I understand that the slowly-varying envelope approximation means that we can write an electromagnetic wave as

$$ E(x,t)=V(x,t)e^{i(k_0x-\omega_0 t)},$$

where $$ \left \vert \frac{dV}{dx} \right \vert \ll |k_0 V(x,t)| \quad \text{and} \quad \left \lvert \frac{dV}{dt} \right \rvert \ll |\omega_0 V(x,t)| \, .$$

I have two questions about this approximation:

  1. Why this implies that the band of the wave is narrow?
  2. What is the maximum value of $Δx$, compared to $\lambda$, for which $V(x+Δx)≈V(x)$ still holds?
  • $\begingroup$ Hint for the question about the narrow band: Use the theorem for the Fourier transform of products and the information about $V$ to constrain it's Fourier transform. I don't understand the part with respect to "how many $\lambda$" since wavelengths are a continuous a number of $\lambda$s does not really make sense. $\endgroup$ Commented Dec 30, 2018 at 18:56
  • $\begingroup$ @SebastianRiese thank you for your comment! I already thought about Fourier transforming the above relationship, but what I obtain is $\omega \tilde V \ll \omega_0 \tilde V$, but this is just telling me that the frequency of variation of $V$ is smaller than $\omega_0$, but I already knew that. My second question can be reformulated as follows: what is the maximum value of $\Delta x$ for which $V(x+\Delta x)\approx V(x)$ still holds? In my opinion is useful to use $\lambda$ as a scale for $\Delta x$. $\endgroup$ Commented Dec 30, 2018 at 19:03
  • $\begingroup$ The second one can be answered by Taylor expansion. $\endgroup$ Commented Dec 30, 2018 at 19:10
  • $\begingroup$ @SebastianRiese to answer the second question using a Taylor expansion I should know a priori what is the exact analytical form of $V(x,t)$, but in my understanding this approximation is made before, because without this approximation the function $V(x,t)$ cannot even be defined. Reformulating again: quantitavely speaking when can I use the Slowly varying envelope approximation? When it is justified? $\endgroup$ Commented Dec 30, 2018 at 19:14
  • $\begingroup$ I think you may be doing things a little upside-down. The decomposition is always exactly possible and not an approximation in it self (factoring out the plane wave of a certain frequency – usually the mean of the Fourier transform of $E$), and if $V$ fulfils the given smallness conditions, then you have a slowly varying envelope, and various things get easy. (As usual in physics we assume $E$ is sufficiently non-singular.) $\endgroup$ Commented Dec 30, 2018 at 19:30

2 Answers 2


1. Narrow-band-ness

Consider, as JEB's answer does, the pure-space dependence $$ E(x,0)=V(x,0)e^{ik_0x},$$ and let's look at the Fourier transform of this (dropping the $t=0$'s for simplicity): \begin{align} \tilde E(k) & = \int E(x)e^{-ikx}\mathrm dx = \int V(x)e^{-i(k-k_0)x}\mathrm dx = \tilde{V}(k-k_0), \end{align} i.e. the frequency-space relationship is, as noted by JEB, the convolution of $\tilde V(k)$ and the Dirac delta that results from Fourier transforming $e^{ik_0x}$. Let's look at $\tilde V(k)$ more closely, though, by first considering the Fourier transform of its derivative $\frac{\mathrm dV}{\mathrm dx}$: \begin{align} \int \frac{\mathrm dV}{\mathrm dx}(x) e^{-ikx}\mathrm dx & = ik \int V(x) e^{-ikx}\mathrm dx =ik\tilde V(k), \end{align} assuming that the boundary terms in $V(x)|_{-\infty}^\infty$ vanish. This means, therefore, that \begin{align} |k\tilde V(k)| & = \left| \int \frac{\mathrm dV}{\mathrm dx}(x) e^{-ikx}\mathrm dx \right| \\ & \ll \left| \int k_0 V(x) e^{-ikx}\mathrm dx \right| = k_0|\tilde V(k)| . \end{align} Taken at face value, that's one heck of a strange statement, but the way to read it is this: within the band in which $|\tilde V(k)|$ is appreciable, you must have $|k|\ll k_0$.

