# Are non-zero-area pulses of electromagnetic radiation possible?

This is a bit of a long-standing question / bone of contention that I've seen floating around, and which I would like to ask here in the hopes of getting some outside perspectives from a broader community.

My neck of the woods centers around pulsed lasers: these can be long or short pulses, at a variety of wavelengths, but in essence we always have some light source which produces radiation that is focused down into some interaction region so that it can do things to atoms or molecules or whatnot. And, in this community, there is a bit of folklore that essentially says something like

You cannot have nonzero-area pulses

where a nonzero-area pulse is a pulse of radiation with electric field $\mathbf E(\mathbf r,t)$ with the property that $$\int_{t_\mathrm{i}}^{t_\mathrm{f}} \mathbf E(\mathbf r,t)\mathrm dt \neq 0,$$ where $t_\mathrm{i}$ and $t_\mathrm{f}$ are pulses well before, and well after, the pulse has started and finished, respectively, and the integral is taken at some fixed position in space in your interaction region, i.e. in the radiation regime kind of far away from the charges and currents that make up the light source that produces that electric field.

This principle has a pretty wide applicability, because it impacts what kinds of vector potentials are of interest: since we normally define the vector potential as $\mathbf A(\mathbf r,t) = -\int_{t_\mathrm{i}}^t \mathbf E(\mathbf r,\tau)\mathrm d\tau$, if all pulses must have zero area then you should only work with vector potentials that go to zero at $t\to\infty$. This is nice because it simplifies things, but it also restricts the kinds of pulses you might hope to shine on your target, if you want to do something like attosecond streaking or whatnot.

Unfortunately, however, this bit of folklore is very rarely backed up with any kind of solid justification (even in e.g. course notes from researchers that I hold in very high regard), and as I will argue below it is on very shaky ground. I would like some help establishing whether it is true or not.

On the justifications front, a friend recently found a resource that offers some foundations for this principle, in the book chapter

from which the relevant passage runs as follows:

### 2.4 The "zero area" pulse

An interesting feature of a travelling electromagnetic wave is that it must have equal amounts of positive and negative field, meaning the shortest optical pulse that could be generated would be one full cycle. It is not possible to generate and transmit a unipolar electromagnetic pulse that consists of only 1/2 cycle. A plot of both cases is shown in fig. 3. This can be easily understood from the Fourier analysis of the propagating wave. The Fourier spectrum of a propagating unipolar pulse (right-hand side of fig. 3) will have a DC term. To launch a unipolar pulse, it would be necessary to transmit a transverse DC potential with the rest of the waves. But it is impossible to establish a transverse DC potential in space, thus it is impossible to create such a pulse.

Because no propagating wave can have a DC component, pulses must have equal amounts of positive and negative field. Propagating fields will always have an oscillatory structure with at least one cycle. Another way to look at this is to recognize that all physically realizable pulses must have an average area of zero - i.e. there is just as much negative as positive field in a total optical pulse.

This is, then, a formal appearance of the principle in the literature - and in a resource that otherwise looks pretty solid, too -, but I find it deeply problematic. My problems are most easily condensed around the phrase

it would be necessary to transmit a transverse DC potential with the rest of the waves

because there is no such thing as a 'transverse DC potential', since DC fields don't have a propagation and the notion of 'transverse' is therefore meaningless. More generally, I'm really troubled by the mixture of a Fourier decomposition (which looks at the whole experiment globally, i.e. you need to treat the entire time interval as a unified whole for the analysis to make sense) with talk of 'launching' those Fourier components.

More practically, though, I find this very weird because it is possible to exhibit radiative solutions of the Maxwell equations that include zero-area pulses, at least in the plane-wave regime.

