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I have been looking into tensor calculus (from a component perspective) and how it can be used to simplify Maxwell's Equations. When the EM Field tensor, $F$ is introduced, it is usually assumed it is a tensor. How could this be shown (and is it possible without going into the lorentz transformations or is it only a tensor with respect to them)?

In this answer to a similar question, it was said how the derivative of a tensor is a tensor of one higher covariant rank, from which followed that $F$ was a tensor, as it is defined as: $$F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$$ And since $A$ is a tensor. In this case what sort of derivative does $\partial_i$ represent and what is it with respect to?

In addition to this, is the significance that it is a tensor that it becomes true in all reference frames (and therefore accounts for SR) or is there some other significance to it?

Thanks very much for any help!

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  • $\begingroup$ What do you mean by 'significance that it is a tensor that it becomes true in all reference frames' ? $\endgroup$
    – Avantgarde
    Commented Jun 25, 2017 at 18:24
  • $\begingroup$ @Avantgarde I mean that since it is a tensor it is independent of the choice of coordinates used, so someone moving close to the speed of light or in a strong gravitational field would use the same equation as someone standing still relative to the system being measured or not in a gravitational field $\endgroup$
    – aidangallagher4
    Commented Jun 25, 2017 at 18:27
  • $\begingroup$ Okay, yes. This is Lorentz covariance. Also, in the equation you wrote, it should be $F_{\mu \nu}$ and not $F^{\mu \nu}$. Both sides of the equation must transform in the same manner. $\endgroup$
    – Avantgarde
    Commented Jun 25, 2017 at 18:38
  • $\begingroup$ Something that you might consider. $F_{\mu\nu}$ has significance also in relation to the potentials involved in creating it (by diffing the potentials, as in your question you ask what the derivatives represent. What do you get when you diff a potential? ) . So although different observers will measure different E and B fields, what is it that remains invariant to all observers? $\endgroup$
    – user154420
    Commented Jun 25, 2017 at 19:14
  • $\begingroup$ en.m.wikipedia.org/wiki/Electromagnetic_tensor $\endgroup$
    – user154420
    Commented Jun 25, 2017 at 19:22

2 Answers 2

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When the EM Field tensor, F is introduced, it is usually assumed it is a tensor. How could this be shown?

You can define $F$ in terms of $A$, or $A$ in terms of $F$. It's a matter of taste.

If you take $A$ as fundamental, and assume that $A$ is a tensor, then $F$ is automatically a tensor because it's a derivative of a tensor. The index notation is defined so that any set of symbols that obeys the grammar of the notation is guaranteed to be a tensor. (In GR, the derivative has to be a covariant derivative, not a partial derivative. I assume whatever book you're referring to either isn't considering curved spacetime or uses $\partial$ as a notation for the covariant derivative.)

If you take $F$ as fundamental, then I think it's a lot more obvious why it's a tensor. The typical way to define it is that the four-acceleration of a charged particle is given by $a_\mu=(q/m)F_{\mu\nu}v^\nu$, where $v$ is the velocity four-vector. Thus $F$ is a linear operator that takes a four-vector as input and gives a four-vector as output, and that's one way of defining what we mean by a rank-2 tensor.

(and is it possible without going into the lorentz transformations or is it only a tensor with respect to them)?

When we say something is a tensor, we normally mean that it transforms (or, according to taste, its components transform) according to the standard transformation rules under any smooth change of coordinates, not just Lorentz transformations. This applies to $F$. (It is possible to have things like psuedotensors or tensor densities that have slightly different transformation properties, but $F$ isn't one of those.)

In addition to this, is the significance that it is a tensor that it becomes true in all reference frames (and therefore accounts for SR) or is there some other significance to it?

Your wording seems a little off to me, but basically yes, the reason we like to deal with tensors is because any tensorial relation remains valid under any smooth change of coordinates, which includes a Lorentz transformation.

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    $\begingroup$ If the connection is symmetric the expression above (the difference of the two derivatives) is the same for partial derivatives and for covariant derivatives (the terms with the Christofell symbols cancel out). $\endgroup$
    – MBN
    Commented Jun 26, 2017 at 8:45
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To directly answer your question, $A$ is the connection one-form on a $U(1)$-bundle and $d$ is the exterior derivatives, therefore $F=dA$. In components, given the definition and the properties of such exterior derivatives, one can show that only the anti-symmetric part is left alive.

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