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Is any matrix a tensor in special relativity?

My question is inspired by the definition of the electromagnetic field tensor in Carroll's Spacetime and Geometry book. In equation (1.69), he defines a matrix which he says is the (0,2) electromagnetic field tensor.

In section 1.8, he states the Maxwell equations in tensor notation, and additionally states that in this form they manifestly transform as tensors. I assume that this is due to the tensorial nature of the electromagnetic field tensor; however, I don't understand why it is a tensor.

Would we have to prove that the electromagnetic field tensor transforms as a tensor by brute calculation? Alternatively, is there some clever and more obvious shorthand that you could use to come to the same realization?

physics
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Dargscisyhp
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    If you like this question you may also enjoy reading this and this Phys.SE posts. – Qmechanic Sep 22 '14 at 19:42
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    All together now: "When it transforms like a tensor." Now, hopefully some of our relativists will explain what "transforms like a tensor", means and we'll be squared away. – dmckee --- ex-moderator kitten Sep 22 '14 at 19:42
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    Could you include Carroll's definition of the tensor here? We need to know where we are starting from. – ACuriousMind Sep 22 '14 at 19:44
  • @ACuriousMind: It doesn't matter what tensor we're talking about. The answer is what dmckee said. –  Sep 22 '14 at 19:57
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    @BenCrowell: Yes, dmckee is certainly correct. But to show of a given object that it is a tensor, we need its definition to examine its transform. – ACuriousMind Sep 22 '14 at 20:02
  • @ACuriousMind: You're right. The title of the question misled me into thinking that the questioner didn't even know what criterion to apply at all. Actually the final sentence makes it clear that s/he does know what the criterion is. I'll edit the title. –  Sep 22 '14 at 21:40

3 Answers3

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Strictly speaking, a matrix is not a tensor, it is a representation of a tensor in a particular basis. You can't tell whether a given matrix is a tensor using only its components. You would have to know how it transforms to different reference frames.

For the electromagnetic field tensor, for example, you could write the equations for some physical configuration of electromagnetic fields, and then write the equations that describe the same physical configuration in a different reference frame, and show that applying the corresponding Lorentz transformation twice converts from one to the other.

$$F^{\mu\nu}_\text{frame 2} = [\Lambda(1,2)]_\alpha^\mu [\Lambda(1,2)]_\beta^\nu F^{\alpha\beta}_\text{frame 1}$$

where I've used $\Lambda(1,2)$ to denote the Lorentz transformation that transforms from frame 1 to frame 2. Doing this calculation out in full gets somewhat tedious, which is why many textbooks don't go through it in full detail, they just make it seem plausible. (But at least Einstein had to do it to show that the theory worked this way.)

Anyway, the point is that the tensorial nature of a quantity is really a consequence of the transformation law, not the representation (the matrix).

David Z
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  • You have explained what it means that $F$ is a tensor (in short, $F'=\Lambda^tF\Lambda$) but you do not address the title of the question: "How to tell that the electromagnetic field tensor transforms as a tensor?" – Filippo Sep 09 '22 at 11:52
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The answer to your question depends quite a bit on what you consider to be fundamental and what you consider to be derived.

A modern, manifestly relativistic treatment of E&M would define the electromagnetic field tensor as

$$ F^a = q\mathcal{F}^a_bv^b, $$

where $\mathcal{F}$ is the field tensor and $F$ is the four-force acting on a particle. By this definition, it's manifestly clear that $\mathcal{F}$ is a tensor; it's a linear operator that takes one vector and makes another. In this approach, the work you have left to do is to show how the components of $\mathcal{F}$ relate to the usual $\mathbf{E}$ and $\mathbf{B}$ three-vectors. For a treatment in this style, see my SR book, http://www.lightandmatter.com/sr/ , section 10.3.

A more lowbrow approach would be to define $\mathbf{E}$ and $\mathbf{B}$ as usual in a freshman physics class, then find their transformation laws. (This is done, e.g., in the classic 1965 book by Purcell, Electricity and Magnetism. The first edition can be found for free online because it was developed under an NSF grant. There is also a nice modern edition from Cambridge University Press.) Having done this, you then have to put them together as the components of $\mathcal{F}$ and show that these transformation laws make $\mathcal{F}$ transform as a tensor.

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Would we have to prove that the electromagnetic field tensor transforms as a tensor by brute calculation?

Not if you accept that the derivative of a tensor is a tensor with one higher covariant rank.

In SR, the scalar electric and vector magnetic potential are components of a four-vector (a rank 1 tensor), the electromagnetic four-potential $A_{\mu}$.

Then, it is the case that the following is a rank 2 covariant anti-symmetric tensor, the electromagnetic tensor:

$$F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$$

Alternatively, is there some clever and more obvious shorthand that you could use to come to the same realization?

In component free notation, we have the tensor equation

$$\mathbf F \stackrel{\mathrm{def}}{=} \mathbf{dA}$$

which defines the electromagnetic tensor as the exterior derivative of the electromagnetic four-potential and leaves no possible doubt that $\mathbf F$ is a tensor.

Alfred Centauri
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