Find the fourier transformation of the electric potential

Question

as part of a homework assignment I have to solve the following problem:

The electric field can be expressed as the gradient of the scalar potential $$\Phi(\vec{x}): \nabla\cdot \vec{E}(\vec{x})= -\nabla ^2 \Phi(\vec{x})= \frac{\rho(\vec{x})}{\epsilon_0}=\frac{q\delta(\vec{x})}{\epsilon_0}$$ with $\rho(\vec{x})$ the charge density and $q$ a single charge and $\delta(\vec{x})$ the dirac delta function. How do I calculate the Fourier transformation $\hat{\Phi}(\vec{k})$ of the potential ?

So far I have tried the following:

$$ \frac{q\delta(\vec{x})}{\varepsilon_0} =-\frac{\nabla^2}{8\pi^3}\int_{\mathbb{R}^3}{\hat{\Phi}(\vec{k}e^{i\vec{k}\cdot \vec{x}} d^3 \vec{k}} = -\nabla^2 \Phi(\vec{x}) $$

$$\iff \frac{q\delta(\vec{x})}{\varepsilon_0} = \frac{-\vec{\nabla}}{8\pi^3}\hat{\Phi}(\vec{k})e^{i\vec{k} \cdot \vec{x}}$$

$$ \iff -\int_{\mathbb{R}^3}{\frac{q\delta(\vec{x})}{\varepsilon_0}8\pi^3 e^{-i \vec{k} \cdot \vec{x}} d^3 \vec{x}} = \hat{\Phi}(\vec{x}) $$

$$ \iff \frac{-q8\pi^3}{\varepsilon_0} \mathfrak{F}_3(\delta(\vec{x})) = \frac{-q8\pi^3}{\varepsilon_0} = \hat{\Phi}(\vec{x}) $$

I believe I made a mistake but I don’t see how to solve it the right way. Any help would be very much appreciated.

Answer 1

In this answer, I am using the physicist's Fourier transform in three dimensions (as this is a physics problem). This is

$$\mathcal{F}(f)(\vec{p})=\int_{\mathbb{R}^3}f(\vec{x})e^{-i\vec{p}\cdot \vec{x}}dx_1dx_2dx_3$$

with inverse transform

$$\mathcal{F}^{-1}(f)(\vec{x})=\int_{\mathbb{R}^3}f(\vec{p})e^{i\vec{p}\cdot \vec{x}}\frac{dp_1dp_2dp_3}{(2\pi)^3}.$$

The mathematician's Fourier transform adds a factor $(2\pi)^{-3/2}$ to the FT and a factor $(2\pi)^{+3/2}$ to the inverse FT, making them more symmetrical.

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Start with

$$-\nabla^2\phi(\vec{x})=\frac{q}{\varepsilon}\delta(\vec{x})$$

and Fourier transform both sides. The LHS becomes $|\vec{p}|^2\tilde{\phi}(\vec{p})$ and the RHS is just $\frac{q}{\varepsilon}$. Hence

$$\tilde{\phi}(\vec{p})=\frac{q}{\varepsilon |\vec{p}|^2}$$

This is an oft-used trick, namely transforming derivative operators into functions via the Fourier transform. See for instance Airy's differential equation.

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This calculation uses the Fourier transform of the derivative, and the Fourier transform of Dirac's delta. We have $\mathcal{F}(\delta)(\vec{p})=1$ straight from the definition of the delta and the FT. The Fourier transform of $\frac{\partial \phi}{\partial x_i}$ is $ip_i\tilde{\phi}$ by integration by parts (where we assume that function $\phi(\vec{x})\to 0$ as $\vec{x}\to \infty$)

$$\int_{\mathbb{R}^3} \frac{\partial \phi(\vec{x})}{\partial x_i} e^{-i\vec{p}\cdot\vec{x}}\, dx_1dx_2dx_3=-\int \phi(\vec{x}) (-ip_i)e^{-i\vec{p}\cdot\vec{x}}\, dx_1dx_2dx_3$$

Answer 2

By Fourier tansform in $\mathbb R^3$ and Fubini, you get the product formula for the triple integral on a complete basis of integrable functions $f(\vec x) =f_1(x)f_2(y)f_3(z)$ as a tensor product of three equal function spaces $$\mathit F(-\Delta)(\vec k) =\text{const}\ {\vec k}^2 $$ because the triple integral factors over the one-dimensional integrals, the constant depending on normalization conventions.

This yields $$\mathit F(-\Delta \Phi)(\vec k) = \frac{q}{\epsilon_0}\mathit F(\delta) * \frac{1}{{\vec k}^2 }$$ because the differential can be applied to the exponential by an integration by parts.

$$\int_\mathbb R \partial_\xi f(\xi) e^{i k \xi} = -i k \int_\mathbb R f(\xi) e^{i k \xi}$$

The inverse Fourier transform of $1/k^2$ is trivially done in cylinder or spherical coordinates, yielding the $1/r$ potential.

$$\int_ 0^{2\pi}\left(\int_ 0^\infty k^2\ \left(\int_ 0^\pi\frac {\sin (\theta) e^{i k r\cos (\theta)}} {k^2}\, d\theta \right)\, dk \right)\, d\phi$$

For simplicity and error avoidance in table lookup, its always better to determine the free constants in linear equations by insertion of the solution form into the original equation.

Answer 3

Skipping some constant factors you have $\nabla^2\Phi=\delta.$ Taking the Fourier transform of both sides gives $-k^2\hat\Phi=1.$ Therefore $$ \hat\Phi = -\frac{1}{k^2}. $$

Actually that is not completely true. The equation $-k^2\hat\Phi=1$ has more solution, namely $$ \hat\Phi = -\frac{1}{k^2} + \vec{A}\cdot\nabla \delta + B \delta. $$ Because of spherical symmetry the $\nabla\delta$ term must vanish, i.e. $\vec{A}=\vec{0}.$ So the Fourier transform has the form $$ \hat\Phi = -\frac{1}{k^2} + B \delta $$ for some constant $B.$