# Some tensor confusion

AxiomOfChoice
I don't think this should be a very difficult question for people who are used to working with tensors, but I'm new to it, so I'm confused. The Wikipedia article on the electromagnetic field tensor $F^{\mu \nu}$ asserts that

$$F_{\mu \nu} F^{\mu \nu} = 2 \left( B^2 - \frac{E^2}{c^2} \right).$$

But if you look at the way they've written out $F_{\mu \nu}$ and $F^{\mu \nu}$ in matrix form, there is no way you get that when you just simply multiply the two matrices together. I mean, how can one obtain a constant from multiplying two square matrices (that aren't one-by-one, obviously) together? What am I missing here?

(The relevant Wikipedia article is http://en.wikipedia.org/wiki/Electromagnetic_tensor" [Broken].)

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Because this is not matrix multiplication! The fact that mu and nu are used on both tensors means they are summed over, so this is an inner product of two tensors, not the product of two matrices!

Homework Helper
Of course you're missing. Taking the minus trace on the resulting matrix. In tensor notation

$$F_{\mu\nu}F^{\mu\nu} = -\delta_{\mu}^{\lambda} F_{\lambda\nu}F^{\nu\mu}$$

Brock Skylar
R u guys taking antifrction into account? I think ur overcomplicating things

Staff Emeritus
Gold Member
What am I missing here?
I answered this question quite recently (for arbitrary matrices), so I'll just quote myself.

If you want to treat this expression as something involving a product of matrices, this is what you need:

Definition of matrix multiplication: (XY)ij=XikYkj
Definition of transpose: (XT)ij=Xji
Definition of trace: Tr X=Xii

These definitions tell us that

$$A_{ij}S_{ij}=(A^T)_{ji}S_{ij}=(A^TS)_{jj}=\operatorname{Tr}(A^TS)$$

Alternatively,

$$A_{ij}S_{ij}=A_{ij}(S^T)_{ji}=(AS^T)_{ii}=\operatorname{Tr}(AS^T)$$

You can also reverse the order of the matrices in the trace, because Tr(XY)=Tr(YX) for all matrices X and Y. (This is very easy to prove using the definitions above).

AxiomOfChoice
Thanks a lot, guys. I think this has cleared up a lot. Can I argue as follows?

$$F_{\mu \nu} F^{\mu \nu} = (\partial_\mu A_\nu - \partial_\nu A_\mu)(\partial^\mu A^\nu - \partial^\nu A^\mu),$$

from which one obtains

$$F_{\mu \nu} F^{\mu \nu} = 2 (\partial_\mu \partial^\mu A_\nu A^\nu - \partial_\nu A^\nu \partial_\mu A^\mu)$$

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Homework Helper
Thanks a lot, guys. I think this has cleared up a lot. Can I argue as follows?

$$F_{\mu \nu} F^{\mu \nu} = (\partial_\mu A_\nu - \partial_\nu A_\mu)(\partial^\mu A^\nu - \partial^\nu A^\mu),$$

from which one obtains

$$F_{\mu \nu} F^{\mu \nu} = 2 (\partial_\mu \partial^\mu A_\nu A^\nu - \partial_\nu A^\nu \partial_\mu A^\mu) [/tex You maye want to correct the sloppy notation and LaTex formatting. And what you wrote, assumingly correct, has nothing to do with the initial question to which answers were already formulated. AxiomOfChoice You maye want to correct the sloppy notation and LaTex formatting. Done. Sorry; I temporarily lost my wireless connection while trying to correct. And what you wrote, assumingly correct, has nothing to do with the initial question to which answers were already formulated. ...I guess I don't see why. The key ingredient I was missing was the fact that repeated indices implied a sum. I put this together with the identity $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ to produce the above. Science Advisor Homework Helper The LaTeX needs further fixing. Well, your question was about the matrices of the e-m tensor. The elements of these matrices are made up of E and B. There's no help you can get on your question by bringing potentials into the discussion, because the E and B are enough (your lagrangian is a quadratic function of E and B, after all) to formulate the simplest possible correct answer you could get (post 3). AxiomOfChoice The LaTeX needs further fixing. The only thing I can see that you might be talking about is the fact that I have written $\partial_\mu A_\nu \partial^\nu A^\mu$ as $\partial_\nu A^\nu \partial_\mu A^\mu$. But what is the difference between $\partial^\nu A_\nu$ and $\partial_\mu A^\mu$? Staff Emeritus Science Advisor Gold Member Thanks a lot, guys. I think this has cleared up a lot. Can I argue as follows? [tex] F_{\mu \nu} F^{\mu \nu} = (\partial_\mu A_\nu - \partial_\nu A_\mu)(\partial^\mu A^\nu - \partial^\nu A^\mu),$$

from which one obtains

$$F_{\mu \nu} F^{\mu \nu} = 2 (\partial_\mu \partial^\mu A_\nu A^\nu - \partial_\nu A^\nu \partial_\mu A^\mu)$$
I'm also wondering what this has to do with your original question. What you overlooked was, as you said yourself, that repeated indices are summed over. Now that you know that, you should see that it's simply a matter of using the definitions of matrix multiplication, transpose and trace. The trace of a matrix is a number. There's no need to use any knowledge of what the components of F are. I'll go one step further and say that it doesn't help at all to know what the components are.

