# Simplifying this equation?

1. May 23, 2015

### beyondthemaths

Hello, I am very new to tensors and GR and would like to ask for guidance to understand how tensor simplification works.

If we have this term $$\frac{1}{2}N_{IJ}F_{\mu\nu}^I\tilde{F}^{J\mu\nu}$$ and I want to derive w.r.t $F^{\rho\sigma I}$

where
- $N_{IJ}$ is a symmetric complex matrix
- $\tilde{F}^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$

- How to derive w.r.t $F^{\rho\sigma I}$? (I have no $\rho\sigma$ in the term above--- I know this how it goes but I have no idea how).

That is to say, had the derivation were w.r.t $F^{\mu\nu}$ and I had instead of $\frac{1}{2}N_{IJ}F_{\mu\nu}^I\tilde{F}^{J\mu\nu}$ a $1/2F^{\mu\nu}F_{\mu\nu}$, the derivation would give me an $F_{\mu\nu}$, but in this simple example I presented now, there is no F with indices ${\rho\sigma I}$ instead of ${\mu\nu}$.

2. May 23, 2015

### pervect

Staff Emeritus
If F is a rank 3 tensor i i.e you have $F^{I}{}_{\mu\nu}$, which is how I interpreted your post (I am not quit clear why the I index is roman and the others are greek), then it's hodges dual, $\tilde{F}$ will be a rank 1 tensor, assuming you're working in the usual 4 dimensonal space. You write the hodges dual operator of a rank 2 tenor (which you also call F) as another rank 2 tensor $\tilde{F}$ which makes sense, but I am not understanding the rank 3 tensor version of your question. I suspect it's just wrong, but I could be misunderstanding something.

add - if the issue is the hodges dual, I'd recommend reading a bit about clifford algebras - the hodge dual operator didn't make any sense for me until I got that background. See for instance "

3. May 23, 2015

### beyondthemaths

No, here "I" is related to the "I" in $N_{IJ}$ which is a matrix. F is a rank 2 tensor. Suppose there were no "I" or "J" to start with, how can I derive $1/2 N_{IJ}F_{\mu\nu}\tilde{F}^{\mu\nu}$ wrt $F^{\rho\sigma}$ meanwhile it has no $\rho$ nor $\sigma$ in it?

4. May 23, 2015

### stedwards

I'm confused as well, but I think what we're going for here, is that the Greek indices are antisymmetric with each other.

Then the Latin indices count through an array of two index antisymmetric tensors and $\tilde{F}^{J\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}{F^{J}}_{\rho\sigma}$.

Substitute the left hand side into your equation.

Last edited: May 23, 2015
5. May 23, 2015

### beyondthemaths

I guess, the question is not clear enough.

My confusion is in the following, if in the least usage of words: If $L = \frac{1}{2}N_{IJ}F_{\mu\nu}^I\tilde{F}^{J\mu\nu}$, what is $\delta L/\delta F^{\rho\sigma J}$?

6. May 23, 2015

### fzero

What you're doing here is called taking a functional derivative. We have to start with the definition

$$\frac{\delta F^I_{\mu\nu}(y)}{\delta F^J_{\rho\sigma}(x)} = \delta^I_J \delta^\rho_\mu \delta^\sigma_\nu \delta(y-x).$$

The Dirac delta is there because we usually take functional derivatives of integrated quantities like the action, rather than just the Lagrangian. The next step would be to take the expression you're differentiating and organize the indices to facilitate using the definition of the functional derivative:

\begin{align} S & = \int d^4y \frac{1}{2}N_{IJ}F_{\mu\nu}^I(y)\tilde{F}^{J\mu\nu}(y) \\ &= \int d^4y \frac{1}{4}N_{IJ}\epsilon^{\mu\nu\rho\sigma} F_{\mu\nu}^I(y) F^J_{\rho\sigma}(y).\end{align}

Now we can compute

$$\frac{\delta S}{\delta F^K_{\lambda\tau}(x)} = \int d^4y \frac{1}{4}N_{IJ}\epsilon^{\mu\nu\rho\sigma} \left( \frac{\delta F_{\mu\nu}^I(y) }{\delta F^K_{\lambda\tau}(x)} F^J_{\rho\sigma}(y) + F_{\mu\nu}^I(y)\frac{\delta F^J_{\rho\sigma}(y) }{\delta F^K_{\lambda\tau}(x)} \right).$$

I'll leave the simplification of that as an exercise for you to try, but post back if you still have trouble.

