Timeline for Multivariable Dirac Delta and Faddeev-Popov Determinant
Current License: CC BY-SA 3.0
14 events
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| Oct 29, 2019 at 21:11 | comment | added | TMS | @AltLHC After scratching my head for some time, I realized how we can use diagonolized $J$ during this proof: We actually assume this at first. Then, when we get the result (before integration), we note that this result is actually rotation invariant, since the determinants (that represents volume) are rotation invariant, and that n-dimensional Dirac delta, is also rotation invariant (see Wikipedia), and thus the result is true for any Jacobean. | |
| Apr 13, 2018 at 15:36 | comment | added | Blind Miner | @JeffDror How do we know that the Jacobian $J$ is unitarily diagonalizable? The statement that the product of diagonal elements is equal to the determinant seems to suggest this. Thanks for your help! | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Oct 9, 2015 at 11:10 | comment | added | AltLHC | So, after a bit of consideration the whole discussion seems silly. Would it not suffice to write $1 = \int d{\mathbf{x}}\;\delta^{(n)}(\mathbf{x}-\mathbf{x}_{0})$ and do a variable change to deduce the transformation properies of the delta function? | |
| Oct 8, 2015 at 17:44 | comment | added | AltLHC | Hm, guess the delta rewriting works for $J_{1,1}=J_{2,1}=...$ as well. | |
| Oct 8, 2015 at 17:41 | comment | added | AltLHC | @JulioParra: Just make $m$ copies of the function such that $f_{j} = f$ - then it is a matrix and the proof is fine. Additionally, $\frac{\partial f_{i}}{\partial x^{i}}$ is not a diagonal matrix but the diagonal elements of the matrix $\frac{\partial f_{j}}{\partial x^{i}}$. | |
| Oct 8, 2015 at 17:23 | comment | added | AltLHC | Nice answer but it seems only to be valid for diagonal Jacobians. In the step after the statement "We choose to isolate each delta function in the equation above for a different $a_j$" you quickly choose $J_{1,1}$ to be the prefactor to the $a_{1}-a_{0,1}$ zero but you could just as well have chosen $J_{2,1}$. For non-diagonal Jacobians this doesn't hold up. Another indication of this problem is in the identification $J_{1,1}J_{2,2}... = \det J$ which only works for diagonal J. | |
| Feb 28, 2014 at 16:47 | comment | added | jpm | I think you have to be more careful here since the function we are considering is $f:\mathbb{R}^n \to \mathbb{R}$ and not $f:\mathbb{R}^n \to \mathbb{R}^m$ then $\frac{\partial f(x^i)}{\partial x^i}$ is simply the normal partial derivative $\frac{\partial f}{\partial x^i}$ and keeping the argument is just misleading. This is different from $\frac{\partial f_i}{\partial x^i}$, which indeed would be a diagonal matrix to the matrix of partial derivatives of all the functions $f_j$ which form an $f:\mathbb{R}^m \to \mathbb{R}^n$ function as $f=(f_1(x^i),f_2(x^i),\cdots,f_m(x^i))$. | |
| Feb 28, 2014 at 15:19 | vote | accept | Physics_maths | ||
| Feb 28, 2014 at 14:35 | comment | added | Physics_maths | Makes much more sense to me now :) thanks for your great answer. | |
| Feb 28, 2014 at 14:30 | comment | added | JeffDror | Whoops, got to exciting to post an answer... My mistake. I updated it now. It is indeed a vector. | |
| Feb 28, 2014 at 14:30 | history | edited | JeffDror | CC BY-SA 3.0 |
added 45 characters in body
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| Feb 28, 2014 at 14:07 | comment | added | Physics_maths | Hi, thanks for the answer, I have some question though. The notation $J _{ ij} \equiv \frac{ \partial g _{ i} }{ \partial a _j } \big|_{ a _0 }$ makes sense because $g_i$ is a component of the vector function $\bf g$. But in the first paragraph you state that \begin{equation} J _{ ij} = \frac{ \partial f ( x ^i ) }{ \partial x ^j } \end{equation} which doesn't make sense or at least does not agree with the one involving g. I get confused by the notation. Your derivation seems correct and I get that but the first part is a little unclear, which was my original question. Thanks. | |
| Feb 28, 2014 at 12:37 | history | answered | JeffDror | CC BY-SA 3.0 |