Bianchi identity with F^ab F^cd

[JustinLevy]

The Reimann curvature tensor has the following symmetry resulting from a Bianchi identity$$R_{abcd}+R_{acdb}+R_{adbc}=0$$
The derivative of the electromagnetic field tensor also yields some of Maxwell's equations from a Bianchi identity$$\partial_\gamma F_{ \alpha \beta } + \partial_\alpha F_{ \beta \gamma } + \partial_\beta F_{ \gamma \alpha } = 0$$
Unfortunately, while I can do tensor manipulations at the index level and calculate results in some coordinate system, when articles start manipulating things index free as "forms" and deriving results, I get lost. So while I recognize the above as coming from Bianchi identities, it is only because I have heard them referred to as such, but beyond checking by hand (painfully to verify the above) I cannot actually "derive" them from some Bianchi identity.

Now the question:

Does $F^{ab} F^{cd}$ have some kind of index permuting symmetry? And if so, is it due to a Bianchi identity?
(Below, there is some evidence a result can be obtained with the Levi-Civita permutation symbol, so the answer may be yes, but I am unsure how to derive it.)

Also, while it may not be pertinent, this object does have some of the symmetries of the Riemann tensor.
Defining $$X^{abcd} = F^{ab} F^{cd}$$ it has the symmetries $$X^{abcd} = X^{cdab}$$ $$X^{abcd} = - X^{bacd} = - X^{abdc}$$

I saw in a discussion on science2.0 with someone trying to derive $X^{abcd} X_{bcda}$ and they found
$$X^{abcd} X_{bcda} = \frac{1}{2}(F^{ab}F_{ab})^2 + \frac{1}{4}(F^{ab}G_{ab})^2$$
Where $G_{ab}$ is the dual electromagnetic field tensor
$$G_{ab}=\frac{1}{2}\epsilon_{abcd}F^{cd}$$

The way in which it was derived was pretty messy, but it looks okay to me. (If you are curious its in the comments of a blog post by Doug Sweetser who used to post here a lot. http://www.science20.com/comments/173209/Some_notes_simplify ) I'm wondering if there is some Bianchi like identity that would make the above "obvious" somehow.

 

[robphy]

This might be useful:
en.wikipedia.org/wiki/Kulkarni%E2%80%93Nomizu_product

 

[WannabeNewton]

This might be useful:
en.wikipedia.org/wiki/Kulkarni%E2%80%93Nomizu_product

Hey robphy (long time no see!). In the link it says that tensor fields formed using the Kulkarni–Nomizu product share the same algebraic symmetries as the Riemann tensor. Does this include the algebraic Bianchi identity, which is what the OP is particularly after? If so do you know of a proof or a link to one?

For $R_{a[bcd]} = 0$ one generally uses the fact that $R^{a}_{bcd}\alpha_a = 2\nabla_{[c}\nabla_{d]}\alpha_b$ and the fact that at any given point we can write $\alpha_a = \nabla_a \alpha$ for some $\alpha$ to immediately conclude $R^{a}_{[bcd]}\alpha_a = 2\nabla_{[c}\nabla_d \nabla_{b]}\alpha = 0$ if $\nabla_a$ is torsion-free hence $R_{a[bcd]} = 0$. But here we made explicit use of the definition $R^{a}_{bcd}\xi^b = -2\nabla_{[c}\nabla_{d]}\xi^a$. How does the proof go for arbitrary tensor fields formed from the Kulkarni–Nomizu product?

 

[robphy]

Hi.
I don't know the answer to the OP's question.
Note that the relation to Bianchi in the above product involves symmetric tensors.

This paper is probably relevant... but I've only skimmed it in the past.

On the Bianchi Identities.
Ravindra S. Kulkarni
Mathematische Annalen (1972)
Volume: 199, page 175-204
https://eudml.org/doc/162323

 

[Bill_K]

Does $F^{ab} F^{cd}$ have some kind of index permuting symmetry? And if so, is it due to a Bianchi identity?
(Below, there is some evidence a result can be obtained with the Levi-Civita permutation symbol, so the answer may be yes, but I am unsure how to derive it.)