To show a specific example, consider the fields \begin{align} \mathbf E(\mathbf r,t) & = E_0 \hat{\mathbf y} f(|x|-ct) \\ \mathbf B(\mathbf r,t) & = \operatorname{sgn}(x)B_0 \hat{\mathbf z} f(|x|-ct), \end{align} where $B_0 = E_0/c$, driven by a sheet of current in the $y,z$ plane with uniform surface current density $$\mathbf K(\mathbf r,t) = -\frac{2B_0}{\mu_0}\hat{\mathbf y} f(-ct). \qquad\$$ These are full solutions of the Maxwell equations, including the boundary terms, irrespective of what form $f$ has, so long as it is differentiable. To disprove the zero-area principle, then, one can simply take the waveform to be any nonzero-area pulse, which can be e.g. $$f(x) = \operatorname{sech}^2(x/\sigma), \qquad\quad$$ among a myriad of other examples.

As you can see, then, there is some significant tension here. Are CR Pollock's arguments simply not applicable to the example above? Or is Pollock's argument just completely flawed and the zero-area principle is not true at all?

On the other hand, though, the passage above does suggest that my plane-wave counter-example is not quite strong enough, because it requires infinite energy and an extended source, so really one should find examples of pulses with finite energy and localized fields, and find explicit sources that will produce them (to really defuse the "you can't 'launch' the DC component of those fields" argument).

And, on the opposite end of the spectrum, the focus on the pulse area as a DC component also raises some interesting flags, because normally we like to work in the paraxial approximation to describe the focusing of a laser beam, but those DC components are very much not describable by those dynamics. Indeed, in the paraxial approximation, the Gouy phase shift of $\pi$ over the focus suggests that if you manage to make a cosine-like pulse before the focus, then it will become a minus-cosine-like pulse after the focus, and that is some pretty odd behaviour when seen through the lens of a DC-like field.

A bit more on the practical and experimental side, we're getting close to the point where we can reliably and consistently produce and measure pulses that look like part (c) of this figure (from this paper), where you really start to concentrate the energy in the central half-cycle of the pulse, tilting the pulse area towards a positive balance. Now, you can argue that the area of the central pulse must always be balanced to zero by all the little negative bits in the leading and trailing edges of the pulse, but frankly I find that perspective on the argument to look very weak.

Which brings me, then, to the formal statement of my question. I've ranted a bit to show the tensions between multiple different statements, facts, and perspectives, and the resolution doesn't look that easy to me. The way I see it, what's required is either

• a counterexample showing pulses with nonzero pulse area, in the radiatively-far regime, with finite energy in the fields, and with the sources shown explicitly; or
• an actual solid proof of the zero-area principle, with none of the fuzzy language in Pollock's passage, which explicitly delimits its validity and its hypotheses, and where the proof and its validity limits clearly show why it is not applicable to the plane-wave example I posted above.

On the other hand, this isn't quite rocket science, either, so hopefully there are already better explorations of this topic in the literature. Has anyone seen such a text? Or if not, can someone show conclusively, as above that the principle is true or false?

• In theory there is no reason why pulses of non-zero area under plot of electric field could not be produced. The argument by Scrinzi makes no sense, travelling pulse in the shape of half wavelength is a possible solution to Maxwell's equations in free space (and thus also solution of the wave equation). One could use a fast shutter to allow only half wavelength of a linearly polarized EM radiation through. In practice this is probably the more difficult the higher the frequency of the primary radiation and when it comes to optical and higher frequencies, may be practically impossible. – Ján Lalinský Apr 14 '17 at 18:58
• A pulse of this kind could also be generated (in principle for anz frequency) by positively charged particle that is subjected to unidirectional constant force for finite period of time. The particle accelerates for this period of time and moves rectilinearly afterwards. The field it produces is a travelling pulse whose electric field points in the direction of acceleration. – Ján Lalinský Apr 14 '17 at 19:02
• @JánLalinský That's my feeling, but if it is indeed possible then one should be able to write down an explicit example, which is what I'm really asking for (if the principle is invalid). Shutters are not practical but conceptually they are just a current density so one could work with that. (On a practical footing, of course, mechanical shutters won't work in the optical regime, but we can get pretty close with optical pulse synthesis as described in the Hassan et al. paper linked in the question.) – Emilio Pisanty Apr 14 '17 at 19:08
• I agree with @JánLalinský in that a "baseband" pulse can be transmitted. As it is a single pulse it has no dc component to be radiated. The Fourier Transform of a baseband pulse of finite support has no discrete $\delta(f)$ content, only continuous spectrum. It is another question what antenna could support such transmission but I think a coupled induction coils where the primary is the transmitting and the secondary is the receiving will do. I am skeptical, though, that such induction arrangement is practical at optical frequencies. – hyportnex Apr 14 '17 at 23:50

