Question about the Dirac delta function

In summary, the conversation revolves around transforming an interval on the x-axis, defined by the parameter L, into a Dirac delta function using integration. The suggestion is to use the indicator function, I(x), instead of the original function, f(x), with L represented by the integral. There is confusion about the usage and representation of the Dirac delta function and its relation to the parameter L.
  • #1
SeM
Hi, if I have an interval on the x-axis, defined by the parameter L, can this, interval be transformed to a Dirac delta function instead, on the x-axis?

Thanks!
 
Physics news on Phys.org
  • #2
Wouldn't it simply become a point ? Like in: the interval ##[x,x+\epsilon]## becomes the point ##x## for ##\epsilon \downarrow 0##
 
  • #3
BvU said:
Wouldn't it simply become a point ? Like in: the interval ##[x,x+\epsilon]## becomes the point ##x## for ##\epsilon \downarrow 0##

Good "point"! Thanks, I will give it a thought!
Cheers
 
  • #4
You can do it with integration, if that is ok. An indicator of the interval (a,b) would be f(x) = ∫-∞xδ(t-a)dt - ∫-∞xδ(t-b)dt.

PS. I'm not sure about the value at the interval endpoints, a and b.
 
  • #5
Thank you! This is indeed a great suggestion. However, L is included in a function I have, f(x), and currently it is a parameter. It would be useful to use your form somehow, as you write:

\begin{equation}
\int_{-\infty}^x \delta(t-a)dt - \int_{-\infty}^x \delta(t-b)dt
\end{equation}

Is L actually a-b here, and t is the variable of my function, f(x)?

If e^(ix)+L is my function, would the equivalent be :

\begin{equation}
\int_{-\infty}^x \delta(e^{it}-a)dt - \int_{-\infty}^x \delta(e^{it}-b)dt
\end{equation} ?

Thanks!
 
  • #6
SeM said:
Thank you! This is indeed a great suggestion. However, L is included in a function I have, f(x), and currently it is a parameter. It would be useful to use your form somehow, as you write:

\begin{equation}
\int_{-\infty}^x \delta(t-a)dt - \int_{-\infty}^x \delta(t-b)dt
\end{equation}

Is L actually a-b here,
For an interval (a,b) where a<b, that would make L negative. If 'L' stands for length, do you mean L = b-a?
If so, you could change the definition of the indicator function (call it I(x) to distinguish it from your f(x)). Then
\begin{equation}
I(x) = \int_{-\infty}^x \delta(t-a)dt - \int_{-\infty}^x \delta(t-(a+L))dt
\end{equation}
and t is the variable of my function, f(x)?
t is just a dummy variable for the integration, so it does not appear as an input or parameter of the indicator function.
If e^(ix)+L is my function, would the equivalent be :

\begin{equation}
\int_{-\infty}^x \delta(e^{it}-a)dt - \int_{-\infty}^x \delta(e^{it}-b)dt
\end{equation} ?
I'm confused about what you are trying to do here.
 
  • #7
[QUOTE\I'm confused about what you are trying to do here.[/QUOTE]Me too. I am unfamiliar with the Dirac delta function. It just says \delta and it gives therefore no idea to me on how it looks like a function.

Nevertheless, you say the dirac f is a dummy function. The thing I am not sure about is that because L is part of the existing function, f(x), say:

\begin{equation}
f(x) = e^{-ipk}/L
\end{equation}

how would the Dirac variant of this look like, where L is represented by the integral?

Thanks
 
  • #8
SeM said:
It just says \delta and it gives therefore no idea to me on how it looks like a function.
'It just says'? Where are you seeing this reference to the delta function? You need to back up and explain more about what your original problem is and why you want to use the delta function.
Nevertheless, you say the dirac f is a dummy function.
Who said that it is a 'dummy' function? I never said that.
The thing I am not sure about is that because L is part of the existing function, f(x), say:

\begin{equation}
f(x) = e^{-ipk}/L
\end{equation}

how would the Dirac variant of this look like, where L is represented by the integral?
Why would you want to replace L with a delta function? Is L a constant?
 
  • #9
FactChecker said:
Why would you want to replace L with a delta function? Is L a constant?
L is a parameter, the width of the space the particle is in.
 

1. What is the Dirac delta function?

The Dirac delta function, also known as the unit impulse function, is a mathematical function that is defined as zero everywhere except at a single point, where its value is infinite. It is often used in engineering and physics to represent a concentrated point of mass, charge, or energy.

2. How is the Dirac delta function represented?

The Dirac delta function is typically represented by the symbol δ or δ(x). It is a continuous function that has a value of 0 for all values of x except at x = 0, where it has a value of infinity.

3. What is the purpose of the Dirac delta function?

The Dirac delta function is used to model idealized point sources in physics and engineering problems. It allows for the simplification of mathematical equations by representing a point source as a function rather than a distribution of values.

4. How is the Dirac delta function used in calculus?

The Dirac delta function is commonly used in calculus to represent discontinuous functions, such as step functions. It is also used to define the derivative of a step function, which is called the Heaviside step function.

5. Can the Dirac delta function be integrated?

Yes, the Dirac delta function can be integrated, but it requires the use of a special type of integration called the Dirac integral. The Dirac integral is defined as the limit of a sequence of approximating functions, and it allows for the integration of the Dirac delta function over a specific interval.

Similar threads

  • Calculus
Replies
25
Views
841
  • Calculus
Replies
2
Views
2K
  • Calculus
Replies
2
Views
1K
Replies
3
Views
2K
Replies
24
Views
2K
Replies
2
Views
829
Replies
1
Views
793
Replies
4
Views
961
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top