# Tag Info

## Hot answers tagged photons

8

The photon couples to all particles with electric charge or magnetic moment. This includes all of the quarks, the charged leptons $e,\mu,\tau$, and their antiparticles. It also includes particles composed of quarks and charged leptons: the proton and neutron (though the neutron only magnetically), the charged mesons, etc. Many electrically neutral mesons, ...

7

If you take an isolated hydrogen atom then the electron sits in well defined atomic orbitals that are eigenfunctions of the Schrodinger equation. This is a stable system that doesn't change with time. If you now introduce an oscillating electromagnetic field (i.e. light) then this changes the potential term in the Schrodinger equation and the hydrogen ...

6

At a basic level: The universe, in the beginning was very hot. So hot in fact that there were no atoms, only electrons and protons and neutrons and photons flying around. The photons were scatting off of the electrons and protons, as they interacted strongly because the electrons and protons are charged. The universe was much like the plasma you find in ...

6

Great question - partial answer. A lot of the opacity of most objects is due to a combination of scatter and absorption. At discontinuities in refractive index, light has a probability of either refracting or reflecting. When you add for example titanium dioxide particles to paint, you create many tiny scatter points. This is what makes white paint (and ...

5

Just think about it. The wavelength is the inverse of the frequency, so the shorter the wavelength, the higher the frequency. Photons in the visible electromagnetic spectrum are generated by an oscillation of charge currents. Consequently, because these oscillations are proportional to the frequency of the light wave, more oscillations yield a higher energy. ...

5

To properly understand what is going on you need to understand general relativity. Massless particles, like photons, travel on null geodesics and mass bends spacetime so the null geodesics are not straight lines. The problem is that neither you nor your teacher understand general relativity so this isn't a very convincing argument. But here is an argument to ...

4

No. Photons and electrons are fundamentally different and you cannot convert one into the other. it is incorrect to state that 'electrons and light are just electromagnetic radiation'; only light is electromagnetic radiation. Photons can be emitted and absorbed, and their number is not conserved. Electron number, on the other hand, is conserved, so ...

3

Since the photon reflects, its momentum changes: $p_{ph}'=-p_{ph}$. But total momentum of the system is conserved: $p_m+p_{ph}=p_m'+p_{ph}'$. Thus, the mirror will change its momentum. But, if the mirror has large mass, then it'll get very small energy from the collision. For zero-mass particle (photon) falling onto the mirror with mass $m_2$, the energy of ...

3

Actually you can argue that circular polarization is the "more natural" basis for a single photon. The photon carries one unit $\hbar$ of angular momentum, and circularly polarized light carries real angular momentum (an opportunity for me to mention one of my favorite experiments ever, using photon polarization to drive a pendulum). A single photon that's ...

3

I'm not very sure what the situation is with harder x-rays, but in the soft x-ray and extreme UV range doing this sort of thing is very hard, and as far as I'm aware there are no polarizing beam splitters in this range. In fact, starting with VUV at about 150nm, it is usually quite hard to get the light to do anything other than propagate in a straight(ish) ...

3

I will reply to Why isn't the CMB at the edges of the universe? Why is it flying around in the middle? The occurrence of space time and matter after the Big Bang happened to all points in our universe. The expansion of space happened at the same rate outwards for all points of the universe. All points of the universe 380.000 years ago had ...

3

Actually the framework where one can describe electromagnetic fields is a classical framework. When one is talking of photons phonons etc one is in the quantum mechanical regime where the concept field, is different. A classical field in physics: A field can be classified as a scalar field, a vector field, a spinor field or a tensor field according ...

3

Newton's law does predict the bending of light. However it predicts a value that is a factor of two smaller than actually observed. The Newtonian equation for gravity produces a force: $$F = \frac{GMm}{r^2}$$ so the acceleration of the smaller mass, $m$, is: $$a = \frac{F}{m} = \frac{GM}{r^2}\frac{m}{m}$$ If the particle is massless then $m/m = 0/0$ ...

2

If you are not familiar with the subject this link (wiki) can give you a basic knowledge of what happens in a simple atom of H, and then you can expand through related articles. All your 4 questions are related, and I added [0-] to the statement in your premise: Photons in all frequencies hitting an object are absorbed in different ways (absorbed, ...

2

Photons coming from changes in the energy level of an electron in a bound state ( atom,molecule,lattice) come in discrete energy slices. emission spectrum of iron Photons coming from the radiation emitted by accelerating or decelerating charged particles are coming in a continuum spectrum. Bremsstrahlung radiation Spectrum of the X-rays emitted ...

