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Relativistic force and electromagnetic tensor follow up

  1. May 31, 2012 #1

    Wox

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    A while back (thread) you guys helped me understand why [itex]\tilde{F}=m\frac{d\gamma\tilde{v}}{dt}[/itex] (3-vectors) as it follows from [itex]\bar{F}=q\Psi\bar{v}[/itex] (4-vectors) and [itex]\tilde{F}=q(\tilde{E}+\tilde{v}\times \tilde{B})[/itex] (3-vectors). However, I had the impression that one also uses [itex]\tilde{F}=m\frac{d\gamma\tilde{v}}{dt}[/itex] in relativistic mechanics, making abstracting of the nature of the force and therefore not necessarily being electromagnetic as in its derivation. Is that true?

    Secondly, there appears to be a chicken-egg problem here. Where does [itex]\bar{F}=q\Psi\bar{v}[/itex] come from? I assume it follows from the classical Lorentz force, but what are the extra's? [itex]\tilde{F}=m\frac{d\gamma\tilde{v}}{dt}[/itex] can't be part of it because then we do have a chicken-egg problem. I've been trying to find some comprehensive reference that builds this up from classical electrodynamics and the properties of Minkowski space, but without success.
     
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  3. May 31, 2012 #2

    Meir Achuz

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    I am not sure what [tex\Psi[/tex] stands for.
    The concept of 'force' is unnecessary in relativity or in QM, and can lead to confusion sometimes. Even with Newton's F=ma there is a chicken-egg problem--Does it define force or mass?
     
  4. Jun 1, 2012 #3

    Wox

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    Sorry, [itex]\Psi[/itex] is the electromagnetic tensor.
     
  5. Jun 1, 2012 #4

    pervect

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    Basically F=m (d/dt) (gamma * acceleration) is not correct. So you'll have a hard time justifying it. See the discussion in wikki about the rather outdated "transverse" and "longitudinal" masses.

    http://en.wikipedia.org/w/index.php...id=491515263#Transverse_and_longitudinal_mass

    Specifically, if an object is moving in the x direction with velocity v:

    f_x = m gamma^3 a_x
    f_y = m gamma a_y
    f_z = m gamma a_z

    Note that the power of gamma is 3 in the direction of motion, and unity transverse to the direction of motion.

    I'd also argue that the electromagnetic Lagrangian and/or Hamiltonian is a better and more fundamental description of the electromagnetic field than one in terms of forces. If you're not familiar with Lagrangians or Hamiltonians yet I'd suggest reading about them, a fairly standard reference is Goldstein's "Classical Mechanics".
     
  6. Jun 1, 2012 #5
    Check out the first few pages of Chapter 3 in MTW. The quantity [itex]\bar{F}=q\Psi\bar{v}[/itex] is equal to the 4-force divided by γ, and [itex]\tilde{F}=m\frac{d\gamma\tilde{v}}{dt}[/itex] is the 4D version of Newton's second law divided by γ. As reckoned from any secific inertial frame of reference, the γ's cancel. However, you should understand that, prior to the cancelation of the γ's, the equations were originally expressed in terms of the 4 force, the 4 velocity, and the 4 acceleration.

    Chet
     
  7. Jun 1, 2012 #6

    Wox

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    Ok, now I'm confused.

    [itex]\bar{F}[/itex]: four force
    [itex]\bar{v}[/itex]: four velocity
    [itex]\bar{p}[/itex]: four momentum
    [itex]\tau[/itex]: proper time
    [itex]\Psi[/itex]: electromagnetic tensor
    [itex]\tilde{F}[/itex]: three force
    [itex]\tilde{v}[/itex]: three velocity
    [itex]\tilde{E}[/itex]: electric field
    [itex]\tilde{B}[/itex]: magnetic field
    [itex]q[/itex]: charge

    [itex]\bar{F}=\frac{d\bar{p}}{d\tau}=q\Psi\bar{v}[/itex]

    [itex]\Leftrightarrow \gamma \frac{d\bar{p}}{dt}=\gamma\Psi(c,\tilde{v})[/itex]
    [itex]\Leftrightarrow \frac{d\bar{p}}{dt}=\Psi(c,\tilde{v})[/itex]
    [itex]\Leftrightarrow (mc\frac{d\gamma}{dt},m\frac{d\gamma\tilde{v}}{dt})=\Psi(c,\tilde{v})[/itex]
    (take only the spatial part)
    [itex]\Leftrightarrow m\frac{d\gamma\tilde{v}}{dt}=q(\tilde{E}+\tilde{v}\times\tilde{B})[/itex]
    [itex]\Leftrightarrow m\frac{d\gamma\tilde{v}}{dt}=\tilde{F}[/itex]

    the relativistic version of Newton's second law of motion in 3D. Now the question is, what follows from what? From classical electrodynamics we have [itex]\tilde{F}=q(\tilde{E}+\tilde{v}\times\tilde{B})[/itex]. So how do we end up at [itex]\bar{F}=q\Psi\bar{v}[/itex]? Or better, where is our starting point: on the top or on the bottom of the above scheme?
     
    Last edited: Jun 1, 2012
  8. Jun 1, 2012 #7

    Wox

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    In equation 3.2a they use [itex]\frac{dp}{dt}=e(E+v\times B)[/itex] where [itex]p[/itex] the four momentum. How is that possible, the left hand side is in 4D while the right hand side is in 3D.
     
  9. Jun 1, 2012 #8

    George Jones

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    On page 73, note carefully the difference in font for p for 3-dimensional notation and for 4-dimensional notation.
     
  10. Jun 1, 2012 #9

    Meir Achuz

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    If you take the time derivative of [tex]m\gamma{\bf a}[/tex],
    you get [tex]d/dt(m\gamma{\bf a})=m\gamma^3[{\bf a+v\times(v\times a})[/tex],
    which is just what you wrote.
     
    Last edited: Jun 1, 2012
  11. Jun 1, 2012 #10

    Meir Achuz

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    If you just substitute for the EM tensor and the 4-velocity, you get the Lorentz force for the vector part.
     
  12. Jun 1, 2012 #11
    This is all explained very effectively and in detail in the first few pages of Chapter 3 in MTW. Study it carefully. You'll get the idea.

    Chet
     
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