Consider some basic cosine waves of the form ${E_i} = {E_0}\cos ({\omega _i}t - {k_i}z)$ with different amplitudes, frequencies and phases. We know a combination of such waves could result in a wave which has an envelope that is traveling at a group velocity and also has a phase velocity.

When there are some basic waves with equal amplitudes propagating in the same direction of propagation, deriving the expression for the resultant wave is easily possible using some simple trigonometric identities. say when there are two waves of the form: ${E_1} = {E_0}\cos ({\omega _1}t - {k_1}z)$ and ${E_2} = {E_0}\cos ({\omega _2}t - {k_2}z)$ we'll have:

$$\begin{align} {E_1} + {E_2} &= {E_0}\bigl(\cos ({\omega _1}t - {k_1}z) + \cos ({\omega _2}t - {k_2}z)\bigr) \\ &= 2{E_0}\cos\biggl(\frac{\omega_1 - \omega_2}{2}t - \frac{k_1 - k_2}{2}z\biggr)\cos\biggl(\frac{\omega_1 + \omega_2}{2}t - \frac{k_1 + k_2}{2}z\biggr) \\ &= \biggl\{ 2{E_0}\cos\biggl(\frac{\omega_1 - \omega_2}{2}t - \frac{k_1 - k_2}{2}z\biggr)\biggr\} \cos(\frac{\omega_1 + \omega_2}{2}t - \frac{k_1 + k_2}{2}z\biggr) \end{align}$$

First part refers to envelope of the resultant wave and group velocity will be derived from this part and the other part deals with the phase velocity.

Deriving such a expression gives a better physical viewpoint of what we are dealing with than when we represent it in the form of a summation of some basic cosine waves. Plus such a relation for the wave function could make it so easy to find out many charactristics of the wave in a brief look at the expression.

Therefore it's so important and helpful to derive an expression for the wave which is caused by superposition of some basic propagating waves. But it was a simple situation.My question is that what if basic waves amplitudes were not the same?

When we are aware of propagation of some basic waves with unequal amplitudes and frequencies, how can we derive the relation for the resultant wave (which actually represent the only changing field in the medium)? And of course I have the same question about the expression for the group velocity.

Like what we did in relation to equal amplitude waves example, is it possible to derive the expressions only by use of simple trigonometric identities?

  • 1
    $\begingroup$ I'd definitely suggest you read the homework policy about that. But the gist of it is that you need to show your work and ask about the specific physics concept that you're getting stuck on. $\endgroup$
    – David Z
    Commented Oct 3, 2013 at 16:20
  • $\begingroup$ @david-z:I've already read that page. My question is not a homework or a homework like question. It's an attempt to find a simple solution to derive and for representing the expression of a wave caused by superposition of some basic waves with unequal amplitudes. It's a try to make the wave and the wave propagation phenomenon tangible as easy as possible. It's dealing with a more general case of how adding some different waves could result in a unitary wave which also has a group and a phase velocity.T am surprised why you think it's of those homework like questions!Please reopen it.Thanks $\endgroup$
    – 2physics
    Commented Oct 3, 2013 at 18:03
  • $\begingroup$ The way you asked it before, it was a homework-like question. But now that you've removed the specific problems and are just asking about a physical concept, it's fine. (It might be mathematical in nature but that's not a big deal; the community will just migrate it to Mathematics if that is the case.) $\endgroup$
    – David Z
    Commented Oct 3, 2013 at 18:14
  • $\begingroup$ What is the point of writing the summation of sine waves as a product? Say you have a summation of 5 waves with different frequencies and amplitudes, what is the benefit of writing them as a product?? $\endgroup$ Commented Oct 3, 2013 at 22:32
  • $\begingroup$ @Gotaquestion: One of the reasons is to derive the relation for the group velocity of the packet in the same procedure as we followed when the waves have same amplitudes and to find a relation for the envelope... $\endgroup$
    – 2physics
    Commented Oct 3, 2013 at 22:52

3 Answers 3


I'm pretty sure that we can't do any better with trig identities; it looks like any such expression would have a term vaguely like $\sqrt{1 + r \cos \theta}$, and then the square root ruins the nice intuition.

In lieu of that I'll offer a derivation of the group velocity that uses no fancy math at all! Hopefully that's what you wanted, even if you didn't directly ask it.

First off, let's figure out what group velocity is. Apparently, group velocity is the velocity of the envelope of a wave. But what is the envelope of a wave? I can create a wave with any initial position and velocity, so it can be an arbitrarily weird shape. There might be no discernible envelope at all.

