Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Delta squared

  1. Nov 10, 2015 #1
    I consider the Dirac delta.

    In physics the delta squared has an infinite norm : $$\int\delta (x)^2=\infty $$

    But if i look at delta being a functional i could write : $$\delta [f]=f (0) $$ hence $$\delta^2 [f]=\delta [\delta [f]]=\delta [\underbrace {f (0)}_{constant function}]=f (0)$$

    Thus in this view $$\delta^2=\delta $$ ?
     
  2. jcsd
  3. Nov 10, 2015 #2
    Dirac delta is defined by the integral
    ##\int \delta(c) g(x) dx = g(c)##
    If you plug in ##g(x) = \delta(x)## then you get
    ##\int \delta(c) \delta(x) dx = \delta(c)## which isn't finite. The integral fails.

    You can think of the integral as a functional, but it doesn't make sense to think of delta as a functional.
     
  4. Nov 11, 2015 #3

    pwsnafu

    User Avatar
    Science Advisor

    Dirac delta is indeed a linear functional. Specifically it is the linear functional such that ##f \mapsto f(0)##. Schwartz distributions are our most developed theory of generalized functions, and there all Schwartz distributions are linear functionals.

    Now to OP. If you want ##\delta^2 = \delta \cdot \delta## to exist as a Schwartz distribution as well, then it turns out that ##\delta^2 = c\delta## for some constant ##c##, but not for the reason you posted. The notation ##\delta^2## is reserved for the product of Dirac with itself, not the composition of Dirac with itself. As to what ##c## is, there's a lot of disagreement. If the product defined is a "normal product" then ##c=0##. Some mathematicians have argued that non-zero c has physical meaning. Some argue ##c = \infty##. Others define it to be ##c = \delta(0)## and not define what that means (as long as it cancels in the end they are happy with it). So yeah a lot of disagreement.

    That being said, if you aren't working with Schwartz distributions, such as working in Colombeu algebra, then it is the operator ##f(x) \mapsto f(-x)^2## (technically the equivalence class of such operators), which in turn corresponds to the non-linear functional ##f \mapsto f(0)^2##

    Edit: One last thing. There are people who write ##\delta^2## for ##\delta(x)\delta(y)##, i.e. the two dimensional Dirac delta. But then you wouldn't have ##\int \delta^2 = \infty## at the top of your post, so I'm ignoring that situation.
     
    Last edited: Nov 11, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Delta squared
  1. Delta Z (Replies: 12)

  2. Square roots and square (Replies: 14)

  3. Dirac delta (Replies: 2)

  4. Delta Squared? (Replies: 2)

  5. Delta question? (Replies: 1)

Loading...