Tag Info

Hot answers tagged

8

The cosmic microwave background has a redshift of about $1100$, see here for instance. Keep in mind that the "surface of last scattering" that gives rise to the CMB in fact existed everywhere in space, it's just that the photons currently reaching us have $z\sim1100$.


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

Unfortunately X-ray and gamma mirrors are impossible to build the way you think - mainly because there is much less interaction with the matter comparing to UV - it will go through all materials commonly used for making mirrors. Even for EUV light (wavelength of 13.5nm) building effective mirrors is a royal pain. As wavelength of X-Rays is very small (down ...


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

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

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 ...


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

Yes it is possible but as BarsMonster points out it isn't like an optical mirror. X-ray reflectors are used in the construction of nuclear weapons and are critical to increasing the yield. How they work is the initial fusion reaction releases high energy radiation, this is then reflected back into the reaction mass increasing the energy levels of the ...


3

That's right, running away from a gamma source fast enough would shift them into the visible portion of the spectrum. It goes without saying that he'd have to run quite fast: $$\frac{\lambda_{\rm obs}}{\lambda_{\rm emit}} = \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$$ Picking rough round numbers for gamma radiation at $\lambda=10\;\rm pm$ and visible at ...


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 $m/m = 0/0$ ...


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

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 ...


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

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

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

For now I will only give you an overview of the ideas involved and show you how you should interpret the idea of a "local realistic theory" that cannot exist at the microscopic scale. Once you've read it, and if you feel you need more mathematical rigor to be convinced, then I will draw you step by step the proof of Bell's inequality (it is not the only one ...


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 ...


1

$E$ = $mc^2$ A much better expression is $E^2 = (mc^2)^2 + (pc)^2$, where $m$ is the "mass" (also known as "intrinsic mass", also known "rest mass", but most physicists nowadays just use "mass") of the particle and $p$ is the particle's momentum. This reduces to $E=mc^2$ in the special case of a particle with zero momentum, but it also reduces to $E=pc$ ...


1

It's all linked, so answering one will elucidate the rest. They don't stay in that same state, if that were the case then you have the problem you mention in point 3). After being excited, an electron loses it's energy by several mechanisms. Some of them radiate again "light" or more generally electromagnetic waves in other frequencies outside the visible ...


1

Transparent media are transparent because the incoming photon does not match any of the available energy levels to transfer its energy to the atom , or molecule, or crystal. A classical analogy is thinking of energy levels as various size sieve holes which allow only certain size of particles to go through. Not any matter of knowing or adjusting, but ...


1

A particle's interaction (with anything it can interact with) can be thought of as it making a measurement of the physical quantity associated with the interaction, This isn't true. All measurements are interactions, but not all interactions are measurements. When we talk about a Lorentz frame in special relativity, we're talking about a very ...


1

I don't know of polarizing x-ray beam splitters, and I suspect they don't exist, but I'm always very interested to be shown wrong. It appears from a cursory search that working x-ray interferometers have been built, and are conceptually more similar to neutron interferometers than to optical interferometers. For neutrons there are polarizing supermirrors, ...


1

The Faraday effect occurs because the phase speed of left-circularly polarized waves can be different than right-circularly polarized waves in certain media. You can think of a linearly polarized electromagnetic wave as a superposition of left and right hand circularly polarized waves. You can visualize how a left and right hand circularly polarized ...


1

Does it need to "slow down" to the new speed of light? Or does a new photon get generated? One has to keep clear in this case the difference between a photon and an electromagnetic wave. An individual photon is an elementary particle, its wavefunction is given by the quantum mechanical form of Maxwell's equations and the square of this wave function ...



Only top voted, non community-wiki answers of a minimum length are eligible