5
$\begingroup$

I know light is created when an electron moves to a lower energy orbital in an atom.

Is this original source of all light we see (sun / light bulbs / fire) or is there another way to produce light other than an electron changing orbital?

The Wikipedia article on light mentions light can originate from black body radiation. Is this a different method that produces light, or does it still just boil down to electrons and orbitals?

$\endgroup$
  • $\begingroup$ Ultimately, EM radiation is the result when you accelerate (or deaccelerate) electric charge, though sometimes this may involve changes in the magnetic distributions of a source. But since magnetism is a consequence of charge in motion, this still holds. $\endgroup$ – jim Aug 29 '19 at 20:08
  • $\begingroup$ So if I accelerate something that has a charge (a battery? an ion? an electric current through a cable?) fast enough, could I make light? And if so, is this ever a common source of light I could see in the real world? $\endgroup$ – default Aug 29 '19 at 20:56
  • $\begingroup$ @default You might be interested in en.wikipedia.org/wiki/Bremsstrahlung and en.wikipedia.org/wiki/Cherenkov_radiation and en.wikipedia.org/wiki/Sonoluminescence. $\endgroup$ – Chiral Anomaly Aug 30 '19 at 3:35
  • $\begingroup$ Check out the answer to this question physics.stackexchange.com/questions/5046/… $\endgroup$ – jim Aug 30 '19 at 5:59
5
$\begingroup$

There are two frames where the electric and magnetic fields and interactions are modeled, the classical one with Maxwell's equations describing the behavior of charged matter, and quantum mechanical ones where classical light equations can be shown to emerge from the superposition of photons(which are quantum entities)

In the classical frame, all accelerating charges radiate light. This carried out in the quantum frame says that all accelerating charged particles will emit photons. The energy for this is given by the force that induces the acceleration.

In addition, quantum mechanically, the transition of electrons to lower energy levels goes through the emission of photons , as you have stated, but not only electrons. In solids and liquids there are other energy levels involved that will allow in general charged particles moving in the electric fields of each other , to produce a plethora of photons from these accelerations, which may be just vibrations and rotations, i.e. changing energy levels in much more complicated solutions.

The temperature of the material follows from a radiation curve in most cases approximately fitted by a black body radiation curve, and this radiation is given off from the complex quantum mechanical interaction of the atoms and molecules involved.

So in lamp light, for example, there are both effects, the atomic and molecular spectrum transitions , and the tail of the black body radiation curve where a continuum of optical frequencies exist, and which dominates.

$\endgroup$
2
$\begingroup$

In atoms, the emission of photons is cause by an electron dropping to a lower energy level. In molecules, on the other hand, there are more mechanisms. A molecule usually have a variety of vibrational modes that can produce radiated photons. The situation becomes even complicated when you add rotations.

Then one can also have vibrations in a solid material among the constituents (atoms or molecules) of the material, such as crystal vibrations. These give rise to phonons, which can sometimes couple to photons (see Brillouin scattering) .

Black body radiation is generally not associated with electrons jumping between energy levels. It is generated by the thermal vibrations in the material. This is perhaps the most common source of light.

$\endgroup$
2
$\begingroup$

Electrons moving to a lower energy orbital is certainly one way that light can be produced - but as I see it, there certainly is another.

For example, the collision of matter and antimatter will produce energy, as defined by the equation $E=mc^2$ - or to be more specific, $E=mc^2/\sqrt{1-v^2/c^2}$. When the matter and antimatter collide - let's say a proton and antiproton - they will release energy that often times takes the form of a photon. It certainly isn't common, and most occurrences we observe are in labs, but it is a possibility.

And this is just one example - there may be others, but I guess the point stands that electrons moving into lower energy orbitals are not the only producers of EM radiation/light.

$\endgroup$
1
$\begingroup$

On a fundamental quantum level light can be emitted when fundamental particles that couple to the electromagnetic field move to a lower energy state. These fundamental particles are leptons (including electrons) and quarks (which make up protons and neutrons), and 'moving to a lower energy state' includes annihilation of particle-anti particle pairs. The state transitions are not limited to only orbital changes, although those will be the majority for everyday life. Most light you will see in everyday life comes from electron transitions. To see light from quark energy transitions you will need to look at the core of a nuclear reaction or at a cosmic ray. To see light from non-orbital electron transitions you will need to look at something containing free electrons, i.e. a plasma such as a lightning bolt or an arc discharge lamp. (I'm not sure how much of the light of those comes from bremsstrahlung, but probably some of it.)

