Dirac delta function in spherical cordinates

[Pushoam]

Homework Statement

Calculate
$\int_{r=0}^\inf δ_r (r -r_0)\,dr$

Homework Equations

$\int_V \delta^3(\vec{r} - \vec{r}') d\tau = 1$

The Attempt at a Solution

$$\int_V \delta^3(\vec{r} - \vec{r}') d\tau =
\int_V \frac {1}{r^2 sinθ}\delta_r(r-r_0) \delta_θ (θ-θ_0) \delta_Φ (Φ-Φ_0) r^2 sinθ dr dθ dΦ = 1
$$
$$\int_{r=0}^ \inf \delta_r(r-r_0) dr \int_{θ=0}^{π/2}\delta_θ (θ-θ_0) dθ\int_{Φ=0}^{2π}\delta_Φ (Φ-Φ_0) dΦ = 1$$
$$\int_{r=0}^ \inf \delta_r(r-r_0) dr = \int_{θ=0}^{π/2}\delta_θ (θ-θ_0) dθ = \int_{Φ=0}^{2π}\delta_Φ (Φ-Φ_0) dΦ = 1$$

Is this correct?

 

[marcusl]

I don't think it is, because you have assumed specific values \theta_0 and \Phi_0 but your problem is one-dimensional and independent of angle. I would approach it as a 1D problem and use the definition of a delta as the limit of a sequence of functions of unit area. Have you covered that in your class?

 

[Pushoam]

Have you covered that in your class?

No

$$\int_{r=0}^ \inf \delta_r(r-r_0) dr \int_{θ=0}^{π/2}\delta_θ (θ-θ_0) dθ\int_{Φ=0}^{2π}\delta_Φ (Φ-Φ_0) dΦ = 1$$

​

$$\int_{r=0}^ \inf \delta_r(r-r_0) dr = \int_{θ=0}^{π/2}\delta_θ (θ-θ_0) dθ = \int_{Φ=0}^{2π}\delta_Φ (Φ-Φ_0) dΦ = 1$$

Are the above two steps correct even if they are not related to the current problem?

 

[marcusl]

Only if \theta_0 and \Phi_0 lie within the limits specified in the integrals. In fact, I've not previously seen the form you used for \delta^3 in the first line under #3 of your original post.

There is lots of information on the web about the delta as a sequence of functions. Page 4 of this link, for instance,
https://redirect.viglink.com/?format=go&jsonp=vglnk_149694239503512&key=6afc78eea2339e9c047ab6748b0d37e7&libId=j3ooulsj010009we000DAatnkkpxm&loc=https%3A%2F%2Fwww.physicsforums.com%2Fthreads%2Fderivative-of-dirac-delta-function.372548%2F&v=1&out=http%3A%2F%2Flinks.uwaterloo.ca%2Famath731docs%2Fdelta.pdf&ref=https%3A%2F%2Fwww.google.com%2F&title=Derivative%20of%20dirac%20delta%20function%20%7C%20Physics%20Forums%20-%20The%20Fusion%20of%20Science%20and%20Community&txt=http%3A%2F%2Flinks.uwaterloo.ca%2Famath731docs%2Fdelta.pdf
contains a concise derivation of the 1D delta function that you may find helpful.

 

[rude man]

Not sure what there is to compute. The definition of the dirac distribution (I still call it function but we must humor the pure mathematicians ) is ∫ f(x) δ(x - x₀) dx over all x space = f(x₀) in cartesian coordinates.

In the case of cartesian coordinates, x ranges over -∞ < x < ∞ whereas in spherical coordinates, r ranges over 0 < r < ∞. So ∫f(r - r₀) dr over 0 to ∞ = f(r₀) and since f(r) = 1 here we simply get 1 for the answer.

Unless the sub-r in δ_r is supposed to connote differentiation, in which case I plead ignorance.

 

[marcusl]

You have quoted the answer, which is already given in the question. I assume that he needs to show how to get the answer.

 

[rude man]

You have quoted the answer, which is already given in the question.

