7
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ I haven't seen so many bad answers to a question! By the charter of this forum, if you can't speak with authority to answer a question, do not answer it. $\endgroup$ – garyp May 4 '16 at 15:49
  • 1
    $\begingroup$ @garyp Sad, but true. It is puzzling that, so far, nobody in the answers even mentioned the Poynting vector. $\endgroup$ – dominecf May 4 '16 at 19:05
  • 2
    $\begingroup$ @garyp Would one of you please answer the question. As you see, many people thinks that EM waves carry the energy through photons. If this is not true, I'd like to know why it isn't in a scientific explanation rather than just saying 'no it is not!' $\endgroup$ – Alper91 May 4 '16 at 20:41
  • $\begingroup$ @Alex, you know the form every light-wave takes, right? Oscillations of an electric and magnetic field repeated over an axis. Every oscillation is exactly the same length along the axis, from the front of the wave all the way to its end. That's the wavelngth. Now imagine that you inhabit an energy mine and your job is to package very specific amounts of energy and send it out of the mine. Your tools are simple: for each amount of energy a train of attached containers, in which the containers are the same size throughout, but the size is picked by you, and you can use any number of them. (cont) $\endgroup$ – David Reishi May 4 '16 at 21:33
  • $\begingroup$ @Alex, (2 of 2). Now imagine that this is the way you choose to package the energy: For every specific amount of energy requested, from smaller to larger, you package the energy in a train of smaller containers, but more of them, in such a way that every container on the train is perfectly full, and there's no energy or container remaining. $\endgroup$ – David Reishi May 5 '16 at 0:04
4
$\begingroup$

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".

Electrostatic and magnetostatic energies

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.

Application to EM waves

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.

A microscopic interpretation in terms of photons?

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.

Conclusion

  • "Its said that electromagnetic waves carry energy. Is this because these waves are made up of electric and magnetic field which can cause changes to the stuff that falls with in their range ?"

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.

  • "Why some electromagnetic waves like gamma or x-rays carry more energy ?"

It's true that one gamma photon individually carries more energy than a visible photon because of the formula $E=\hbar\omega$. BUT...

  • "why more frequency of electromagnetic waves will have more energy "

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).

$\endgroup$
0
$\begingroup$

I like the following explanation. Although mostly mathematical, it illustrates a point Feynman tried to convey in his lectures. Energy is always conserved, and whenever it seems like it isn't you just need to look harder.

Let's recall that in the context of classical particle mechanics one defines the energy of the system of $n$ particles as the sum of the sum of the individual kinetic energies and the sum of the potential energy produced by every interaction (making sure one does not count an interaction twice) between the particles in the system. With such a definition one can prove that for an energy that evolves in the interval $[t_i,t_f]$ we have $E(t_f)-E(t_i)=W_{\text{non_con}}$, where the term on the right hand side is the work done by non-conservative forces.

Now, as far as classical physics goes, non-conservative forces all arise from the difficulty of calculating the energy interactions precisely. That observation allows us to confirm energy conservation for at least forces one may consider as "natural". Non the less, once one starts studying electromagnetism, one of the most common phenomena in our everyday experience one quickly realize the electric and magnetic forces are not conservative! In particular, there is a maxwell equation that states $\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}$, prohibiting the existence of a function $U$ such that $\vec{E}=\nabla U$ in the presence of non-static magnetic fields. Therefore one reaches to the conclusion that in the presence of arbitrary electromagnetic fields energy is not conserved!

