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Maxwell tensor's antisymmetry

  1. Jul 8, 2010 #1
    i am reading a book written by malcolm ludvigsen and i have difficulty in understanding the following:

    he introduces the maxwell tensor via

    m[tex]\ddot{x}[/tex] = eF(v)

    where v is the four-velocity and [tex]\ddot{x}[/tex] the four-acceleration of a charged partice.

    he then states that F(a,b) = aF(b) is "clearly" antisymmetric, i.e. F(a,b)=-F(b,a).

    well, i know that it is antisymmetric. but if i wouldn't i then i'd find it quite hard to reach that conclusion having only the given information.

    can someone please tell me how to easily see the obvious antisymmetry by only these information?
  2. jcsd
  3. Jul 8, 2010 #2


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    Hard to say what the author had in mind based on the information you've posted.

    One way to see that it has to be antisymmetric is that it has to have the right number of degrees of freedom. It should have 6 d.f., corresponding to the components of E and B.

    From the equation of motion of the charged particle, you also get unphysical results if, for example, F has only a nonvanishing time-time component and everything else is zero. Then a particle initially at rest would start to change its energy, without changing its three-velocity.

    More generally, you want [itex]v^2=1[/itex] always (with a +--- metric). This can only happen if the derivative of this quantity is zero, which means [itex]<v,\dot{v}>=0[/itex]. To ensure that [itex]v^TFv=0[/itex] for all v, I think F has to be antisymmetric. (I could be wrong about this -- my linear algebra is rusty.)
  4. Jul 9, 2010 #3
    We have F(a,b)+F(b,a)=F(a+b,a+b) by linearity, which equals (a+b) dot F(a+b) by definition. Since F(v) is orthogonal to v for all v because the acceleration is orthogonal to the velocity, this equals zero. So F(a,b)=-F(b,a).

    Last edited: Jul 9, 2010
  5. Jul 9, 2010 #4
    thanks @all

    Aha. That's great. Thank you very much. I'm happy now.
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