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This question already has an answer here:

In my high school physics class, I was taught that the energy of light is dependent only on the frequency, as demonstrated in the equation $E = h \cdot \nu$.

My question is, why is amplitude part of the equation? As the amplitude of the light increases, it gets more intense, i.e. brighter (ignoring that more light also makes things brighter), so wouldn't it make sense that light that has a greater amplitude also has more energy? And as an extension of that assumption: that lower amplitudes have less energy?

EDIT: not a duplicate, because I'm asking why amplitude is not part of the energy equation, not if a particle has amplitude.

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marked as duplicate by John Rennie energy Jul 6 '18 at 3:58

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As you've said, that's the energy of a light particle, not an arbitrary light wave. As you point out, an electromagnetic plane wave with electric field amplitude $E_0$ has total average energy density $u=\varepsilon_0E_0^2/2$. The formula of energy you have given actually relates this to the number density $n$ of light particles per unit volume, by $u=nh\nu$. As such, $$n=\frac{\varepsilon_0E_0^2}{2h\nu}.$$

EDIT The thing to be considered here is, while the amplitude of the wave does matter for its total energy, it has no bearing on the energy of the particles that make it up.

What increases the amplitude of an electromagnetic wave is an increase in the number of photons (light particles, each with energy $E=h\nu$) that are part of this entire wave.

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  • $\begingroup$ Can you simplify this? I don't really understand what those equations mean. Remember, this was high school physics, not college or career-field level. $\endgroup$ – The Eye Jul 6 '18 at 3:47
  • $\begingroup$ So, basically, light is a particle in a wave, and the particle has an energy separate from the wave's energy, similar to water molecules in relation to actual waves? $\endgroup$ – The Eye Jul 6 '18 at 5:11
  • $\begingroup$ yes water molecules carry the water wave dynamics. The difference lies in that photons do not add up to light by a simple addition of vectors , as with water waves, but by a superposition of their quantum mechanical wave function . $\endgroup$ – anna v Jul 6 '18 at 11:02
  • $\begingroup$ Actually no. To the both of you. These photons are not localized particles that can vibrate. They actually are the units of vibration. What oscillates here is the electromagnetic field, and the minimum energy for it to vibrate with a certain frequency $\nu$ is $h\nu$. It has nothing to do with what makes up the vacuum. $\endgroup$ – Gabriel Golfetti Jul 6 '18 at 14:45
  • $\begingroup$ The water molecules don't play the same role as the photons. They are more akin to the EM field itself. And by the way, the wave function of the EM field is a vector. So the total wave is, indeed, the vector sum of constituent waves. $\endgroup$ – Gabriel Golfetti Jul 6 '18 at 14:47
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In simple terms:

A building is made up by bricks. Bricks are not a building.

Photons are the elementary particles/constituents of light, but they are not light. What we call light is an ensemble of photons in a mathematical superposition, and the classical theory of light emerges from the underlying quantum mechanical level of photons.

The classical theory, is a wave theory which does have amplitude, the more light, the more energy. We have experimentally found that this is built up by photons which are massless particles carrying energy and obeying special relativity. The wave nature of the photon is in the quantum mechanical wave function, and its amplitude is connected with the probability of detection of the photon, and not its energy.

One can estimate the number of photons in the classical light wave by using the formula given by Gabriel.

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  • $\begingroup$ "Light is described by a classical theory" ─ are you claiming that quantum theory is invalid for the description of light? $\endgroup$ – Emilio Pisanty Jul 6 '18 at 10:49
  • $\begingroup$ @EmilioPisanty You misunderstand; classical theory emerges from the underlying quantum field theory, in a mathematically smooth manner. Classical electromagnetic theory is best for large ensembles of photons the way thermodynamics (also an emergent theory) is best for a large ensemble of particles/molecules. $\endgroup$ – anna v Jul 6 '18 at 11:02
  • $\begingroup$ As it stands, your text reads as the claim I described above. If that's not what you meant, then it might be a good idea to edit it. Or leave it as is with the claim's implications as they stand, too. $\endgroup$ – Emilio Pisanty Jul 6 '18 at 11:06

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