# Kronecker delta symbol in calculation question

1. Jul 9, 2014

### CAF123

As part of a physics calculation, I have the following integral $$\int d \bar x a^{\sigma} \left[-\partial_{\mu}\left(\frac{\delta x^{\nu}}{\delta a^{\sigma}}\right) (\partial_{\nu}\Phi )\frac{\partial L}{\partial (\partial_{\mu}\Phi)} + \partial_{\mu}\left(\frac{\delta x^{\mu}}{\delta a^{\sigma}}\right)L\right],$$ with Einstein summation convention understood.

Using the fact that $x'^{\mu} = x^{\mu} + a^{\mu}\,\,\,(1)$, this reduces to

$$\int d \bar x a^{\sigma} \left[-\partial_{\mu}\delta^{\nu}_{\sigma} (\partial_{\nu} \Phi) \frac{\partial L}{\partial (\partial_{\mu}\Phi)} + \partial_{\mu} \delta^{\mu}_{\sigma}L\right]$$

In the first term in the first integral the term $\partial_{\mu}$ only acts on the bracketed term shown. When I use the result (1) this bracketed term changes to a delta, which then consequently changes the index on the following $\partial_{\nu}$ term. My question is, what does the term $\partial_{\mu}$ now act on? Before it acted only on the bracketed term, but now that that is replaced with a delta and using the fact that $\delta^{\nu}_{\sigma}\partial_{\nu} = \partial_{\sigma}$ it seems that the only possibility would be for it to act on the new $\partial_{\sigma}\Phi$ term. Is that correct?

2. Jul 9, 2014

### Fredrik

Staff Emeritus
That's not a possibility. A product like f'(x)g(x) isn't going to change into cg'(x) where c is some number, just because f is something simple like a constant function.

3. Jul 9, 2014

### CAF123

Hi Fredrik,
I agree, that is why I asked the question. I don't see what else the resulting expression would be - the $\partial_{\mu}$, being a differential operator, has to act on something, so what would it be? I don't see what else it would be that would keep the summation convention sensible.

Thanks.

Last edited: Jul 9, 2014
4. Jul 9, 2014

### Fredrik

Staff Emeritus
You seem to expect that there's some fancy mathematics at play here, but there isn't. The key here is the trivial observation that if it acts on a function f, and f=g, then it acts on g.

This is a situation where it may help to be careful with the notation. For example, don't simplify notations like f'(x) to f'. If D takes a function to its derivative, be careful to write (Df)(x) instead of something like D(f(x)). If you know what everything acts on at the start of the calculation, you will know it at the end of the calculation too.

5. Jul 9, 2014

### CAF123

Ok, so in this example, $$\partial_{\mu} \left(\frac{\delta x^{\nu}}{\delta a^{\sigma}}\right) (\partial_{\nu}\Phi) = \partial_{\mu} \delta^{\nu}_{\sigma} \partial_{\nu}\Phi$$ Now at this point, I have my difficulty. By making use of the Kronecker symbol, thereby changing the $\nu$ index on the subsequent partial to a $\sigma$, I get rid of it. So I understand your last post in that $\partial_{\mu}$ still acts on the bracketed term throughout, but given what it reduces to and the properties of the Kronecker symbol, I don't really see how to write down the next step above.

$\partial_{\mu} (\partial_{\sigma}\Phi)$ would suggest the $\partial_{\mu}$ acts on the $\partial_{\sigma}\Phi$ term but I know that not to be true. It is really the derivative acting on the Kronecker symbol that I can't quite see how to express.

Thanks.

6. Jul 9, 2014

### Fredrik

Staff Emeritus
I think you're going to have to explain the $\delta(\text{something})/\delta(\text{something})$ notation. Is this a calculation from a book?

You say that $\frac{\delta x^\nu}{\delta a^\sigma}=\delta^\nu_\sigma$ (or at least that's what I thought you where saying). I don't know what that could mean if not that $\frac{\delta x^\nu}{\delta a^\sigma}$ is equal to the constant function $x\mapsto\delta^\nu_\sigma$. All partial derivatives of that function (and all other constant functions) are of course zero. If that's not what you want, then something is wrong at an earlier stage.

7. Jul 10, 2014

### CAF123

Sort of, I'm trying to obtain the result in Di Francesco's book on CFT where he derives the Noether current. The integral above is exactly the second term in the very last integral of my other thread here:https://www.physicsforums.com/showthread.php?t=760137 I could not find a way to make this second term go to zero generally, nor could my professor (can you see any way? ;) ) so we thought we would try it out on various symmetry transformations to test its validity. As you can see from that thread, strangerep already got the same result and indeed I have verified the term vanishes for a Lorentz transformation, but not quite translation (the case in this thread) or dilations. The notation is just a derivative wrt the symmetry parameter a. $x^{\nu}$ and $a^{\mu}$ are contravariant components of 4 vectors, for example if the space is (3+1) dim Minkowskian.

Yes

For a translation, $x'^{\mu} = x^{\mu} + a^{\mu} \Rightarrow x'^{\mu} - x^{\mu} = \delta x^{\mu} = a^{\mu}$ so $\delta x^{\mu}/\delta a^{\sigma} = \delta^{\mu}_{\sigma}$. That is just a mathematical re writing of the fact that the change of the vector $\mathbf x$ in the $\mu$ direction is only dependent on the component of $\mathbf a$ in the $\mu$ direction.

Last edited: Jul 10, 2014
8. Jul 10, 2014

### Fredrik

Staff Emeritus
Wait, so you need this integral to be zero, and you've found that each term contains a derivative of a constant function? That solves the problem, doesn't it?

I'm looking at the book right now. I really hate its presentation of these things.

9. Jul 10, 2014

### CAF123

Do you mean to say that $\partial_{\mu}\delta^{\nu}_{\sigma} = 0$ since the Kronecker delta can be represented as the unit matrix, which is constant.
If that is all there is to it, then thanks - I got distracted by the property of the Kronecker symbol.
I see, it appears strangerep and vanhees71 shared that opinion.

10. Jul 10, 2014

### Fredrik

Staff Emeritus
It's even simpler than that. For all $\nu,\sigma\in\{0,1,2,3\}$, we have $\delta^\nu_\sigma\in\{0,1\}$, so $\partial_\mu$ acts on one of the two constant functions $x\mapsto 0$ or $x\mapsto 1$.

But I have to say that I still don't understand the notations. So I still can't be 100% sure that the calculations must be interpreted this way.