1
$\begingroup$

I am having troubles in understanding how to correctly perform the continuum limit of a double sum containing a Kronecker delta.

Imagine to integrate a function depending on $t$ and $t'$, both ranging from $0$ (initial time) to $T$ (final time):

$$I:=\int_0^Tdt_1\int_0^Tdt_2 f(t_1,t_2).\tag{1}$$

The corresponding Riemann sums, dividing the time intervals in slices of width $\epsilon=T/N$ is:

$$I_{disc}:=\sum_{j,j'=1}^N \epsilon^2 f(j\epsilon,j'\epsilon).\tag{2}$$

Obviosly lim$_{N\rightarrow\infty}I_{disc}=I$. Now consider the case when only the diagonal elements of the double integral are different from zero i.e.

$$J_{disc}:=\sum_{j,j'=1}^N \delta_{j,j'}\epsilon^2 f(j\epsilon,j'\epsilon).\tag{3}$$

I would expect that, in the continuum limit $N\rightarrow \infty$ ($\epsilon\rightarrow 0$) it becomes

$$J:=\int_0^Tdt\int_0^Tdt'\delta(t-t') f(t,t'),\tag{4}$$

where $\delta(t-t')$ is a Dirac delta.

Here is the problem: the Kronecker delta is adimensional, while the Dirac Delta has the dimensions of seconds$^{-1}$. This implies that $J_{disc}$ and $J$ have different dimensions, which does not make any sense. Therefore, there must be some mistake I am doing in going from the discrete to the continuum version of $J$. Could you help me spotting it and, more important, suggest the way to do this limit correctly?

$\endgroup$
  • $\begingroup$ Maybe $\delta\left(\frac{t-t'}{T}\right)$? $\endgroup$ – Cryo May 14 at 0:06
1
$\begingroup$

The translation between the Kronecker delta function and the Dirac delta distribution is

$$ \frac{1}{\epsilon}\delta_{j,j^{\prime}}\qquad\longrightarrow\qquad\delta(t-t^{\prime}),\tag{A}$$

where $\epsilon$ is the "volume" of a unit-cell in the discretization. See e.g. this related Phys.SE post.

In particular, the rhs. of OP's eq. (3) should be divided with $\epsilon$ to have a finite continuum limit.

$\endgroup$
  • $\begingroup$ Thank you very much, I think this clarifies! To summarize, if I just take (3) as it is and perform the limit I get zero, which is also consistent of doing an integral of a finite quantity (the function f) on a set with null measure (the line s=s'). $\endgroup$ – Sandro May 15 at 14:01
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic May 15 at 16:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.