1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dirac delta

  1. Oct 16, 2009 #1
    hi,

    may someone help me to clarify my doubts...

    in my work, i encounter diracdelta square [tex]\delta(x-x_1)\delta(x-x_2)[/tex] i am not sure what it means... it seems if i integrate it

    [tex]\int dx \;\delta(x-x_1)\delta(x-x_2) = \delta(x_1-x_2)[/tex] is either zero of infinity.

    is this correct?

    thanks
     
  2. jcsd
  3. Oct 17, 2009 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Technically, saying that it has "value zero or infinity" doesn't make sense. Any Dirac delta only makes sense under an integral sign (although in physics, we tend to think of it as being an "infinite spike with a finite area").

    It is correct that
    [tex]
    \int dx \;\delta(x-x_1)\delta(x-x_2) = \delta(x_1-x_2)
    [/tex]

    So, again, this expression again only makes sense inside an integral, like
    [tex]\int dx_1 \int dx \; \delta(x - x_1) \delta(x - x_2) = \int dx_1 \; \delta(x_1 - x_2)[/tex]
    which is one or zero (depending on whether or not x2 lies in the integration interval of the x1 integral).
     
  4. Oct 18, 2009 #3
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Dirac delta
  1. Dirac delta function (Replies: 2)

Loading...