Talk:Kronecker delta/Archive 1

Archive 1

Unit Impulse Function

The unit impulse function is defined as a combination of two formulas. One defining where the function is zero:

$\delta (x-y)=0,x\neq y$

And one stating that the entire integral must be equal to one:

$\int _{-\infty }^{\infty }\delta (x-y)f(x)dx=1,$

Maybe it should be pointed out that the unit impulse is the same one as the Dirac delta. Anyway I would move the whole treatment of the unit impulse function to the appropriate article (instead of giving overhere two definitions of this other mathematical object). Andy 22:36, 20 September 2006 (UTC)

$\int _{-\infty }^{\infty }\delta (x-y)f(x)dx=f(y),$

--Bob K 13:51, 21 September 2006 (UTC)

TeX cases

I don't like the ", if" in the definition, just \begin{cases} 1 & x = 0 \\ ... usually seems sufficient. --136.152.170.83 (talk) 04:16, 15 August 2008 (UTC)

Earlier comments

To claim as the article does that the Kronecker delta is a discrete analogue of the Dirac delta is very funny. The Dirac delta came afterwards, was named delta by analogy with Kronecker's and is not even an integral kernel, while the Kronecker delta is a bona fide matrix. — Miguel 22:12, 2004 Nov 21 (UTC)

There's another definition of the generalized Kronecker delta that is defined by the sign of the permutation that maps the contravariant indices to the covariant ones. It's equal to and sometimes defined by a determinant of regular Kronecker deltas. Do a search on Planetmath for more info. Jason Quinn 19:57, 11 Mar 2005 (UTC)

I don't get why it references an OR gate in the end. That doesn't make sense to me. Peter (talk) 08:10, 28 August 2008 (UTC)

Punctuation

In the article, there is first a comma between the indices, and later there is no comma. Are both forms of punctuation equally correct? Ezekiel 33:32 (talk) 08:35, 6 June 2010 (UTC)

I have seen both used. Without comma is more usual but can occasionally cause confusion. (Eg if i=222 and j=2 without comma one would write \delta_{2222} which also might mean i=22 and j=22.) MathHisSci (talk) 09:50, 9 August 2010 (UTC)

New symbol

I am wondering if a symbol which behaves as the opposite of Kronecker delta would be useful. I need it for my research. I want to propose $!\delta _{nk}$ to that end. So $!\delta _{nk}=0$ when n=k and $!\delta _{nk}=1$ when $n\neq k$

Wilkn (talk) 01:22, 31 October 2010 (UTC)

You might want to try $1-\delta _{nk}$. Tkuvho (talk) 05:22, 31 October 2010 (UTC)

Thanks !

Wilkn (talk) 22:55, 8 November 2010 (UTC)

Clarification: is the Kronecker delta a function, or "notational shorthand"?

The statement that the Kronecker delta "is treated as a notational shorthand rather than as a function" has been in the lead section of this article since its inception in 2003. I've tagged this as needing clarification because the lead also states that "the Kronecker delta... is a function." Which is it then? Or is it both? Either way, we need to clarify. --TSchwenn (talk) 19:24, 18 May 2012 (UTC)

Good points. As far as I know its the identity tensor (which amounts to the shorthand for 0 or 1) in descrete contexts, or the delta function in the continuous case (descrete indices are replaced by continuous variables). It might help to define both at the beginning and keep the descrete and continuous versions seperate through the article, preferably the dscrete case first since its easier to understand, then the continuous. Anyone can resection things, but I'm no expert on distribution theory though... F = q(E+v×B) ⇄ ∑_ic_i 17:40, 21 May 2012 (UTC)

By delta function, I guess you mean Dirac delta function, which could be used to define the Kronecker delta like this:

$\delta _{ij}=\int _{j}^{j}\delta (i)di$

Does that work? --TSchwenn (talk) 21:56, 21 May 2012 (UTC)

Not quite. If you are integrating w.r.t. j then it cannot appear as a free variable on the left hand side. You need something like

$\delta _{ij}=\int _{j}^{j}\delta (x-i)\,dx$

Gandalf61 (talk) 07:46, 22 May 2012 (UTC)

The Kronecker delta is a function (of discrete variables). The Dirac delta function, which acts on continuous variables, is not a function, but it is a generalized function. --Trovatore (talk) 21:41, 21 May 2012 (UTC)

I will delete the "notational shorthand" bit in the article then. Thanks for clarifying! --TSchwenn (talk) 22:02, 21 May 2012 (UTC)

Generalized version

I'm not a mathematician, but it seems to me the generalized Kronecker delta should be -1 for odd index permutations, unless there is more than one kind of generalized version? Reference: http://books.google.com/books?id=Gu3-ciFTbQgC&pg=PA277#v=onepage&q&f=false — Preceding unsigned comment added by 70.95.65.244 (talk) 11:34, 7 June 2012 (UTC)

