# Paradox of wave energy

Imagine that a wave with energy $E$ is given. as we know $E$ is relevant to the $A^2$($A$ is amplitude) now consider another wave (as same as the first one) and these two wave having a constructive interference(it's global we don't have destructive interference anywhere else) now we have a new wave with $A' = 2A$ now if we apply our energy equation we might conclude that the energy should $4E$but energy conservation will fail because we expect that the total energy should be $2E$.why??

• Duplicate: physics.stackexchange.com/q/7707 – lemon Mar 7 '15 at 10:08
• @jack see my answer to the question about destructive interference of waves – Sofia Mar 7 '15 at 10:26
• thanks @lemon but what if the constructiveness of our interference is global?in the post which you have mentioned it's not global and we have both constructiveness and destructiveness together – jack Mar 7 '15 at 10:32
• physics.stackexchange.com/q/47105 – mmesser314 Mar 7 '15 at 10:59

## 2 Answers

Let me restate the observation a little: Energy in waves are proportional to the square of the amplitude so when say three idential waves are added together the result has $$3^2$$ the energy instead of the expected triple of the energy.

The method that we use to describe waves is for our convenience, not nature's. Being human, we like linear things so we use the amplitude. For example, with a lot of waves the wave equation is linear so we can take two solutions and add them together to get another solution. This is very convenient for calculations so our tendency is to use the amplitude to describe waves.

However, using the amplitude to describe waves is NOT so convenient for describing energy as energy is proportional to the SQUARE of the amplitude, hence the paradox.

So what happens is that if you already have a particular wave, it's not as easy as it looks to add a second identical wave. That is, the second wave requires more energy than the first. In fact, when you take three waves and add then one by one the energy required is going to be proportional to:

1st wave: $$A^2$$

2nd wave: $$(4-1)A^2 = 3A^2$$

3rd wave: $$(9-4) = 5A^2$$

From a physical point of view, why is it that the later waves require more energy?

The reason is that the later added waves start in places where it is more difficult to go further. They have to go higher on top of the old highs and lower below the old lows.

Let's suppose you were digging a hole. Assuming dirt doesn't require breaking but just needs to be lifted, how should you charge for energy? Should you charge proportional to the depth (amplitude)? Or to the square of the depth (amplitude squared or energy)?

It's easy to see that the deeper the hole, the more the AVERAGE gravitational energy that is needed to pull some of the dirt up to the top. And with a little calculus you will quickly discover that the gravitational energy in digging a hole is proportional to the square of the depth. If you want to make this into a wave problem, let the "hole" be the amplitude of the height in a plowed field. As the amplitude of the plowing triples, the energy requirement goes up nine times.

The same calculation applies to sound waves. Here, the depth of the hole becomes the amplitude of the pressure.

Assume for simplicity a cavity of length $L$ in which two waves propagate in opposite directions

$$y_{+}(x, t) = Ae^{i(kx - \omega t)} \tag{i}$$

and

$$y\_(x, t) = Ae^{i(-kx - \omega t)}, \tag{ii}$$

with $k = \pi/L$, i.e. the fundamental mode of vibration. It's obvious that the sum of the intensities of the two waves is $2|A|^2$.

Now let's find their superposition

$$y(x) = y_{+}(x, t) + e^{2i\phi} y\_(x, t) = 2A e^{i\phi} cos \left(\frac {\pi x}{L} - \phi \right) e^{-i\omega t}. \tag{iii}$$

It's clear that the condition $y(0) = y(L) = 0$ imposes $\phi = \pi /2$. In this case we have constructive superposition. Now, let's calculate the average intensity over the chord.

$$\langle I \rangle = \frac {1}{L} 4|A|^2\int_0^L sin^2 \left( \frac {\pi x}{L} \right) dx = \frac {4|A|^2}{2L}\int_0^L \left[1 - cos\left(\frac {2\pi x}{L} \right) \right] dx = 2|A|^2. \tag{iv}$$

So, is there any problem?

• @jack I have replied you right now. Hope that my reply answers your problem with constructive superposition. – Sofia Mar 7 '15 at 13:39