So let's back up and find some examples. When we talk about the envelope of a wave, we mean some curve you draw around an oscillation, like this.

enter image description here

In order to do this, there must be a well defined oscillation to draw the envelope around. That means that our wave must be made up of individual frequencies that are close to one central frequency. To make things convenient, let's write that schematically as $$\text{wave} = \sum_{k'} \sin(k'x - \omega(k') t) \text{ for a bunch of } k' \approx k$$ However, if we just have one frequency, the wave is just an infinite sinusoid $\sin(kx)$. This doesn't have an envelope, strictly speaking, because it just goes on forever at the same amplitude. We must have waves of other frequencies, which will constructively and destructively interfere with each other, to actually get an envelope.

So we've concluded that the envelope is defined by where a bunch of sinusoids making up our wave constructively or destructively interfere. Their phases are, as a function of space and time, $$\phi(k') = k'x - \omega(k') t$$ Now let's assume for simplicity that the top of the envelope is at $x = 0$ at time $t = 0$. That means that the waves must constructively interfere there, so all the $\phi(k')$ are about the same.

As time goes on, the envelope will move, but the peak will still be where the phases of the component waves are the same. That means $$\text{peak of envelope satisfies } \frac{d\phi(k')}{dk'} = 0$$ Performing the differentiation, we have $$x - \frac{d\omega(k')}{dk'} t = 0$$ Since we said $k' \approx k$, let's drop the primes and rearrange for $$\frac{x}{t} = \frac{d \omega(k)}{dk}$$ But $x/t$ is exactly the speed of the peak of the envelope, so this is the group velocity.

  • $\begingroup$ Thank you and yes, this is the general way of explaining the group and phase velocities relations.:)I also have been disappointed to find a trig solution. I've noticed that in some books(such as Field and wave Electromagnetics by David Cheng)the author has tried to explain the group velocity relation using the special case of summation of cosine waves of equal magnitudes, which is not a general way of explaining the relation.That's why I thought may be we have to consider a general case to explain the issue in the same way.Thanks for your time & clear & easy to understand explanation $\endgroup$
    – 2physics
    Commented Feb 12, 2016 at 8:47

I think there is a misconception here. The point of writing the sum of two signals as product is NOT to compute the phase velocity or the group velocity. It is to show the difference between them only. It doesn't add any information to write the signals as a sum or a product.

Having two signals with equal amplitudes is a particular nice example because:

1.The mathematics is so easy.
2.The graph of this function has a clearly defined envelope and carrier.
3.Since the graph has a clear envelope, it means it is so easy to distinguish what is the difference between the phase velocity and group velocity. Yet you can’t tell what the value of any of them is.

The general definitions of phase and group velocities are: $$ v_{phase}=\frac{\omega}{k} $$ $$ v_{group}=\frac{\partial \omega}{\partial k}\bigg|_{\omega=\omega_0}=\left(\frac{\partial k}{\partial \omega}\right)^{-1} $$

One has always to keep in mind that phase velocity and group velocity are CHARACTERISTICS OF THE MEDIUM OF PROPAGATION. They don’t depend on the propagating signal; they depend on the medium of propagation. For every medium there is what is called a “dispersion relation” which is basically a relation relating the frequency to wave vector, or in another words the wave vector as a function of the frequency. For free space the dispersion relation is: $$ \omega=\omega_{(k)}=ck $$ A typical dispersion relation in plasma is: $$ \omega^2=\omega^2_p+3v^2_{th}k^2 $$ All the parameters in the previous equation aside from omega and k are medium parameters.

Having the dispersion relation, one can compute an expression for the group and phase velocity without knowing the signal. It is enough to know the frequency of the signal to know the wave vector. If a signal is sinusoidal or a summation of sinusoidal, their frequencies are well defined, so the phase and group velocity of every signal can be determined by substituting the frequency in the expressions of the phase and group velocity.

Now mathematically speaking, there is no general way of re-writing the summation of arbitrary sinusoidal signals as a product. The only practical way is to plot the summation as function of the phase which will show the periodicity of the signal (if it is periodic in the first place).

