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Formal definition of Dirac's delta and charge densities

  1. May 29, 2013 #1
    Hi, i'm new to this forum but i've been aware of its existence for a while and it's pretty cool. I finally came up with a question to post so here i am :)

    I've read a few nice posts on this forum about this topic, but I couldn't find the answer to what i'm looking for. I'm familiar with the Dirac delta function in the context of distributions, as it is used in electrical engineering, but i don't get how to apply it to mass or charge densities. The thing is that charge densitiy, for example, is defined as the function that verifies the following property:

    $$ \int_{V} \rho (x,y,z) dV = Q $$Where, ##\rho## is the charge density, Q is the net charge inside the volume V, and dV is the differential volume element.

    This definition is not only integral, which already is a problem for me, but it also involves a 3D volume integral and hence, 3 coordinates. The problem in the formalisation arises because i've no idea how to integrate distributions in one variable, let alone 3d integrals. Just to be clear, the formalisation in the distribution context of the dirac delta function that i know is:

    $$ < \delta (x-x_0), \varphi (x)> = \varphi (x_0) $$

    Where ##\varphi(x)## is a infinitely differentiable function and has a compact support. Now, i've done some reading and found out that it can be generalized to ##X = (x_1,x_2,...,x_n)\in \mathbb{R}^n## in the following way:

    $$ < \delta (X), \varphi (X)> = \delta (x_1) · \delta(x_2) ··· \delta (x_n) $$

    And i would like to say that it equals ##\varphi (X_0)##, but this is not clear at all, since there is no definition for the product between distributions that i'm aware of... so it's like a dead end. On the other hand it even feels natural to define it like ##< \delta (X-X_0), \varphi (X) > = \varphi (X_0)##. What is defined for distributions is tensor product and convolution product, maybe that's the way to go?

    In any case, even assuming the multi dimensional delta is well defined over a proper space ##D^{'} = \{ T / T:D\rightarrow \mathbb{C} \}## where ##D = \{ \varphi / \varphi : \mathbb{R}^n \rightarrow \mathbb{C}, \text{smooth and of compact support} \} ##, there's still a bigger problem: what am i to do with this "practical definition" of the delta:

    $$ \int_{-\infty}^{+\infty} f(x) \delta (x - x_0) dx = f(x_0) $$

    It just doesn't fit in the distribution theory.. Is there a way to define integration and make this work? I've read that the mass density of point particles is just ##\rho = \sum m_i \delta (\vec{r}-\vec{r}_i )## which actually makes sense using the above definition, it is clear than when integrating the total mass will be the sum of all the ##m_i## inside the volume ##V##. But if things start to get messy, without the formalisms i start to get lost. For example, how would you define a linear or surface density?

    I'll try to summarize my questions here because i feel i made quite a mess trying to fully explain the problem.

    1. What's the formal definition of the delta in the context of charge or mass densities? Is it a distribution or something else entirely? Even if it's not, is it possible to interpet it that way?

    2. If so, how do you define the integral of a distribution?

    3. How do you manage to work out a distribution associated with a density function given that the definition is integral in its nature, and not differential.

    4. I know it's possible to define it as ##\rho = \frac{dQ}{dV}##, is this in any way more connected to distributions than the integral form?

    All in all i guess i'm looking for the connection between the theory and delta i already know and
    this use of the delta in other fields of physics.

    I guess that's it. Anyway, any kind of input is fully appreciated, I'd certainly would like an example on how to calculate linear and surface densities.

    Thanks for your time!
  2. jcsd
  3. May 29, 2013 #2

    Simon Bridge

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    1. the definition is exactly the same as it is for anything - in terms of a distributional limit.
    2. don't understand the question
    3. you apply it's known properties
    4. No.

    I suspect you are over-thinking things: a point charge Q at the origin would have the charge density: ##\rho(r)=Q\delta(r)##
    So any integral of the density that does not include the origin will return zero, and any that do include the origin returns Q.

    What's the problem?
  4. May 29, 2013 #3
    hi simon, thanks for taking the time to answer. The thing is that if the delta is a distributional limit, it's a limit to a succession of other distributions, thus, it's a distribution (i'm using the word "distribution" in the mathematical sense of a linear functional, just so that we are on the same page).

    The problem is that you can't integrate a distribution... there's no such thing as ##\int \delta(x) dx## (not formally at least) since the delta is this: ## <\delta (x),\varphi (x) > = \varphi (0)##. It's a function that takes another function and returns a number. The delta is a "functional", not a function in the traditional way, and as such you can't integrate it in a Riemann integral. That's why I ask how do you define the integral of a distribution, given that it's not a function.

    I know it's over thinking, you can just learn the properties and go along with it, but i'd like to know what i'm doing... otherwise I never learn it and I keep forgetting the rules. As an example, I understand the case of a point charge or mass, but i just can't extrapolate this to a linear charge density... I lack the insight that a formal definition provides. How do you write, in terms of a delta, the charge density of a ring? I just don't see how to apply the property that the delta integrates 0 everywhere but in a single point, to a ring or any other curve.
  5. May 30, 2013 #4

    Simon Bridge

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  6. May 30, 2013 #5
    That bracket notation is taken to be an integral in most cases (hilbert spaces, what not).
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