Prove: EM Waves Addition Superposition Equivalence

[fluidistic]

Homework Statement

Show that the superposition of three waves with frequencies \omega _c, \omega _c + \omega _m and \omega _c - \omega _ m and same amplitude are equivalent to another wave of frequency \omega _c which is modulated by a sinusoidal wave with frequency \omega _m, i.e. E=E_0 \left [ 1+a \cos (\omega _m t) \right ] \cos (\omega _c t).

Homework Equations

Not sure, but I started with E_1=E_0 e^{i (\omega _c t + k_1 x)}, E_2=E_0 e^{i \left [ (\omega _c + \omega _m )t + k_2 x \right ]} and E_3=E_0 e^{i \left [(\omega _c + \omega _m)t + k_1 x \right ]}.

The Attempt at a Solution

Using the equations in the Relevant part, I just summed them up and factorized by E_0. I'm wondering if I started with the right equations. What do you think? I'm asking this question because I'm unsure and further I don't really see how to reach the answer from the equations I've put.

 

[gabbagabbahey]

I'd assume that the speed of the wave is the same for all 3 constituent waves, allowing you to express k for each one.

Also, I'd imagine that you are supposed to assume the fields are real; so you will either want to use cosines/sines or take the real part of complex exponentials. For example, the electric field \textbf{E} of a plane wave is often taken to be the real part of a complimentary complex-valued field:

\tilde{\mathbf{E}}=\tilde{\mathbf{E}}_0{\rm{e}}^{{\rm{i}}(\textbf{k}\cdot\textbf{r}-\omega t)}

Where the tildes denote complex-valued quantities. This is done since exponentials are often easier to do calculations with.

 

[fluidistic]

Thanks gabba hey. Yes I would take the real part when I'm done with the simplifications.
So I wasn't wrong take a wave of the form E=E_0 \cos (\ometa t + \vec k \codt \vec r)?

 

[gabbagabbahey]

Yes, \LaTeX typos aside (where in the Greek alphabet is "\omet[/color]a" located? ), that assumed form should be fine. You could even include a phase factor if you want (i.e. \textbf{E}=\textbf{E}_0\cos(\textbf{k}\cdot\textbf{r}-\omega t+\delta)) as long as you assume that all 3 waves are in phase (have the same \delta).

 

[fluidistic]

Yes, \LaTeX typos aside (where in the Greek alphabet is "\omet[/color]a" located? ), that assumed form should be fine. You could even include a phase factor if you want (i.e. \textbf{E}=\textbf{E}_0\cos(\textbf{k}\cdot\textbf{r}-\omega t+\delta)) as long as you assume that all 3 waves are in phase (have the same \delta).

Oops, ometa and codt instead of cdot...
Now I notice that you put a minus sign in "-\omega t". Is it important? I didn't use any minus sign, should I? Isn't just the direction of propagation that changes?

 

[gabbagabbahey]

Yes, the negative sign simply means that the wave will propagate in the same direction as \textbf{k} instead of opposite to it. It is convention to assume that k=||\textbf{k}|| is positive and so in order for the waves to travel with positive speed, forwards in time, you use the negative sign.

 

[fluidistic]

Ok thanks, I'll use the negative sign.

 

[fluidistic]

I just summed up E_1 with E_2 and it's already very messy, I can't believe that adding E_3 will simplificate anything.
I reach E_1+E_2=2E_0 \left [ \cos \left ( \frac{(k_1+k_2)x-(2 \omega _c + \omega _m)t + 2 \alpha}{2} \right) \cdot \cos \left ( \frac{(k_1-k_2)x+ \omega _m t}{2} \right ) \right ].
If I sum the third wave I'll get a function of the form \cos ( \cos (...)) which doesn't seem to simplificate and I won't reach the answer. Am I missing something?

 

[gabbagabbahey]

If I were you, I'd start with adding E_2 and E_3 instead

 

[fluidistic]

Indeed, I should have thought more before heading to the arithmetics.
I now reach E_2+E_3=2E_0 \left [ \cos \left ( \left ( \frac{k_2+ k_3}{2} \right ) x + \alpha - \omega _c t \right ) \cos \left ( \left ( \frac{k_2 - k_3}{2} \right ) x - \omega _m t \right ) \right ]. I'm sure this simplifies more and I should do it before adding E_1. I'm not sure which trigonometric relation to use here. What do you think?

