Dummy index and renaming if not a tensor

[binbagsss]

Homework Statement

see below

Relevant Equations

see below

Apologies it's been a little while since I've done this, but I believe the rule is, that if the object is not a tensor you can not rename the dummy index?

For example, i have the action

##\int d^3 x \epsilon^{uvp} A_u \partial v A_p ## and I want to write this in terms of the ##i## and ##0## components (so that I can take ##\frac{\delta S}{\delta A_0}## and so later on, I imagine, once I've intergreated by parts etc, it will become handy to rename the dummy indicies, but this can not be done for the levi-civita symbol...is this correct?

many thanks

 

[strangerep]

[...] but I believe the rule is, that if the object is not a tensor you can not rename the dummy index?

I don't think there's any such rule. If an index is a (repeated) dummy summation index, you can rename it.

For example, i have the action
##\int d^3 x \epsilon^{uvp} A_u \partial v A_p ## and I want to write this in terms of the ##i## and ##0## components (so that I can take ##\frac{\delta S}{\delta A_0}## and so later on,[...]

You might not need to do that. E.g., $$\epsilon^{\mu\nu\rho} \frac{\delta A_\rho}{\delta A_0} ~=~ \epsilon^{\mu\nu\rho} \delta^0_\rho ~=~ \epsilon^{\mu\nu 0} $$

 

[binbagsss]

I don't think there's any such rule. If an index is a (repeated) dummy summation index, you can rename it.

You might not need to do that. E.g., $$\epsilon^{\mu\nu\rho} \frac{\delta A_\rho}{\delta A_0} ~=~ \epsilon^{\mu\nu\rho} \delta^0_\rho ~=~ \epsilon^{\mu\nu 0} $$

Sorry , well I meant like if you have ##A^a B_a A_b B^b - A^aA_pB_aB^p## you can rename ##p## in the second term to ##b##say. (The reason I want to rename is to cancel the variation##\delta A_{\mu}##but in the two terms I have got they have different indices in this variation).. (but I will also read through what you said in a bit)

 

[binbagsss]

I don't think there's any such rule. If an index is a (repeated) dummy summation index, you can rename it.

You might not need to do that. E.g., $$\epsilon^{\mu\nu\rho} \frac{\delta A_\rho}{\delta A_0} ~=~ \epsilon^{\mu\nu\rho} \delta^0_\rho ~=~ \epsilon^{\mu\nu 0} $$

okay that looks simple enough, so also using the chain rule on the variation ##\frac{\delta S}{\delta A_0}= \frac{\delta S}{\delta A_{\rho}}. \frac{\delta A_\rho}{\delta A_0}##?

Actually in this case, it was a variation wrt ##A_{i}## and the final answer is a expression involving ##\epsilon^{ij}## with only two-indices and not three, I'm really confused how one managed to go from an expression involving a three index levi-civta tensor to an expression involving only two.

Many thanks

 

[strangerep]

Actually in this case, it was a variation wrt ##A_{i}## and the final answer is a expression involving ##\epsilon^{ij}## with only two-indices and not three, I'm really confused how one managed to go from an expression involving a three index levi-civta tensor to an expression involving only two.

Well, I can't be sure without seeing the full original context. Maybe because one of ##\epsilon## indices is ##0##?

(This is why we normally insist on a more complete "homework statement". The one in your opening post is not adequate for helping you properly.)

 

[binbagsss]

Well, I can't be sure without seeing the full original context. Maybe because one of ##\epsilon## indices is ##0##?

(This is why we normally insist on a more complete "homework statement". The one in your opening post is not adequate for helping you properly.)

okay but in answer to the original question , say i have ##\epsilon_{abc}A^bB^{ac} ## can i rename say ## b \to d## as:##\epsilon_{adc}A^dB^{ac}##?, for the levi-civita symbol, as it's not a tensor, was my initial question

 

[strangerep]

okay but in answer to the original question , say i have ##\epsilon_{abc}A^bB^{ac} ## can i rename say ## b \to d## as:##\epsilon_{adc}A^dB^{ac}##?, for the levi-civita symbol, as it's not a tensor, was my initial question

Yes. The summation convention doesn't care whether it's a tensor or a goose.

 

[binbagsss]

Well, I can't be sure without seeing the full original context. Maybe because one of ##\epsilon## indices is ##0##?

(This is why we normally insist on a more complete "homework statement". The one in your opening post is not adequate for helping you properly.)

Hi okay, apologies I am now stuck on this (this wasn't the origianl question nor intention of my post and hence why I did not state the full problem at the start, but I think it makes more sense to post it here since we went that way anyway..)So i have ## \epsilon^{uvp} a_u \partial_v a_p ## (as the integrand under a 3d integral) and I wantt o find the variation w.r.t ##a_i##. where ##i = 1,2 ## running only over the spatial indicies.

So, so far I have

## \epsilon^{nup}(a_u + \delta_u)\partial_v(a_p+\delta a_p) ##and so to order ##\delta a## after integrating the by parts the first term to order :

##\epsilon^{nup} (- \partial_v a_u \delta_p + \partial_v a_p \delta a_u ##.

And now this is where my confusion is as I can't factorise out ##\delta a ## to get ##\delta S / \delta a^c ## since both terms have different indicies on delta a .

So ## \delta S / \delta a^i= \delta S/ \delta a^ c . \delta c/ \delta a^i ##, however , where ##c## runs over all 3 indices, and ##i## only the spatial ones

but I'm stuck because i don't have ## \delta S/ \delta a^ c ## to even apply the chain rule since i can't factorise out ##a## as said above

thanks

 

[strangerep]

## \epsilon^{nup}(a_u + \delta_u)\partial_v(a_p+\delta a_p) ##

Where did the "n" index on ##\epsilon## come from? A typo? I'm guessing what you meant was: $$\epsilon^{uvp}(a_u + \delta_u)\partial_v(a_p+\delta a_p) ~~~?$$

and so to order ##\delta a## after integrating the by parts the first term to order : ##\epsilon^{nup} (- \partial_v a_u \delta_p + \partial_v a_p \delta a_u ##.

And here there seem to be more typos? I guess you meant $$\epsilon^{uvp} (- \partial_v a_u \delta a_p + \partial_v a_p \delta a_u) ~~~~~?$$

And now this is where my confusion is as I can't factorise out ##\delta a ## to get ##\delta S / \delta a^c ## since both terms have different indicies on delta a .

So ## \delta S / \delta a^i= \delta S/ \delta a^ c . \delta c/ \delta a^i ##, however , where ##c## runs over all 3 indices, and ##i## only the spatial ones

but I'm stuck because i don't have ## \delta S/ \delta a^ c ## to even apply the chain rule since i can't factorise out ##a## as said above

OK. In baby steps, here's what to do. First split it into 2 pieces:
$$\epsilon^{uvp} \partial_v a_p \delta a_u ~-~ \epsilon^{uvp}\partial_v a_u \delta a_p $$
Now rename the indices in one of the terms, so that ##\delta a_p## appears in both terms. (I'll choose the 1st term and interchange ##u \leftrightarrow p##).
$$\epsilon^{pvu} \partial_v a_u \delta a_p ~-~ \epsilon^{uvp}\partial_v a_u \delta a_p $$
Now remember that you can interchange any 2 indices on the ##\epsilon## provided you also change the sign. So the above can become
$$-\epsilon^{uvp} \partial_v a_u \delta a_p ~-~ \epsilon^{uvp}\partial_v a_u \delta a_p $$
which is just $$-2 \epsilon^{uvp} \partial_v a_u \delta a_p ~~.$$ Can you continue from there?