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I know that wavelength $\times$ amplitude is equal to the speed of light $c$ when looking at an electromagnetic wave, but does a higher aplitude mean greate energy carried by the wave? And if so, is there a formula related to energy and aplitude?

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In simple terms - energy of a photon associated with light is related to frequency (Plank's constant multiplied by frequency). Intensity of light - energy crossing unit area per unit time - is directly proportional to square of amplitude of em wave.

Let me clarify a few concepts

  1. Wave-particle duality also applies to electromagnetic waves. That means, electromagnetic waves behaves as waves in many situations however it also behaves as particles in some other situation - for example interference is a wave phenomena whereas light's interaction with mettar - photo electric emission is a particle phenomena.

  2. Now the particle associated with electromagnetic waves are called photons (Photon https://g.co/kgs/JYG1Fr)

  3. Energy of a photo is geven by Plank's constant (h) multiplied by frequency of the EM wave.

  4. Another analogous to energy of em wave is intensity it is defined as the total energy carried by em wave through unit surface area in unit time.

  5. This intensief of light depends on the amplitude of em waves - which is closely related to number of photons crossing unit area in unit time. It is proportional to square of amplitude http://www.acoustics.salford.ac.uk/feschools/waves/propagation.php
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In cgs units, the energy density carried by an electromagnetic wave in the vacuum is $$u = {1\over2}(\mathbf{E}^2+\mathbf{B}^2),$$ so yes, greater amplitude means more energy (per unit volume) is carried by the electromagnetic wave.

The formula for the electromagnetic energy density can be generalized to any system of charged particles and electromagnetic fields. For more information you can refer e.g. to Section 6.7 of the Jackson.

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  • $\begingroup$ Why minus before $\mathbf B^2$? $\endgroup$ – Ruslan Nov 7 '18 at 12:43
  • $\begingroup$ It should be a plus: Energy per unit volume is $\mathcal E= \frac 12(\epsilon_0 |{\bf E}|^2 + |{\bf B}|^2/\mu_0)$ $\endgroup$ – mike stone Nov 7 '18 at 13:32
  • $\begingroup$ The $-$ sign gives a Lorentz-invariant Lagrangian, but the energy, a Hamiltonian, shouldn't be invariant because of the privileged role time derivatives have in the usual Legendre transform. As with the classical example $L=\frac{p^2}{2m}-V,\,H=\frac{p^2}{2m}+V$, the end result of the Legendre transform is a sign change. $\endgroup$ – J.G. Nov 7 '18 at 15:31
  • $\begingroup$ right it's with a plus. my bad $\endgroup$ – TotoPasta Nov 8 '18 at 14:08

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