How to evaluate the coordinate transformation of Dirac delta function?

Question

I have come across the following calculation while evaluating path integrals.How do one determine the coordinate transformation of delta function? For instance,If the delta function $\delta(X_{N}-X_{f})$ (where $X_{N}=X(t_{N})$ and $X_{f}=X(t_{f})$) is transformed to to another set variables $Z_{N}$ and $Z_{f}$ from X coordinate. How does $\delta(X_{N}-X_{f})$ transforms to $\delta(Z_{N}-Z_{f})$. Also how does the following transforms transform to $Z$ coordinate? $\int dX_{N} \delta(X_{N}-X{f})f(X_{N})$. The transformation of $X$ to $Z$ is defined to be $X(t)=Z(t)exp(-\omega t)$

Answer 1

Of course we can evaluate the given integral without transforming the Dirac $\delta-$functions since : \begin{equation} \int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left(x_{_{\rm{N}}}\!-\!x_{_{\rm{F}}}\right) f\left(x_{_{\rm{N}}}\right)\mathrm dx_{_{\rm{N}}}= f\left(x_{_{\rm{F}}}\right)=f\left[e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z\left(t_{_{\rm{F}}}\right)\right]= f\left(e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z_{_{\rm{F}}}\right) \tag{01} \end{equation} But if for other purposes we want to see what is happening via transformation of the Dirac $\delta-$functions, we could proceed as follows. Let the transformation between $\:x,z\:$
\begin{equation} x\left(t\right)=e^{\boldsymbol{-}\omega t}z\left(t\right) \tag{02} \end{equation} We express the variables under the integral in (01) from $x$-coordinate to $z$-coordinate \begin{align} x_{_{\rm{N}}} & =x\left(t_{_{\rm{N}}}\right)=e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z\left(t_{_{\rm{N}}}\right)=e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}} \tag{03a}\\ x_{_{\rm{F}}} & =x\left(t_{_{\rm{F}}}\right)=e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z\left(t_{_{\rm{F}}}\right)=e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z_{_{\rm{F}}} \tag{03b}\\ \mathrm dx_{_{\rm{N}}} & = e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}\mathrm d z_{_{\rm{N}}} \tag{03c} \end{align} and so \begin{align} \int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left(x_{_{\rm{N}}}\!-\!x_{_{\rm{F}}}\right) f\left(x_{_{\rm{N}}}\right)\mathrm dx_{_{\rm{N}}} & = \int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left(e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}}\!-\!e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z_{_{\rm{F}}}\right) f\left(e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}}\right) e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}\mathrm d z_{_{\rm{N}}} \nonumber\\ & =e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left(e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}}\!-\!e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z_{_{\rm{F}}}\right) f\left(e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}}\right) \mathrm d z_{_{\rm{N}}} \nonumber\\ & =\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left[z_{_{\rm{N}}}\!-\!e^{\omega \left(t_{_{\rm{N}}}\boldsymbol{-}t_{_{\rm{F}}}\right)} z_{_{\rm{F}}}\right] f\left(e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}}\right) \mathrm d z_{_{\rm{N}}} \tag{04} \end{align} The last equality in (04) is valid since⁽¹⁾ \begin{equation} \int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\delta\left(\alpha x+\beta\right)\mathrm{h}\left(x\right)\mathrm dx=\dfrac{1}{\boldsymbol{\vert}\alpha\boldsymbol{\vert}} \int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\delta\left(x\boldsymbol{+}\frac{\,\beta\,}{\alpha}\right)\mathrm{h}\left(x\right)\mathrm dx \tag{05} \end{equation} written symbolically \begin{equation} \delta\left(\alpha x+\beta\right) \doteq \dfrac{1}{\boldsymbol{\vert}\alpha\boldsymbol{\vert}} \delta\left(x\boldsymbol{+}\frac{\,\beta\,}{\alpha}\right) \tag{06} \end{equation} The equality is proved by making in the 2nd term of (04) the following substitutions \begin{align} x & \quad \boldsymbol{-\!\!\!-\!\!\!-\!\!\!-\!\!\!\longrightarrow} \quad z_{_{\rm{N}}} \tag{07a}\\ \alpha & \quad \boldsymbol{-\!\!\!-\!\!\!-\!\!\!-\!\!\!\longrightarrow} \quad e^{\boldsymbol{-}\omega t_{_{\rm{N}}}} \tag{07b}\\ \beta & \quad \boldsymbol{-\!\!\!-\!\!\!-\!\!\!-\!\!\!\longrightarrow} \quad \boldsymbol{-}e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z_{_{\rm{F}}} \tag{07c} \end{align} So \begin{equation} \int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left(x_{_{\rm{N}}}\!-\!x_{_{\rm{F}}}\right) f\left(x_{_{\rm{N}}}\right)\mathrm dx_{_{\rm{N}}} =\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left[z_{_{\rm{N}}}\!-\!e^{\omega \left(t_{_{\rm{N}}}\boldsymbol{-}t_{_{\rm{F}}}\right)} z_{_{\rm{F}}}\right] f\left(e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}}\right) \mathrm d z_{_{\rm{N}}}= f\left(e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z_{_{\rm{F}}}\right) \tag{08} \end{equation} a result identical to that given in (01).

Now, I don't think it would be useful to write (08) symbolically
\begin{equation} \delta\left(x_{_{\rm{N}}}\!-\!x_{_{\rm{F}}}\right) \doteq \delta\left[z_{_{\rm{N}}}\!-\!e^{\omega \left(t_{_{\rm{N}}}\boldsymbol{-}t_{_{\rm{F}}}\right)} z_{_{\rm{F}}}\right] \tag{09} \end{equation} since this is valid in the special case of transformation (02).

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^{(1) Equation (05) is proved as equation (A-07) in Example A of my answer here : Physical meaning of the Jacobian in relation to Dirac delta function.}