Variable transformation of a Dirac delta function

Question

I am struggling to understand the variable transformation of a Dirac delta function. More specifically, a transformation of the following type, $$\delta(a\chi(z)-b) \rightarrow \delta(z-c)$$ Here, $a, b$ and $c$ are constants. The specific relationship between $\chi(z)$ and $z$ is, $$\chi(z)=\int{\frac{1}{H(z)}dz}$$ where $H(z)$ is a non-zero, positive and smoothly varying function of $z$. In the context of my Physics problem, for the sake of the interested audience, $H(z)$ is the Hubble parameter of the Universe while $\chi(z)$ is the comoving distance and $z$ is the redshift of any time in the past.

So, let me add what I have done so far:

I start by defining the Dirac delta function in the form a unit step function $\Theta (a\chi(z)-b)$as, $$\frac{d}{d(a\chi(z))}\Theta(a\chi(z)-b)=\delta(a\chi(z)-b)$$ Then converting this in the form of z using the chain rule as, $$\frac{d}{dz}\Theta(a\chi(z)-b)\frac{dz}{d(a\chi(z))}$$ Using the relation with $H(z)$, we can write the equation as, $$\frac{d}{dz}\Theta(a\chi(z)-b)\frac{H(z)}{a}$$ Also, $\chi(z)$ can also be replaced by the integral as well, so that left side containing the unit step function can be written as, $$\frac{H(z)}{a}\frac{d}{dz}\Theta(\ a\int_0^z{\frac{1}{H(z')}dz'}-b)$$.

After this, I am not sure what else to try. Hopefully, this addition helps. Also, please point out an error if you see one.

Answer 1

Generally, $\delta(f(x))$ can be written as a sum of $\delta$'s at the zeros of $f$: $$ \delta(f(x)) = \sum_{f(x_i)=0} \frac{1}{|f'(x_i)|} \delta(x-x_i) . $$

Using this we get $$ \delta(a\chi(z)-b) = \sum_{a\chi(z_i)-b=0} \frac{1}{|a\chi'(z_i)|} \delta(z-z_i) = \sum_{a\chi(z_i)-b=0} \frac{H(z_i)}{|a|} \delta(z-z_i) . $$

If $a\chi(z)-b$ has only one zero $z_0$ this reduces to $$ \delta(a\chi(z)-b) = \frac{H(z_0)}{|a|} \delta(z-z_0) . $$