1

Formally, if I want to measure the length of a closed curve $f(x,y) = 0$, I presumed I could write:

$$ L = \int^\infty_{-\infty}\int^\infty_{-\infty} \delta( f(x,y) )\, dx\, dy, $$

but trying this out I don't think this works. What is wrong with this formula?

Edit: Am I missing a measure like a Jacobian or something? How can you prove this?

zooby
  • 4,343
  • 1
    It's a little unclear what you mean by "measure a closed curve". You mean arc length? – Adrian Keister May 31 '18 at 19:36
  • yes. Measuring the length of a curve. – zooby May 31 '18 at 20:16
  • The limits on the integral are all $+\infty$. – Benedict W. J. Irwin May 31 '18 at 21:15
  • If your curve was a single point, then it's 'length' would be $1$ according to this. – Benedict W. J. Irwin May 31 '18 at 21:18
  • Let's check what happens for a circle. Let $R$ be the radius. Then we can take $f(x, y) = x^2 + y^2 - R^2,$ so we get $$ L = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \delta(x^2 + y^2 - R^2) , dx , dy \ = { \text{ polar coordinates } } \ = \int_0^{2\pi} \left( \int_0^\infty \delta(r^2 - R^2) , r , dr \right) , d\theta \ = { u = r^2 } \ = \int_0^{2\pi} \left( \int_0^\infty \delta(u-R^2) , \frac12 du \right) , d\theta \ = \int_0^{2\pi} \frac12 , d\theta = \pi $$ @BenedictWilliamJohnIrwin. Wouldn't we get the same result for a single point where $R = 0$? – md2perpe May 31 '18 at 21:45
  • @md2perpe That is somewhat perplexing...

    Well the content of this question might help: https://math.stackexchange.com/questions/619083/dirac-delta-function-of-non-linear-multivariable-arguments

     – Benedict W. J. Irwin May 31 '18 at 21:59
  • I believe the real formula for the length of the line would be $\int\int \delta(f(x,y)) \sqrt{ (\partial_x f )^2 + (\partial_y f )^2 } dx dy $ – zooby Jun 01 '18 at 16:05

1 Answers1

2

For the integral $L := \iint_{\mathbb R^2} \delta(f(x,y)) \, dx \, dy$ to measure the length of the curve $f(x,y) = 0$ we must have $\|\nabla f(x,y)\| = 1$ where $f(x,y) = 0.$ This is related to the scaling property.

So in the case of the circle with center in $(0,0)$ and radius $R,$ we could take $f(x,y) = \frac{1}{2R} (x^2+y^2-R^2).$ This would then give $$ L = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \delta\left( \frac{1}{2R}(x^2 + y^2 - R^2) \right) \, dx \, dy \\ = \{ \text{ polar coordinates } \} \\ = \int_0^{2\pi} \left( \int_0^\infty \delta\left( \frac{1}{2R}(r^2 - R^2) \right) \, r \, dr \right) \, d\theta \\ = \left\{ u = \frac{1}{2R}r^2 \right\} \\ = \int_0^{2\pi} \left( \int_0^\infty \delta\left( u-R/2 \right) \, R \, du \right) \, d\theta \\ = \int_0^{2\pi} R \, d\theta = 2\pi \, R. $$

md2perpe
  • 26,770