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

  1. Jun 22, 2012 #1
    1 = 0 <-- What goes fundamentally wrong? Delta distribution

    [itex]2 \pi a = \iint \delta(a^2 - (x^2+y^2)) \mathrm d x \mathrm d y [/itex]

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

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

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

    [itex]\frac{\pi}{a} =2 \pi \int \delta'(a^2 - r^2) \cdot r \cdot \mathrm d r [/itex]

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

    [itex]\frac{1}{a} = - \int \frac{\partial}{\partial r} \delta(a^2 - r^2) \mathrm d r[/itex]

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

    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.
     
  2. jcsd
  3. Jun 22, 2012 #2
    Re: 1 = 0 <-- What goes fundamentally wrong? Delta distribution

    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.
     
  4. Jun 22, 2012 #3
    Re: 1 = 0 <-- What goes fundamentally wrong? Delta distribution

    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.
     
  5. Jun 22, 2012 #4
    Re: 1 = 0 <-- What goes fundamentally wrong? Delta distribution

    [itex]\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[/itex]

    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...
     
  6. Jun 22, 2012 #5
    Re: 1 = 0 <-- What goes fundamentally wrong? Delta distribution

    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.
     
  7. Jun 22, 2012 #6
    Re: 1 = 0 <-- What goes fundamentally wrong? Delta distribution

    Of course, thank you :)
     
  8. Jun 22, 2012 #7
    Re: 1 = 0 <-- What goes fundamentally wrong? Delta distribution

    But... say we reinterpret the correct equation [itex]\pi = \int \delta(x^2+p^2-E) \mathrm d x \mathrm d p[/itex]. 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?
     
  9. Jun 22, 2012 #8

    micromass

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

    I'm not an expert on distributions. But one of my favorite books "Functional Analysis" by Lax says that [itex]\delta(x^2)[/itex] doesn't even have a sensible definition. Maybe that is what goes wrong?
     
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