For the record, that's Tensor Calculus and Riemannian Geometry 22nd edition (2007) by D. C. Agarwal, published by Krishna Prakashan Media. No ISBN to be found. --TSchwenn (talk) 06:07, 8 June 2012 (UTC)

Theodore Frankel (2012) The Geometry of Physics: An Introduction 3rd edition, published by Cambridge University Press, ISBN 9781107602601 has the same definition as is suggested here. In fact, doing a google book search on "generalized kronecker delta" appears to find only this definition. Thus, this seems to be more useful and certainly more notable than what appears to be shorthand for a repeated product of Kronecker deltas, so I'll replace the generalization presently in the article. — Quondum^☏ 13:27, 12 June 2012 (UTC)

Error in recursive definition of generalized Kronecker delta

くま兄やん (talk · contribs) noticed that my recursive definition of the generalized Kronecker delta was wrong and gave a correction for it. Due to over-confidence in my own work, repeating the same error, and disliking the format of his alternative, I reverted to my version. Quondum‎ (talk · contribs) gave another corrected version; and simultaneously, I saw the error of my version. However, I still do not like the format of the offered alternatives. How about this:

$\delta _{\nu \sigma _{1}\cdots \sigma _{n}}^{\mu \rho _{1}\cdots \rho _{n}}=\delta _{\nu }^{\mu }\delta _{\sigma _{1}\cdots \sigma _{n}}^{\rho _{1}\cdots \rho _{n}}-\sum _{k=1}^{n}\delta _{\sigma _{k}}^{\mu }\delta _{\sigma _{1}\cdots \sigma _{k-1}\;\nu \;\sigma _{k+1}\cdots \sigma _{n}}^{\rho _{1}\cdots \rho _{n}}\,$

version? JRSpriggs (talk) 13:37, 2 November 2012 (UTC)

This expression suggests something special about the first term, whereas all terms are actually interchangeable by symmetry. Some simple pattern is preferable. Your suggestion has an unswapped index, then that index in turn swapped with another for the remaining terms. I haven't checked the signs. My two patterns were first a cyclic shift, and second a list with one index removed and placed on the first delta. I find the pattern of one index being treated specially the least intuitive (the most difficult to infer a pattern from). The one source (Lovelock & Rund) gives the following formula (and there is some merit in using a sourced formula):

$\delta _{h_{1}\cdots h_{r}}^{j_{1}\cdots j_{r}}={\delta _{h_{r}}^{j_{r}}\delta _{h_{1}\cdots h_{r-1}}^{j_{1}\cdots j_{r-1}}-\delta _{h_{r}}^{j_{r-1}}\delta _{h_{1}\cdots h_{r-1}}^{j_{1}\cdots j_{r-2}j_{r}}+\delta _{h_{r}}^{j_{r-2}}\delta _{h_{1}\cdots h_{r-1}}^{j_{1}\cdots j_{r-3}j_{r-1}j_{r}}\cdots +(-1)^{r+1}\delta _{h_{r}}^{j_{1}}\delta _{h_{1}\cdots h_{r-1}}^{j_{2}\cdots j_{r}}}$

— Quondum 21:46, 2 November 2012 (UTC)

Either the permutation leaves the first index where it was ($\mu =\nu \,$) or it moves it to another location ($\mu =\sigma _{k}\,$). In the first case, one can simply apply the previous generalized Kronecker delta to handle the rest of the permutation. In the second case, do a single transposition (getting a factor of −1) to reduce this to the first case. JRSpriggs (talk) 07:29, 3 November 2012 (UTC)

I prefer that the expression on the left dictates the index naming and that we use the variable p, as for the other expressions. This would suggest one of the following expressions.

Yours, with indices renamed accordingly would be:

$\delta _{\nu _{1}\cdots \nu _{p}}^{\mu _{1}\cdots \mu _{p}}=\delta _{\nu _{p}}^{\mu _{p}}\delta _{\nu _{1}\cdots \nu _{p-1}}^{\mu _{1}\cdots \mu _{p-1}}-\sum _{k=1}^{p-1}\delta _{\nu _{k}}^{\mu _{p}}\delta _{\nu _{1}\cdots \nu _{k-1}\;\nu _{p}\;\nu _{k+1}\cdots \nu _{p-1}}^{\mu _{1}\cdots \mu _{p-1}}.$

An expression derived from that which I gave above would be (with the hat indicating an omitted index):

$\delta _{\nu _{1}\cdots \nu _{p}}^{\mu _{1}\cdots \mu _{p}}=\sum _{k=1}^{p}(-1)^{p-k}\delta _{\nu _{k}}^{\mu _{p}}\delta _{\nu _{1}\cdots {\hat {\nu }}_{k}\cdots \nu _{p}}^{\mu _{1}\cdots \mu _{p-1}}.$

These are compact enough that we could include both; they each have their merits. — Quondum 13:02, 3 November 2012 (UTC)

OK. JRSpriggs (talk) 09:41, 4 November 2012 (UTC)

Cool. I've replaced the list expressions with these general expressions. — Quondum 10:21, 4 November 2012 (UTC)