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I ran across the identity: $$ \delta(f(x)) = \sum_i \frac{\delta(x - x_i)}{|f^\prime(x_i)|} $$ where $\delta(x)$ is the Dirac delta and $x_i$ are the roots of some function $f(x)$. Now, I have seen in physics textbooks people use $$ \delta(\vec{r} - \vec{r}_0) = \frac{1}{r^2\sin\theta}\delta(r-r_0)\delta(\theta - \theta_0)\delta(\phi - \phi_0) $$ as the Dirac delta in spherical coordinates. Where I am deeply confused is that some replace the $\theta$ part with $\delta(\theta - \theta_0)/\sin\theta \rightarrow \delta(\cos\theta - \cos\theta_0)$. I thought I was comfortable with this transformation because of the above rule, but I not so sure anymore. Take for example, a line defined as $\delta(\cos^2\theta - 1)$. By the above identity, I should be able to transform this to $$ \delta(\cos^2\theta - 1) = \frac{\delta(\theta)}{|2\sin(0)|} + \frac{\delta(\theta - \pi)}{|2\sin\pi|}, $$ but as you can see, this function blows up at the poles. So this means we cannot really write $$ \frac{\delta(\theta - \theta_0)}{\sin\theta} = \delta(\cos\theta - \cos\theta_0), $$ correct? Is this one of those common physics habits of fudging up the math? I see the $\delta(\cos\theta)$ so much, I just feel that I am grossly misunderstanding something, but it is not easy to Google. Any insight is greatly appreciated! Thanks.

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  • $\begingroup$ Which physics textbooks? Which pages? $\endgroup$
    – Qmechanic
    Commented Apr 7, 2023 at 19:44
  • $\begingroup$ @Qmechanic, I remember seeing it in undergraduate E&M, but I cannot readily find it. I know it is in Barton's "Elements of Green’s Functions and Propagation". Also, Eq. (19) in this this paper and Eq. (13) in this web book $\endgroup$
    – Marco
    Commented Apr 7, 2023 at 21:57
  • $\begingroup$ In your original setting you need to avoid $\theta_0$ being a multiple of $\pi$ I guess. Which makes sense anyway because this causes a division by 0 either way. $\endgroup$
    – Ian
    Commented Apr 8, 2023 at 3:51

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The definition of the composition of the Dirac Delta, $\delta \circ f$, with a function $f$ assumes that $(1)$ $f$ is continuously differentiable and $(2)$ $f'$ is nowhere zero. Under such assumptions then if $f(x_i)=0$ for $i=1, \cdots N$ we define $\delta\circ f$ by

$$\delta \circ f=\sum_{i=1}^N \frac{\delta_{x_i}}{|f'(x_i)|}$$

where $\langle \delta_{x_i}, \phi \rangle =\phi(x_i)$ for any $\phi\in C_C^\infty$.

Then, we have

$$\langle \delta\circ f, \phi\rangle =\sum_{i=1}^N \frac{\phi(x_i)}{| f'(x_i)|}$$

Now, suppose $f(x)=\cos^2(x)-1$. Note that $f(0)=f(\pi)=0$. Note also $f'(x)=\sin(2x)$ so that $f'(0)=f'(\pi)=0$. Hence, $f(x)=\cos^2(x)-1$ fails to satisfy the assumptions of the definition.

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  • $\begingroup$ Thank you very much for this answer. Would you be able to explain to me why so many substitute $\delta(\theta - \theta_0) / \sin\theta$ with $\delta(\cos\theta - \cos\theta_0)$? If you look at my comments above, I cite a couple references where this is used. $\endgroup$
    – Marco
    Commented Apr 8, 2023 at 14:26
  • $\begingroup$ You're welcome. My pleasure. Let $f(\theta)=\cos(\theta)-\cos(\theta_0)$. Then, $f(\theta_0)=0$ and $f'(\theta)=\sin(\theta)$. Now use the definition for the composite Dirac Delta. $\endgroup$
    – Mark Viola
    Commented Apr 8, 2023 at 15:31
  • $\begingroup$ Thanks! I think you are nearing the end of my confusion. One last question, why do you use $f^\prime(\theta)$ instead of $f^\prime(\theta_0)$ if $\theta_0$ is the zero of $f(\theta)$ and the composite Dirac delta is defined only at the zeros? $\endgroup$
    – Marco
    Commented Apr 8, 2023 at 16:24
  • $\begingroup$ I think this is the main source of my confusion. Basically, $\delta(\cos\theta - \cos\theta_0) = \delta(\theta - \theta_0)/|-\sin\theta_0|$ based on the definition of composite Dirac delta, no? So where did the non-constant $\sin\theta$ go? In my mind, $\delta(\theta - \theta_0)/\sin\theta = \delta(\cos\theta - \cos\theta_0)\sin\theta_0/\sin\theta$ based on the definitions provided. $\endgroup$
    – Marco
    Commented Apr 8, 2023 at 16:37
  • $\begingroup$ Note that $f(x)\delta(x-x_0)=f(x_0)\delta(x-x_0)$ $\endgroup$
    – Mark Viola
    Commented Apr 9, 2023 at 15:57

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