3
$\begingroup$

I am having some issues understanding Dirac's delta function/distribution behaviour under change of coordinates. There is a statement, if $(x_1,\ldots,x_n)$ are cartesian coordinates and $y_1,\ldots,y_n$ are some general coordinates in Euclidean $\mathbb{E}^n$ space, then: $$ x_1=x_1(y_1,\ldots,y_n),\\ \vdots \\ x_n=x_n(y_1,\ldots,y_n)$$ and the Jacobian: $$ J(y_1,\ldots,y_n) = \begin{vmatrix} \frac{\partial x_1}{\partial y_1} & \cdots & \frac{\partial x_1}{\partial y_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial x_n}{\partial y_1} & \cdots & \frac{\partial x_n}{\partial y_n} \end{vmatrix} $$

define the transformation. Provided that $J(y_1^{(P)},\ldots,y_n^{(P)})\ne0$ for $(y_1^{(P)},\ldots,y_n^{(P)})$ coordinates of some point $P$ coresponding to cartesian coordinates $(x_1^{(P)},\ldots,x_n^{(P)})$, then:

$$\delta(x_1-x_1^{(P)})\cdots\delta(x_n-x_n^{(P)}) = \frac{1}{|J(y_1^{(P)},\ldots,y_n^{(P)})|} \delta(y_1-x_1^{(P)})\cdots\delta(x_n-y_n^{(P)})$$.

I wonder how to show this fact? If I consider $\delta$ in the original idea by Dirac, then:

$$f(x_1^{(P)},\ldots,x_n^{(P)})=\int_{\Omega}\delta(x_1-x_1^{(P)})\cdots\delta(x_n-x_n^{(P)}) f(x_1,\ldots,x_n)\,{\rm d}x_1\ldots{\rm d}x_n=\\= \int\ldots\int\frac{\delta(y_1-y_1^{(P)})\cdots\delta(y_n-y_n^{(P)})}{|J(y_1^{(P)},\ldots,y_n^{(P)})|} f(y_1,\ldots,y_n) |J(y_1^{(P)},\ldots,y_n^{(P)})|\,{\rm d}y_1\ldots{\rm d}y_n=\\ = \begin{cases} f(y_1^{(P)},\ldots,y_n^{(P)}) & \quad \text{all integration bounds contain the point $P$}\\ 0 & \quad \text{Any one of the integration bounds does not contain the point $P$} \end{cases} $$

And almost all seems fine but I am not sure if such proof is sufficient for this interpretation.

When I consider $\delta$ as a distribution, I do not really understand how to show that this holds not even whether it is true. I remember that for a distribution, one can define a multiplication with a $C^\infty$ function, but I still cannot see how to show that the above holds, since in terms of distributions $\delta$ is defined only as a continuous linear functional that does $<\delta,\phi>=\phi(x)$ to any $C^\infty$ test function with compact support in some set (there is really no integration involved).

$\endgroup$
  • $\begingroup$ Hint: The space of distributions comes equipped with a topology in which you can approximate the $\delta$ distribution by integration against smooth functions. See what happens if you do that. $\endgroup$ – Chris Janjigian Aug 31 '18 at 15:11
  • $\begingroup$ I suggest you take $n=1$ first, and see what happens. All the ideas of the general proof are already present in that simpler case, but you will not have to deal with cluttered expressions. $\endgroup$ – Antoine Aug 31 '18 at 15:29
0
$\begingroup$

Suppose that we have some diffeomorphism/parameterization $x = p(y)$. If $u$ is a continuous function and $\phi$ is a test function, then we have $$ \left< u \circ p, \phi \right> = \int u(p(y)) \, \phi(y) \, dy = \int u(x) \, \phi(p^{-1}(x)) \, \left|\det\frac{\partial y}{\partial x}\right| \, dx = \left< \left|\det\frac{\partial y}{\partial x}\right| u, \phi \circ p^{-1} \right> , $$ where $\frac{\partial y}{\partial x}$ is the Jabobian matrix. This result can a bit sloppy be written $$ u(y) = \left|\det\frac{\partial y}{\partial x}\right| u(x). $$

We then define $u \circ p$ for a distribution $u$ by the above result.

Note that $$ \left| \det\frac{\partial y}{\partial x} \right| = \left| \det \left( \left( \frac{\partial x}{\partial y} \right)^{-1} \right) \right| = \left| \left( \det\frac{\partial x}{\partial y} \right)^{-1} \right| = \left| \det\frac{\partial x}{\partial y} \right|^{-1} = \left| \det J \right|^{-1} , $$ where $J$ is as in your post.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.