When applied to the full electric field, this then tells you that within the band in which $|\tilde E(k_0 + \Delta k)|$ is appreciable, you must have $|\Delta k| \ll k_0$, and this is the precise meaning that we assign to $E(x)$ being a narrowband wavepacket.

2. $V(x+Δx)≈V(x)$

Your second question,

[given that $|\frac{dV}{dx}|\ll |k_0 V(x,t)|$,] what is the maximum value of $Δx$, compared to $\lambda$, for which $V(x+Δx)≈V(x)$ still holds?

is rather harder, because neither of the statements involved, $|\frac{dV}{dx}|\ll |k_0 V(x,t)|$ and $V(x+Δx)≈V(x)$, is a hard quantitative statement.

Nevertheless, the way to tackle this is via a Taylor expansion, \begin{align} | V(x+\Delta x) - V(x) | & = |\Delta x \frac{\mathrm dV}{\mathrm dx}(x)| + \mathcal O(\Delta x^2) \\ & \ll |\Delta x \: k_0 V(x)| + \mathcal O(\Delta x^2) \\ & = 2\pi \left|\frac{\Delta x}{\lambda} V(x)\right| + \mathcal O(\Delta x^2) . \end{align} If you have a quantitative statement for what that "$\ll$" means, then it will directly give you a statement for the fractional change in terms of $\Delta x/\lambda$ displacements that you will observe in $V(x)$.

  • $\begingroup$ Indeed my second question is the real core of my doubts. I know that we do not know enough about $V(x,t)$ to be really quantitative about its behaviour. In truth I was hoping in the existence of a some kind of criterion to decide when this approximation can be applied and when not. In particular I was trying to solve another doubt I had, closely related. In case you have time and will I would like to ask you kindly if you would like to have a look to this other question: physics.stackexchange.com/questions/451004/… $\endgroup$ Commented Jan 1, 2019 at 17:36

Let's look at fixed time:

$$ E(x) = V(x)e^{-ik_0x} \equiv V(x)\phi_{k_0}(x) $$

I wrote $\phi(x)$ to highlight the fact that it is a fast phase modulation of a slowly varying function.

What does slowly varying mean? It means the Fourier transform:

$$ \tilde V(k) = \int{e^{ikx}V(x)dx} $$

has its power below some $k_V$ that satisfies:

$$ ||k_V|| \ll k_0$$

It also has some bandwidth $\Delta k \approx 2 k_V$ (there a positive and negative components for a real $V(x)$).

All this condition means is that the wavelength of $V(x)$ is much longer than the carrier wavelength, $1/k_0$. How much depends on your system. If it's microwaves in a circuit, then you have to look at the hardware. If you are solving a boundary value problem, you have to look at the conditions of the problem. A factor of "100" is certainly good, "10" is OK, and "2" is pushing it.

As pointed out in the comments, the FT of a product is a convolution of the multiplicands' FTs, and:

$$ \tilde \phi(k) = \delta(k_0) $$

Now the convolution with a delta functions is a shift operation, hence:

$$ \tilde E(k) $$

looks like

$$ \tilde V(k) $$

shifted from 0 to that carrier frequency $k_0$.

Thus, the signal has some bandwidth $\Delta k$ centered around $k_0$.

Now you can revisit the "slow" question by saying $\Delta k/2 \ll k_0$--meaning the all the power is in the positive frequencies, which is the same as the earlier condition but view slightly differenlty. They are equivalent to the conditions on the derivative stated in the OP.

Adding the time-coordinate back in doesn't change anything, unless you have a difficult dispersion relation.

  • $\begingroup$ You definitely nailed the first question, but I still have a few doubts about the second one. I am particularly interested in this question because is closely related to another one I still have not solved. If you are interested I would like to ask you kindly if you would like to have a look to this other one: physics.stackexchange.com/questions/451004/… $\endgroup$ Commented Jan 1, 2019 at 17:38

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