# Why the frequency component is not included when the energy of a wave is described?

Energy of a wave is directly proportional to its Amplitude squared. There is no inclusion of frequency in it. But by intuition if one thinks about it, isn't it obvious that a wave with higher frequency should have higher energy? According to particle theory, the energy of a photon depends on frequency which makes sense. But in classical theory, how is it included? I am not seeing it. Please explain.

• Why do you expect the "energy" of the EM wave be frequency dependent? Does the energy of an elastic or an acoustic wave or of an LC resonator depend its frequency? Note that in the equation $\mathcal E =hf$ is not the energy of the EM wave whose frequency is $f$. Commented Jul 16 at 14:04
• wouldn't a person require more energy to vibrate a rope faster(i.e. higher frequency)? Thankyou verymuch for replying. Commented Jul 16 at 14:36
• I think you are confusing the concepts of the rate of working that is power with energy. When you pull a spring with a mass on it you expend work of the amount by which your muscle's chemical energy decreases (friction ignored). This amount of work depends how far out you have pulled the spring and the stored elastic energy is equal to the work you have done on it. When you let it go that energy will oscillate back and forth between kinetic and elastic so that their sum stays constant (friction ignored) independently of its amplitude, $f$ depends on the mass and the spring's elasticity. Commented Jul 16 at 15:52

This really depends on the wave. If it's a wave in rope, there is kinetic energy associated with the velocity of the rope moving up and down, and in the linear limit:

$$v \propto A\omega$$

which has momentum (density):

$$p = \rho v$$

and energy, ofc.

But if it's an electromagnetic wave, the energy density goes as:

$$u = \frac 1 2 \Big(\vec E \cdot \vec D + \vec B \cdot \vec H \Big)$$

so basically amplitude squared.

Unlike the changing displacement of a rope, there is no kinetic energy in the changing of the electric field.

Now you might think: "But a changing E field produces a B fields, so the faster it changes, the stronger the B field, so the more energy and vice versa).

This is the wrong premise. A changing E field doesn't create anything, but it does match the gradient of the B field (and vice versa), so the higher frequency is just compensated by a shorter wavelength--which agrees with out understanding of waves.

• Thankyou. So, what you are saying is, if we want a higher frequency EM wave, hence we start oscillating the electrons faster(lets say in an antenna), so the energy required would be same? It won't increase because we have increased the frequency of the electrons to make then oscillate faster? Commented Jul 16 at 14:56

According to particle theory, the energy of a photon depends on frequency which makes sense. But in classical theory, how is it included? I am not seeing it.

Energy density vs. energy flux
One has to distinguish here energy density and energy flux. Energy density is not explicitly dependent on energy, but the energy flux, i.e., how fast the energy propagates from one place to another (the energy "carried" by the wave), is indeed dependent on frequency (i.e., on the rate of change in time), as is described by the Poynting theorem: $$\nabla \cdot\mathbf{S} = \frac{\partial u}{\partial t}$$

Quantum vs. classical
Trying to draw analogy with the photons however touches upon the crucial difference between the classical and quantum waves. In classical theory one can excite waves of any energy at any frequency - for high frequency we can always excite a wave of sufficiently weak amplitude, so that it has low energy density. In Quantum theory the excitations are quantized, so that a wave of frequency $$\omega$$ cannot have energy lower than $$\hbar\omega$$ (forgetting for the moment the energy of the vacuum fluctuations.) This was the insight behind the Planck's law.

• Thankyou verymuch Commented Jul 17 at 2:50