0
$\begingroup$

To be clear here, My question arises from studying Electromagnetic induction and polarisation and photoelectric effect. From here I've gained the knowledge that creation of a photon involves:

  • Electromagnetic Induction - The fact that a time varying spatial magnetic field which creates a closed loop electric field [Super Confusing as to why universe does that]
  • Polarisation - In the books, they show that light is oscillating electric field vectors and magnetic field vectors
  • Photoelectric Effect - Creation of light has something to do with an electron returning to a lower energy state, not saying that it's the only way.

So how does light get produced? I'm seeking in-depth explanation.


Some Context

The question arose when I was talking to my friend about a hypothesis of mine where I was saying that If I released a photon from , say a torch. I could, under some known error margin, say at what instant the photon was created but according to what I've heard multiple times, time stops at light speed. So, any photon at 'c' would basically never had experienced any time. So, from the perspective of the photon, its birth can't be defined. We seem to reach a paradox here, if I observe a time at which the photon was created but the photon has no notion of time, it has no notion of a beginning or an end. Then, I tried to expand the concept of the hypothesis to say that maybe in a similar sense, the universe doesn't hold a concept of beginning or an end. As we experience it's perspective, we can't determine its birth time using our maths which has been cultivate throughout this universe, and we also know that maths do break down when we try to solve equations for the t=0 time of Big Bang. NOTE: This talk was just to have fun, I just like physics and like to talk about the universe, however, I understand that I might have just annoyed someone by ignoring a dozen of laws and theorems that I'm yet to find out about. APOLOGIZES!


I'm expecting a somewhat dumbed down explaination, now don't get too carried away by that. I'm a high school student but I do love a bit of technical mumbo jumbo but I'm just asking to keep in mind that I'm in high school. Example:

Say we have an equation: x3+x2-x+44=0 Instead of saying the degree of equation is 3 maybe say the highest power in the equation is 3.

$\endgroup$
5
$\begingroup$

You are rambling in the text of the questions, so I will just address the title, which might attract a google search.

How is light produced?

The underlying framework of nature from which all classical theories emerge is quantum mechanical , based on special relativity and for large distances General Relativity, though gravity has not yet been definitively quantized ( only effective theories exist).

Light is a classical physics concept it is mathematically beautifully described by Maxwell's equations, and it arises from changes in electric or magnetic fields.

Photons are elementary particles in the particle standard model, and classical light emerges from a confluence of innumerable photons.That electromagnetic radiation (light) emerges from a superposition of photons can be shown mathematically, for those interested in quantum electrodynamics.

To understand how light is produced one has to understand the underlying quantum mechanical processes, which are many.

One is in transitions from excited energy levels of atomic or molecular bound states to a lower energy level, when there is an emission of a photon. The excitation energy to start with can be single other photons, or to energy supplied by the temperature of a wire, for example, where the tail of the black body radiation can have visible frequencies.This is the light coming from incandescent lamps, where the temperature of the wire is increased by the applied voltage to the point of incandescence.

A continuous spectrum of photons is supplied by the plasma of the sun, where a large part of the black body radiation, due to motion of electrons and ions generates photons in the visible range. These include compton scattering, i.e. the scattering of a photon on a charged particle and entering the visible part of the spectrum.

A fire has a combination of energy level change photons with plasma induce photons, etc.

The way these photons built up one by one the light that we see with our eyes is not a summation , as a summation of bricks make up a wall. It is a superposition of the quantum mechanical wave functions of the photons

photwvf

which builds up the classical electromagnetic field with its electric and magnetic field properties. The complex conjugate square of the superposed photon wavefunctions gives the probability of a photon interacting at (x,y,z,t) including in the retina of your eye to give the impression of "light".

To address a misunderstanding, you say:

We seem to reach a paradox here, if I observe a time at which the photon was created but the photon has no notion of time, it has no notion of a beginning or an end

The photon has no brain that can contain notions. It is always mathematically possible to define coordinate transformations, but one has to keep consistency, not mix up coordinate systems, as you are doing introducing your observations in your coordinate system (at rest) observing a photon that has a beginning and an end , with a framework going with the photon speed c, where due to the form of the Lorenz transformations there is no meaning in distance or time intervals because of the infinities introduced by transforming to such a coordinate system. I will copy this answer:

When we're travelling at the speed of light, or at the speed very very very close to the light, there's NOTHING USEFUL to talk about distance and time any more, and thus there's also nothing useful to attach any rest frame to it, because basically they (distance and time) don't exist any more. They are zero and not useful.