??

 

[haruspex]

Homework Statement

Calculate
$\int_{r=0}^\inf δ_r (r -r_0)\,dr$

Not sure what the expression means.
As written, this is purely a 1-dimensional integral. The answer will depend whether r₀ is positive or negative.
If you mean it as a volume integral, $\int_{\infty}\delta(\vec r -\vec {r_0}).\vec {dr}$, then as @rude man says, it is by definition 1.

 

[rude man]

Not sure what the expression means.
As written, this is purely a 1-dimensional integral. The answer will depend whether r₀ is positive or negative.
If you mean it as a volume integral, $\int_{\infty}\delta(\vec r -\vec {r_0}).\vec {dr}$, then as @rude man says, it is by definition 1.

Hm, I didn't necessarily mean it as a volume integral and I don't see that it matters whether r₀ is + or -.
So, confusion all around!

 

[haruspex]

I didn't necessarily mean it as a volume integral

You were interpreting it as r in spherical coordinates. If not as a volume integral then it must be a linear integral in some arbitrary but unspecified direction, and δ must be defined in terms of line integrals in that direction. In that case, it is a semi-infinite line and r₀ could be negative, making the result zero.

 

[rude man]

You were interpreting it as r in spherical coordinates. If not as a volume integral then it must be a linear integral in some arbitrary but unspecified direction, and δ must be defined in terms of line integrals in that direction. In that case, it is a semi-infinite line and r₀ could be negative, making the result zero.

If a semi-infinite line then a negative r₀ is not defined.

 

[haruspex]

If a semi-infinite line then a negative r₀ is not defined.

The integral range is a semi-infinite line. Just like x from 0 to ∞ in Cartesian.

 

[rude man]

The integral range is a semi-infinite line. Just like x from 0 to ∞ in Cartesian.

Right. So there are no negative numbers, by definition.

 

[haruspex]

Right. So there are no negative numbers, by definition.

Eh? There are no negative values of r in the range, but there is nothing to stop r₀ being negative.

 

[rude man]

Eh? There are no negative values of r in the range, but there is nothing to stop r₀ being negative.

r₀ must be in the range of r which in turn is in the range 0 to ∞, hence r₀ cannot be <0.

 

[haruspex]

r₀ must be in the range of r which in turn is in the range 0 to ∞, hence r₀ cannot be <0.

I guess it depends on your view of the conventions.
By convention, we represent points in the plane or 3-space in polars using only non-negative values of r, but that is merely a canonical form. In the same way, we pick ranges for the angles. But in plane (r, θ) coordinates I would say the point (-1, 0) exists, and it is the same as the points (1, π) and (1, 3π).
Anyway, the whole discussion is academic. We need to know what the question intends. It makes no sense to me as it stands.

 

[rude man]

Anyway, the whole discussion is academic. We need to know what the question intends. It makes no sense to me as it stands.

Agreed.

 

[rude man]

But in plane (r, θ) coordinates I would say the point (-1, 0) exists, and it is the same as the points (1, π) and (1, 3π).

Don't agree. Different rules in Australia?

 

[Pushoam]

Are the above two steps correct even if they are not related to the current problem?

Only if $\theta_0 and \Phi_0$ lie within the limits specified in the integrals

Sorry for being late in replying,
Thank you all for the replies,
I asked this question because I thought that the rules for calculating integration in spherical coordinates are different but I was wrong.
The above steps are correct by the definition of δ- function (Only if $\theta_0 and \Phi_0$ lie within the limits specified in the integrals).

 

[rude man]

There is actually a small issue here still:
Consider: ∫f(x)δ(x - a)dx = f(a)/2 if a is either the upper or lower limit of integration. I.e. not f(a).
So ∫δ(r - r₀)dr with limits of integration 0 and ∞ = 1/2 if r₀ = 0.

 

[Pushoam]

Consider: ∫f(x)δ(x - a)dx = f(a)/2 if a is either the upper or lower limit of integration. I.e. not f(a).

