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

[quietrain]

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!

 

[homology]

What are your integration variables in each case?

 

[quietrain]

there is like no limits? i don't know. on the 2nd line, it picks out r...

 

[homology]

Ah, I'm almost certain you've got Griffiths there. If I had to hazard a guess I'd say that $$d\tau$$ 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:

$$\int \rho(x')\delta(x-x')dx' = \rho(x)$$

 

[quietrain]

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?

 

[homology]

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,

$$\int_{b_1}^{b_2} f(x)\delta(x-a)dx=f(a)$$

if $$a\in[b_1,b_2]$$

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:

$$<br /> \int_{\mathbb{R}}f(x)\delta(x+a)dx=f(-a)<br />$$

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 $$\vec{r}-\vec{r'}=\vec{0}$$ same time that $$\vec{r'}-\vec{r}=\vec{0}$$

 

[quietrain]

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

$$\int_{b_1}^{b_2} f(x)\delta(x-a)dx=f(a)$$

if $$a\in[b_1,b_2]$$


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:

$$<br /> \int_{\mathbb{R}}f(x)\delta(x+a)dx=f(-a)<br />$$

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 $$\vec{r}-\vec{r'}=\vec{0}$$ same time that $$\vec{r'}-\vec{r}=\vec{0}$$

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!