Understanding the Antisymmetry of the Maxwell Tensor

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Discussion Overview

The discussion centers on the antisymmetry of the Maxwell tensor as introduced in the context of the motion of charged particles. Participants explore the implications of the tensor's properties and seek clarity on how to derive its antisymmetry from the given equations and definitions.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in understanding the antisymmetry of the Maxwell tensor based solely on the information provided in the book by Malcolm Ludvigsen.
  • Another participant suggests that the antisymmetry can be inferred from the requirement of having the correct number of degrees of freedom, specifically 6, which correspond to the components of the electric and magnetic fields.
  • A further argument is made that unphysical results would arise if the tensor had only a nonvanishing time-time component, implying that the tensor must be antisymmetric to maintain physical consistency.
  • A participant presents a linearity argument, stating that the relationship F(a,b) + F(b,a) = F(a+b,a+b) leads to the conclusion that F(a,b) = -F(b,a) due to orthogonality conditions.
  • Another participant acknowledges the clarity provided by the linearity argument and expresses satisfaction with the explanation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a single method to demonstrate the antisymmetry of the Maxwell tensor, with multiple viewpoints and reasoning presented. Some participants agree on the necessity of antisymmetry based on physical principles, while others focus on mathematical arguments.

Contextual Notes

There are unresolved assumptions regarding the definitions and properties of the Maxwell tensor, as well as the implications of the equations of motion for charged particles. The discussion reflects varying levels of familiarity with the underlying mathematics and physics.

Who May Find This Useful

This discussion may be of interest to those studying electromagnetism, tensor calculus, or the dynamics of charged particles in physics, particularly in relation to the properties of the Maxwell tensor.

M. Kohlhaas
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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?
 
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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.)
 
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).

-Matt
 
Last edited:
thanks @all

dodelson said:
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).
Aha. That's great. Thank you very much. I'm happy now.
 

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