Also, while it may not be pertinent, this object does have some of the symmetries of the Riemann tensor.
Defining $$X^{abcd} = F^{ab} F^{cd}$$ it has the symmetries $$X^{abcd} = X^{cdab}$$ $$X^{abcd} = - X^{bacd} = - X^{abdc}$$

I saw in a discussion on science2.0 with someone trying to derive $X^{abcd} X_{bcda}$ and they found
$$X^{abcd} X_{bcda} = \frac{1}{2}(F^{ab}F_{ab})^2 + \frac{1}{4}(F^{ab}G_{ab})^2$$
Where $G_{ab}$ is the dual electromagnetic field tensor
$$G_{ab}=\frac{1}{2}\epsilon_{abcd}F^{cd}$$

The way in which it was derived was pretty messy, but it looks okay to me. (If you are curious its in the comments of a blog post by Doug Sweetser who used to post here a lot. http://www.science20.com/comments/173209/Some_notes_simplify ) I'm wondering if there is some Bianchi like identity that would make the above "obvious" somehow.

Not only is it not "obvious", it's not correct!

What they're trying to evaluate is $F^{ab}F_{bc}F^{cd}F_{da}$. Since $F^{ab}$ has only two algebraically independent invariants, $F^{ab}F_{ab} = G^{ab}G_{ab} = 2(E^2 - B^2)$ and $F^{ab}G_{ab} = 2 E \cdot B$, it must be a linear combination of $(F^{ab}F_{ab})^2$, $(F^{ab}F_{ab})(F^{ab}G_{ab})$ and $(F^{ab}G_{ab})^2$, and all we need to do is determine the coefficients. The easiest way to do this is to consider special cases.

Take the case in which $B = 0$. Then $F^{ab}F_{bc}F^{cd}F_{da} = 2 F^{0i}F_{i0}F^{0j}F_{j0} = 2 E^4$. That determines the first coefficient, $(1/2) (F^{ab}F_{ab})^2$. They got that right.

Next, take the case in which $E$ and $B$ are parallel. Then all the terms in $F^{ab}F_{bc}F^{cd}F_{da}$ involving both $E$ and $B$ vanish. The best you can do is start out $F^{0i}F_{ij}\cdots \sim E \times B$. There is no way to form $E \cdot B$. Consequently the second and third coefficients vanish.

The correct answer is $F^{ab}F_{bc}F^{cd}F_{da} = (1/2) (F^{ab}F_{ab})^2$

 

[JustinLevy]

Since $F^{ab}$ has only two algebraically independent invariants, $F^{ab}F_{ab} = G^{ab}G_{ab} = 2(E^2 - B^2)$ and $F^{ab}G_{ab} = 2 E \cdot B$, it must be a linear combination of $(F^{ab}F_{ab})^2$, $(F^{ab}F_{ab})(F^{ab}G_{ab})$ and $(F^{ab}G_{ab})^2$, and all we need to do is determine the coefficients.

Can you expand on this?
How do you know there are only two algebraically independent invariants?
I might agree that there are only two algebraically independent quadratic invariants. But I'm not sure how to actually prove that, let alone leap to there not being any higher order invariants.

Next, take the case in which $E$ and $B$ are parallel. Then all the terms in $F^{ab}F_{bc}F^{cd}F_{da}$ involving both $E$ and $B$ vanish. The best you can do is start out $F^{0i}F_{ij}\cdots \sim E \times B$. There is no way to form $E \cdot B$. Consequently the second and third coefficients vanish.

In the way they worked it out, the $(E \cdot B)$ comes about indirectly. For example, if we take your scenario with E and B parallel in x direction, and then if it turns out the invariant is
$2(E_x)^4 + 2(B_x)^4$
Then this gives
$2(E_x)^4 + 2(B_x)^4 = 2((E_x)^2 - (B_x)^2)^2 + 4 (E_x B_x)^2 = \frac{1}{2}(F^{ab}F_{ab})^2 + \frac{1}{4}(F^{ab}G_{ab})^2$

I'm sorry if I'm just having trouble following your thought process, but it seems like some of your leaps are not justified. Can you help me fill in the gaps?

 

[JustinLevy]

For $R_{a[bcd]} = 0$ one generally uses the fact that $R^{a}_{bcd}\alpha_a = 2\nabla_{[c}\nabla_{d]}\alpha_b$ and the fact that at any given point we can write $\alpha_a = \nabla_a \alpha$ for some $\alpha$ to immediately conclude $R^{a}_{[bcd]}\alpha_a = 2\nabla_{[c}\nabla_d \nabla_{b]}\alpha = 0$ if $\nabla_a$ is torsion-free hence $R_{a[bcd]} = 0$. But here we made explicit use of the definition $R^{a}_{bcd}\xi^b = -2\nabla_{[c}\nabla_{d]}\xi^a$. How does the proof go for arbitrary tensor fields formed from the Kulkarni–Nomizu product?

