary changes of coordinates to ensure that the velocity in the Lorentz Formula always be that unique "absolute velocity".
     
    Last edited: May 25, 2009
  11. martillo Registered Senior Member

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    andbna:
    I don't know if I undertood your point. I take:
    The problem is that then always a total null force woulod exist and a current wire would never affect a moving electron around wich I think is not the case.
     
  12. tsmid Registered Senior Member

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    As indicated above already, if there is a current in the wire (i.e. if there is a net motion of the electrons with regard to the ions in the wire) then this produces a magnetic field B around the wire which is independent of the reference frame (as the current in the wire is independent of the reference frame). The magnetic force on a test charge is then given by F=q*vxB where v is the velocity of the test charge relative to the wire (or rather to its center of mass).

    Thomas
     
    Last edited: May 25, 2009
  13. martillo Registered Senior Member

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    tsmid:
    Again: In the first case the electron has v=0 relative to the wire and so zero Magnetic Force. In the problem both the wire and the electron increment their velocity such that their relative velocity is always the same so you would have that always would be no Magnetic Force in the isolated electron which I think is not the case...
     
  14. martillo Registered Senior Member

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    przyk:
    You are right in the point that actually the relativistic prediction do agree with the classical prediction. I was wrong in that point but the inconsistence rises then from the other point of view: Relativity states there are no privileged frames of reference and so in the presented problem there's a change in the real movement of the electron just because a change in the frame of observation (the observer)!

    I must rewrite the problem.

    Once again thanks very much for your logical, rational and mathematical refutations that shows me the real wrong points in my propositions...
    But then what do you think of my point now: Relativity would have to state that a change in the observer would change the movement of the electron???
     
  15. tsmid Registered Senior Member

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    In F=q*vxB, v is the relative velocity between the test charge and the wire, so it is by definition independent of the reference frame (and B is independent of the reference frame as well as it depends on the difference of the velocities of the electrons and ions in the wire).

    Thomas
     
  16. martillo Registered Senior Member

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    So you say the Magnetic Force is walways zero on the isolated electron???
     
  17. tsmid Registered Senior Member

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    If the relative velocity between the electron and the wire is zero, then the magnetic force will be zero in all inertial reference frames (all forces are by definition identical in inertial reference frames).

    Thomas
     
  18. przyk squishy Valued Senior Member

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    You're confusing two parameters here: the velocity of the current and the velocity of the charge the magnetic field is acting upon are two different things. The Lorentz force depends linearly on each and quadratically if you vary both.

    This is an unneccessary ad-hoc patch that doesn't generalize to anything beyond the special case you happen to be worrying about. For example, how does it apply to martillo's scenario, where there's no wire at all?

    Note that, in anything but a wire's rest frame, there's a net flow of positive charge (slight excess of protons over bound electrons) which will contribute to the magnetic field around the wire in that frame. If the unbound electrons in a neutral wire (those contributing to the current \(I_{\text{e}} = \lambda v_{\text{e}}\)) have a charge density \(\lambda\), this leaves a charge density \(-\lambda\) of protons and unbound electrons which will constitute a current \(I_{0} = - \lambda v_{0}\), so the net magnetic field can be expressed as:
    \( \begin{align} B \;&=\; \frac{\mu_{0}}{2 \pi r} \bigl( I_{\text{e}} \,+\, I_{0} \bigr) \\ \\ \\ \\ \;&=\; \frac{\mu_{0}}{2 \pi r} \bigl( v_{\text{e}} \,-\, v_{0} \bigr) \lambda \end{align} \)​

    I don't know if this is what you meant, but the point is there's no need to make special assumptions that don't generalize to the rest of electrodynamics.

    No it wouldn't. For a finite wire, the entire wire would be length contracted and the total charge remains the same. Note that this wouldn't actually make much difference to the calculation: if the wire is long enough, the extremities would have a negligible effect on the fields near the middle of the wire.

    As far as relativity and charge conservation go, what's usually meant by charge conservation is that the charge density and current obey the local continuity equation everywhere in space and time:
    \(\frac{\partial \rho}{\partial t} \,-\, \bar{\nabla} \cdot \bar{j} \;=\; 0\)​

    This can be expressed as \(\part_{\mu} j^{\mu} \,=\, 0\), which is manifestly covariant. In other words, if it holds in one frame, it holds if you Lorentz-transform the four-divergence and the four-current, and therefore automatically holds in every frame.
     
    Last edited: May 25, 2009
  19. przyk squishy Valued Senior Member

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    He calculated that the forces exactly cancel at the speed of light.

    No they're not. In relativity the four-force, which contains the 3-force, has the transformation law of a four-vector. You could deduce how the force transforms between reference frames based on how accelerations transform and \(\bar{F} \,=\, m \bar{a}\)

    As a general rule, all vector quantities are frame-dependent in relativity. Strictly-speaking, they're frame dependent even if you don't consider relativity, since they're also variant under rotations.
     
  20. przyk squishy Valued Senior Member

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    It already does: when you switch between reference frames, trajectories transform under the Lorentz transformation, whether you're dealing with electrons or not. The change in the movement of the electron is exactly the change you expect due to time dilation.
     
  21. martillo Registered Senior Member

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    przyk,
    Sorry przyk but I cannot "swallow" that. Do you realize you are admiting that a change in the selected frame of observation of a "system" would alter its dynamics!
    No, no way...
    This is absurd.
     
  22. martillo Registered Senior Member

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    So the isolated electron would not be affected no matter the current that could pass through it??? No way. Just review some basic Physics books about the Magnetic Field and Force of current carrying wire where the magnetic does not depend on the velocity of any wire just the linear current.

    If not convinced in principle the experiment could be done to see what really happens.
     
    Last edited: May 25, 2009
  23. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Maxwells equations are manifestly Lorentz covariant---in fact, this is one of the ways that Einstein discovered special relativity. So any confusion about some experiment which puts the two at odds is a pretty outrageous claim.
     
    Last edited: May 25, 2009

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