2

And so photons have mass No - photons don't have mass - they have momentum. And energy. But just because energy is equivalent to mass, doesn't mean they have mass. And they can only travel at the speed of light. A photon cannot travel at any other speed - so you cannot apply the Lorentz transformation to it. The Lorentz transformation applies to "rest ...

2

There are several things in play here. A photon emitted as a result of a transition of electrons between two well defined orbitals in principle has a well defined state However, the uncertainty principle limits how well that state can be known. You have shown yourself familiar with the energy-time formulation of the uncertainty principle; a transition ...

2

As my answer to How do photons know they can or can't excite electrons? explains, when a photon interacts with matter it is no longer just a photon. The photon/matter system has to be described by a wavefunction that describes both and that isn't separable into a photon bit and a matter bit. When the interaction is strong, e.g. in Bose-Einstein ...

2

One cannot ignore quantum mechanics when one is talking of photons. Photons, in addition to giving rise to transitions of bound states or being emitted by changes in energy levels of bound states ( atom, molecule,lattice) can also interact elastically with electrons and the fields that collectively build up around matter. Now a liquid which has a solute and ...

2

Yes,the photoelectric effect can be explained without photons! One can read it in L. Mandel and E. Wolf, Optical Coherence and Quantum Optics, Cambridge University Press, 1995, a standard reference for quantum optics. Sections 9.1-9.5 show that the electron field responds to a classical external electromagnetic radiation field by emitting ...

2

You'll need two things, the solar zenith angle $\theta$ and the air mass $AM$. The zenith angle is given by $$\cos\theta = \cos\phi \cos \delta + \sin\phi \sin\delta \sin h$$ where $\phi$ is your latitude, $\delta$ is the declination of the Sun, and $h$ is the hour angle. You should know your own latitude. The declination of the Sun can be found on a ...

2

The classical electromagnetic field given mathematically by Maxwell's equations can be proven to emerge from a confluence of individual photons, which photons are described by the Quantum Mechanical form of Maxwell's equations. Thus the classical wave is made up by zillions of photons with energy $h\nu$, where $\nu$ is the frequency of the classical wave. ...

2

In case you "run out of photons", you must switch to probabilistic description of quantum mechanics. Let's consider an extreme case: You have an emitter of spherical waves which radiates just one single photon. You place a lot of detectors some meters apart (with same distance) from the emitter. QM says that the photon propagates as a probabilistic wave to ...

2

Briefly, the formula $E=mc^2$ applies only particles at rest in an inertial frame of reference. Since there is no rest frame for a photon, no inertial reference frame in which a photon is at rest, one cannot apply the formula $E=mc^2$ to a photon. In more detail, the four-momentum of a particle has components $(\frac{E}{c}, \vec p)$ The four-momentum for ...

1

What you read is correct. I am not sure if those were the exact words of your teacher but according to the general theory of relativity, sun doesn't "attract" the photon (or any other body for that matter). In fact gravity is not even a real force. Let me briefly state what the theory of relativity has to say about gravity without going into the complicated ...

1

Absorption isn't an instant event. At the level of simple quantum mechanics, this system can be described as follows. Evolution of electron in crystal is governed by Schrödinger's equation. External electromagnetic field, namely the light which we shine on the crystal, is a periodic addition to the Hamiltonian. When you start shining light at the crystal, ...

1

'The mirror is given a momentum twice that of the incoming photon. As a mirror is typically quit heavy, lets say one gram. Its kinetic energy due to momentum it received will be extremely small. However, the photon will actually change its energy by the same amount, thus its wavelength changes, but not much.

1

The correct general equation is $E^2=m^2c^4+p^2c^2$, since $m=0$, $E=pc$. A photon's momentum is $p=\hbar k$, where k is the wavenumber ($k=\frac{2\pi}{\lambda}$). This would be consistent with $E=h\nu$ where $\nu$ is the frequency.

1

@Emilio Pisanty mentioned the process where "if you emit photons at an atom it will gain electrons": a photon creates an electron-positron pair in the Coulomb field of the ion, the electron is captured by the atom, and the positron goes free. I just wanted to give a reference (it is theoretical, and I don't know if this process was observed experimentally: ...

1

I think you have invented the zone plate, a kind of specialized flat circular diffraction grating that acts as a lens. It consists of a set of concentric transparent zones of decreasing widths. The width you have derived is the diameter of the first zone. Subsequent zones interfere constructively by allowing paths differing by integer multiples of the ...

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