After such a disturbing conclusion one must try to analyze the matter more closely. In particular, lets try to calculate the work done by the electromagnetic field on a region of space $\Omega$ with a charge density $\rho$. We have $$W_{\text{EM}}=\int_\Omega dV\rho\int_\mathcal{C}(\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{r}=\int_\Omega dV\rho\int_{t_i}^{t_f}dt(\vec{E}+\vec{v}\times\vec{B})\cdot \vec{v}= \int_\Omega dV\rho\int_{t_i}^{t_f}dt\vec{E}\cdot \vec{v}= \int_\Omega dV\int_{t_i}^{t_f}dt\vec{E}\cdot \rho\vec{v}=\int_{t_i}^{t_f}dt\int_\Omega dV\vec{E}\cdot\vec{J}=\int_{t_i}^{t_f}dt\int_\Omega dV\vec{E}\cdot\frac{1}{\mu_0}\left(\nabla\times\vec{B}-\mu_0\epsilon_0\frac{\partial\vec{E}}{\partial t} \right)=\\ \int_{t_i}^{t_f}dt\int_\Omega dV\left(\frac{1}{\mu_0}\vec{E}\cdot\nabla\times\vec{B}-\epsilon_0\vec{E}\cdot\frac{\partial\vec{E}}{\partial t} \right)=\\ \int_{t_i}^{t_f}dt\int_\Omega dV\left\{\frac{1}{\mu_0}\left[(\nabla\times\vec{E})\cdot\vec{B}-\nabla\cdot(\vec{E}\times\vec{B})\right]-\epsilon_0\frac{1}{2}\frac{\partial E^2}{\partial t} \right\}=\\ \int_{t_i}^{t_f}dt\int_\Omega dV\left( -\frac{1}{\mu_0}\frac{\partial\vec{B}}{\partial t}\cdot\vec{B}-\frac{1}{\mu_0}\nabla\cdot(\vec{E}\times\vec{B})-\epsilon_0\frac{1}{2}\frac{\partial E^2}{\partial t} \right)=\\ -\int_{t_i}^{t_f}dt\int_\Omega dV\frac{1}{\mu_0}\nabla\cdot(\vec{E}\times\vec{B})-\int_\Omega dV\int_{t_i}^{t_f}dt\left( \frac{1}{2\mu_0} \frac{\partial B^2}{\partial t}+\frac{1}{2}\epsilon_0\frac{\partial E^2}{\partial t}\right)=\\ -\int_{t_i}^{t_f}dt\int_\Omega dV\frac{1}{\mu_0}\nabla\cdot(\vec{E}\times\vec{B})-\int_\Omega dV\left( \frac{1}{2\mu_0} B^2(t_f)+\frac{1}{2}\epsilon_0E^2(t_f)\right)+\int_\Omega dV\left( \frac{1}{2\mu_0} B^2(t_i)+\frac{1}{2}\epsilon_0E^2(t_i)\right)=\\ -\int_{t_i}^{t_f}dt\int_{\partial\Omega} d\vec{A}\cdot\frac{1}{\mu_0}(\vec{E}\times\vec{B})-\int_\Omega dV\left( \frac{1}{2\mu_0} B^2(t_f)+\frac{1}{2}\epsilon_0E^2(t_f)\right)+\int_\Omega dV\left( -\frac{1}{2\mu_0} B^2(t_i)+\frac{1}{2}\epsilon_0E^2(t_i)\right) $$

To better understand this result let's define the electromagnetic energy (a very suggestive name) at time $t$ as $E_{\text{EM}}(t)=\int_\Omega dV\left( -\frac{1}{2\mu_0} B^2(t)+\frac{1}{2}\epsilon_0E^2(t)\right) $ and the Pointyng vector as $\vec{S}=\frac{1}{\mu_0}(\vec{E}\times\vec{B})$ we have that the work done by the electromagnetic force is $$W_{\text{EM}}=E_{\text{EM}}(t_i)-E_{\text{EM}}(t_f)-\int_{t_i}^{t_f}dt\int_{\partial\Omega} d\vec{A}\cdot\vec{S}$$

Now, one can write the mechanical result as $E(t_f)-E(t_i)=W_{\text{non_con}}+W_{\text{EM}}$ where now the electromagnetic force is not taken into account when calculating the work done by non conservative forces. By plugging in the previous result one has $E(t_f)-E(t_i)+E_{\text{EM}}(t_f)-E_{\text{EM}}(t_i)=W_{\text{non_con}}-\int_{t_i}^{t_f}dt\int_{\partial\Omega} d\vec{A}\cdot\vec{S}$ and we get to the exciting conclusion that all we needed was to redefine energy taking into account the one due to the electromagnetic field to recover $E(t_f)-E(t_i)=W_{\text{non_con}}-\int_{t_i}^{t_f}dt\int_{\partial\Omega} d\vec{A}\cdot\vec{S}$. It is now that we have redefined energy that we can interpret the term $\int_{t_i}^{t_f}dt\int_{\partial\Omega} d\vec{A}\cdot\vec{S}$ as energy escaping from the system. This is the key of light waves transfering energy. They do it through the flux of the Pointyng vector.