$\endgroup$
  • $\begingroup$ Just to clear up, the Quantum Mechanical Complex Wave Function you have shown, contains 2 other functions E<sub>T</sub> and B<sub>T</sub> which are in terms of r vector and t. It represents Time varying Electric and Magnetic fields right? $\endgroup$ – Alpha Mineron Oct 3 '17 at 17:17
  • $\begingroup$ Plus, that special relativity stuff you mentioned in the end, cleared up why I’m wrong but Jeez! I have a lot to study in the coming years. $\endgroup$ – Alpha Mineron Oct 3 '17 at 17:18
  • $\begingroup$ Yes, the complex function has an average E and an average B as a function of the four vectro (r,t) , and that is why E and B appear when a lot of photons are superposed and the complex conjugate squared of the total wave function is taken. $\endgroup$ – anna v Oct 3 '17 at 17:56
1
$\begingroup$

I will try to answer the main question.

What is light

Classicaly, light is considered to be an electromagnetic wave, which means that it has an electric and a magnetic component. Each component is perpendicular to the other and to the direction of propagation (of light), as you can see here!

Quantum mechanical nature of light

It is important to understand that a key feature of quantum mechanics is quantization. Energy is quantized, it has a discrete nature, it is not continuous. The quanta of the electromagnetic field is the photon. In a quantum system (e.g. an atom), the energy levels are also quantized. An atom cannot have any energy, it can have only specific amount of energy. Have a look at this! Now let's take two energy levels of an atom, $|1>$ and $|2>$. This first energy level has associated a lower energy state, let's call that $E_{1}$, and the second one has a higher energy state, $E_{2}$. Let's say the atom is first in the $|1>$ state, now to excite the atom means to force the transition from $|1>$ to $|2>$. For this we have give the atom an amount of energy equal to the difference between the two states ($E_{2}-E_{1}$). Because now the atom has a higher energy state it will eventually go back to the lower one because every physical system tends toward the lowest energy state possible. In this de-excitation, the atom will release the extra energy as a photon of light (if the transition is radiative). The energy of the photon is given by the difference between the two energy levels, as you can see here. $$h\nu=E_{2}-E_{1}$$ An excited nucleus will also emit light as gamma radiation. A different way to produce light is by annihilation of matter with antimatter. For example, if an electron encounters a positron, they will a form an unstable system called positronium and then they will eventually annihilate emitting two gamma photons. Also if charged particles are accelerated, they emit electromagnetic radiation. In particular, when the charged particle is decelerated, the radiation that is emitted is called Bremsstrahlung radiation. More on this here and here.


Edit

Electric and magnetic fields are really two aspects of the same thing. A magnetic field seen from an inertial reference frame can be seen as a combination of electric and magnetic fields from another inertial reference frame. You can make this transformation from an inertial frame to another using Lorentz transformation.

The polarization of light is a property that tells you how the field components oscillate. This image might help you visualize it.

The photoelectric effect happens when sufficiently energetic photons are incident on a material. The effect consists in the emission of electrons from the material. What happens when a photoelectron is emitted is that it absorbs all of the energy of the photon and escapes from the atom. The kinetic energy of the photoelectron is given by the difference between the energy of the photon and the work that had to be done in order to remove the photoelectron from the atom: $$K=h\nu-\phi$$

$\endgroup$
  • $\begingroup$ @annaV has already tackled the core of the question(The Complex Quantum Wave Function) but you have only provided an insight towards the concepts mentioned in the question which could still be helpful to others. $\endgroup$ – Alpha Mineron Oct 3 '17 at 17:23
  • 1
    $\begingroup$ I would suggest that you actually insert the images into your post and provide a summary of the main point you are trying to convey in a blockquote when you provide a link to external websites. Links tend to expire and become obsolete, so a quick summary while providing one will both give the read an insight as to what your point is and also secure your point in case the link expires. $\endgroup$ – Alpha Mineron Oct 3 '17 at 17:26

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.