# EM tensor invariants

1. Mar 4, 2013

### BruceW

Hey everyone

I was looking up the EM tensor on wikipedia, and they mention two invariants. There is the obvious one - The Lorentz invariant
$$B^2- \frac{E^2}{c^2}$$
And there is also the product of the EM tensor with its dual, which they call the pseudoscalar invariant:
$$\frac{1}{2} \epsilon_{\alpha \beta \gamma \delta} F^{\alpha \beta} F^{\gamma \delta} = - \frac{4}{c^2} ( \mathbf{E} \cdot \mathbf{B} )$$
And they also mention the determinant of the EM tensor, but it is just the square of the above invariant, so it does not give us any more information.

Anyway, so my question is: are there any other invariants? And if these two are the only two, then can this be shown using the properties of the metric and the fact that the EM tensor is anti-symmetric?

2. Mar 4, 2013

### Bill_K

The only quantities you can form from the components of the EM field that are invariant under rotations are E2, B2 and E·B. Well so E2 + B2 is the energy density, T00, so that leaves just the two combinations you mentioned.

3. Mar 4, 2013

### WannabeNewton

While not the EM tensor itself, it might be worth noting that the electromagnetic stress energy tensor as constructed from the electromagnetic field tensor, $T_{ab} = \frac{1}{4\pi }(F_{ac}F_{b}^{c} - \frac{1}{4}g_{ab}F_{de}F^{de})$, is itself invariant under duality rotations $\tilde{F_{ab}} = F_{ab}cos\alpha + *F_{ab}sin\alpha$. I wonder if there is something similar for the Bel - Robinson tensor (not exactly analogous in the sense of duality of course since the Bel - Robinson tensor is a totally symmetric tensor), which is constructed in a manner very similar to the EM stress tensor by using the Weyl curvature tensor in place of $F_{ab}$. Anyways, very nice question Bruce!

4. Mar 4, 2013

### BruceW

aha! I just found this wikipedia page, which says that the two invariants I mentioned are the only two "fundamental invariants of the electromagnetic field"
http://en.wikipedia.org/wiki/Classification_of_electromagnetic_fields#Physical_interpretation

But they don't say why these are the only two. They do talk a bit about linear operators. I will have a read of it when I get time. I was also thinking that maybe we can consider
$$F^{\alpha \beta} F^{\gamma \delta}$$
and think of all the possible ways of contracting the indices, (and also use the fact of antisymmetry) to show that there are only two different invariants.

Edit: but then what about using more outer products, like FFF instead of just FF... Maybe if I just think about F (as a linear operator), and how it maps 4-vectors, then maybe that is the way to get the answer..

5. Mar 4, 2013

6. Mar 4, 2013

### bcrowell

Staff Emeritus
In addition to these quantities that are invariant at a point in spacetime, there is also at least one interesting nonlocal invariant quantity. The total power radiated by a point charge is Lorentz invariant if there's symmetry in the rest frame.

GR also has similar quantities for gravitational waves. As with E&M, you can use them to test whether a certain field is a radiation field. This is the basic idea behind Petrov types.

7. Mar 4, 2013

### BruceW

I don't really get what you meant. You said that E^2, B^2 and $\vec{E} \cdot \vec{B}$ are the 3 quantities that are invariant under rotations. I'm guessing you mean that these are the 3 quantities that are invariant under spatial rotations. But I don't see how this helps in understanding the invariants of the EM tensor. Maybe I wasn't specific enough about what I wanted to find. I wanted to find the electromagnetic invariants of special relativity. In other words, the Poincare invariant quantities associated with the EM field. I don't think we need to use the mathematics of groups, but that would be cool if someone had an explanation using groups. I think we just need to use the idea of the EM tensor as being a skew-symmetric linear operator.

Also, bcrowell, that is pretty interesting. So am I right in thinking that the thing you mentioned is a global quantity that is invariant under Lorentz transforms? That is unusual, right? We usually talk about local (tensor) scalars as having Lorentz invariance. So we are lucky that there is this global quantity that is Lorentz invariant.

