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I am trying to determine the envelope function of a superposition of two waves. I will give a concrete example:

Let
$$ f(x, t_1) = \cos(2x + t_1) + \cos(2x + 2t_1) $$

Using a trigonometric identity, this can be rewritten as:
$$ f(x, t_1) = 2 \cos\left( \frac{t_1}{2} \right) \cdot \cos\left( 2x + \frac{3t_1}{2} \right) $$

By graphing the wave, I found that the actual bounding function appears to be:
$$ \pm 2 \cos\left( 2x + \frac{3t_1}{2} \right) $$

I am not sure of generally how to find the envelope. Furthermore, i need a clear definition of what the envelope is. I have found terminology such as spatial and temporal envelope, however it is not clear what exactly these terms means.

Envelope definition

I am working with the following definition of the envelope (written by me...):

Let
$ u(x, t) = f(x, t) \cdot \text{carrier}(x, t) $

Then $ f(x, t) $ is called an envelope function if it satisfies the following two conditions:

  • Bound condition:
    $$ |u(x, t)| \leq |f(x, t)| \quad \text{for all } x, t $$

  • Attainability condition:
    $$ \text{For all } x, \text{ there exists some } t \text{ such that } |u(x, t)| = |f(x, t)| $$

Additionally, I define:
$$ E_{\text{upper}}(x, t) := \max_t |f(x, t)|, \quad E_{\text{lower}}(x, t) := \min_t |f(x, t)| $$

In my example, I believe the correct envelope is:
$$ f(x, t_1) = 2 \cos\left( 2x + \frac{3t_1}{2} \right) $$

Phase velocity

This envelope has phase:
$$ \phi(x, t) = 2x + \frac{3t}{2} \Rightarrow v_{\text{phase}} = -\frac{3}{4} $$

Interestingly, this is also the phase velocity of the full summed wave.

So I’m confused: does this mean that in this case the phase velocity equals the group velocity?
Is this a coincidence or a misunderstanding on my part?

Would appreciate clarification — especially if there's a flaw in how I defined the envelope or interpreted its velocity.

Thanks in advance!

Mad
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1 Answers1

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To meaningfully talk about an envelope (wave packet) in physical sense, the frequencies should be clustered around a central frequency in Fourier space. In other words, the width of the spectrum, $\Delta\omega$ should be much smaller than the central ("carrying") frequency: $\omega_0\gg\Delta\omega$.

In case of only two waves, this means that the difference of frequencies is much smaller than their average (or any of the frequencies): $$ |\omega_1-\omega_2|\ll\omega_1,\omega_2,\frac{\omega_1+\omega_2}{2}. $$ This is clearly not the case in the example given in the Q ($\omega_1=1, \omega_2=2$). Mathematically, one could still apply the same definition that one uses for adding any two waves (see the first two references below for more detailed math), but physically it just doesn't make much sense.

Remark: It is also worth noting that even on the formal level, the discussion in the Q misidentifies the smaller ("envelope") frequency and the higher ("carrying") frequencies - a mistake that wouldn't probably occur, if the mathematical inequality above was satisfied.

Stationary beats
Final frequency with two waves combined
Interpreting group velocity of free particle wave packet
Understanding group velocity

Roger V.
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