Consider the problem of projectile motion in 2 dimensions. Launch angle is constant. Range of projectile, $x$, then depends only on launch speed, $v$, and is given by \begin{equation} x=v^2, \quad v\in [0,1] \tag{1} \end{equation} Above equation has been non-dimensionalised (by taking maximum range as our length scale, and maximum launch speed as our velocity scale), so all quantities are dimensionless. Probability density function for launch speed is assumed uniform over the interval $[0,1]$: \begin{equation} f(v)=1, \quad \textrm{if}~v\in [0,1]\tag{2} \end{equation} and zero otherwise. I want to find p.d.f for range of projectile, $x$. An easy way of doing this \begin{equation} f(x)=\left| \frac{dv}{dx}\right|f(v)=\frac{1}{2\sqrt{x}}, \quad x\in [0,1]\tag{3} \end{equation}

However I wanted to solve the same problem using Dirac delta function: \begin{align} f(x) & =\int_0^1 dv~f(x|v)~f(v) \\ & = \int_0^1 dv~f(x|v) \\ & = \int_0^1 dv~\delta(v^2-x)\tag{4} \end{align} Here $f(~|~)$ denotes conditional p.d.f.. Last line was arrived at because for given value of $v$, it is certain that we shall obtain that value of $x$ that satisfies the equation $v^2-x=0$. Now I make use of the identity for delta function \begin{align} \delta(g(x))=\sum_i \frac{\delta(x-x_i)}{|g'(x_i)|}\tag{5} \end{align} Here $x_i$ are roots of function $g(x)$, and $g'\equiv \dfrac{dg}{dx}$. Now $g(v)=v^2-x$, whose roots are $\pm \sqrt{x}$. We reject the negative root because $v\geq 0$. $g'=2v$. Hence \begin{align} f(x) & =\int_0^1 dv~\delta(v^2-x) \\ & = \int_0^1 dv~\frac{1}{2\sqrt{x}}\delta(v-\sqrt{x}) \\ & = \frac{1}{2\sqrt{x}}\tag{6} \end{align} which is correct.

However instead of $f(x|v)=\delta(v^2-x)$, we could equally well have begun with the equation $f(x|v)=\delta(v-\sqrt{x})$, because at least according to me, physical content of both equations is identical. However the last choice yields a completely different p.d.f.: \begin{align} f(x) & =\int_0^1 dv~\delta(v-\sqrt{x})=1\tag{7} \end{align} I don't think I have done anything wrong mathematically (if I have, please point out). To a mathematician of course the two functions are different, and so the fact that they yielded different p.d.f.s is not surprising. But when the equations are put in their proper physical context, both have identical physical content (as far as I can see). This example makes me wonder if Dirac Delta function may be used unambiguously in solving physical problems. While this was a simple problem where a second method of solution was available and so we could compare, what does one do in more complicated situations where such a comparison is not possible?

  • $\begingroup$ I found this link - physics.stackexchange.com/q/54778 - since I posted this question, yet I am not sure how it answers my question. $\endgroup$ – Deep Oct 31 '16 at 8:00

The ambiguity is resolved once you think about the dimension of the $\delta$-function. It's actually your method of non-dimensionalization that led you astray here. Given $\delta(f(x))$, the $\delta$-function is a density with inverse dimensions of $f$. That is, $\delta(v^2 - x)$ has dimensions of inverse length while $\delta(v-\sqrt{x})$ has dimensions of inverse square-root-of-length. Looking at the equation $$ f(x) = \int f(x\vert v)f(v)\mathrm{d}v,$$ we observe that $f(x)$ is supposed to have dimensions of inverse length, and $f(v)$ has dimensions of inverse velocity. Since $\mathrm{d}v$ has units of velocity, it follows that $f(x\vert v)$ must have units of inverse length for the equation to be consistent. Therefore, $\delta(v^2-x)$ is consistent while $\delta(v-\sqrt{x})$ is not.