8. Mar 4, 2013

### bcrowell

Staff Emeritus
I'll see if I can fill in some of what Bill left to the imagination. Any Poincare invariant must be both an invariant under Lorentz boosts and under rotations. Given the two vectors E and B, the only way to form rotational invariants is to form dot products, which gives E2, B2, and $E\cdot B$. Once you've formed these, you can do any arithmetic operations with them that you like and still get a rotational invariant, although it's not guaranteed to be a Lorentz invariant. However, that doesn't change the number of degrees of freedom, it just rearranges them. In particular, we can rearrange these into the combinations $A_1=E^2+B^2$, $A_2=E^2-B^2$, and $A_3=E\cdot B$. (Please get hip to relativistic units with c=1 -- it's a total waste of your time to keep writing all those factors of c, and it makes your equations harder for others to read.) Of these three new combinations, we know that although $A_1$ is rotationally invariant, it's not a Lorentz invariant, since it's the 00 component of the stress-energy tensor. Therefore there are only two independent degrees of freedom that give us invariants, $A_2$ and $A_3$.

We can of course form all kinds of things like $A_4=(A_3)^{17}+\sin(A_2/A_3)$, and these are also invariants. However, they aren't independent degrees of fredom. That is, it's not possible to have fields whose values of $A_2$ and $A_3$ are the same but whose values of $A_4$ are unequal.

9. Mar 4, 2013

### Staff: Mentor

10. Mar 4, 2013

### robphy

If I am not mistaken, (someone check my math)
you can calculate the principal invariants of a tensor:
$$F^a{}_a, F^a{}_{[a} F^b{}_{b]}, F^a{}_{[a} F^b{}_{b}F^c{}_{c]}, F^a{}_{[a} F^b{}_{b} F^c{}_{c}F^d{}_{d]}$$ in 4-dimensions.
You'd get the trace (sum of the eigenvalues), sum of products-of-pairs of eigenvalues, sum of products-of-triples, and finally, the product-of-the-4-eigenvalues (the determinant).

For a real antisymmetric matrix, the eigenvalues are imaginary.
So, only those with sums of products of even-numbers-of-eigenvalues will be nonzero.

11. Mar 4, 2013

### Staff: Mentor

In MTW there is a discussion of a representation of a generic EM field using two wedge products; IIRC (don't have my copy handy to check right now), it looks something like this:

$$F = E dt \wedge dz + B dx \wedge dy$$

This would imply that there can only be two independent invariants, since there can only be two independent coefficients of the wedge products.

12. Mar 4, 2013

### WannabeNewton

Last edited by a moderator: May 6, 2017
13. Mar 5, 2013

### Bill_K

Ai. Because in order for a quantity to be invariant under the entire Lorentz group, it must first be invariant under spatial rotations. Si?

So the only quantities that can possibly be Lorentz invariant are a subset of the rotational invariants. There are three independent rotational invariants, and one of them turns out to be T00, leaving two.

14. Mar 5, 2013

### bcrowell

Staff Emeritus
The determinant is $(B\cdot E)^2$, and $F^{ab}F_{ab}=2(B^2-E^2)$ (presumably corresponding to your sum of products of pairs). Seems to make sense.

I suppose you can also form various scalars by differentiating the fields. E.g., you could make $(\nabla_a F^{ab})( \nabla^c F_{cb})$, which equals the squared norm of the current four-vector. Some such scalars would vanish identically by Maxwell's equations.

15. Mar 6, 2013

### samalkhaiat

The number of independent invariants constructed from rank-2 antisymmetric tensor depends on the number of dimensions of space-time. You can show that, for $SO(2k)$ and $SO(2k + 1)$, that number of independent invariants is $k$.

Sam

16. Mar 10, 2013

### BruceW

I see now, yeah that is a pretty clever way to get the answer. And that does answer my question. Thanks to you and Bill_K.

Ah, wow, a manifestly relativistic answer. This is what I was hoping for. These 'principle invariants' Am I right in thinking that they are the coefficients of the characteristic polynomial of the matrix? And the paper which WannabeNewton gave link to says that you can instead use the eigenvalues themselves as the invariants of the matrix. I guess that if the eigenvalues are the invariants, then it is equivalent to instead say that the coefficients are the invariants.

When we transform the coordinates by using a transformation matrix, the eigenvalues of any matrix stay the same. So the eigenvalues of our 'electromagnetic matrix' will be the same, even if we do any kind of linear transform of the coordinates. (This can be a rotation, or boost or whatever). So it makes sense to me that the eigenvalues of our 'electromagnetic matrix' are invariants. But why should we believe that these are the only invariants? There is nothing that I can think of which limits the invariants to these ones. I am hoping that there is some kind of theorem which says that the only invariants of a tensor under a linear coordinate transformation are the eigenvalues of the associated matrix. Has anyone heard of such a theorem?

Note: I am trying to be good, and keep the two notions of 'matrix' and 'tensor' separate. I understand that a matrix is really a tensor in a particular coordinate system, while a tensor is a more abstract mathematical object. But often the two concepts are used interchangeably when they really shouldn't be.