But what is the difference between $\partial^\nu A_\nu$ and $\partial_\mu A^\mu$?

$$\partial^\nu A_\nu=\eta^{\nu\rho}\partial_\rho\eta_{\nu\sigma}A^\sigma=\delta^\rho_\sigma\partial_\rho A^\sigma=\partial_\mu A^\mu$$

Homework Helper
[...]$$F_{\mu \nu} F^{\mu \nu} = 2 (\partial_\mu \partial^\mu A_\nu A^\nu - \partial_\nu A^\nu \partial_\mu A^\mu)$$

You wrote that, which is wrong.

$$\partial_\mu \partial^\mu A_\nu A^\nu \neq \partial_{\mu} A^{\nu}\partial_{\nu} A^{\mu}$$

Gold Member
I'm also wondering what this has to do with your original question. What you overlooked was, as you said yourself, that repeated indices are summed over. Now that you know that, you should see that it's simply a matter of using the definitions of matrix multiplication, transpose and trace. The trace of a matrix is a number. There's no need to use any knowledge of what the components of F are. I'll go one step further and say that it doesn't help at all to know what the components are.

$$\partial^\nu A_\nu=\eta^{\nu\rho}\partial_\rho\eta_{\nu\sigma}A^\sigma=\delta^\rho_\sigma\partial_\rho A^\sigma=\partial_\mu A^\mu$$

Did you just use eta to denote the metric tensor? Can I formally protest because I'm much more used to seeing it as g? >_>

Did you just use eta to denote the metric tensor? Can I formally protest because I'm much more used to seeing it as g? >_>

Well presumably the OP is referring to electromagnetism in flat, aka minkowski, spacetime, rather than a general spacetime... In which case we usually write eta instead of g.

Gold Member
Oh, even in minkowski spacetime, I usually see g=diag(1,-1,-1,-1). This is the convention I am familiar with haha.

Staff Emeritus
Gold Member
I use g for the metric tensor and also for its matrix of components in a coordinate system, but η is the specific matrix you mentioned (sometimes defined with the opposite sign), so Minkowski spacetime is defined by the choice "g=η, in an inertial coordinate system". In a coordinate-independent expression, I would always use g for the metric tensor, even if we're dealing with Minkowski spacetime.

AxiomOfChoice
You wrote that, which is wrong.

$$\partial_\mu \partial^\mu A_\nu A^\nu \neq \partial_{\mu} A^{\nu}\partial_{\nu} A^{\mu}$$

Sorry...I don't see what the issue is. Obviously,

$$(\partial_\mu A_\nu - \partial_\nu A_\mu)(\partial^\mu A^\nu - \partial^\nu A^\mu) = \partial_\mu A_\nu \partial^\mu A^\nu - \partial_\mu A_\nu \partial^\nu A^\mu - \partial_\nu A_\mu \partial^\mu A^\nu + \partial_\nu A_\mu \partial^\nu A^\mu.$$

I don't see how one gets the $\partial_{\mu} A^{\nu}\partial_{\nu} A^{\mu}$ you mention from that. As best I can tell, the mistake I have made is commuting some of the derivatives; i.e., that $\partial_\mu \partial^\mu A_\nu A^\nu \neq \partial_\mu A_\nu \partial^\mu A^\nu$. But that raises another issue: What can I commute in this business, and what can't I?

Thanks for your help, and for being patient. I'm in a Intro GR class where the professor assumes familiarity with this notation. Which I lack.

AxiomOfChoice
There's no need to use any knowledge of what the components of F are. I'll go one step further and say that it doesn't help at all to know what the components are.

Really? I don't see how I'm supposed to eventually get $E$ and $B$ into this business without referring to what the components of $F$ are.

Staff Emeritus
Gold Member
Really? I don't see how I'm supposed to eventually get $E$ and $B$ into this business without referring to what the components of $F$ are.
I thought the original question was "how can this be a number and not a matrix?", but if you also want to know why the right-hand side is that specific combination of E and B, you obviously have to use the equalities that describe how A is related to B,E and F.

As best I can tell, the mistake I have made is commuting some of the derivatives; i.e., that $\partial_\mu \partial^\mu A_\nu A^\nu \neq \partial_\mu A_\nu \partial^\mu A^\nu$. But that raises another issue: What can I commute in this business, and what can't I?
This isn't just a matter of commuting derivative operators. You need to keep track of when they act on one of the factors and when they act on a product of two of them. Use the product rule when necessary.

AxiomOfChoice
This isn't just a matter of commuting derivative operators. You need to keep track of when they act on one of the factors and when they act on a product of two of them. Use the product rule when necessary.