There may be other mechanisms involving strongly curved space-time, i.e. black holes, which I don't understand but these are currently only conjectures AFAIK. And as long as we know that we don't have a complete theory of physics due to gravity not being part of the standard model, we also know that we may not have a complete picture of all the mechanisms that could generate light.

$\endgroup$
0
$\begingroup$

The primary motivation for this question was to understand the physical principles behind most of the light I see. After researching some of the material from the other answers, is seems that there are two physical sources for basically all light I might encounter in my lifetime:

  • Incandescence (Black body radiation)
  • Luminescence (electrons releasing light when moving to a state of lower energy)

https://en.wikipedia.org/wiki/List_of_light_sources

$\endgroup$
  • $\begingroup$ While this is a very useful distinction of light sources for practical purposes, on a theoretical level incandescence is also an example of luminescence. The impacts of atoms and molecules bumping into each other from their thermal motion excites electrons into a higher energy state. When the electron falls back to a lower state it releases radiation. $\endgroup$ – JanKanis Aug 30 '19 at 18:44
  • $\begingroup$ @JanKanis ok, that's what my original question was tyring to answer. Can you point me to a source that specifies that is what happening in incandescence / black body radiation? I was having trouble finding something that explicitly says that. $\endgroup$ – default Aug 30 '19 at 18:47
  • $\begingroup$ Good question. This is from my general physics background knowledge. For such basic stuff it is not always easy to find a proper reference quickly. I usually start from Wikipedia, the Thermal Radiation page has some of this, but doesn't include a real quantum level explanation and also doesn't have many references in the intro and overview sections. Unfortunately the problem of thermal radiation was an important driver in the development of quantum mechanics so Google gives me mostly that instead of a quantum explanation of thermal radiation. $\endgroup$ – JanKanis Aug 31 '19 at 10:12
  • $\begingroup$ So I'm afraid I can't help you with a good reference. $\endgroup$ – JanKanis Aug 31 '19 at 10:14
  • $\begingroup$ Also see this question which you may find interesting. Though it doesn't seem to have the reference you are looking for. $\endgroup$ – JanKanis Aug 31 '19 at 10:15
0
$\begingroup$

There is a spectrum of equilibrium radiation described by Planck $$u_{\omega}=\frac{\hbar \omega_{nm}^3}{\pi^2 c^3}\frac{1}{exp(\frac{\hbar \omega_{nm}}{kT})-1}$$ where the magnitude of the frequency $\omega_{nm}$ is determined by the formula $\omega_{nm}=E_n-E_m$, $E_n=-\frac{me^4}{2n^2\hbar^2}$ From the formula for equilibrium radiation, by integrating over frequency, the formula for the radiation of an absolutely black body is obtained $S=\sigma T^4$ The formula $E_n=-\frac{me^4}{2n^2\hbar^2}$ is general in nature and describes the classical radiation of an electromagnetic wave of an accelerated moving electron at any frequency. I do not agree Calc-You-Later with the fact that during anihilation it is impossible to apply the formula for the energy of an electron in an atom. Anihilation can be considered as a collision of two particles, and the Rutherford collision formula is valid for them. According to Rutherford's formula, the energy in a collision is determined by $E_n=-\frac{1}{2n^2}$ The proportionality coefficient is determined from the following considerations. The electron is located at a distance $a_0=\frac{\hbar}{m e^2}$ from the nucleus and the energy of transition to a free state $E_{\infty}-E_1=\frac{m c^2}{2*137^2}=\frac{e^2}{2a_0}$. During anihilation, the electron and positron approach each other at a distance $r_e=\frac{e^2}{m c^2}$ and the energy of transition to a free state $E_{\infty}-E_1=m c^2=\frac{e^2}{r_e}$ The origin of vibration is also a formula for the energy of an electron in an atom. The formula for vibration of molecules can be obtained from the ratio $$E_{n+1}-E_n=\frac{me^4(n+1/2)}{\hbar^2 n^2(n+1)^2}=\hbar \omega(n+1/2)$$ This formula is of fundamental importance. It reflects the fact of the absence of oscillations with a zero quantum number, since in the case of a zero quantum number the frequency tends to infinity. All these formulas have a common basis, they are due to the electromagnetic potential of Coulomb.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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