I don't see how the Kulkarni–Nomizu product will help us here. Maybe I'm missing something.

But your notes on the Bianchi identity gives me some ideas.
$F_{ab} = \nabla_a A_b - \nabla_b A_a = 2 \nabla_{[a} A_{b]}$
So this gives at a point for some $\alpha$:
$F^{ab}F^{cd} = 4 (\nabla^{[a} A^{b]})(\nabla^{[c} A^{d]}) = 4 (\nabla^{[a} \nabla^{b]}\alpha)(\nabla^{[c} \nabla^{d]} \alpha)$

Hmm... I guess I'm not as familiar with index manipulations as I hoped. How exactly do I "combine" brackets?
Trying to use the Bianchi identity to give me some of Maxwell's equations, can I do the following?
At a point for some $\alpha$
$\begin{align*} \nabla^{[a}F^{bc]} &= \nabla^{[a} 2 \nabla^{[b} A^{c]]} \\ &= \nabla^{[a} \nabla^{b} A^{c]} + \nabla^{[a} \nabla^{c} A^{b]} \\ &= \nabla^{[a} \nabla^{b} A^{c]} - \nabla^{[a} \nabla^{b} A^{c]} = 0 \end{align*}$
But that seemed to give me zero before I got to even use $\nabla^{[a} \nabla^{b} \nabla^{c]}\alpha = 0$
(Am I misusing this here?)So if I consider the object
$Q^{abcd} = \nabla^{[a}\nabla^b \nabla^c \nabla^{d]} \alpha^2$
any (permuted) third derivative on $\alpha$ will be zero, so is this equivalent to the following?
$Q^{abcd} = (\nabla^{[a}\nabla^b \alpha)(\nabla^c \nabla^{d]} \alpha)$

This seems to be getting much closer, but I'm just thinking aloud here. Hopefully this isn't all gibberish.

 

[Bill_K]

How do you know there are only two algebraically independent invariants?
I might agree that there are only two algebraically independent quadratic invariants. But I'm not sure how to actually prove that, let alone leap to there not being any higher order invariants.

The fact that the electromagnetic field tensor has just two algebraically independent invariants is a standard theorem. As usual, it's mentioned only tangentially in Wikipedia,

"every other invariant can be expressed in terms of these two"

The invariants are used to give the algebraic classification of the electromagnetic field at a point. The field can either be null (both invariants are zero, such as for a plane wave), or nonnull (at least one of the invariants is nonzero).

Similarly, the Riemann tensor (EDIT: in vacuum) has four algebraic invariants, two of them quadratic, two of them cubic.

In the way they worked it out, the $(E \cdot B)$ comes about indirectly. For example, if we take your scenario with E and B parallel in x direction, and then if it turns out the invariant is
$2(E_x)^4 + 2(B_x)^4$
Then this gives
$2(E_x)^4 + 2(B_x)^4 = 2((E_x)^2 - (B_x)^2)^2 + 4 (E_x B_x)^2 = \frac{1}{2}(F^{ab}F_{ab})^2 + \frac{1}{4}(F^{ab}G_{ab})^2$

Sorry, you're right about this. (Eating my words. )

 

[WannabeNewton]

I don't see how the Kulkarni–Nomizu product will help us here.

I just saw robphy's comment that it only applies to products of symmetric tensors so I agree with you there.

(Am I misusing this here?)

You're not misusing it but as you've seen it isn't necessary. If you have $F^{ab} = 2\nabla^{[a}A^{b]}$ then $\nabla^{[a}F^{bc]} = 0$ is immediate because, in more transparent notation, what you've written is $F = dA$ and $dF = d^2 A = 0$ since $d^2$ is nilpotent meaning the second power always vanishes. All this falls out because $F_{ab}$ is an exact 2-form. We don't have that same luxury with $R_{abcd}$ so we make do by other means when showing $R_{a[bcd]} = 0$ (such as the above) and $\nabla_{[a}R_{bc]de}= 0$.

So if I consider the object
$Q^{abcd} = \nabla^{[a}\nabla^b \nabla^c \nabla^{d]} \alpha^2$
...

But $\nabla^{[a}\nabla^b \nabla^c \nabla^{d]} \alpha^2 = \nabla^{[a}\nabla^b \nabla^{[c} \nabla^{d]]} \alpha^2 = 0$ so I'm not immediately seeing how useful it will be to you.

