Its said that electromagnetic waves carry energy. Is this because these waves are made up of electric and magnetic fields which can cause changes to the stuff that falls with in their range? Is that why its said that electromagnetic waves carry energy? Why do some electromagnetic waves like gamma or x-rays carry more energy? I read that that is because they have high frequency but I don't get how and why higher frequency electromagnetic waves have more energy. What is the reason for that?
[Physics] How do electromagnetic waves carry energy
electromagnetic-radiationelectromagnetismenergyfrequencyphotons
Best Answer
You should remember one thing : electromagnetic field is just a spatial representation of how electric charges interact with each other, and by "interact" I actually mean "exchange some energy".
Lets imagine that we want to build "from scratch" a given charge distribution $\rho(\textbf{x})$. That means that we have to bring close to each over different kind of charges, and we know that, by doing so, charges will interact with each other following the Coulomb's law. It very basically states that same signe charges repell each other wherease opposite signe charges attracts each other.
Just by saying this, we actually already know that their is some energy stored EM field since we can induce movement (repulsion/attraction) just by making charges interact with each other. That means that some kind of potential energy has been converted in kinetic energy in order to produce movement.
Without going too deep into the calculations, one can compute the total work $W_e$ one has to provide in order to build such distribution $\rho(\textbf{x})$ (i.e. to bring the separated charges together from the infinity): $$ W_e=\frac{1}{2}\int\mathrm{d}\textbf{x}_1\,\mathrm{d}\textbf{x}_2\,\frac{\rho(\textbf{x}_1)\rho(\textbf{x}_2)}{4\pi\epsilon_0|\textbf{x}_1-\textbf{x}_2|} $$ where $\epsilon_0$ is the vacuum permittivity.
Using Maxwell's equations, it is possible to express $W_e$ in term of electric field $\textbf{E}$ radiated by the charge distribution $\rho$ : $$ W_e=\int\mathrm{d}\textbf{x}\,\frac{\epsilon_0\textbf{E}^2}{2} $$
Actually, the same goes if you have some current distribution $\textbf{j}(\textbf{x})$ (i.e. some charges moving and flowing through space), the total work $W_m$ needed in order to produce such current can be expressed in terms of radiated magnetic field $\textbf{B}$ : $$ W_m=\int\mathrm{d}\textbf{x}\,\frac{\textbf{B}^2}{2\mu_0} $$ where $\mu_0$ is the vacuum permeability.
What I wanted to show you is that EM field is radiated by charges/currents distributions and that this EM field is storing some energy that comes from the interactions between the charges.
If you concede that EM field can carry some energy, just like in the previous case, then you should have no problem to see why EM waves carry some energy as well. EM waves are just a particular case of EM field that can propagate through space and time. For simplicity, we can considere a monochromatic plane wave of frequency $\omega$, where : $$ \textbf{E}(\textbf{x},t)=\textbf{E}_0\,e^{\mathrm{i}(kx-\omega t)}\quad\text{and}\quad\textbf{B}(\textbf{x},t)=\frac{1}{\omega}\textbf{k}\times\textbf{E}(\textbf{x},t) $$ where $\textbf{k}=k\,\hat{x}$ is the wave vector.
In this case now, it means that the quantity $\mathcal{E}=W_e+W_m$ varies in respect of time. Without going into to much details, you can play with the maths and show that the time derivative of $\mathcal{E}$ can be interpreted as an energy flux density $\mathbf{\Pi}$ called Poynting vector such that : $$ \int\mathrm{d}\textbf{x}\,\frac{\partial\mathcal{E}}{\partial t}=-\oint\mathrm{d}\textbf{S}\cdot\mathbf{\Pi}\quad\text{with}\quad\mathbf{\Pi}=\frac{1}{\mu_0}\,\textbf{E}\times\textbf{B} $$ where $\textbf{S}$ is a surface enclosing your charge distribution.
Units of $\mathbf{\Pi}$ are $\text{W.m}^{-2}$ so it tells you how much energy is radiated out by your charges/currents distribution by unit of time and surface.
You can compute $\mathbf{\Pi}$ for EM plane wave and find : $$ \mathbf{\Pi}=\frac{\textbf{E}^2_0}{\mu_0 c}\cos^2(kx-\omega t)\,\hat{x} $$ where $c=1/\sqrt{\epsilon_0\mu_0}$ is the light velocity in the vacuum.
As stated before, the Poynting vector tells you how much energy is carried by the EM wave. But does not explain why "higher frequency means more energy" since the amplitude of $\mathbf{\Pi}$ is independent on $\omega$.
The electromagnetic power carried by the EM wave is simply defined as : $$ \mathcal{P}=\int\mathrm{d}\textbf{S}\cdot\mathbf{\Pi} $$ which is expressed in Watt, i.e. in Joules per second. This quantity then may be interpreted as a photon flux such that : $$ \mathcal{P}=\Phi\hbar\omega $$ where $\hbar\omega$ is the energy carried by one photon, and $\Phi$ is the number of photon by unit time passing through the surface $\textbf{S}$. This formula tells you that if you want your wave to carry 1W, then you will need $2.5\times10^{18}$ photons per second of 2.5eV energy (i.e. 500nm wavelength). Or you could also have 1W with $6.2\times10^{12}$ photons per second of 1MeV energy (typically gamma ray). You see you need much much less gamma photons than visible photons for your wave to carry 1W.
EM field, and in particular EM waves carry some energy because they are representations of how charges interact with each other through Coulomb's potential energy.
It's true that one gamma photon individually carries more energy than a visible photon because of the formula $E=\hbar\omega$. BUT...
This is not necessary true because we just seen that the amplitude of the energy density of an EM wave does not depend on $\omega$. The reason is that you have also to take into account how many photons can represent your wave (the $\Phi$ here).