# I Gradient of Dirac delta

1. Jul 29, 2016

### IanBerkman

Dear all,

I have a quick question, is the following statement true?
$$\nabla_\textbf{x'} \delta(\textbf{x}-\textbf{x'}) = \nabla_\textbf{x} \delta(\textbf{x}-\textbf{x'})?$$

I thought I have seen this somewhere before, but I could not remember where and why.
I know the identity $d/dx \delta(x) = \delta(x)/x$ but I do not see how to implement this into the above equation.

Ian

2. Jul 29, 2016

### BvU

3. Jul 29, 2016

### Orodruin

Staff Emeritus
I suggest you think about the more general question when you replace the delta by an arbitrary function of a vector, i.e., how is the gradient of f(x-x') wrt x related to its derivative wrt x'?

4. Jul 29, 2016

### IanBerkman

.For a general function the answer is $\nabla'f(x-x')=-\nabla f(x-x')$. I was too distracted with the delta function itself, this also explains the minus sign

5. Jul 29, 2016

### vanhees71

To check, what's right, you need to remember the meaning of the $\delta$ distribution. It's a functional acting on an appropriate set of test functions (e.g., the smooth functions with compact support, $C_0^{\infty}(\mathbb{R}^3)$. Then you have by definition [corrected due to #7 and #8]
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \vec{\nabla}_x \delta^{(3)}(\vec{x}-\vec{x}') f(\vec{x}')=\vec{\nabla}_x \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \delta^{(3)}(\vec{x}-\vec{x}') f(\vec{x}')=\vec{\nabla}_x f(\vec{x}).$$
On the other hand, via integration by parts,
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \vec{\nabla}_{x'} \delta^{(3)}(\vec{x}-\vec{x}') f(\vec{x}')=-\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \delta^{(3)}(\vec{x}-\vec{x}') \vec{\nabla}_{x'} f(\vec{x}')=-\vec{\nabla}_{x} f(\vec{x}).$$
Thus comparing the two formulae tells you
$$\vec{\nabla}_x \delta^{(3)}(\vec{x}-\vec{x}')=-\vec{\nabla}_{x'} \delta^{(3)}(\vec{x}-\vec{x}').$$

Last edited: Sep 14, 2016
6. Jul 29, 2016

### IanBerkman

It is clear now, thank you.

7. Sep 14, 2016

### Orodruin

Staff Emeritus
Both cases should be f(x').

8. Sep 14, 2016

### vanhees71

Of course, it's a stupid typo. I'll correct it right now.