7. May 23, 2015

### beyondthemaths

Hello!! Thank youuu, that is what I was meaning. So, if I want to carry on from your last step,

$$\frac{\delta S}{\delta F_{\lambda \tau}^K(x)}=\int{d^4y\frac{1}{4}N_{IJ}\epsilon^{\mu\nu\rho\sigma}(\delta^I_K\delta^{\lambda}_{\mu}\delta^{\tau_\nu}\delta(y-x)F_{\rho\sigma}^J(y)+F_{\mu\nu}^I(y)\delta^J_K\delta^{\lambda}_{\rho}\delta^{\tau}_{\sigma}\delta(y-x))}$$
I suppose $y=x$ here in my action, so $\delta(y-x)=\delta(0)=1$, isn't it?
So if I am on the right track, then in my case, the $\lambda = \rho, \tau=\sigma, K=J$ so applying the formula of kronecker delta, my final answer would leave $$\int{d^4y\frac{1}{4}N_{IJ}\epsilon^{\mu\nu\rho\sigma}(F_{\mu\nu}^I)}$$

Is that correct?

8. May 23, 2015

### fzero

No, this is the Diract delta, so it satisfies

$$\int d^4y f(y) \delta(y-x) = f(x).$$

No. Let's go step by step, first using the Dirac delta

$$\begin{split} \frac{\delta S}{\delta F_{\lambda \tau}^K(x)} & =\int{d^4y\frac{1}{4}N_{IJ}\epsilon^{\mu\nu\rho\sigma}(\delta^I_K\delta^{\lambda}_{\mu}\delta^\tau_\nu\delta(y-x)F_{\rho\sigma}^J(y)+F_{\mu\nu}^I(y)\delta^J_K\delta^{\lambda}_{\rho}\delta^{\tau}_{\sigma}\delta(y-x))} \\ & =\frac{1}{4}N_{IJ}\epsilon^{\mu\nu\rho\sigma}\left( \delta^I_K\delta^{\lambda}_{\mu}\delta^\tau_\nu F_{\rho\sigma}^J(x)+F_{\mu\nu}^I(x)\delta^J_K\delta^{\lambda}_{\rho}\delta^{\tau}_{\sigma} \right)\\ & =\frac{1}{4} \left( N_{KJ}\epsilon^{\lambda\tau \rho\sigma} F_{\rho\sigma}^J(x)+ N_{IK}\epsilon^{\mu\nu\lambda\tau} F_{\mu\nu}^I(x)\right)\\ & = \frac{1}{4} \epsilon^{\lambda\tau \rho\sigma} \left( N_{KI}F_{\rho\sigma}^I(x)+ N_{IK} F_{\rho\sigma}^I(x)\right) \\ & =\frac{1}{2} \left( N_{KI} \tilde{F}^{I\lambda\tau}(x)+ N_{IK} \tilde{F}^{I \lambda\tau}(x) \right). \end{split}$$

In the second to last line we have used the symmetries of the $\epsilon$ tensor and changes of summed indices.

9. May 23, 2015

### beyondthemaths

Oh my, what I did was terrible! I now understand this, thank you very much!

10. May 23, 2015

### beyondthemaths

I have one question, as I said above $N_{IJ}$ is a complex symmetric matrix, that is I do not know if we should consider $I$ and $J$ to be functional here and embed them in a Kronecker delta or should I consider them as if they were not there because those are things that span the space and almost are present in SUGRA stuff (Especially special geometry sections)?

That is one hand, one the other, I would like to ask you about something, if it were $$N_{IJ}\tilde{F}^I_{\mu\nu}\tilde{F}^{J\mu\nu}$$Does this mean that this will be $$\frac{N_{IJ}}{4}(\epsilon_{\mu\nu\rho\sigma}F^{I\rho\sigma}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}^J)$$ and thus using the relation$$\epsilon^{\rho\sigma\mu\nu}\epsilon_{\mu\nu\rho'\sigma'}=-2(\delta^{\rho}_{\rho'}\delta^{\sigma}_{\sigma'}-\delta^\rho_{\sigma'}\delta^{\sigma}_{\rho'})$$ so we get $$N_{IJ}(-2)(\delta^{\rho}_{\mu}\delta^{\sigma}_{\nu}-\delta^{\rho}_{\nu}\delta^{\sigma}_{\mu})F^{I\rho\sigma}F^J_{\rho\sigma}$$
then if I followed what you guided me through above, I get $$-2(F^{I\mu\nu}F^J_{\mu\nu}-F^{I\nu\mu}F^J_{\nu\mu})$$ and since at last, F is an antisymmetric tensor, we get a zero? Is that correct? (The zero typically scares me and this is my first attempt after your above guidance so I would like to make sure from you, if you want I could copy this in a new question)

Thank you @fzero and sorry since I extended this to a further question, please let me know if it is better to write it in a new question.

11. May 23, 2015

### fzero

Usually the $I,J$ should be treated as numerical indices referring to some internal symmetry. We use a notation to distinguish them from the spacetime indices $\mu,\nu,\ldots$. I didn't use the symmetry of $N_{IJ}$ above, but that can be used to simplify the expression to one term.