I hope now that it is clear that writing the sum as a product won’t add any information

  • $\begingroup$ Thanks; I already knew and of course writing the product form doesn't add any information, I was just trying to find a better representation for appearance of the wave function. An appearance by which you can derive some of those variables more easier and faster at a glance(like what we saw in the example of equal amplitudes). How/why are you so sure that there is no general method of rewriting it that way? and yes,phase velocity and group velocity are CHARACTERISTICS OF THE MEDIUM OF PROPAGATION but also representation of basic signals in the medium shows these characteristics. $\endgroup$
    – 2physics
    Commented Oct 4, 2013 at 22:43
  • $\begingroup$ If you looked for trigonometric identities anywhere you find that there are some formulas for converting sums to products and vice-versa. But those are for very special cases like same amplitude, same frequency, or integer multiple of frequencies. If there is a simple way of doing it why would one bother by using Fourier transform?? And I don't know what you mean by "representation of basic signals in the medium shows these characteristics". Can you tell me what is the group velocity of the expression you wrote above? $\endgroup$ Commented Oct 4, 2013 at 23:27
  • $\begingroup$ @Gotaquestion I'm not quite sure I understand the statement "One has always to keep in mind that phase velocity and group velocity are CHARACTERISTICS OF THE MEDIUM OF PROPAGATION. They don’t depend on the propagating signal; they depend on the medium of propagation." It seems like a counter example to this is a Stokes wave in water, which has a dispersion relation that depends on the amplitude of the wave, so without knowledge of the signal (i.e. the amplitude), one would improperly diagnose the dispersion relation. $\endgroup$
    – Nick P
    Commented Jan 2, 2014 at 23:25
  • $\begingroup$ @NickP I should have made it clear that I was referring to electromagnetic waves (I don't know a lot about water waves). This question was a continuation to another question about EM waves, so I kept speaking on that topic $\endgroup$ Commented Jan 2, 2014 at 23:40
  • $\begingroup$ @NickP and Gotaquestion. Nicks counterexample is about nonlinear waves. For many waves we study just simple linear waves of smallish amplitude then the dispersion relation depends on medium and not on wave amplitude.. though velocities do depend on wave frequencies. Water is a fantastic neglected in school simple case especially readily observed in daily life with dispersion w^2=g.k and phase (crest) velocity turns out to be double the group (packet) velocity. Curiously just the opposite in quantum wave mechanics , Schrodinger nonrelativistic version. $\endgroup$
    – blanci
    Commented Oct 7, 2020 at 2:43

(EDITED adding a solution to the original question taken from this source)

A general expression using any number of tones, of any frequency, and any propagation constant, could be written as follows:

\begin{equation} s(x,t) = \sum_i A_i \cos(\omega_i t - k_ix) \end{equation}

Where the frequencies and propagation constants can be defined with respect to the respective values at the center frequency (ωc and kc) as follows:

\begin{equation} \omega_i = \omega_c + \Delta\omega_i \end{equation} \begin{equation} k_i = k_c + \Delta k_i \end{equation}

After reordering terms and using basic trigonometrics:

\begin{equation} \begin{split} s(x,t) &=\sum_i A_i \cos(\Delta\omega_i t - \Delta k_ix)\cos(\omega_c t - k_cx)\\ &-\sum_i A_i \sin(\Delta\omega_i t - \Delta k_ix)\sin(\omega_c t - k_cx) \end{split} \end{equation}

Now using the equation of a sum of two tones of equal frequency but different amplitudes we can conclude:

\begin{equation} s(x,t) = \underbrace{e(x,t)}_{\text{envelope}}\underbrace{\cos[\omega_c t - k_cx - \varphi(x,t)]}_{\text{carrier}} \end{equation}

\begin{equation} e(x,t) = \sqrt{\left[\sum_i A_i \cos(\Delta\omega_i t - \Delta k_ix)\right]^2+\left[\sum_i A_i \sin(\Delta\omega_i t - \Delta k_ix)\right]^2} \end{equation}

\begin{equation} \varphi(x,t) = \arctan\left(\cfrac{-\sum_i A_i \sin(\Delta\omega_i t - \Delta k_ix)}{\sum_i A_i \cos(\Delta\omega_i t - \Delta k_ix)}\right) \end{equation}

From e(x,t) it can be concluded that the envelope moves according to the typical definition of group velocity (dw/dk). This demonstrates that the generalization asked by the OP original question is possible.

Actually we can get another conclusion: what it is more difficult to generalize is the phase velocity due to the complexity of φ(x,t). (Note that φ(x,t) is actually zero in the typical example that only considers two tones of equal amplitudes in s(x,t) ).

  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$ Commented Nov 12, 2022 at 13:50

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