Notice that this example exemplifies how energy is not a concept which is obvious from observing nature. We believe it is because we have used it so much during our academic experience that we try to naturalize it. Non the less, energy is actually a quantity that we define as we want and the surprising fact about the law of conservation of energy is not that a number is conserved, but that humans have managed to find ways to redefine a number so that it does!

Now note that the definition of the Pointyng vector, whose flux determines the amount of energy light carries is proportional to the amplitudes of the electromagnetic field, not the frequency. But you pointed out that the energy of an electromagnetic wave depends on frequency not amplitud. This is the beginning of quantum mechanics! Your observation along with the proof above show classical mechanics is an incomplete theory of physics!

$\endgroup$
-2
$\begingroup$

The energy is carried in individual photons. A photon with twice the frequency has twice the energy. x-rays are made of photons with higher frequencies. The energy is transferred as kinetic energy as with the photoelectric effect. The energy of a photon is calculated as E=hf (Energy=Plank's constant x the frequency).

$\endgroup$
  • $\begingroup$ "kinetic energy as with the photoelectric effect". you mean to say that electric and magnetic field cause photon in the surrounding because of photoelectric effect and then these photon travel as kinetic energy ? $\endgroup$ – Alex May 4 '16 at 9:49
  • 1
    $\begingroup$ Photons are local phenomena that only happen when we perform a measurement on a quantum field. The energy is being carried by the field. $\endgroup$ – CuriousOne May 4 '16 at 10:02
  • 1
    $\begingroup$ do we get electromagnetic waves or photon when charges are accelerated ? photons cause electromagnetic waves or electromagnetic waves cause photons ? @CuriousOne $\endgroup$ – Alex May 4 '16 at 10:54
  • $\begingroup$ @Alex: In the far field you get a chance to detect photons, in the near field it seems to be complicated. I don't have a good physical intuition to offer for that. $\endgroup$ – CuriousOne May 4 '16 at 20:25
-2
$\begingroup$

EM Waves are basically spinnings of photons. To create a EM Wave, you basically move some electrons in a directional way(think of an antenna). Electron is a charged particle as well as protons. Charged particles emit photons, and if you emit photons in an ordered way such as this:

Antenna emitting photons

You are seeing a dipol antenna. When you apply negative voltage(intense electron charge referenced to antenna) to antenna input electrons move towards endpoints of the antenna and reflect in opposite if you apply positive voltage electrons will move toward the source so if you apply the voltage(reference electron charge) to antenna in a frequency (and if VSWR of the antenna suits the frequency) you will emit bunch of photons in an order (in a wavelength) in a direction.

And if the photons hit an antenna in the similar order and characteristic like the emitting antenna, photons will absorbed by the antenna and they will push the electrons to higher electron bands and finally electrons will be free due to high energy state and free electrons will create a charge that way. Of course, if you want to carry a meaningful signal(meaningful energy) you have to take care about emitting and receiving antenna characteristics.

EM waves just photon movements in a wave-like order. And photons carry energy. It is simple as that.

$\endgroup$
  • 2
    $\begingroup$ EM wave are very decidedly not spinning photons. What does "emit a bunch of photons" mean? Some of this answer has some validity, but there's some nonsense here, too. And none of it answers the question. $\endgroup$ – garyp May 4 '16 at 15:47
  • 1
    $\begingroup$ So you do not agree that energy is carried by photons? $\endgroup$ – Alper91 May 4 '16 at 15:56
  • 2
    $\begingroup$ The question is made in terms of classical E&M where there are no photons. $\endgroup$ – garyp May 4 '16 at 15:58
  • 1
    $\begingroup$ Garyp is correct, EM waves are like acoustic waves with the exception that they are generally transverse, whereas acoustic waves are longitudinal, and that they propagate as disturbances in the electric and magnetic fields, rather than through materials. Neglecting this, the formalism is the same. Photons are not EM waves. They are part of a particle physics formalism of electromagnetism which is necessarily equivalent but distinct. $\endgroup$ – user113857 May 4 '16 at 19:06
  • 2
    $\begingroup$ You both are absolutely mistaken. Electromagnetic waves carry energy through photons, virtual photons. If there was no charge there were no EM waves at all. And if there is an EM wave in the environment, there is an absolute change in intensity of photons in the environment on each point through the way EM wave propagates. $\endgroup$ – Alper91 May 4 '16 at 19:32

protected by Qmechanic May 4 '16 at 18:53

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.