When you are using the Einstein summation convention where repeated indices are summed over, you have to be careful not to use the same index to refer to different sums. A rule of thumb is that a given index should never appear more than twice in a single term, otherwise it's too easy to get confused. So you should write

$$\frac{N_{IJ}}{4}(\epsilon_{\mu\nu\rho'\sigma'}F^{I\rho'\sigma'}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}^J)$$

You've gotten indices confused. We have

$$-\frac{2N_{IJ}}{4}(\delta^{\rho}_{\rho'}\delta^{\sigma}_{\sigma'}-\delta^\rho_{\sigma'}\delta^{\sigma}_{\rho'})(F^{I\rho'\sigma'}F_{\rho\sigma}^J) = -\frac{N_{IJ}}{2} \left( F^{I\rho\sigma}F_{\rho\sigma}^J-F^{I\sigma\rho}F_{\rho\sigma}^J\right)$$

Using the antisymmetry of $F$ we see that the 2 terms add rather than cancel.

As long as the questions are related I don't think it breaks any rules. Also I will see notifications for activity in this thread.

12. May 23, 2015

### beyondthemaths

Thank you for being very helpful. I appreciate it a lot @fzero!

13. May 23, 2015

### beyondthemaths

Then it is okay to consider the I and J and place them in a kronecker delta and this has nothing to do with the fact that they are functional or not.

14. May 23, 2015

### fzero

I should have said that in the context of supergravity and related topics it is typical to see an index like that referring to some discrete label $I,J=1,2,\ldots ,N$ so you do sums over them rather than integration.

At the risk of confusing you, I would mention that I have seen a "supercondensed" notation in the context of functionals where an index could refer to a continuous variable, but I tend to doubt your text is using it. It would be mentioned or otherwise clear from the context.

15. May 23, 2015

### beyondthemaths

Indeed, those are what I am talking about so to cut the story short, I will continue with what we started $\delta^{()}_{()}\delta^I_J$, safe no?

16. May 23, 2015

### fzero

In the functional derivative, yes.

17. May 24, 2015

### beyondthemaths

Hello @fzero ! Sorry I am reviving the thread again but as I was reviewing what I learned yesterday, I am wondering about something:

In this one, now in order to proceed like we did in our first case, I should lower indices in $F^{I\rho\sigma}$ right? So I can also use the $\epsilon$ for that purpose? That is to say $F^{I\rho\sigma}=\epsilon^{\rho\sigma\mu\nu}F^I_{\mu\nu}$ and then I continue. Is that correct?

18. May 24, 2015

### physciencer

@fzero I have a question, when @beyondthemaths wrote at the very beginning $1/2N_{IJ}F^I_{\mu\nu}\tilde{F}{J\mu\nu}$ and then the functional derivative as you have defined it will be something like $\frac{\delta S}{\delta F_{\lambda \tau}^K}$, what I notice was that $F_{\lambda \tau}^K$ had the indices lowered. Now my question is, were it $1/2 N_{IJ}\tilde{F}^I_{\mu\nu}F^{J\mu\nu}$ instead, the duality of electromagnetic tensor equation will lead us to write $1/4N_{IJ}\epsilon_{\mu\nu\rho\sigma}F^{I \rho\sigma}F^{J\mu\nu}$, now the functional derivative will be $$\frac{\delta S}{\delta F_{\lambda \tau}^K} = \int{... \epsilon_{\mu\nu\rho\sigma}\frac{N_{IJ}}{4}[\frac{\delta F^{I \rho\sigma}}{\delta F_{\lambda \tau}^K}F^{J\mu\nu} + ...] }$$
does the equation you wrote for the directional derivative in your very first post remain the same if you have indices like that?

Thank you.

19. May 24, 2015

### fzero

No, $\epsilon^{\rho\sigma\mu\nu}F^I_{\mu\nu} = 2 \tilde{F}^I_{\mu\nu}$. To raise and lower indices you must use a metric $g_{\mu\nu}$. In fact, whenever we have an expression like $A^a A_a$, there is a hidden factor of the metric since $A^a A_ = g_{ab} A^a A^b$. In the formula you quoted, in order to take the functional derivative, it would be simpler to write $F^{I\rho\sigma} F^J_{\rho\sigma} = g^{\rho\rho'} g^{\sigma\sigma'} F^I_{\rho\sigma}F^J_{\rho'\sigma'}$ first.

20. May 24, 2015

### fzero

You make a similar point to beyondthemaths' last post. In this case it is simplest to introduce the metric to put the indices in the lowered positions used in the definiton of the derivative. Alternatively, one can use the metric to formulate the proper rule to take the derivatives with the indices in mixed position. I think it's best to simplify in the Lagrangian while learning this and then learn the shortcuts by experience. This way you're not trying to memorize lots of things at the outset.

Also if $F$ is a field strength and the potential $A$ appears elsewhere in the action, then you would have to vary with respect to the potential instead of $F$ to get consistent results.