 

[gabbagabbahey]

Usually waves of different frequency traveling in the same medium will have the same speed, so you expect

k_1=\frac{\omega_c}{c}, \;\;\;\;\;\; k_2=\frac{\omega_c+\omega_m}{c}\;\;\;\;\;\; \text{and}\;\;\;\;\;\;k_3=\frac{\omega_c-\omega_m}{c}

 

[fluidistic]

Usually waves of different frequency traveling in the same medium will have the same speed, so you expect

k_1=\frac{\omega_c}{c}, \;\;\;\;\;\; k_2=\frac{\omega_c+\omega_m}{c}\;\;\;\;\;\; \text{and}\;\;\;\;\;\;k_3=\frac{\omega_c-\omega_m}{c}

Ah right. I reach E_2+E_3=2E_0 \cos \left [ \omega _c \left ( \frac{x}{c}-t \right ) + \alpha \right ] \cos \left [ \omega _m \left ( \frac{x}{c} -t \right ) \right ].
The "alpha" term seems bothering, it's the phase of any of the 3 waves. Now I must add this expression to the first wave... I guess it'll be complicated. Let's see.

Edit: I just summed the first wave to the last expression. I get that E_1+E_2+E_3=3 E_0 \cos \left [ \omega _c \left ( \frac{x}{c}-t \right ) +\alpha \right ] + 2 E_0 \cos \left [ \omega _m \left ( \frac{x}{c}-t \right ) \right ].
I really don't see how to reach the result.

 

[gabbagabbahey]

Adding E_1 should be a piece of cake since E_1=E_0 \cos \left [ \omega _c \left ( \frac{x}{c}-t \right ) + \alpha \right ][/tex] <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" />

 

[fluidistic]

Adding E_1 should be a piece of cake since E_1=E_0 \cos \left [ \omega _c \left ( \frac{x}{c}-t \right ) + \alpha \right ][/tex] <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" />

<br /> <br /> Well yes. I&#039;ve edited my last post to include where I headed. <br /> Now I believe I should use a trigonometric trick, but I don&#039;t see how to get rid of alpha.

 

[gabbagabbahey]

Ah right. I reach E_2+E_3=2E_0 \cos \left [ \omega _c \left ( \frac{x}{c}-t \right ) + \alpha \right ] \cos \left [ \omega _m \left ( \frac{x}{c} -t \right ) \right ].
The "alpha" term seems bothering, it's the phase of any of the 3 waves. Now I must add this expression to the first wave... I guess it'll be complicated. Let's see.

Edit: I just summed the first wave to the last expression. I get that E_1+E_2+E_3=3 E_0 \cos \left [ \omega _c \left ( \frac{x}{c}-t \right ) +\alpha \right ] + 2 E_0 \cos \left [ \omega _m \left ( \frac{x}{c}-t \right ) \right ].
I really don't see how to reach the result.

I don't see how you ended up with that

\begin{aligned}E_1+E_2+E_3 &amp;= E_0 \cos \left [ \omega _c \left ( \frac{x}{c}-t \right ) + \alpha \right ] +2E_0 \cos \left [ \omega _c \left ( \frac{x}{c}-t \right ) + \alpha \right ] \cos \left [ \omega _m \left ( \frac{x}{c} -t \right ) \right ] \\ &amp;= E_0 \cos \left [ \omega _c \left ( \frac{x}{c}-t \right ) + \alpha \right ] \left(1+2\cos \left [ \omega _m \left ( \frac{x}{c} -t \right ) \right ]\right)\end{aligned}

Which is your desired result

 

[fluidistic]

Ok thank you. I made an arithmetics error, I was too fast.
So the answer is slightly different from the one provided but I think we've done a more general case (i.e. considering a phase) although I'm not 100% sure.
Problem solved. Thanks for all.

 

[gabbagabbahey]

The point is that E_0 \cos \left [ \omega _c \left ( \frac{x}{c}-t \right ) + \alpha \right ] is a wave of frequency \omega_c, and \cos \left [ \omega _m \left ( \frac{x}{c}-t \right )\right ] is a sinusoidal wave of frequency \omega_m, so

E_{\text{total}}=E_0 \cos \left [ \omega _c \left ( \frac{x}{c}-t \right ) + \alpha \right ] \left(1+2\cos \left [ \omega _m \left ( \frac{x}{c} -t \right ) \right ]\right)

is a wave of frequency \omega_c, modulated by a sinusoidal wave of frequency \omega_m...which is exactly what the question asked you to show.