Dirac delta function confusion

In summary, the conversation revolves around trying to evaluate the D'Alambertian of ##\epsilon(t)\delta(t^2-x^2-y^2-z^2)##, where ##\epsilon(t)## is a sign function. The goal is to prove that it solves the Klein-Gordon equation, but there are difficulties in simplifying the expression further. The problem becomes more complicated when considering the case where ##x=y=z=0## and further clarification is needed to proceed with the simplification. There is also a question about the setting of the problem and whether the function at the origin is zero or not.
  • #1
Antarres
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As a part of a bigger problem, I was trying to evaluate the D'Alambertian of ##\epsilon(t)\delta(t^2-x^2-y^2-z^2)##, where ##\epsilon(t)## is a sign function. This term appears in covariant commutator function, so I was checking whether I can prove it solves Klein-Gordon equation. Since there's no mass involved in the term, I would assume that D'Alambertian of the function should be proportional to the function itself(unit mass case).

However after some calculation(product + chain rule), using that ##(\epsilon(t))' = 2\delta(t)## in the sense of distributions, I get after some algebra:
$$\square(\epsilon(t)\delta(\sigma)) = 2\delta'(t)\delta(\sigma) + 8\epsilon(t)\delta'(\sigma)+8t\delta(t)\delta'(\sigma) + 4\epsilon(t)\sigma\delta''(\sigma)$$
where ##\sigma = t^2-x^2-y^2-z^2##.
Here, however, I run into problems. For example, the third term in the equation comes from the middle term of product expansion:
$$\square(f(t)g(\sigma)) = \left(\frac{\partial^2}{\partial t^2}f(t)\right)g(\sigma) + 2\frac{\partial}{\partial t}f(t)\frac{\partial}{\partial t}g(\sigma) + f(t)\square g(\sigma)$$
This middle term in our particular case looks like:
$$2\frac{\partial}{\partial t}\epsilon(t)\frac{\partial}{\partial t}\delta(\sigma) =4\delta(t)\frac{\partial \sigma}{\partial t}\frac{d}{d\sigma}\delta(\sigma) = 8t\delta(t)\delta'(\sigma)$$
So the prime means that it's the derivative with respect to ##\sigma##, or equivalently, it's just the derivative of the delta function composed to ##\sigma##. But then here I run into problems simplifying this, because I feel like I can't employ the normal derivative rule for distributions:
$$\delta'(x)f(x) = -f'(x)\delta(x)$$
I feel like I can pass the derivative this way only for normal partial derivatives(with respect to ##t##, ##x##, ##y##,##z##), but not the ##\sigma## derivatives. So I am unsure how to proceed with simplifying this expression further. It seems to me that if I multiply delta function with some expression, it changes, and that usual rules we have for delta function do not hold if multiplied by some factor.
For example in the factor above I have ##t\delta(t)##, which should be evaluated as zero, but I'm unsure if it will be evaluated as zero if it's multiplied by this delta function derivative. And the same goes for the other factors.

I feel like going in circles trying to figure it out, so some clarification would be very helpful.
 
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  • #2
Since
[tex]\delta(t^2-l^2)=\frac{1}{|l|}(\delta(t+l)+\delta(t-l))[/tex]
Except the case x=y=z=0, 8tδ(t)δ′(σ)=0 because points of anomaly, 0, -l, and l with coefficient zero do not coincide.
Then you should focus on the case x=y=z=0. For t=0 8tδ(t)δ′(t^2) is an odd function of t so it should be zero or not defined there sandwitched with divergences of ##\pm \infty## as function ##\frac{1}{t}##.

I have a question on the setting of the problem.
Antarres said:
As a part of a bigger problem, I was trying to evaluate the D'Alambertian of ϵ(t)δ(t2−x2−y2−z2), where ϵ(t) is a sign function.
Your setting at the origin
[tex]\epsilon(t)\delta(t^2)[/tex]
seems not clear to me. Fot t=0, this odd function of t is zero or anything else?
 
Last edited:
  • #3
anuttarasammyak said:
[tex]\delta(t^2-l^2)=\frac{1}{|l|}(\delta(t+l)+\delta(t-l))[/tex]
I think you mean
[tex]\delta(t^2-l^2)=\frac{1}{2|l|}(\delta(t+l)+\delta(t-l))[/tex]

jason
 
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1. What is the Dirac delta function?

The Dirac delta function, also known as the Dirac delta distribution, is a mathematical function that is used to represent a point mass or impulse in the context of calculus and signal processing. It is defined as a function that is zero everywhere except at the origin, where it is infinite, and its integral over the entire real line is equal to one.

2. How is the Dirac delta function different from a regular function?

The Dirac delta function is not a regular function in the traditional sense, as it is not defined at every point on the real line. It is considered a distribution, which is a more general concept than a function. It is often used as a mathematical tool to simplify calculations and describe physical phenomena, but it does not have a specific value at any point other than the origin.

3. Can the Dirac delta function be graphed?

No, the Dirac delta function cannot be graphed in the traditional sense because it is not a continuous function. However, it can be represented visually as a spike or impulse at the origin, with an area of one under the curve. This representation is often used in engineering and physics to illustrate the concept of a point mass or impulse.

4. How is the Dirac delta function used in practical applications?

The Dirac delta function has many practical applications in various fields, including physics, engineering, and signal processing. It is commonly used to model point masses, impulses, and other discontinuities in physical systems. It is also used in the study of differential equations, Fourier analysis, and the convolution integral.

5. What are some common misconceptions about the Dirac delta function?

One common misconception about the Dirac delta function is that it represents an actual function that exists at the origin. In reality, it is a mathematical construct that is used to simplify calculations and describe physical phenomena. Another misconception is that the Dirac delta function is equal to infinity at the origin, when in fact it is undefined at that point. Additionally, some people may mistakenly think that the Dirac delta function is only applicable in physics, when it is actually used in a wide range of fields and applications.

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