What is the purpose of the Dirac delta function in three dimensions?

In summary, the conversation discusses the use of the Dirac delta function in 3D and its properties. The speaker asks for clarification on the use of the function in integrals and how it picks out certain variables. The expert explains that the delta function should only be used within integrals and clarifies its properties in 1D and 3D. The conversation ends with the speaker expressing their understanding of the function.
  • #1
quietrain
655
2
i don't really understand the dirac delta function in 3D.

is it right that integral of f(r)d3(r-a)dt = f(a)

where a = constant ,r is like variable x in 1D dirac delta function?




so why when i have f(r')d3(r-r') , it picks out f(r)?

where r is now a constant and r' is a variable

shouldn't it be f(r')d3(r'-r), then it picks out f(r) ?

it is as though saying that (r-r') = (r'-r) when we use dirac delta function?

thanks!
 
Physics news on Phys.org
  • #2
What are your integration variables in each case?
 
  • #3
there is like no limits? i don't know. on the 2nd line, it picks out r...

dirac.jpg
 
  • #4
Ah, I'm almost certain you've got Griffiths there. If I had to hazard a guess I'd say that [tex]d\tau[/tex] is ranging over the r' though I'm not familiar with the notation. But it would make sense if 'r' is the position we're interested in knowing the potential at we'd want to add up (i.e. integrate) all the contributions from other points (the r').

So a 1-d version would look like:

[tex]
\int \rho(x')\delta(x-x')dx' = \rho(x)
[/tex]
 
  • #5
but i thought the dirac delta function works like this

f(x)d(x-a), it picks out f(a)

so from your 1d example, you mean it also works like this

f(x)d(a-x) = f(a)

so mathematically, the (x-a) or (a-x) are equal in the dirac delta function?

(x-a) is just a shift of the axis right? where a is constant, meaning it shifts the x-axis towards +a right?

so (a-x) is also a shift of axis? so now it shifts the axis in +a or -a?
 
  • #6
quietrain said:
but i thought the dirac delta function works like this

f(x)d(x-a), it picks out f(a)

It doesn't make sense to write the delta function outside of an integral. It only has meaning inside of an integral. So yes,

[tex]
\int_{b_1}^{b_2} f(x)\delta(x-a)dx=f(a)
[/tex]

if [tex]a\in[b_1,b_2][/tex]

so from your 1d example, you mean it also works like this

f(x)d(a-x) = f(a)

so mathematically, the (x-a) or (a-x) are equal in the dirac delta function?
In both cases the delta function is zero unless x=a
(x-a) is just a shift of the axis right? where a is constant, meaning it shifts the x-axis towards +a right?

so (a-x) is also a shift of axis? so now it shifts the axis in +a or -a?

x-a is a shift right by a, and a-x=-x+a is a reflection of the x-axis and then a shift left (which means a shift right). But just try to see it this way, the integral is zero unless the argument of the delta function is zero. That happens when (above) x=a. But notice that:

[tex]

\int_{\mathbb{R}}f(x)\delta(x+a)dx=f(-a)

[/tex]

I can do that without thinking about how the function's been shifted but rather just seeing where the delta function is equal to zero. Its the same in 3D. When is [tex]\vec{r}-\vec{r'}=\vec{0}[/tex] same time that [tex]\vec{r'}-\vec{r}=\vec{0}[/tex]
 
  • #7
homology said:
It doesn't make sense to write the delta function outside of an integral. It only has meaning inside of an integral. So yes,

[tex]
\int_{b_1}^{b_2} f(x)\delta(x-a)dx=f(a)
[/tex]

if [tex]a\in[b_1,b_2][/tex]


In both cases the delta function is zero unless x=a


x-a is a shift right by a, and a-x=-x+a is a reflection of the x-axis and then a shift left (which means a shift right). But just try to see it this way, the integral is zero unless the argument of the delta function is zero. That happens when (above) x=a. But notice that:

[tex]

\int_{\mathbb{R}}f(x)\delta(x+a)dx=f(-a)

[/tex]

I can do that without thinking about how the function's been shifted but rather just seeing where the delta function is equal to zero. Its the same in 3D. When is [tex]\vec{r}-\vec{r'}=\vec{0}[/tex] same time that [tex]\vec{r'}-\vec{r}=\vec{0}[/tex]



OH i see . so the crux is just to make the dirac delta function 0 so that it satisfy the definition that it is infinity at that point and the integral over all space is thus 1.

thank you very much!
 

1. What is the 3D Dirac Delta Function?

The 3D Dirac Delta Function, also known as the three-dimensional delta function, is a mathematical function used in physics and engineering to model point sources or point masses in three-dimensional space. It is defined as:

δ(x,y,z) = 0 for all (x,y,z) ≠ (0,0,0)

δ(x,y,z) = ∞ at (x,y,z) = (0,0,0)

The function is zero everywhere except at the origin, where it is infinite.

2. How is the 3D Dirac Delta Function used in physics?

The 3D Dirac Delta Function is used to represent point-like objects in three-dimensional space, such as particles with mass, charge, or magnetic moment. It is also used in electromagnetism to model point charges or point dipoles. In quantum mechanics, it is used to represent the position of a particle in three-dimensional space.

3. What are the properties of the 3D Dirac Delta Function?

Some important properties of the 3D Dirac Delta Function include:

  • Normalization: ∫δ(x,y,z) dV = 1, where the integral is taken over all space.
  • Shift invariance: δ(x-a, y-b, z-c) = δ(x,y,z) for any constants a, b, and c.
  • Scale invariance: δ(kx, ky, kz) = δ(x,y,z)/|k| for any nonzero constant k.
  • Symmetry: δ(x,y,z) = δ(z,y,x) = δ(x,z,y) = ...
  • Convolution: The 3D Dirac Delta Function can be convolved with any function f(x,y,z) to yield f(0,0,0). In other words, ∫f(x,y,z)δ(x,y,z) dV = f(0,0,0).

4. Can the 3D Dirac Delta Function be graphed?

No, the 3D Dirac Delta Function cannot be graphed in the traditional sense because it is a point-like function with infinite value at the origin and zero everywhere else. However, it can be visualized as a spike at the origin in a 3D plot, with the height of the spike representing the infinite value of the function at that point.

5. How is the 3D Dirac Delta Function related to the 1D and 2D Dirac Delta Functions?

The 3D Dirac Delta Function is the three-dimensional analog of the 1D and 2D Dirac Delta Functions. Just like the 3D function, the 1D and 2D functions are zero everywhere except at their respective origins, where they are infinite. The 1D and 2D functions can be obtained by taking cross-sections of the 3D function in the x-y and x-y-z planes, respectively.

Similar threads

  • Calculus
Replies
25
Views
840
Replies
3
Views
2K
Replies
18
Views
1K
Replies
2
Views
829
Replies
4
Views
961
Replies
4
Views
10K
Replies
1
Views
793
  • Advanced Physics Homework Help
Replies
5
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
12
Views
3K
Back
Top