Ok. I'm trying to work this out. I'm guessing that the expression

$$\partial_\mu A_\nu \partial^\mu A^\nu,$$

corresponds to a sum that contains 16 summands, and that the (0,0) summand looks like

$$\underbrace{\frac{1}{c} \frac{\partial}{\partial t} \frac{\phi}{c}}_{\partial_0 A_0} \underbrace{\frac{1}{c} \frac{\partial}{\partial t} \frac{\phi}{c}}_{\partial^0 A^0}.$$

Are you saying that this is actually

$$\frac{1}{c^4} \frac{\partial}{\partial t} \left( \phi \frac{\partial \phi}{\partial t} \right)$$

Staff Emeritus
Gold Member
No, what I'm saying is that if you want to move the derivative operators around in that middle expression, you have to do it with the product rule in mind: Df Dg = D(Df g)-D2f g.

AxiomOfChoice
No, what I'm saying is that if you want to move the derivative operators around in that middle expression, you have to do it with the product rule in mind: Df Dg = D(Df g)-D2f g.

Well, that's what I thought I was saying...so is it wrong to say

$$\partial_0 A_0 \partial^0 A^0 = \frac{1}{c^4} \left( \left( \frac{\partial \phi}{\partial t} \right)^2 + \phi \frac{\partial^2 \phi}{\partial t^2}\right)$$

?

Staff Emeritus
Gold Member
Yes it's wrong, because the left-hand side is equal to the first term (give or take minus signs and factors of c...I didn't check those details).

AxiomOfChoice
Yes it's wrong, because the left-hand side is equal to the first term (give or take minus signs and factors of c...I didn't check those details).

Wow... ... I'm starting to feel like a complete idiot...so how do I know that in the expression

$$\underbrace{\frac{1}{c} \frac{\partial}{\partial t} \frac{\phi}{c}}_{\partial_0 A_0} \underbrace{\frac{1}{c} \frac{\partial}{\partial t} \frac{\phi}{c}}_{\partial^0 A^0},$$

that first partial derivative with respect to $t$ only acts on the first $\phi$ and not the second one?

Staff Emeritus
Gold Member
Because the product of the two real numbers f'(x) and g'(x) is f'(x)g'(x), not D(Df·g)(x). It really is as simple as that.

AxiomOfChoice
Because the product of the two real numbers f'(x) and g'(x) is f'(x)g'(x), not D(Df·g)(x). It really is as simple as that.

Ok. Well, first of all, thanks for all of your help. Second of all, are there any good online resources you could recommend for learning this stuff systematically, as opposed to bugging you guys incessantly?

Staff Emeritus
Gold Member
I don't know any, but it might help you to look at some older threads about expressions with lots of indices. One thing that will definitely help is to simply decide to make sure at every step that you're using the definitions and not throwing in assumptions of your own. I think you're finding this difficult just because you expect it to be much more difficult than it really is. These expression don't involve anything more complicated than sums and products of real numbers, and partial derivatives.

AxiomOfChoice
I don't know any, but it might help you to look at some older threads about expressions with lots of indices. One thing that will definitely help is to simply decide to make sure at every step that you're using the definitions and not throwing in assumptions of your own. I think you're finding this difficult just because you expect it to be much more difficult than it really is. These expression don't involve anything more complicated than sums and products of real numbers, and partial derivatives.

You're probably right. But I'm trying not to throw in any assumptions. See, for example, the $\partial_0 A_0 \partial^0 A^0$ confusion with the first partial derivative and what it acts on...there is some convention that says the first partial acts ONLY on the first $\phi$ and not on the second. What is this convention? Because, if one were to just write out

$$\partial_0 A_0 \partial^0 A^0 = \frac{1}{c} \frac{\partial}{\partial t} \frac{\phi}{c} \frac{1}{c} \frac{\partial}{\partial t} \frac{\phi}{c},$$

it seems to require an assumption not to let that first $\partial_t$ act through to the second occurrence of $\phi$...

Staff Emeritus
Gold Member
What you're saying about the notation is true. It's not 100% obvious what it means, if you only look at the notation. But in this case, you can also look at the expression you obtained it from. You started with

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

$$F_{\mu\nu}F^{\mu\nu}=(\partial_\mu A_\nu - \partial_\nu A_\mu)(\partial^\mu A^\nu - \partial^\nu A^\mu)$$

Let's look at the expression $\partial_\mu A_\nu$. This is just a function from $\mathbb R^4$ into $\mathbb R$. Alternatively, you can interpret it as meaning $\partial_\mu A_\nu(x)$ with the x suppressed (i.e. not typed because you're trying to save time). This is just a real number. Let's go with the latter interpretation. Then the expression above is of the form (A+B)(C+D) where A,B,C,D are real numbers. You know that this is equal to AC+AD+BC+BD, and somehow you still managed to convince yourself that the product of A and C isn't AC.

If you find the notation confusing, just insert parentheses where you think they're needed. Most authors seem to use the convention that a derivative operator only acts on the first thing that's immediately to its right, and if you can remember that, you can avoid having to type a lot of parentheses.

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