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I Gradient of Dirac delta

  1. Jul 29, 2016 #1
    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.

    Thanks in advance,

    Ian
     
  2. jcsd
  3. Jul 29, 2016 #2

    BvU

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  4. Jul 29, 2016 #3

    Orodruin

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    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'?
     
  5. Jul 29, 2016 #4
    .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
     
  6. Jul 29, 2016 #5

    vanhees71

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    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
  7. Jul 29, 2016 #6
    It is clear now, thank you.
     
  8. Sep 14, 2016 #7

    Orodruin

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    Both cases should be f(x').
     
  9. Sep 14, 2016 #8

    vanhees71

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