The Dirac Delta Function in Three Dimensions

The three-dimensional delta function must satisfy: \begin{equation} \int\limits_{\hbox{$\scriptstyle all space$}} \delta^3(\rr-\rr_0)\,d\tau=1 \end{equation} where $\rr=x\,\xhat +y \,\yhat +z\,\zhat$ is the position vector and $\rr_0=x_0\,\xhat +y_0 \,\yhat +z_0\,\zhat$ is the position at which the “peak” of the delta function occurs. In rectangular coordinates, it is just the product of three one-dimensional delta functions: \begin{equation} \delta^3(\rr-\rr_0)=\delta(x-x_0)\,\delta(y-y_0)\,\delta(z-z_0) \end{equation} so that: \begin{equation} \int\limits_{\hbox{$\scriptstyle all space$}} \delta^3(\rr)\,d\tau = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \delta(x)\,\delta(y)\,\delta(z)\, dx\, dy\, dz = 1 \end{equation}

But in curvilinear coordinates, with $d\tau=h_u h_v h_w\,du\,dv\,dw$, it has a Jacobian in it. Thus, \begin{equation} \delta^3(\rr-\rr_0)\ne\delta(u-u_0)\,\delta(v-v_0)\,\delta(w-w_0) \end{equation} but instead: \begin{equation} \delta^3(\rr-\rr_0) = {1\over h_u h_v h_w} \delta(u-u_0)\,\delta(v-v_0)\,\delta(w-w_0) \end{equation}

In particular, in cylindrical coordinates: \begin{equation} \delta^3(\rr-\rr_0) = {1\over r} \delta(r-r_0)\,\delta(\theta-\theta_0)\,\delta(z-z_0) \end{equation} and in spherical coordinates: \begin{equation} \delta^3(\rr-\rr_0) = {1\over r^2\sin(\theta)} \delta(r-r_0)\,\delta(\theta-\theta_0)\,\delta(\phi-\phi_0) \end{equation}

Just as with the delta function in one dimension, when the three-dimensional delta function is part of an integrand, the integral just picks out the value of the rest of the integrand at the point where the delta function has its peak. \begin{equation} \int\limits_{\hbox{$\scriptstyle all space$}} f(\rr)\,\delta^3(\rr-\rr_0) \,d\tau = f(\rr_0) \end{equation}

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