Simplifying this equation?

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}##.
 

pervect

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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 "
[PDF]Imaginary Numbers are not Real — the Geometric Algebra of space-time", http://geometry.mrao.cam.ac.uk/wp-content/uploads/2015/02/ImagNumbersArentReal.pdf
[
 
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?
 
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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 [itex]\tilde{F}^{J\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}{F^{J}}_{\rho\sigma}[/itex].

Substitute the left hand side into your equation.
 
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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}##?
 

fzero

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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}##?
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.
 
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?
 

fzero

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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?
No, this is the Diract delta, so it satisfies

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

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

fzero

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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)?
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.

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

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

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)
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.

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.
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.
 
Thank you for being very helpful. I appreciate it a lot @fzero!
 
Usually the I,JI,J should be treated as numerical indices referring to some internal symmetry. We use a notation to distinguish them from the spacetime indices μ,ν,…\mu,\nu,\ldots. I didn't use the symmetry of NIJN_{IJ} above, but that can be used to simplify the expression to one term.

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.
 

fzero

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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.
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.
 
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,…,NI,J=1,2,\ldots ,N
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?
 

fzero

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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?
In the functional derivative, yes.
 
Hello @fzero ! Sorry I am reviving the thread again but as I was reviewing what I learned yesterday, I am wondering about something:

##−N_{IJ}/2(F^{I\rho\sigma}F^J_{\rho\sigma}−F^{I\sigma\rho}F^J_{\rho\sigma})##
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?
 
@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.
 

fzero

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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?
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.
 

fzero

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@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.
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.
 
Also if FF is a field strength and the potential AA appears elsewhere in the action, then you would have to vary with respect to the potential instead of FF to get consistent results.
Do you mean that I should have derived wrt A? Well, because what I am trying to find is ##G_{\mu\nu}=\delta S/\delta F^{\mu\nu}##. The formula that I have come across in the web says so, maybe in few resources it says so, in some others, it says ##G_{\mu\nu}=\epsilon_{\mu\nu\rho\sigma}\delta S/\delta F## (I don't know the difference and which to use) but in any case, it never said ##\delta S/\delta A##. I am a little confused here.
 

fzero

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Do you mean that I should have derived wrt A? Well, because what I am trying to find is ##G_{\mu\nu}=\delta S/\delta F^{\mu\nu}##. The formula that I have come across in the web says so, maybe in few resources it says so, in some others, it says ##G_{\mu\nu}=\epsilon_{\mu\nu\rho\sigma}\delta S/\delta F## (I don't know the difference and which to use) but in any case, it never said ##\delta S/\delta A##. I am a little confused here.
How to vary the action comes down to context. If the action only depends on the field strength and you have some physical reason to derive a quantity related to the variation wrt the field strength then there should be no problem. I just pointed that out because if you were facing some different problem, like the EM field coupled to an electron, then the EM potential would appear explicitly in the action and it wouldn't generally be appropriate to only vary wrt to the field strength anymore. It's just something to keep in mind that is bound to crop up in your future studies.
 
Ok thank you a lot. I now got what you meant.
 
Ah @fzero so you mean ##\delta S/\delta F^{\lambda \tau}= \frac{\delta S}{g^{\lambda \lambda'}g^{\tau \tau'}\delta F_{\lambda'\tau'}}##?
 

fzero

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Ah @fzero so you mean ##\delta S/\delta F^{\lambda \tau}= \frac{\delta S}{g^{\lambda \lambda'}g^{\tau \tau'}\delta F_{\lambda'\tau'}}##?
Well it's probably clearer to write

$$\delta S/\delta F^{\lambda \tau}= g_{\lambda \lambda'}g_{\tau \tau'} \frac{\delta S}{delta F_{\lambda'\tau'}},$$

but yes you can derive rules for how the functional derivatives are related by raising and lowering indices (provided that the metric itself doesn't depend on the field that you're varying). I would recommend that beginners not do this and start by organizing indices in the action like

$$ F^{\lambda \tau} F_{\lambda \tau} = g^{\lambda \lambda'}g^{\tau \tau'}F_{\lambda \tau} F_{\lambda' \tau'}.$$

Then you're taking the same type of derivative everywhere. If you are already comfortable doing it the other way, that is fine too.
 

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