Edit: Also, bcrowell makes a good point about differentiating the EM tensor to get other invariants. I was just considering the actual tensor, but differentiation is a natural extension.

17. Mar 11, 2013

### robphy

I always strive for manifestly relativistic invariants.

Yes, these are the coefficients of the characteristic polynomial.
These might be useful references
http://www.encyclopediaofmath.org/index.php/Invariants,_theory_of
and
http://www.math.unibas.ch/~kraft/Papers/KP-Primer.pdf (near p 16)
(Maybe someone more algebraically-inclined can translate.)

I think it is more correct to say that the set of "elementary symmetric functions of the eigenvalues" are invariants.

See above.
I think the situation is that the set of principal invariants can generate the ring of invariants.
http://en.wikipedia.org/wiki/Ring_(mathematics)

My above reply generates algebraic invariants.
Including derivatives of tensors would generate differential invariants.
One could generalize the formulas I first posted to tuples of tensors (e.g. invariants generated from a pair of tensors).

18. Mar 16, 2013

### BruceW

Awesome. These are some good links for me to get my teeth into. And I can do some google searches, since I know what to look for now. Thank you! and thanks to everyone who replied to my thread. Some interesting posts.

19. Mar 16, 2013

### Bill_K

Given a choice, I prefer simple one-line answers to long ones, where they exist. But if complexity is what turns you on, you might next try working out the invariants of the Riemann tensor.

20. Mar 17, 2013

### BruceW

haha, true. what 'turns me on' is principles which can be applied to a whole bunch of stuff in physics, which might seem to be completely disconnected at first sight. Which is why I am happy to have learned a bit about the 'principle invariants' of a general matrix.

On the other hand, I also like simple principles. I think 'simplicity' and 'generality' are what I like to see in theories.

Edit: I can see why you answered my question with the short answer though. I didn't fully explain what I was hoping to find. And it is better to answer with the short answer first, since I might not have been interested in the long answer.

Last edited: Mar 17, 2013
21. Mar 20, 2013

### Bill_K

Early on in this thread, WannaBeNewton mentioned duality rotations. Duality is a symmetry of the Maxwell Equations that makes the discussion of invariants even simpler.

Define the complex Faraday tensor Pμν = Fμν + i *Fμν. Then Maxwell's Equations written in terms of Pμν become just a single equation, Pμν = 0, invariant under the duality rotation Pμν → e Pμν.

Equivalently, the field can be expressed in terms of a complex 3-vector: P = E + iB, and one can write the 3-d Maxwell Equations in terms of P.

Relevance of this to the present discussion is that Lorentz invariants must also conform to this symmetry: if you apply a duality rotation to an invariant, you'll get another invariant. Consequently, there's only one complex invariant, whose real and imaginary parts are the two invariants we know about, namely P·P = (E + iB)·(E + iB) = (E·E - B·B) + i(2E·B).

I'll answer my own question. Vacuum General Relativity also has a duality. Define a complex Riemann tensor Pμνστ = Rμνστ + i *Rμνστ. Then as before, Pμνστ = 0 (the Bianchi identities) and Pμνστ → e Pμνστ (the symmetry).

Equivalently the Riemann tensor (10 independent components in vacuum) is represented in 3-d terms by two symmetric traceless rank two tensors with 5 components each, Eij = R0i0j and Bij = ½ εjmn R0imn. The complex version is a single complex 3-d traceless tensor, P = E + iB.

The Riemann invariants? Now they're easy to write down.

The first one is quadratic, similar to the Maxwell invariant except E and B are now tensors instead of vectors: Tr(P·P) = Tr(E·E - B·B) + 2i Tr(E·B).

The second one is cubic: Tr(P·P·P) = Tr((E + iB)·(E + iB)·(E + iB)) = Tr (E·E·E - 3 E·B·B) + i Tr(3 E·E·B - B·B·B). Splitting them into real and imaginary parts, we have the four Riemann invariants.

EDIT: moved the phrase "in 3-d terms", for clarity.

Last edited: Mar 20, 2013
22. Mar 20, 2013

### TrickyDicky

The Riemann tensor in 3D has a total of 6 algebraic independent components. You must be thinking of the 4D Riemann tensor that has a total of 20, 10 in vacuum.

23. Mar 20, 2013

### Bill_K

That's correct. The 4-d vacuum Riemann tensor, which I'm splitting into its 3-d parts. Just as we do for the 4-d Faraday tensor.

24. Mar 20, 2013

### TrickyDicky

I see, I interpreted "in 3d terms" out of context in that sentence.