In summary, the answer to the question contains a delta function. You can find the delta function by applying the divergence theorem.
  • #1
UVCatastrophe
37
8
For a research project, I have to take multiple derivatives of a Yukawa potential, e.g.

## \partial_i \partial_j ( \frac{e^{-m r}}{r} ) ##

or another example is

## \partial_i \partial_j \partial_k \partial_\ell ( e^{-mr} ) ##

I know that, at least in the first example above, there will be a Dirac delta function somewhere in the answer,

## \partial_i \partial_j \{ \ldots \} \supset - 4\pi \delta^{i j} \delta^{(3)} (\mathbf{x} ) ##

If you do this by naively applying the chain rule and product rule to the functions, unit vectors, and their derivatives, you will miss it. The way to see that the answer contains a delta function is to apply the divergence theorem, realize there's a contradiction, and then add the 3D Dirac delta ad hoc. (Refer to Chapter 1.5 of Griffiths Electrodynamics for instance.)

This is a subtle point: easy to miss, and worrisome that I have to put some of the answer in by hand. Does anyone know if there are other functions, or combinations of functions and unit vectors, that straight up chain/product rule will give incomplete answers? Or can someone assure me that the delta function is the only case? Thanks!
 
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  • #2
I wouldn't blame the chain and product rules. 1/r diverges at r=0 and r does not have a continuous derivative there, so you have to decide how to interpret this singularity and the derivatives. E.g., you could replace 1/r by something smooth and finite like ##1/\sqrt(r^2+a^2)## and take the limit a to 0 after you did all the derivatives.
 
  • #3
You can try the "test function" method. If ## \partial_i \partial_j ( \frac{e^{-m r}}{r} ) ## is something involving Dirac deltas and their derivatives, Examine it's action by for example

$$\int d^3x f(\mathbf{r}) \partial_i \partial_j ( \frac{e^{-m r}}{r} ) $$

And then appliMaybe that will be helpful. (Oh I see you need see this about the divergence theorem, nevermind)...
 

1. What are derivatives in 3D and Dirac Delta?

Derivatives in 3D refer to the process of calculating the rate of change of a function in three-dimensional space. This involves finding the partial derivatives with respect to each variable. The Dirac Delta, also known as the Dirac function, is a mathematical function that is used to represent point masses or impulses in physics and engineering.

2. How are derivatives in 3D and Dirac Delta related?

The Dirac Delta is often used in the context of derivatives in 3D because it allows for the representation of point masses or impulses in three-dimensional space. This is useful in physics and engineering applications where point masses or impulses may occur.

3. What is the relationship between the Dirac Delta and the delta function in calculus?

The Dirac Delta is a specialized version of the delta function in calculus. The Dirac Delta is a continuous function while the delta function is a discrete function. Additionally, the Dirac Delta is only non-zero at a single point, while the delta function may have multiple non-zero points.

4. How is the Dirac Delta used in solving physical problems?

The Dirac Delta is used to represent point masses or impulses in physical problems. This allows for the simplification of equations and the solution of problems involving point masses or impulses. The Dirac Delta is also used in the study of quantum mechanics and signal processing.

5. Are there any limitations or challenges when working with derivatives in 3D and Dirac Delta?

One limitation of working with derivatives in 3D and Dirac Delta is that they can be difficult to visualize and understand due to their abstract nature. Additionally, the Dirac Delta is not a true function and has some mathematical properties that can make it challenging to work with. However, with proper understanding and application, these tools can be valuable in solving complex physical problems.

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