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Can someone explain what the general concept is behind the EM tensor, as written in component form with the four-gradient? $$F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$$ Is that essentially saying the tensor is defined as the change along vector $\mu$ of $A$, as measured along $\nu$, minus the change along vector $\nu$ of $A$, as measured along $\mu$? I don't quite understand the indices of the RHS of the equation and why they are "opposite" ($\partial^{\mu}A^{\nu}$) instead of "matching" ($\partial^{\mu}A^{\mu}$). I understand the EM four-potential vector $A^{\mu}$ pretty easily on its own, but I don't quite understand what physical relevance the second index represents, especially in relation to the first, except that they both become the indices of the EM tensor.

Forgive me, I am self-taught and I haven't been able to find much of an answer on my own.

Qmechanic
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Garrett Miller
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  • Do you know about the “curl” of a vector field? https://en.m.wikipedia.org/wiki/Curl_(mathematics) – robphy Jun 23 '23 at 00:30
  • I don't quite understand the indices of the RHS of the equation. Free (i.e., non-contracted) indices must be the same on both sides of an equation. – Ghoster Jun 23 '23 at 00:31
  • Do you understand that that equation stands for multiple component equations such $F^{12}=\partial^1A^2-\partial^2A^1$ when $\mu$ is 1 and $\nu$ is 2? – Ghoster Jun 23 '23 at 00:34
  • Oh, wow, yeah. I was thinking about it so physically that I forgot about the literal definition of components. So then, per your example Ghoster, $F^{12}$ is the change along the x-axis of the y-component of $A$, minus the change along the y-axis of the x-component of $A$? – Garrett Miller Jun 23 '23 at 00:41
  • Robphy, I understand curl pretty well, but not fully. I know that the bottom-right portion of the EM tensor is very similar to the cross product of vectors (the magnetic field specifically in this case), I just didn't really understand the significance of the component form. – Garrett Miller Jun 23 '23 at 00:43
  • @GarrettMiller Yes. – Ghoster Jun 23 '23 at 00:44
  • The free indices on the left and on the right must be the same in order to ensure that both sides transform in the same way under Lorentz transformations, making the equality hold in every inertial reference frame. Such an equation is “manifestly covariant”. – Ghoster Jun 23 '23 at 00:50
  • Ah! Ok, that makes a lot of sense too. When you say the same order, just the first term? Or how does the second term, being reversed, fit that concept? Or am I overthinking it? – Garrett Miller Jun 23 '23 at 00:55
  • Not Strictly Related : Vector product in a 4-dimensional Minkowski spacetime. In the end of my answer you could see how the Electromagnetic Field Tensor is produced formally as the 4-curl of the electromagnetic potential 4-vector , equations (28) to (31). – Frobenius Jun 23 '23 at 02:19
  • Parse that as “same in-order-to”, not “same-in-order to”. If there is a free $\mu$ and a free $\nu$ on the left then there must be a free $\mu$ and a free $\nu$ in each term on the right to have Lorentz covariance. The order in which $\mu$ and $\nu$ appear doesn’t matter. – Ghoster Jun 23 '23 at 02:41

1 Answers1

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In coordinates $(x^0,x^1,x^2,x^3) \equiv (ct, \vec r)$, the index notation simply means that e.g.

$$F^{01} = \partial^0 A^1 - \partial^1 A^0$$

where $(A^0,A^1,A^2,A^3) \equiv (c\phi, \vec A)$ is the 4-potential. In Cartesian coordinates, the Minkowski metric takes the form

$$\eta = \pmatrix{-1 & 0 &0 &0 \\ 0 & 1 & 0 & 0 \\ 0 &0 & 1 & 0 \\ 0 & 0 & 0 &1}$$ and so $\partial^0 = -\partial_0\equiv -\frac{\partial}{\partial (ct)}$ whereas $\partial^i = \partial_i \equiv \frac{\partial}{\partial r^i}$ for $i=1,2,3$. As a result, for $i=1,2,3$ we have

$$F^{0i} = \partial^0 A^i - \partial^i A^0 = -\frac{1}{c} \frac{\partial A^i}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial r^i}$$

Recalling that $\vec E = -\frac{\partial \vec A}{\partial t} - \nabla \phi$, the components $F^{0i}$ are simply $E^i/c$. Filling in the rest of the tensor,

$$F = \pmatrix{0& E^1/c & E^2/c & E^3/c \\ -E^1/c & 0 & B^3 & -B^2 \\ -E^2/c & -B^3 & 0 & B^1 \\ -E^3/c & B^2 & -B^1 & 0 }$$

where $\vec B = \nabla \times \vec A$.

So that's how the indicies are defined and how $F$ can be interpreted. Why we would define $F$ in the first place is a deeper and more interesting question.

In the context of special relativity, the fundamental object is $F$, not $\vec E$ or $\vec B$. Given a particular choice of reference frame, we can define 3-vectors $\vec E$ and $\vec B$ by picking out the components of $F$ as shown above, but those components generally mix together when we boost to a different reference frame. In that sense, the splitting of the electromagnetic field into an electric part $\vec E$ and a magnetic part $\vec B$ is artificial and unnatural (though it is of course often useful).

Historically, $\vec E$ and $\vec B$ were understood as distinct objects, not as different aspects of the same thing (i.e. $F$). The Maxwell equations show that despite appearing to be independent, $\vec E$ and $\vec B$ are related to one another very intimately. If you work with the Maxwell equations for a bit, you discover that observers who are in relative motion to one another must observe different electric and magnetic fields - for example, an observer standing next to a stationary charge sees a pure electric field, while an observer moving with respect to the charge sees both electric and magnetic fields.

This naively suggests that the Maxwell equations only work for certain special observers, and that you need to be in the right reference frame to compute the correct fields. However, through an apparent coincidence, it turns out that even though $\vec E$ and $\vec B$ must be different in different frames, they conspire together such that you get the correct trajectory for a particle no matter what frame you work in. The realization that this isn't a coincidence - and that $\vec E$ and $\vec B$ are distinct aspects of a single, fundamental entity - is (at least in part) what fueled Einstein's development of special relativity.

Things get even deeper and more interesting when we move to relativistic quantum mechanics, where electromagnetism arises essentially automatically by imposing something called $\mathrm U(1)$ gauge invariance on the quantum fields. This is far beyond the scope of this question from a technical standpoint, but it is worth mentioning that in this much more advanced (but physically fundamental) context, it is once again $A$ and $F$ which play a fundamental role.

J. Murray
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  • Ok, J.Murray. Great answer! Really illuminating. I know that $\vec{E}$ and $\vec{B}$ are dependent on frame of reference, and that $A$ and $F$ are the fundamental invariants, but this gave a great visual of how those break down into their respective components. – Garrett Miller Jun 23 '23 at 00:52
  • @GarrettMiller Thanks, glad it was helpful. – J. Murray Jun 23 '23 at 00:55