This is indeed a somewhat tricky question. It is related to the issue of laser vacuum acceleration of charges, which is generally believed to be impossible - a claim that goes under the name of Lawson-Woodward theorem. There has been a bit of a discussion a while ago in PRL, see the paper

Coherent Electron Acceleration by Subcycle Laser Pulses. B. Rau, T. Tajima, and H. Hajo, Phys. Rev. Lett. 78, 3310 (1997)

who claim to have found unipolar sub-cycle pulses that can be used for electron acceleration. This has been criticised in an interesting comment on that paper at

Comment on “Coherent Acceleration by Subcycle Laser Pulses”. Kwang-Je Kim, Kirk T. McDonald, Gennady V. Stupakov, and Max S. Zolotorev, Phys. Rev. Lett. 84, 3210 (2000),

which I consider to be both relevant and correct. It is certainly true that unipolar, pulsed plane wave solutions to the 3d wave equation exist mathematically (see e.g. OP), but they do not have finite energy and/or cannot be generated by a source of finite extent as Kim et al. point out (with a reference to Feynman). Put differently, this implies that the question indeed cannot be answered in a physically sensible way without considering the generation of the pulse, i.e. its source together with the realisation of the latter.

• I'm not sure I buy the argument that all reasonable radiation-regime fields must fall off as $1/r$. As a possible counterexample, you can make a reasonable approximation to a gaussian beam whose source is a current sheet confined to a disk. At distances much greater than the Rayleigh range after the focus, once diffraction takes over, then yes, the field falls of as $1/r$, but within the focus itself it's not really true, and you do want to include this region as a relevant one. – Emilio Pisanty Apr 27 '17 at 9:54
• Nevertheless, the Kim et al. paper is the clearest exposition of the grounding for the zero-area principle, so I'm awarding the bounty to this answer. It'll take me a bit to fully digest it, though. – Emilio Pisanty Apr 27 '17 at 9:55
• Thanks - and no worries. Your question is still part of (my) ongoing research. Regarding Gaussian beams: they're quite useful, but only approximate solutions of the wave equation. For instance, there is always a longitudinal electric field component. McDonald has something to say about this here. – Tom Heinzl Apr 27 '17 at 10:06
• @TomHeinzl - Quick question: Is this related to laser wakefield acceleration? If so, I do not think it is considered impossible. Unless I am much mistaken, they routinely do things like this at SLAC (e.g., portal.slac.stanford.edu/sites/ard_public/facet/pages/rpwa.aspx) among other places (e.g., nature.com/articles/srep43910). Did I misunderstand something here? – honeste_vivere Mar 11 '18 at 18:58
• @honeste_vivere Laser wakefield acceleration is caused by the ionized wake of the laser pulse (more specifically, riding on the charge imbalance caused by the fact that electrons are lighter and they move away from the focus faster) and it is longitudinal and not transverse (i.e. along $\vec k$ instead of $\vec E$). It is not caused (directly) by the laser's electric field exerting a force on the charges to be accelerated. – Emilio Pisanty Apr 12 '18 at 17:29

This is indeed a highly topical issue. Consider research papers such as

which tend to support the notion that unipolar radiation is indeed achievable.

• I appreciate the references and I'll follow them up, but if the zero-area principle is false then I do want a concrete counterexample in analytical form. – Emilio Pisanty Apr 16 '17 at 10:26

an actual solid proof of the zero-area principle, with none of the fuzzy language in Pollock's passage, which explicitly delimits its validity and its hypotheses, and where the proof and its validity limits clearly show why it is not applicable to the plane-wave example I posted above.

