## 1 = 0 <-- What goes fundamentally wrong? Delta distribution

$2 \pi a = \iint \delta(a^2 - (x^2+y^2)) \mathrm d x \mathrm d y$

Differentiating both sides w.r.t. "a" (using chain rule on the RHS) gives

$\frac{\pi}{a} = \iint \delta'(a^2 - (x^2+y^2)) \mathrm d x \mathrm d y$

Changing variables (and noting that integrand is independent of theta)

$\frac{\pi}{a} =2 \pi \int \delta'(a^2 - r^2) \cdot r \cdot \mathrm d r$

Eliminate the pi on both sides, but more importantly we can rewrite the RHS as (using chain rule!)

$\frac{1}{a} = - \int \frac{\partial}{\partial r} \delta(a^2 - r^2) \mathrm d r$

However the RHS is obviously zero, explicitly: $0= \delta(a^2-r^2) \big|_{r=0}^{r=+\infty}$.

DISCLAIMER: I know I use the delta as a physicist would, but please be flexible... I would really appreciate knowing where it essentially goes wrong.

 Hmmm ... have you forgotten about the constant of integration at all? I don't know which region you're integrating over in the first equation, so I can't verify it.
 I think the very first equation is wrong. The delta function has units of 1/(units of argument). So delta(a^2 - x^2 - y^2) has units of [length]^-2. Thus the RHS of the first equation is unitless, whereas the LHS has units of length, so the equation can't be correct. I haven't worked it out, but I think if you are more careful you can evaluate the RHS of the first equation and it will be some pure number like 2*pi. Then of course differentiating that wrt a gives zero, as you found.

## 1 = 0 <-- What goes fundamentally wrong? Delta distribution

 Quote by The_Duck I think the very first equation is wrong. The delta function has units of 1/(units of argument). So delta(a^2 - x^2 - y^2) has units of [length]^-2. Thus the RHS of the first equation is unitless, whereas the LHS has units of length, so the equation can't be correct. I haven't worked it out, but I think if you are more careful you can evaluate the RHS of the first equation and it will be some pure number like 2*pi. Then of course differentiating that wrt a gives zero, as you found.
$\iint \delta(a^2-(x^2+y^2) ) \mathrm d x \mathrm d y = \iint \delta(a^2 - r^2) \cdot r \cdot \mathrm d r \mathrm d \theta = \pi \int \delta(r^2 - a^2) \mathrm d (r^2) = \pi$

Err, you seem to be right... I'm totally flabbergasted. Integrating the delta-function on a circle over the plane does not give its circumference? How is that possible...

 I think if you integrated the delta function delta(a - sqrt(x^2 + y^2)) you would indeed get the circumference. You have to recall that if f(x_0) = 0, the integral of delta(f(x)) scales inversely with f'(x_0). For example the integral of delta(r - sqrt(x^2 + y^2)) is not the same as the integral of delta(100*[r - sqrt(x^2 + y^2)]) even though both arguments have zeros on the same circle in the plane.
 Of course, thank you :)
 But... say we reinterpret the correct equation $\pi = \int \delta(x^2+p^2-E) \mathrm d x \mathrm d p$. The RHS is by definition the (exponential of the) phase volume for a one-dimensional harmonic oscillator (2-dimensional phase space) with energy E. So its entropy is independent of its energy? I.e. its temperature is infinite? Am I again overlooking something?
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus I'm not an expert on distributions. But one of my favorite books "Functional Analysis" by Lax says that $\delta(x^2)$ doesn't even have a sensible definition. Maybe that is what goes wrong?