I have the following Maxwell's equations:

$$\nabla \times \mathbf{h} = \mathbf{j} + \epsilon_0 \dfrac{\partial{\mathbf{e}}}{\partial{t}} + \dfrac{\partial{\mathbf{p}}}{\partial{t}},$$

$$\nabla \times \mathbf{e} = - \mu_0 \dfrac{\partial{\mathbf{h}}}{\partial{t}}$$

According to my textbook (provided as a passing comment by the author), the Fourier transform,

$$F(\omega) = \int_{-\infty}^\infty f(t) e^{-j \omega t} \ dt,$$

can be applied to Maxwell's equations to go from the time domain $t$ to the angular frequency domain $\omega$.

My understanding is that this would take us from

$$\nabla \times \mathbf{h} = \mathbf{j} + \epsilon_0 \dfrac{\partial{\mathbf{e}}}{\partial{t}} + \dfrac{\partial{\mathbf{p}}}{\partial{t}}$$


$$\nabla \times \mathbf{H} = \mathbf{J} + j \omega \epsilon_0 \mathbf{E} + j \omega \mathbf{P} = \mathbf{J} + j \omega \mathbf{D}$$


$$\nabla \times \mathbf{e} = -\mu_0 \dfrac{\partial{\mathbf{h}}}{\partial{t}}$$


$$\nabla \times \mathbf{E} = - j \omega \mu_0 \mathbf{H}$$

I want to understand how to do this for the learning experience.

I have experience with the Laplace transform but not the Fourier transform, and I cannot find anything online that goes through the steps of the transformation. Do we just apply the Fourier transform $F(\omega)$ to every term in Maxwell's equations? How do we deal with the presence of vector terms in the context of such an integration?

For instance, we have

$$\nabla \times \mathbf{h} = \mathbf{\hat{i}} (\partial_y h_z - \partial_z h_y) - \mathbf{\hat{j}} (\partial_x h_z - \partial_z h_x) + \mathbf{\hat{k}} (\partial_x h_y - \partial_y h_x)$$

I would greatly appreciate it if people could please take the time to clarify this.

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    $\begingroup$ Are you sure you want a Fourier transform from $t$-space to $\omega$-space, rather than one from $\mathbf{x}$-space to $\mathbf{k}$-space? $\endgroup$
    – J.G.
    Commented May 24, 2019 at 6:41
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    $\begingroup$ @J.G. For instance, my understanding is that going from $\nabla \times \mathbf{h} = \mathbf{j} + \epsilon_0 \dfrac{\partial{\mathbf{e}}}{\partial{t}} + \dfrac{\partial{\mathbf{p}}}{\partial{t}}$ to $\nabla \times \mathbf{H} = \mathbf{J} + j \omega \epsilon_0 \mathbf{E} + j \omega \mathbf{P} = \mathbf{J} + j \omega \mathbf{D}$ is an application of the Fourier transform to go from the time domain to the angular-frequency domain, $\omega$? I hope my analysis of this is accurate. $\endgroup$ Commented May 24, 2019 at 6:44
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    $\begingroup$ Yes. The alternative I'm describing would give $i\mathbf{k}\times\mathbf{H}=\mathbf{J}+\epsilon_0\partial_t\mathbf{E}+\partial_t\mathbf{P}$. $\endgroup$
    – J.G.
    Commented May 24, 2019 at 6:45
  • $\begingroup$ @J.G. Hmm, are you saying that the textbook is incorrect and what I'm trying to do actually can't be achieved? It was given as a passing comment, and I wanted to try and derive it myself for the learning experience. $\endgroup$ Commented May 24, 2019 at 7:55
  • 1
    $\begingroup$ It's a well known problem. These differential operators transform in a nice form under Fourier transform. Check this, then, you can apply it to the Maxwell's equations. $\endgroup$ Commented May 24, 2019 at 11:01