How do we get it?
In Cartesian Coordinates system, if a is either lower or upper limit ,we can easily change the limit such that a is neither lower nor upper limit as the integrand is 0 for x>a or x<a because of δ - function.

But in spherical coordinates , I can't do it for r=0 as a lower limit.
So, is the result valid only for r=0 as a lower limit spherical or cylindrical systems?

 

[rude man]

How do we get it?
In Cartesian Coordinates system, if a is either lower or upper limit ,we can easily change the limit such that a is neither lower nor upper limit as the integrand is 0 for x>a or x<a because of δ - function.

But in spherical coordinates , I can't do it for r=0 as a lower limit.
So, is the result valid only for r=0 as a lower limit spherical or cylindrical systems?

Well, the result is valid only if r₀ > 0. In the cartesian system it is, as you point out, always possible to "wrap around" both sides of the nominal location of the delta function when we integrate it, even at the origin x=y=z=0. In spherical and cylindrical coordinates it is not, simply because negative values of r are impermissible; we can't wrap below r=0 yet we must if the delta function is to have any meaning. So, it has no meaning there. (EDITed out part of last sentence as being irrelevant to this post).BTW I consider this topic belongs in the advanced forum. If you go there, orodruin, others & I have just posted in a similar vein, suggest you look at it: https://www.physicsforums.com/threa...in-spherical-coordinates.765520/#post-5782888

 

[Pushoam]

Thanks for the reply.

What is meant by BTW?

 

[Pushoam]

Consider: ∫f(x)δ(x - a)dx = f(a)/2 if a is either the upper or lower limit of integration. I.e. not f(a).

rude man,
I wanted to know how you got this as I have never come across it.
Can you give me a reference for it?

 

[cnh1995]

What is meant by BTW

By the way.

 

[rude man]

rude man,
I wanted to know how you got this as I have never come across it.
Can you give me a reference for it?

Sorry about the "BTW" Pushoam. A common abbreviation used by the lazy!
I attached a reference you asked for. See at top of page 1.

 

[Pushoam]

rude man,
I wanted to know how you got this as I have never come across it.
Can you give me a reference for it?

The given reference doesn't answer this question.
It simply says that δ- function is defined this way, but the books which I have read till now don't include
"∫f(x)δ(x - a)dx = f(a)/2 if a is either the upper or lower limit of integration. I.e. not f(a) " in definition.
So, does it mean that these books define δ- function incompletely?

Anyway, I am not being able to appreciate the significance of this statement "∫f(x)δ(x - a)dx = f(a)/2 if a is either the upper or lower limit of integration. I.e. not f(a)" .
For Cartesian Coordinates System, it doesn't matter as we can always wrap up the upper or lower limits and for Spherical or Cylindrical , it becomes meaningless (as you mentioned).

Can you please give me an example which will illustrate why it is necessary to know it( ∫f(x)δ(x - a)dx = f(a)/2 if a is either the upper or lower limit of integration. I.e. not f(a)) ?

 

[rude man]

The given reference doesn't answer this question.
It simply says that δ- function is defined this way, but the books which I have read till now don't include
"∫f(x)δ(x - a)dx = f(a)/2 if a is either the upper or lower limit of integration. I.e. not f(a) " in definition.
So, does it mean that these books define δ- function incompletely?

Apparently so.

Can you please give me an example which will illustrate why it is necessary to know it( ∫f(x)δ(x - a)dx = f(a)/2 if a is either the upper or lower limit of integration. I.e. not f(a)) ?

Your problem asked to evaluate an integral with given limits. I have no idea what your problem represents, some real physical situation or whatever. I merely pointed out that the integral, wherever it came from, evaluates to either 1 or 1/2 depending on whether r₀ = 0 or > 0. I myself have never encountered a situation where I needed to invoke that feature (of evaluating to 1/2 instead of 1). The delta function is a mainstay in sampling theory yet I never encountered a situation where the signal being sampled could not be handled by an integral with limits surrounding both sides of the delta function location, i.e. r₀ in your case.