# Find the fourier transformation of the electric potential

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.

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.

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

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.

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 $$\vec{x}\to \infty$$)

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

• Is this rule ($\mathfrak{F}(f^{(n)})(k) = (2\pi i k ) \mathfrak{F} (k)$ also applicable to $\vec{x},\vec{k} \in \mathbb{R}^3$ Thanks again for your explanation.
– T_B
Commented Dec 14, 2023 at 11:27
• Yes, it does. I've rewritten to clarify. Commented Dec 14, 2023 at 13:15

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.

• Thank you Roland for your extremely helpful answer. So I m correct if I came to the result $\mathfrak{F}(\Phi(\vec{x})) = \frac{-q}{\varepsilon_0 4 \pi^2 \vec{k}}$ . And how did you reach the result of the potential $\Phi(\vec{x})$ by using the inverse fouriertransformation? Sorry for the huge follow up question, but now I believe I start understanding it a bit better.
– T_B
Commented Dec 14, 2023 at 13:08
• In your tripple integral, how are you getting to the $k^2$ in front of the inner integral? Since the inner integral doesn’t deppend on k, would the $k^2$ and $\frac{1}{k^2}$ not combine to $1$? Sorry for the stupid question. I think I didn’t quite understand that part just yet.
– T_B
Commented Dec 14, 2023 at 17:22
• The innermost$k^{-2}$ is the potential, the radial measure of integration is $k^2 dk$ in spherical coordinates by the surface measure of spheres. The cancellation is the central trick for the constant surface integral of the gradient over any sphere yieding the charge. For massive waves $-\Delta +m^2$ there is a denominator $(k^2+m^2)^{-1}$ yielding the short range Yukawa. Commented Dec 14, 2023 at 17:43
• Now I see, thank you very much for you time and help. Do you suggest that to slove that inner integral with the residue theorem as a contourintegral?
– T_B
Commented Dec 14, 2023 at 20:42
• No, the inner integral is simply solved by substitution $\cos \theta = z, dz=\sin \theta \ d\theta$ Commented Dec 15, 2023 at 5:48

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 $$B.$$

• Why is $\delta'(k)$ not spherically symmetric? Commented Dec 14, 2023 at 20:18
• @MarkViola. The distributional vector field $\nabla\delta$ is spherically symmetric, but the constant vector field $\vec A$ can not be, unless it vanishes. Commented Dec 14, 2023 at 20:37
• Does the solution $\hat \Phi=-\frac1{k^2}+A \delta'(k) +B\delta(k)$ for any $A$ and any $B$ satisfies $k^2\hat \Phi=-1$, does it not?. Commented Dec 14, 2023 at 21:10
• @MarkViola. We should really write $$\hat\Phi(\vec k) = -\frac{1}{|\vec k|^2} + \vec{A}\cdot\nabla\delta(\vec k) + B\delta(\vec k).$$ This satisfies $$|\vec k|^2\hat\Phi(\vec k)=-1$$ for all $\vec A\in\mathbb C^3$ and $B\in\mathbb C.$ Commented Dec 14, 2023 at 21:22