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I'm having trouble understanding the following. Say I have a quantity of objects arranged spatially, with distribution $dn/dx$. For one reason or another*, I want to sum auto-pairs of spatial points, but only pairs at the same place. Thinking about this discretely, we have

$$ \sum_{i,j} (\bar{n}_i \Delta x) (\bar{n}_j \Delta x) \delta_{ij} = \sum_i (\bar{n}_i \Delta x)^2 $$

Clearly this is dimensionless on both sides. However, moving into the continuous limit, I expect to have

$$ \int\int \frac{dn}{dx_i}\frac{dn}{dx_j}\delta(x_i-x_j)dx_idx_j = \int \left(\frac{dn}{dx}\right)^2 dx. $$

Here both sides have the same dimensions (inverse length), but this is different from the sum, which is dimensionless. What am I doing wrong??

Thanks for any help!


* For those who are interested, this problem arises (or at least I think it does) when finding the variance of a sum of correlated variables. In particular, let the number of objects be a function of $x$ and $y$, and let them be spatially correlated in $y$, but not $x$. Then

$$ {\rm Var}(\sum N_i) = \sum_{x_i}\sum_{x_j}\sum_{y_i}\sum_{y_j} (\bar{n}_{x_i} \Delta x \Delta y) (\bar{n}_{x_j} \Delta x \Delta y)C_{y_i y_j} \delta_{x_i x_j} = \sum_{x_i}\sum_{y_i}\sum_{y_j} (\bar{n}_i \Delta x \Delta y)^2 C_{y_i y_j} .$$

The rest follows as above (note that I have that the spatial covariance in $y$ at a given $x$ is $(\bar{n}_x \Delta x \Delta y)^2 C_{y_i y_j}$).

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  • $\begingroup$ You have to consider the differences of the discrete Dirac-Delta operator and the continuous Dirac-Delta distribution. If your discretization widths $\Delta x_i,\Delta x_j$ shrink your sum converges to zero since $\delta_{ij} = 1$ when $i=j$ and $\Delta x_i,\Delta x_j\rightarrow 0$. You have to replace the discrete Dirac-Delta by a density like $\frac{\delta_{i,j}}{\Delta x_i}$ to get something that converges to the Dirac-Delta distribution. $\endgroup$
    – Tobias
    Commented Nov 18, 2016 at 7:11
  • $\begingroup$ @tobias -- I think this is the correct answer. Would you like to post it as an answer so I can accept? In my case, the discrete version is "correct" while the integrated version is wrong, so insead I'd have to replace the continuous Dirac with something like $\delta(x-k) dx$. $\endgroup$ Commented Nov 21, 2016 at 2:53

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I think that the discrepancy is that your discrete case is not the strict analogue of your continuous case.

If I were summing up just values of $n$, and using the Kronecker delta to pull out one particular value, say $k$, then we'd have $$ \sum_i a_i\delta_{ik} = a_k, $$ and its continuous analogue would be $$ \int a(x)\delta(x-k)\,dx = a(k). $$ In all cases, the units are the same (whatever units $a$ has). Notice that the discrete version lacks the extra $\Delta x$ that you included.

Working backwards from your continuous version, we have \begin{align} \int \frac{dn}{dx_i}\frac{dn}{dx_j}\delta(x_i-x_j)\,dx_i\,dx_j &= \int \left(\frac{dn}{dx}\right)^2\,dx \\ \text{(Going to discrete case)}&\Rightarrow \sum_i \bar{n}^2_i \Delta x \\ &= \left[\sum_{i,j}\bar{n}_i\bar{n}_j \delta_{ij}\right]\Delta x, \end{align} so both versions have units of inverse length.

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  • $\begingroup$ You could be right, but I'm not absolutely convinced. The sum of $a_i$ does not converge to the integral over $a$, rather the sum of $a_i \Delta x$ does. Unless you are saying what @tobias did -- i.e. that $\delta_{ik}$ in the continuous limit literally translates to $\delta(x-k) dx$? $\endgroup$ Commented Nov 21, 2016 at 2:56
  • $\begingroup$ Yep, I think @Tobias is correct too. I think we're ultimately saying the same thing, but I was only working by analogy, and Tobias provided the underlying reason why. $\endgroup$
    – robotopia
    Commented Nov 22, 2016 at 0:04

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