  • $\begingroup$ As I have mentioned in the very beginning, all quantities are dimensionless. For launch angle $\theta\in (0,\pi/2)$, and launch speed $v\in [0,v_0]$, maximum possible range is $x_0=v_0^2/2g~\sin(2\theta)$. Actual equation for range is $x=v^2/2g~\sin(2\theta)$, which when divided by $x_0$ on both sides yields non-dimensional equation mentioned in the beginning. So the problem is not with the dimensions. But thanks for your answer. $\endgroup$ – Deep Nov 1 '16 at 5:35
  • $\begingroup$ +1 Since you gave me the idea that Dirac delta has dimensions, I have used Dirac delta with $\textit{dimensional}$ equations with success. $\endgroup$ – Deep Nov 4 '16 at 5:30
  1. Besides ACuriousMind's suggestion to check dimensions, you could also integrate the wrong distribution $$\delta(\sqrt{x}-v)=2v\delta(x-v^2)$$ over $x$ and see that you don't get the p.d.f. (2) that you started from, but instead a wrong distribution $f(v)=2v$.

  2. The moral is that the Dirac delta distribution $\delta(g(x,v))$ does not just depend on the support $$\overline{\{(x,v)\in [0,1]^2~|~g(x,v)=0\}},$$ but also on the function $g$.

  • $\begingroup$ +1 So using Dirac delta in physical problems can be troublesome. Dimension is not the problem because equation that I am dealing with is dimensionless. Is there is a good reference on Dirac delta and its properties, suitable for a non-mathematician? $\endgroup$ – Deep Nov 1 '16 at 5:41

The dimensionality is one way to describe the problem. Another way is to keep in mind that the joint probability distribution has to be part of a volume form on the space that is invariant under every diffeomophism of the coordinate system. So, while your declaration that "units aren't a factor" is, strictly, true, the joint probability density function (PDF) volume form being: $$\operatorname{d}P(x,v) = \delta\left(v - \sqrt{x}\right)\operatorname{d}v\operatorname{d}x,$$ can only correct when the range, $x$, is uniformly distributed and $\operatorname{d}P(v|x) = \delta\left(v - \sqrt{x}\right)\operatorname{d}v$. Put another way, the conditional form you gave $\operatorname{d}P(x|v) = \delta\left(v - \sqrt{x}\right)\operatorname{d}x$ is not invariant under diffeomophisms of $x$.

Sorry, I didn't see the "non mathematician" remark. The important part is that you can change your coordinate system and the answer that comes out of any integral has to be the same as long as the parts you integrate over correspond. In this example you worked carefully to choose scales where $x$ and $v$ are unitless, but the equations have to work even if you choose different units for $x$ without changing the units of $v$.

In particular, under a simple rescaling $x \rightarrow ax$ delta functions obey: $$\delta(ax) = \frac{\delta(x)}{|a|},$$ and the change of coordinates in the integral yields $\operatorname{d}x \rightarrow a \operatorname{d}x$. So, in order for any average value of some quantity $f(x)$, defined by $$\langle f \rangle \equiv \int f(x) \, \delta( g(x) )\operatorname{d}x,$$ to be invariant under rescaling of $x$, we need $g(x) = x$.

  • $\begingroup$ +1 Thanks for your answer. I am not sure what you mean by "...different units for $x$...". If by units you mean actual measurement units (such as metre, feet, inches, etc.), then dimensionless equations are indeed invariant w.r.t. change in units. But I think that's not what you mean. Any way thanks for pointing out the relevant mathematical concepts. $\endgroup$ – Deep Nov 4 '16 at 9:19

Answer offered by @Qmechanic, that Dirac delta function depends also on the functional form of its argument, is correct in regard to $\textit{dimensionless}$ equations. Answer offered by @ACuriousMind, that dimensions of Dirac delta correctly dictates the functional form of its argument, seems to work when dealing with $\textit{dimensional}$ equations. In dimensional form conditional p.d.f. $f(x|v)$ written as either $\delta (\frac{v^2\sin(2\theta)}{2g}-x)$, or $\delta (\frac{v^2}{2g}-\frac{x}{\sin(2\theta)})$, or $\delta (\frac{v^2}{g}-\frac{2x}{\sin(2\theta)})$ etc., all of them seem to work, so far the argument has dimensions of inverse length. This means that if one is going to use Dirac delta functions, one had better work with dimensional equations. However there is going to be a problem when dealing with intrinsically dimensionless quantities such as angle or function of angle.

To summarize, this means that $\delta(f(x))=\delta(g(x))$ if and only if: (1)$f,g$ have dimensions and their dimensions is identical, (2)$f(x)=0\Leftrightarrow g(x)=0$, i.e. one can obtain equation $g(x)=0$ by manipulation of equation $f(x)=0$ and vice versa. I could not prove it, but examples I have worked out so far seems to support this conclusion.


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