I'll comment on this; I'm reluctant to call the following a solid proof but it can at least be an input for discussion.
I'll use rationalized Gauss units so I'll be missing some $c,\mu_0,\epsilon_0$ factors.

In the Lorenz gauge the 4-potential generated by a source $j$ is given by $$A^{\mu}(\textbf{x},t)= \int d^3\,y \frac{1}{4\pi} \frac{1}{|\textbf{x-y}|} j^{\mu}(t-|\textbf{x-y}|,\textbf{y})$$

Under the assumption that, for all times, the source is spatially bounded ( $j^{\mu}(t,y)=0 \quad\text{if} \quad |y|>l, \forall t$ ) one can write the following expression for $A$ in the radiative limit $r=|x|\rightarrow \infty$ (that is the expansion of $A$ at the order $\mathcal{O}(1/r)$.) \begin{equation} A^{\mu}(x,t)=\frac{1}{4\pi r} \int j^{\mu}(t-r +\frac{\textbf{x$\cdot$y}}{r}, \textbf{y} ) \,d^3\,y \qquad(*) \end{equation} which is obtained simply expanding $|\textbf{x-y}|$ and neglecting terms of order $1/r^2$. One can then show that this $A$ in the radiative limit satisfies the wave relations of the free potential (up to higher order corrections). I'll need one of these relations: definig the 4-vector $n=(1, \frac{\textbf{x}}{r})$ it holds that $\partial_{\mu} A^{\nu} = n_{\mu} \dot A^{\nu}$.

Then the $i$-component of the area of the pulse $$\int E^i (\textbf{x},t) dt = \int -\partial^0 A^i +\partial^i A^0 \,dt= [-A^i(\textbf{x},t) +n^i A^0(\textbf{x},t)]_{t_i}^{t_f} \qquad (§)$$

But we already knew something like this, you wrote it yourself that the area is proportional to the potential $A$ ( this formula here can be made more similar to yours using that $\textbf{E}= \textbf{$n \times n \times\dot A$}$ in lorenz gauge and radiative limit). But we see also two conditions that make our area vanish. We need:

1. a spatially-bounded source (to get the (*) expansion and wave relations)

2. we ask that our source is turned off in the far past and far future ($j^{\mu}(\textbf{y},t)=0 \quad \text{for} \,\, t\sim \pm \infty$) , and that $t_i, t_f$ are sufficiently larger (in modulus) than $r$

Then by using (*) to evaluate the potentials in the expression for the area, they turn out zero because the currents that source them are zero when calculated at those times $t_{i,f}-r +\frac{\textbf{x$\cdot$y}}{r}$.

This is compatible with your example; your source stays "on" for all times, thus assumption #2 is not satisfied.

One could argue that this was just the $\mathcal{O}(1/r)$ order. This is true, I was focusing only on the radiation fields.
One could also argue that it's assumtpion #2 that does all the work; it does, but it sounds reasonable. It means, in my interpretation, that we have to integrate for a time large enough to see the entire pulse.
Moreover, by removing #2, one could get all kinds of behaviour. You already gave an example of a source that gives positive area; one could instead choose a periodic source that gives zero area.
This last paragraph explains why I was reluctant to call this a solid proof; you can say that it is more of an extended comment on what kind of sources give zero area pulses.

Edit: following Emilio's comment, we'd like to remove the requirement that $j^0$ must vanish. Luckily there's a wave relation that helps us.
In the radiative limit and in the Lorenz gauge $\textbf{E}= \textbf{$n \times n \times\dot A$}$. We remember from (*) that $\textbf{A}$ depends only on the spatial components of the source $\textbf{j}$. Putting the two facts together in the expression for the area $\int \textbf{E} \,dt= \textbf{n$\times$n$\times$}[ \textbf{A}]^{t_f}_{t_i}$ we get that only $\textbf{j}$ is required to vanish in order to get zero area.

