Prove: EM Waves Addition Superposition Equivalence

[fluidistic]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

Using the equations in the Relevant part, I just summed them up and factorized by [tex]E_0[/tex]. 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 [itex]k[/itex] 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 [itex]\textbf{E}[/itex] of a plane wave is often taken to be the real part of a complimentary complex-valued field:

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

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 [tex]E=E_0 \cos (\ometa t + \vec k \codt \vec r)[/tex]?

 

[gabbagabbahey]

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

 

[fluidistic]

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

Oops, ometa and codt instead of cdot...
Now I notice that you put a minus sign in "[tex]-\omega t[/tex]". 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 [itex]\textbf{k}[/itex] instead of opposite to it. It is convention to assume that [itex]k=||\textbf{k}||[/itex] 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 [tex]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 ][/tex].
If I sum the third wave I'll get a function of the form [tex]\cos ( \cos (...))[/tex] 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 [tex]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 ][/tex]. I'm sure this simplifies more and I should do it before adding [tex]E_1[/tex]. 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

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

 

[fluidistic]

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

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

Ah right. I reach [tex]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 ][/tex].
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 [tex]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 ][/tex].
I really don't see how to reach the result.

 

[gabbagabbahey]

Adding [itex]E_1[/itex] should be a piece of cake since [itex]E_1=E_0 \cos \left [ \omega _c \left ( \frac{x}{c}-t \right ) + \alpha \right ][/tex]

 

[fluidistic]

Adding [itex]E_1[/itex] should be a piece of cake since [itex]E_1=E_0 \cos \left [ \omega _c \left ( \frac{x}{c}-t \right ) + \alpha \right ][/tex]

Well yes. I've edited my last post to include where I headed.
Now I believe I should use a trigonometric trick, but I don't see how to get rid of alpha.

 

[gabbagabbahey]

Ah right. I reach [tex]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 ][/tex].
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 [tex]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 ][/tex].
I really don't see how to reach the result.

I don't see how you ended up with that

[tex]\begin{aligned}E_1+E_2+E_3 &= 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 ] \\ &= 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}[/tex]

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

[tex]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)[/tex]

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