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I need to evaluate the integral \begin{equation} \int_0^1\mathrm dt\,f\left(t\right)\delta^{\left(3\right)}\left(\vec r\left(t\right)-\vec r_0\right) \end{equation} where there is only one $0\leq t_0\leq 1$ such that $\vec r\left(t_0\right)=\vec r_0$. I need to solve the integral in terms of $t_0$ and $\vec r^{\left(i\right)}\left(t_0\right)$ ($\vec r\left(t\right)$ and its derivatives at $t_0$).

I tried evaluating as a product of three delta functions, but that took me nowhere.

Chetan Vuppulury
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2 Answers2

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Forgive my previous manifestly incorrect answer; this should be correct. $$\begin{align}\int_0^1\mathrm{d}t f(t)\delta^{(3)}(\mathbf{r}(t)-\mathbf{r}_0) & =\int^1_0 \mathrm{d}t f(t)\delta(x(t)-x_0)\delta(y(t)-y_0)\delta(z(t)-z_0) \\ & =\int_0^1 \mathrm{d}tf(t)\frac{\delta(t-t_0)}{|x'(t_0)|}\frac{\delta(t-t_0)}{|y'(t_0)|}\frac{\delta(t-t_0)}{|z'(t_0)|} \\ & = \int_0^1\mathrm{d}t \frac{f(t)}{|x'(t_0)y'(t_0)z'(t_0)|}\delta(t-t_0) \\ & = \frac{f(t_0)}{|x'(t_0)y'(t_0)z'(t_0)|} \end{align}$$

John Dumancic
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    Why did the other $\delta\left(t-t_0\right)$ disappear? I did the same calculation, but ended up with $\frac{f\left(t_0\right)\delta\left(0\right)^2}{\left\lvert\dot x\left(t_0\right)\dot y\left(t_0\right)\dot z\left(t_0\right)\right\rvert}$. – Chetan Vuppulury Nov 04 '19 at 15:25
  • I thought as follows: $\delta(t-t_0)*\delta(t-t_0)$ is zero everywhere, except at $t_0$, so the product would be $\delta(t-t_0)$. I tested this out on a graphing utility by defining $\delta(t-t_0)\approx \frac{1}{|a|\sqrt{\pi}}e^{-\left(\frac{x}{a}\right)^2}$ where a is small, set a = 0.0000000001, and graphing $\delta(x-2)\delta(x-2)$, which centered around 2. I could be incorrect, however; the product of two distribution is normally undefined, after all. – John Dumancic Nov 04 '19 at 20:19
  • How do you go from the second to third lines? That is, how do you get one $\delta(t - t_o)$ where there were three? – Daddy Kropotkin Mar 19 '21 at 12:55
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Not a separate answer, but just as a basic illustration what the mapped delta function means. Since $x(t)=x_0$ only for a single $t_0$ (but you can generalize this to a number of roots), $\delta(x(t)-x_0)$ is certainly zero for almost all $t$. The only thing that interests you is the neighbourhood of $t_0$, because only there the delta function deviates from zero.

Hence you can approximate $x(t)$ by a Taylor expansion around $t_0$: $$x(t)\approx x(t_0)+x^\prime(t_0)\cdot (t-t_0)= x_0+\eta\cdot (t-t_0)$$ Then $$\delta(x(t)-x_0)=\delta(\eta\cdot (t-t_0))$$ Suppose $\eta$ was positive. By a simple linear scaling of the integration variable you can now calculate the behavior of that distribution: $$\int \phi(t)\delta(x(t)-x_0)dt=\int \phi(t)\delta(\eta\cdot (t-t_0))dt=$$ $$=\int \phi(t)\frac{\delta(\eta\cdot (t-t_0))}{\eta}\eta d(t-t_0)=\int \phi(t(\tau))\frac{\delta(\tau)}{\eta}d\tau=\frac{1}{\eta}\phi(t(\tau=0))$$ where $\tau=\eta\cdot (t-t_0)$ has been introduced. Since $\tau=0$ is equivalent to $t=t_0$ it follows that $$\int \phi(t)\delta(x(t)-x_0)dt=\frac{1}{\eta}\phi(t_0)=\int \phi(t)\frac{\delta(t-t_0)}{\eta}dt$$ for any arbitrary test function $\phi(t)$. If $\eta$ is negative, the change of variables flips direction of the integral, so if you want to keep the integration direction (which you normally do for easier notational reasons...), you have to account for the sign before the integral, i.e. $$\int \phi(t)\delta(x(t)-x_0)dt=-\frac{1}{\eta}\phi(t_0)=\int \phi(t)\frac{\delta(t-t_0)}{-\eta}dt$$ With $\eta=x^\prime(t_0)$, the relation to prove is written more compactly as: $$\delta(x(t)-x_0)=\frac{\delta(t-t_0)}{|x^\prime(t_0)|}$$

oliver
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  • The initial problem was how to generalise this to the case where X(t) takes values in higher dimensional spaces, whole this only does the 1-dimensional case – Chetan Vuppulury Mar 19 '21 at 19:05
  • For sure! I just wanted to remark where the transformed delta function generally comes from. If you already know this, perfect, then skip it. However, other readers of this question might not, and it's always good to have such things close by. – oliver Mar 19 '21 at 20:22
  • Right, of course! – Chetan Vuppulury Mar 19 '21 at 20:29