Proof of the wave relations
We remember the formula for $A$ in the radiative limit \begin{equation} A^{\mu}(x,t)=\frac{1}{4\pi r} \int j^{\mu}(t-r +\frac{\textbf{x$\cdot$y}}{r}, \textbf{y} ) \,d^3\,y \qquad(*) \end{equation} and the light-like vector $n=(1,\textbf{x}/r)$. Let's proceed to prove $\partial_{\mu} A^{\nu} = n_{\mu} \dot A^{\nu}$. We'll only consider the contributions to order $1/r$ (for example we won't need to derive the $\frac{1}{4\pi r}$ in front of the integral). We have to note that $$\frac{\partial}{\partial x^{\mu}}(t-r+\textbf{n\cdoty}) = \begin{cases} 1 & \text{if \mu=0} \\ \frac{-x^i}{r} + \mathcal{O}(1/r) = n_i + \mathcal{O}(1/r) & \text{if \mu=i} \end{cases}$$ so that derivative equals $n_{\mu}$ plus higher order terms. Then

\begin{align} \partial_{\mu} A^{\nu} &= \frac{1}{4\pi r} \int \frac{\partial}{\partial x^{\mu}}j^{\nu}(t-r +\textbf{n$\cdot$y}, \textbf{y} ) \,d^3\,y \\ &= \frac{1}{4\pi r} \int \partial_{0}j^{\nu}(...) \frac{\partial}{\partial x^{\mu}}(t-r+\textbf{n$\cdot$y}) \,d^3\,y \\ &=\frac{1}{4\pi r} \int \partial_{0}j^{\nu}(...) (n_{\mu} + \mathcal{O}(1/r) ) \,d^3 \,y \\ &= n_{\mu} \dot A^{\nu} \end{align}

So we have $\partial_{\mu} A^{\nu} = n_{\mu} \dot A^{\nu}$. In the Lorenz gauge: $$\partial_{\mu}A^{\mu}=0 \rightarrow n_{\mu}\dot A^{\mu}=0$$ so $$\dot A^0= \textbf{n \cdot \dot A}$$ Finally: \begin{align} E^{i}&=\partial^{i}A^{0}-\partial^{0}A^{i} \\ &=n^i \dot A^0 - \dot A^i \\ &=n^i (\textbf{n $\cdot \dot A$}) - \dot A^i \\ \rightarrow \textbf{E}&= \textbf{n (n$\cdot \dot A$) - $\dot A$ } \\ &=\textbf{n $\times$n $\times \dot A$} \\ \end{align} to get the last line we remember that $\textbf{n$\cdot$n}=1$.
My reference for these relations is: K. Lechner - "Elettrodinamica Classica".

• I'm not entirely sure that your assumption #2 isn't too strong when applied to $j^\mu$ in full. For the spatial components $j^i$ it's perfectly reasonable, but I suspect it's too strong when applied to the charge density $j^0$; breaking this would describe a charge that moved from A to B in a one-off go, and that sounds plenty reasonable. Wouldn't this make your $n^i A^0|_{t_i}^{t_f}$ term nonzero? – Emilio Pisanty Apr 20 '17 at 18:13
• @EmilioPisanty now that you make me think about it, it seems too strong indeed. In my mind I was imagining $j$ as a temporary dipole that is formed from a neutral distribution and then re-absorbed, so I didn't question the condition on $j^0$. But we're lucky, because the requirement on $j^0$ can be removed. I 've added an edit to the post doing so. – tbt Apr 21 '17 at 10:46
• I'm not sure I buy that radiative-limit formula. Can you point to a reference where it's derived? – Emilio Pisanty Apr 21 '17 at 10:52
• Sadly my reference is in Italian; later today i'll look for a reference in English, if I don't find it I will edit the post adding a derivation. – tbt Apr 21 '17 at 10:56
• Added the derivation to the original post – tbt Apr 21 '17 at 21:10