2 Answers 2


The first thing to remember is that the Fourier transform $\mathcal{F}$ is linear: $$ \mathcal{F}(\alpha f+\beta g)=\alpha \mathcal{F}(f)+\beta \mathcal{F}(g) $$ It also changes differentiation into multiplication. Let $F(\omega)$ be the transform of $f(t)$. Then: $$ \mathcal{F}(D_tf)(\omega)=\int_{-\infty}^{\infty}D_tf(t)e^{-j\omega t}\mathrm{d}t = f(t)e^{-j\omega t}|_{-\infty}^{\infty}+j\omega\int_{-\infty}^{\infty} f(t)e^{-j\omega t}\mathrm{d}t = 0 + j\omega F(\omega) $$ For this to actually be correct, the boundary term from integration by parts has to disappear in the limit. This imposes some technical conditions. Also, if $f$ happens to be periodic, the Fourier transform (likely) won't exist in the standard sense and may require the use of distributions. I won't dwell on these issues here (mostly because I don't actually know a whole lot about all the theory behind these concerns), but perhaps you can make a separate question specifically about when this process works, or maybe someone else can add another answer regarding this.
Anyway, I'll just do the formal manipulations here (as is usually done in physics classes). In your case, we have functions of several variables $({\bf r}, t)$. However, this isn't a problem. For example: $$ \mathcal{F}(\partial_t{\bf e})({\bf r},\omega)=\int_{-\infty}^{\infty}\partial_t{\bf e}({\bf r},t)e^{-j\omega t}\mathrm{d}t= j\omega {\bf E}({\bf r},\omega) $$ Here, the process is exactly the same as the above since we can keep ${\bf r}$ fixed and so you can use integration by parts on $\partial_t$. If you're worried about vector terms, notice that this is no different from integrating any other function ${\bf f}(u)$ with values in $\mathbb{R^3}$ (I'll omit the integration limits because they don't really matter): $$ \int{\bf f}(u)\mathrm{d}u=\left(\int f_x(u)\mathrm{d}u,\int f_y(u)\mathrm{d}u,\int f_z(u)\mathrm{d}u\right) $$ Or, in component notation ($i=x,y,z$):$$ \left(\int{\bf f}(u)\mathrm{d}u\right)_i=\int f_i(u)\mathrm{d}u$$ The curl is also not problematic as the spatial derivatives commmute with time integration: $$ \int\partial_x{\bf f}({\bf r},t)\mathrm{d}t=\partial_x\int{\bf f}({\bf r},t)\mathrm{d}t $$ Because of this, you can immediately conclude that $\mathcal{F}(\partial_x{\bf f})=\partial_x\mathcal{F}({\bf f})$. Writing out only the $x$-component: $$ \left(\mathcal{F}(\nabla\times{\bf f})\right)_x=\mathcal{F}(\partial_yf_z-\partial_zf_y)=\partial_yF_z-\partial_zF_y = (\nabla\times{\bf F})_x $$ Here ${\bf F}=\mathcal{F}({\bf f})$. The other two components work in exactly the same way. Finally, take the equation $$\nabla\times{\bf h}={\bf j}+\epsilon_0\partial_t{\bf e}+\partial_t{\bf p}$$ and apply $\mathcal{F}$ to both sides. By linearity and what we've said about $\nabla\times$, we find $$ \mathcal{F}(\nabla\times{\bf h})=\mathcal{F}({\bf j}+\epsilon_0\partial_t{\bf e}+\partial_t{\bf p})$$ $$ \nabla\times\mathcal{F}({\bf h})=\mathcal{F}({\bf j})+\epsilon_0\mathcal{F}(\partial_t{\bf e})+\mathcal{F}(\partial_t{\bf p}) $$ Keeping in mind what $\mathcal{F}$ does to $\partial_t$, we finally get: $$ \nabla\times{\bf H}={\bf J}+j\omega\epsilon_0{\bf E}+j\omega{\bf P} $$ The other equation follows suit.

As one commenter pointed out, this is a Fourier transform on time. It is also possible to do a Fourier transform on the spatial coordinates. These two approaches are complementary; often, both Fourier transforms are taken and so we change from $({\bf r},t)$-space to $({\bf k},\omega)$-space where ${\bf k}$ is the wave vector.

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    $\begingroup$ Your answer is superb! Thank you very much for taking the time to clarify this. $\endgroup$ Commented May 25, 2019 at 9:27

In general case, these equations don't apply and you need fourier sum for periodic signals and fourier integral for non periodic signals. However, in time harmonic case(sinusoidal time harmonic), the derivatives always operate on $e^{j \omega t}$, so differentiating once means multiplying by $j \omega$, second derivative will become ${-\omega}^2$ etc. Apply these to the Maxwell equations and the result follows immediatly.

Edit for clarification

Instead of considering sine or cosine functions, we consider exponential function, $f(t) = f e^{j \omega t}$. Where $f(t)$ is time harmonic function and $f$ has both real and imaginary part and $f$ is not time dependent: $f = f_r +j f_i$ . From this complex notation, we can get back to the sinusoidal form at any time by taking the real part of $f(t)$; $F(t) = \Re(f e^{j \omega t})$ and using the fact $e^{j \omega t} = cos(\omega t) + j sin(\omega t)$.

We also see that $f'(t) = j \omega f(t)$, $f''(t) = (j \omega)^2f(t) = -\omega^2f(t)$ and so on for higher derivatives.Which means that you replace time derivatives on your equations with multiplication. Note that the functions now depend on only 3 location coordinates $\mathbf{\vec{r}}$, as the time dependency $e^{j \omega t }$ has been dropped out of every term of the Maxwell's equations. These do not work on general case, but the time harmonic case is by far the most important in practice.

  • $\begingroup$ Can you please elaborate? It isn't clear to me what you're saying. $\endgroup$ Commented May 24, 2019 at 7:53
  • $\begingroup$ @ThePointer I edited it a little $\endgroup$
    – J sx
    Commented May 24, 2019 at 8:46
  • $\begingroup$ I've already done gone through the derivation you're describing: math.stackexchange.com/questions/3234770/… But how does any of this relate to the derivation using the Fourier transform? $\endgroup$ Commented May 24, 2019 at 9:08
  • $\begingroup$ @ThePointer If you compute the fourier coefficients of a time harmonic signal, you'll find that the nonzero coefficients correspond to $\Re(f e^{j \omega t})$ $\endgroup$
    – J sx
    Commented May 24, 2019 at 9:47
  • $\begingroup$ Ok, but I don't see how this is relevant to my question. $\endgroup$ Commented May 24, 2019 at 9:49

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