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Is there a coordinate independent Dirac delta function?

  1. Dec 26, 2012 #1
    I have been wondering exactly how one would express the Dirac delta in arbitrary spaces with curvature. And that leads me to ask if the Dirac delta function has a coordinate independent expression. Is there an intrinsic definition of a Dirac delta function free of coordinates and metrics? Or as a distribution does it inherently need a coordinized space on which to distribute its values? Thanks.
     
  2. jcsd
  3. Dec 26, 2012 #2

    pwsnafu

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    The only possibility I can think of would be "Dirac is the identity element for convolution".
     
  4. Dec 28, 2012 #3
    What about the Dirac measure? This is called a "measure", and any measure can be integrated, right? This Dirac delta measure is defined in terms of elements and sets and not on coordinates. So does this qualify as a coordinate independent form of the Dirac delta function? Or would the coordinate free expression have to be in terms of elements/points only and not sets?
     
  5. Dec 28, 2012 #4

    pwsnafu

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    It depends on what "coordinate free" means right? Normally we represent Dirac like
    ##\int_X f(t) \delta(t-a) dt = f(a)##
    or something similar. The problem is we use coordinate systems to express "a" itself, so if you change the coordinate system you change how you write "a". So my interpretation of "coordinate free" would be "changing coordinates doesn't change the expression", at least in the context this question. If you meant something else well you'll need to define it.

    Secondly, you wrote "any measure can be integrated". Well you mean "integrate with respect to the measure" not "integrate the measure itself". And yes there is a concept of "derivative of a measure" but it's very different.

    Now does the Dirac measure satisfy the problem? Well that depends on whether you can pick an isolated point from a set without reference to a coordinate system (remember coordinate system is really just a systematic way of writing elements with respect to each other). Can you pick 0 without knowing where it is? And if you can do this, how is this any different to "normal" way we write Dirac?
     
  6. Jan 3, 2013 #5
    Let me turn this around and ask, if it can be proved that the integral of the dirac delta does not change its value or its appearance (a change of coordinates only changes the variable used in the integral, but otherwise stays the same), would that prove that there does exist a coordinate independent intrinsic definition of the dirac delta?

    Does the measure have to be integrated against a function of compact support? What about integrating against f(x)=1?

    I suppose you could do a formal sum over every point, p, in the set of your manifold, M, without referring to coordinate system. If you did this for every point in M, you'd get 1 for every p [itex]\in[/itex] A [itex]\subset[/itex] M for which δp(A)=1. Then Ʃpδp(A) would give you the size of A, right?
     
  7. Jan 4, 2013 #6

    pwsnafu

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    You need to be more rigorous in what you want from "coordinate independent intrinsic definition". I say this because "Dirac is the identity element of the convolution" is a coordinate independent intrinsic definition. Proving that there really is a distribution that satisfies that definition is long and tedious (it uses a technique called "approximating the identity") but is doable.

    As to your specific question, I guess so?

    Ah. First consider this: suppose we have a smooth function whose domain is ℝ, ##f\in C^\infty(\mathbb{R})##. Now is this function an element of ##C(0,\infty)##? Well it isn't because the domains are different, but we can take f and restrict it to the positives and that restriction is an element of ##C(0,\infty)##.

    Same thing here. The Dirac as a measure can be integrated against f(x)=1, but the Dirac as a Schwartz distribution can't. Nonetheless, you can take the Dirac measure and restrict is to become a Schwartz measure.

    Yes, but in a trivial manner. p is always an element of A, so really you are just evaluating ##\sum_{p\in A}1##.

    Edit: Oh wait you're summing over ##p \in M##? Then that means M is countable.
     
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