I tried a couple of different index manipulations last night to see if $F^{ab}F^{cd}$ obeys the algebraic Bianch identity but I wasn't getting anywhere. It's likely possible I'm missing something entirely obvious but I'm as stuck as you are right now!

 

[JustinLevy]

Thanks for the info on invariants. This is quite interesting.

The fact that the electromagnetic field tensor has just two algebraically independent invariants is a standard theorem.

I'm really curious about this now. Is this discussed in Jackson? Maybe its time I got a copy of Landau-Lifshitz.
The leap from quadratic invariants to all invariants seems almost mystical.
Is this true for any antisymmetric rank 2 tensor? Or do the other symmetries of $F^{ab}$ come into play?

Similarly, the Riemann tensor has four algebraic invariants, two of them quadratic, two of them cubic.

Can you recommend a textbook to read up on this?
And what about the usual Ricci curvature scalar, $R=R^{abcd}g_{ac}g_{bd}$? Does contracting with the metric not count here? So does that mean things like $R^{ab} R_{ab}$ don't count either?

 

[JustinLevy]

You're not misusing it but as you've seen it isn't necessary. If you have $F^{ab} = 2\nabla^{[a}A^{b]}$ then $\nabla^{[a}F^{bc]} = 0$ is immediate...

So in what sense does that vanishing have anything to do with the Bianchi identity?
Maybe it is a bit of a misnomer that those parts of Maxwell's equations come from the Bianchi identity, because now it feels like it is independent of the Bianchi identity.

But $\nabla^{[a}\nabla^b \nabla^c \nabla^{d]} \alpha^2 = \nabla^{[a}\nabla^b \nabla^{[c} \nabla^{d]]} \alpha^2 = 0$ so I'm not immediately seeing how useful it will be to you.

Wait, why is that first term equal to the second (with the extra permutation)?

Is my leap from
$Q^{abcd} = \nabla^{[a}\nabla^b \nabla^c \nabla^{d]} \alpha^2$
to
$Q^{abcd} = (\nabla^{[a}\nabla^b \alpha)(\nabla^c \nabla^{d]} \alpha)$
justified?

I was hoping this was getting closer to
$F^{ab}F^{cd} = 4 (\nabla^{[a} \nabla^{b]}\alpha)(\nabla^{[c} \nabla^{d]} \alpha)$

But yeah, it may not be useful at all.

 

[Bill_K]

Similarly, the Riemann tensor (EDIT: in vacuum) has four algebraic invariants, two of them quadratic, two of them cubic.

And what about the usual Ricci curvature scalar, $R=R^{abcd}g_{ac}g_{bd}$? Does contracting with the metric not count here? So does that mean things like $R^{ab} R_{ab}$ don't count either?

Forgot to say I meant the vacuum Riemann tensor. If you include the Ricci tensor, things get rather more complicated!

Here's the full list of invariants in that case. The four invariants for vacuum are the real and imaginary parts of what they call W₁ and W₂.

 

[WannabeNewton]

So in what sense does that vanishing have anything to do with the Bianchi identity?

See chapter 15 of MTW. By the way the statement $\nabla_{[a}F_{bc]} = 0$ refers to the differential Bianchi identity, which is what is discussed in chapter 15 of MTW. For $R_{abcd}$ this is $\nabla_{[a}R_{bc]de} = 0$. The differential Bianchi identity can be motivated geometrically ("boundary of a boundary is zero") so make sure to keep it distinct from the algebraic Bianchi identity which for $R_{abcd}$ is the usual $R_{a[bcd]} = 0$.

Wait, why is that first term equal to the second (with the extra permutation)?

$T^{[a_1...a_n]} = T^{[a_1...[a_i a_j]...a_n]}$ for any $1 \leq i,j \leq n$ for any tensor $T^{a_1..a_n}$.

Is my leap from
$Q^{abcd} = \nabla^{[a}\nabla^b \nabla^c \nabla^{d]} \alpha^2$
to
$Q^{abcd} = (\nabla^{[a}\nabla^b \alpha)(\nabla^c \nabla^{d]} \alpha)$
justified?

Well yes but it won't really help you because $Q^{abcd} = 0$ identically. This is clear from the first line but you can see this in the second line as well because $(\nabla^{[a}\nabla^b \alpha)(\nabla^c \nabla^{d]} \alpha) = (\nabla^{[a}\nabla^b \alpha)(\nabla^{[c} \nabla^{d]]} \alpha) = 0$