Some tensor confusion

  • #1

Main Question or Discussion Point

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 [itex]F^{\mu \nu}[/itex] asserts that

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

But if you look at the way they've written out [itex]F_{\mu \nu}[/itex] and [itex]F^{\mu \nu}[/itex] 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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Answers and Replies

  • #2
Nabeshin
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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!
 
  • #3
dextercioby
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Of course you're missing. Taking the minus trace on the resulting matrix. In tensor notation

[tex] F_{\mu\nu}F^{\mu\nu} = -\delta_{\mu}^{\lambda} F_{\lambda\nu}F^{\nu\mu} [/tex]
 
  • #4
R u guys taking antifrction into account? I think ur overcomplicating things
 
  • #5
Fredrik
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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

[tex]A_{ij}S_{ij}=(A^T)_{ji}S_{ij}=(A^TS)_{jj}=\operatorname{Tr}(A^TS)[/tex]

Alternatively,

[tex]A_{ij}S_{ij}=A_{ij}(S^T)_{ji}=(AS^T)_{ii}=\operatorname{Tr}(AS^T)[/tex]

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).
 
  • #6
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),
[/tex]

from which one obtains

[tex]
F_{\mu \nu} F^{\mu \nu} = 2 (\partial_\mu \partial^\mu A_\nu A^\nu - \partial_\nu A^\nu \partial_\mu A^\mu)
[/tex]
 
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  • #7
dextercioby
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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),
[/tex]

from which one obtains

[tex]
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.
 
  • #8
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 [itex]F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu[/itex] to produce the above.
 
  • #9
dextercioby
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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).
 
  • #10
The LaTeX needs further fixing.
The only thing I can see that you might be talking about is the fact that I have written [itex]\partial_\mu A_\nu \partial^\nu A^\mu[/itex] as [itex]\partial_\nu A^\nu \partial_\mu A^\mu[/itex]. But what is the difference between [itex]\partial^\nu A_\nu[/itex] and [itex]\partial_\mu A^\mu[/itex]?
 
  • #11
Fredrik
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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),
[/tex]

from which one obtains

[tex]
F_{\mu \nu} F^{\mu \nu} = 2 (\partial_\mu \partial^\mu A_\nu A^\nu - \partial_\nu A^\nu \partial_\mu A^\mu)
[/tex]
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 [itex]\partial^\nu A_\nu[/itex] and [itex]\partial_\mu A^\mu[/itex]?
[tex]\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[/tex]
 
  • #12
dextercioby
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[...][tex]
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 wrote that, which is wrong.

[tex] \partial_\mu \partial^\mu A_\nu A^\nu \neq \partial_{\mu} A^{\nu}\partial_{\nu} A^{\mu} [/tex]
 
  • #13
Matterwave
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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.



[tex]\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[/tex]
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? >_>
 
  • #14
Nabeshin
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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.
 
  • #15
Matterwave
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Oh, even in minkowski spacetime, I usually see g=diag(1,-1,-1,-1). This is the convention I am familiar with haha.
 
  • #16
Fredrik
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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.
 
  • #17
You wrote that, which is wrong.

[tex] \partial_\mu \partial^\mu A_\nu A^\nu \neq \partial_{\mu} A^{\nu}\partial_{\nu} A^{\mu} [/tex]
Sorry...I don't see what the issue is. Obviously,

[tex]
(\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.
[/tex]

I don't see how one gets the [itex]\partial_{\mu} A^{\nu}\partial_{\nu} A^{\mu}[/itex] you mention from that. As best I can tell, the mistake I have made is commuting some of the derivatives; i.e., that [itex]\partial_\mu \partial^\mu A_\nu A^\nu \neq \partial_\mu A_\nu \partial^\mu A^\nu[/itex]. 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.
 
  • #18
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 [itex]E[/itex] and [itex]B[/itex] into this business without referring to what the components of [itex]F[/itex] are.
 
  • #19
Fredrik
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Really? I don't see how I'm supposed to eventually get [itex]E[/itex] and [itex]B[/itex] into this business without referring to what the components of [itex]F[/itex] 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 [itex]\partial_\mu \partial^\mu A_\nu A^\nu \neq \partial_\mu A_\nu \partial^\mu A^\nu[/itex]. 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.
 
  • #20
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

[tex]
\partial_\mu A_\nu \partial^\mu A^\nu,
[/tex]

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

[tex]
\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}.
[/tex]

Are you saying that this is actually

[tex]
\frac{1}{c^4} \frac{\partial}{\partial t} \left( \phi \frac{\partial \phi}{\partial t} \right)
[/tex]
 
  • #21
Fredrik
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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.
 
  • #22
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

[tex]
\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)
[/tex]

?
 
  • #23
Fredrik
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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).
 
  • #24
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... :frown: ... I'm starting to feel like a complete idiot...so how do I know that in the expression

[tex]
\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},
[/tex]

that first partial derivative with respect to [itex]t[/itex] only acts on the first [itex]\phi[/itex] and not the second one?
 
  • #25
Fredrik
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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. :smile:
 

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