The Dirac delta "function" - part II

The relation between δ(x) and dx

After introducing the Dirac delta "function" \(\delta(x)\) in the previous post, I'll try now to explain the relation between it and the differential element \(\mathrm{d}x\). In the process, I'll through all mathematical rigour out the window and aim for an intuitive understanding.

The "integral" of the Dirac delta equals 1, as we can see by taking a constant field \(\mathbf{E}(\mathbf{r}) = \mathbf{E}_0\) in Eq. (4) of post I:

\begin{equation}
\int \delta(\mathbf{r} - \mathbf{r}_0) \mathrm{d}^3 \mathbf{r} = 1
\label{eq:normDirac}
\end{equation}

keeping in mind that the integration sign in \eqref{eq:normDirac} is just a convenient symbol, and not a true Riemann (or Lebesgue) integral. We could even say that the integration is superfluous, since the Dirac delta is only different from zero in \(\mathbf{r}_0\), so why not write directly:

\begin{equation}
\delta(\mathbf{r} - \mathbf{r}_0) \mathrm{d}^3 \mathbf{r} |_{\mathbf{r}_0} = 1
\label{eq:normDirac2}
\end{equation}

or even:

\begin{equation}
\delta(\mathbf{r} - \mathbf{r}_0) = \dfrac{1}{\mathrm{d}^3 \mathbf{r} |_{\mathbf{r}_0}}
\label{eq:normDirac3}
\end{equation}

where the bar indicates that the volume element is to be evaluated at \(\mathbf{r}_0\): we'll soon see what this means.

While Equation \eqref{eq:normDirac} can be seen as shorthand notation for a more complicated concept, \eqref{eq:normDirac2} and \eqref{eq:normDirac3} are mathematically meaningless. They can still remind us of some important properties of the Dirac delta:

• It has units of reciprocal volume, as we have already seen.
• It transforms as the reciprocal of the volume element under a change of coordinates.

Change of coordinates

When switching coordinates from \(\mathbf{r}\) to \(\mathbf{u}\), the volume element changes according to:

\begin{equation}
\mathrm{d}^3 \mathbf{r} |_{\mathbf{r}_0} = \left | J \left [\mathbf{r}(\mathbf{u}) \right ]_{\mathbf{u}_0} \right | \mathrm{d}^3 \mathbf{u} |_{\mathbf{u}_0}
\label{eq:Jacob}
\end{equation}

where \(\mathbf{r}_0 = \mathbf{r}(\mathbf{u}_0)\) and \(J\) is the determinant of the Jacobian matrix that converts from \(\mathbf{u}\) to \(\mathbf{r}\) (to be evaluated in \(\mathbf{u}_0\)). We assume that \(\mathbf{u}(\mathbf{r})\) is a one-to-one correspondence and that \(J\) is always nonzero.

Under the same transformation, the Dirac delta behaves as:

\begin{equation}
\delta(\mathbf{r} - \mathbf{r}_0) = \dfrac{1}{\left | J \left [\mathbf{r}(\mathbf{u}) \right ]_{\mathbf{u}_0} \right |} \delta(\mathbf{u} - \mathbf{u}_0)
\label{eq:Jacobinv}
\end{equation}

and we can conclude that

The Dirac delta behaves as one over the element volume

in the sense discussed above. Finally, note that \eqref{eq:Jacobinv} is an example of the composition of the Dirac delta with a function \(f(x)\) (for simplicity, we will work in one dimension), given by:

\begin{equation}
\delta(f(x)) = \sum_{i} \dfrac{1}{\left |  f'(x_i)\right |} \delta (x - x_i)
\label{eq:comp}
\end{equation}

where \(x_i\) are the solutions of \(f(x) = 0\). They can be multiple, since \(f\) does not generally fulfill the conditions we imposed on the Jacobian transformation.