Is the Fourier Transform of $$\nabla f\cdot \nabla g$$ from $\vec{x}$ to $\vec{k}$ space a convolution?

I know that, for a certain definition of the FT, $$\nabla f\to \vec{k} F $$ where $F(\vec{k})$ is $f(\vec{x})$ after FT.

I suspect the FT of $\nabla f\cdot \nabla g$ involves a convolution though? Something like

$$\int d^3 k' \,\,\vec{k}\, F(\vec{k})\cdot (\vec{k}-\vec{k}')G(\vec{k}-\vec{k}') \,\, ?$$

Sorry for the probably simple question. Also, if anyone has good comprehensive notes on FTs that'd be much appreciated too. Thanks!


Lets try to figure it out. Let $\phi(\vec{k})$ be the Fourier transform of $\nabla f \cdot \nabla g $.

$$ \nabla f \cdot \nabla g = \int \phi(\vec{k}) \ e^{i\vec{k}\cdot\vec{x}} \ \mathrm d^3\vec{k} $$

Now we write $f$ and $g$ in terms of their Fourier transforms.

$$ \int \int \nabla \left(\hat{f}(\vec{k})e^{i\vec{k}\cdot\vec{x}}\right) \cdot \left(\hat{g}(\vec{q})e^{i\vec{q}\cdot\vec{x}}\right)\ \mathrm d^3\vec{k}\ \mathrm d^3\vec{q} = \int \phi(\vec{k}) \ e^{i\vec{k}\cdot\vec{x}} \ \mathrm d^3\vec{k} $$

$$ \int \int \hat{f}(\vec{k})\hat{g}(\vec{q}) \left( \vec{k} \cdot \vec{q}\right) e^{i(\vec{q}+\vec{k})\cdot\vec{x}} \ \mathrm d^3\vec{k}\ \mathrm d^3\vec{q} = \int \phi(\vec{k}) \ e^{i\vec{k}\cdot\vec{x}} \ \mathrm d^3\vec{k} $$

To kill some of the integrals I am going to apply the fourier transform to both sides, i.e., multiply both sides by $\exp(-ip\cdot x)/ (2 \pi)^3$ and integrating with respect to $x$. On the right we will get $\phi(p)$ on the left we will get a delta function which collapses one of the integrals.

$$ \frac{1}{(2\pi)^3} \int\int \int \hat{f}(\vec{k})\hat{g}(\vec{q}) \left( \vec{k} \cdot \vec{q}\right) e^{i(\vec{q}+\vec{k}-\vec{p})\cdot\vec{x}} \ \ \mathrm d^3\vec{x} \mathrm d^3\vec{k}\ \mathrm d^3\vec{q} = \phi(\vec{p}) $$

$$ \frac{1}{(2\pi)^3} \int\int \hat{f}(\vec{k})\hat{g}(\vec{q}) \left( \vec{k} \cdot \vec{q}\right)(2\pi)^3 \delta^{(3)}(\vec{q}+\vec{k}-\vec{p})\ \mathrm d^3\vec{k}\ \mathrm d^3\vec{q} = \phi(\vec{p}) $$

$$ \int \hat{f}(\vec{k})\hat{g}(\vec{p}-\vec{k}) \left( \vec{k} \cdot (\vec{p}-\vec{k})\right) \ \mathrm d^3\vec{k} = \phi(\vec{p}) $$

So it seems that the identity you propose it true, or at least consistent with my result.


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