# Is the intensity of any type of wave proportional to the amplitude squared?

Assume that a wave is any function of a physical quantity $$y$$ that respects the equation:

$$\nabla^2y=\frac{1}{v^2}\frac{\partial^{2} y}{{\partial t}^2}.$$

Is there any way you can prove that the intensity $$I$$ of a wave is proportional to the amplitude squared (for one-dimensional, two-dimensional or three-dimensional waves)? Because I think that to determine $$I$$ we need to know first the physical quantity, then apply the laws to get $$I$$... But as far as I know, that relation exists for every wave.

If there is any way to do so, how can I do it?

• I think that for sinusoidal waves this works, right? You can take for a physical quantity $y$ the power: $$P(y) = \kappa y^2 \therefore P(x, t) = \kappa A^2 cos^2(kx-\omega t + \delta).$$ Then take the mean value in a period: $$I=\frac{1}{2}\kappa A^2.$$ Is that right? If it is, why we have this relation for the power $P$? Sep 18 at 2:15

Intensity refers to the amount of energy carried by a wave through a surface that is perpendicular to the direction of propagation, per unit area per unit time. So in order to define the intensity of a wave, we first need to define the energy carried by the wave. In order to define the energy, we need to know the Lagrangian or the Hamiltonian that governs the wave. It is not enough just to know the equation of motion that the wave obeys.

A wave equation of the form

$$\nabla^2 y = \frac{1}{v^2} \partial_{tt} y$$

arises as a consequence of a Lagrangian density of the form

$$\mathcal{L} = \frac{\mu}{2} (\partial_{t}y)^2 - \frac{\tau}{2} \|\nabla y\|^2$$

(For mechanical waves, $$\mu$$ is the density of the medium, and $$\tau$$ is the "stiffness". However, the above mathematical formulation also applies to other kinds of waves, such as electromagnetic waves.)

You can verify that applying the Euler–Lagrange equations to the above Lagrangian density gives you back the wave equation (where $$v = \sqrt{\tau/\mu}$$).

By applying Noether's theorem to this Lagrangian density, we can derive the conserved energy density,

$$u = \frac{\mu}{2} (\partial_t y)^2 + \frac{\tau}{2} \|\nabla y\|^2$$

From this expression it's evident that if $$y$$ is increased by a factor of $$k$$ then the energy density $$u$$ is increased by a factor of $$k^2$$. The same is then true for the intensity.

The role of measure (in the mathematical sense) is applied to vector fields with (for instance) E-field orientations N-S and E-W. Those two orthogonal orientations result in a sum (in the case of interpenetrating multiple waves) that is added in quadrature because that's the rule for adding orthogonal components' amplitudes to make a vector result's amplitude.

So, for noninteracting waves, in three dimensions, with two polarizations (like all transverse waves), we do expect root-sum-square rules. The breaking of such a convention is possible. Mathematically, one can compute distances with Manhattan-distance rules (sum of absolute values of orthogonal components), though one does the homogeneous-space distance with root-sum-square.

Lightwave interference, analogously, would only follow different rules in a nonlinear medium (and those exist, like the birefringent calcite/iceland spar).