Hot answers tagged photons
47
Individual photons are very small and don't have much energy.
If you put a lot of them together in one place you can hurt somebody - by simply supplying enough power to melt an object (ask any spy on a table underneath a laser beam).
There is another very odd feature of photons. Although lots of them can provide a lot of energy and heat an object, it takes ...
18
I have a somewhat non-physics answer for you. If you allow me to broaden your question a bit to "why doesn't light kill or otherwise make all life on Earth impossible" the answer is that the Earth is in what we call "the habitable zone".
If the Sun produced so much light or light at such high energies that it would kill you, it also would heat the planet ...
10
Cute question!
For a neutrino with mass $m$ and energy $E\gg m$, we have $v=1-\epsilon$, where $\epsilon\approx (1/2)(m/E)^2$ (in units with $c=1$). IceCube has detected neutrinos with energies on the order of 1 PeV, but that's exceptional. For neutrinos with mass 0.1 eV and an energy of 1 PeV, we have $\epsilon\sim10^{-32}$.
The time of flight for ...
10
This question is more interesting than I thought at first. I like it. There are several different parts to an answer to this question; I'll just contribute a couple that have something in common: our bodies (and everything else, it has nothing to do with bodies) also emit photons about as fast as they absorb them.
On the macroscopic/thermal scale, we have ...
10
A general photon isn't too dangerous. Most photons that we encounter have the power to heat our bodies and not much else. The heat we absorb from photons daily isn't that much, so this is rarely a problem.
Now, an interesting thing about photons is that two photons of a lower energy do not make a single photon of higher energy (frequency). So a million ...
6
The word "stimulated" means that the emission of the photon is "encouraged" by the existence of photons in the same state as the state where the new photon may be added. The "same state" is one that has the same frequency, the same polarization, and the same direction of motion. Such a one-photon state may be described by the wave vector and the polarization ...
5
Because you can prepare a state with an arbitrarily long wavelength, hence arbitrarily low energy, photon. That's essentially the definition of a massless particle. If you put in an IR regulator, by putting the system in a box for example, a gap appears since there is now a largest possible wavelength. This can be mimicked by giving the photon a small mass. ...
5
Quoting from my copy of the 2nd edition of Jackson's book on Classical Electrodynamics, section 1.2:
Assume that the force varies as $1/r^{2+\epsilon}$ and quote a value or limit for $\epsilon$. [...] The original experiment with concentric spheres by Cavendish in 1772 gave an upper limit on $\epsilon$ of $\left| \epsilon \right| \le 0.02$.
followed a ...
4
Photons do not have mass, and they do have momentum. The issue is that $p = mv$ is only an approximation. A more accurate equation for the momentum of a relativistic particle like a photon is $pc = \sqrt{E^{2}-m^{2}c^{4}}$. Plugging in 0 for the mass and rearranging, we can see that a photon's momentum is its energy is divided by the speed of light.
4
Regarding your first question. When you are asking it, you should understand that it has an answer only in some model -- there is no universal relation that holds in every imaginable model of electromagnetic interactions. I personally do not know a model that would break the inverse square law in the way you want.
However, if you accept that electromagnetic ...
3
Let us clear some misunderstandings:
if an electron absorbs a photon to get exited to a higher energy level,
It is not the electron that absorbs the photon to go to a higher energy level. It is the whole atom, which is represented by a potential well with energy levels filled by electrons up to a point. A photon with the correct energy, i.e. an ...
3
Photons don't have mass, but they do have energy and momentum. And since they can be absorbed or reflected, they can transfer their momentum to whatever it is that reflects or absorbs. The amount of energy is proportional to the frequency $\nu$ of the light: $E = h\, \nu$, where $h$ is Planck's constant. The momentum is $p = h\, nu / c$, in whatever ...
2
Both you and freude are correct.
The units of Avagadro's constant are atoms / mole, but atoms is just a number and is dimensionless. That's why we write Avagadro's constant as mol$^{-1}$, and why the atoms units disappear from your final equation.
Response to comment:
You can write your expression for $\mu$ in various ways. In your expression:
$$ \mu = ...
2
Jinawee and dmckee have already given answers describing the bounds from the spherical capacitor technique.
A different, and more model-dependent, approach is to build and test empirically a theory in which the photon has nonzero mass. There are some theoretical difficulties involved, e.g., local gauge invariance is broken, and it's not trivial to show that ...
2
As Michael Brown mentioned in the comments, no one will explain this as well as Feynman (at least, no one we know of that's alive). But that doesn't mean your question doesn't deserve at least our attempt. So here is mine and I will try to keep this in the simplest terms I can.
(Aside: to all of the physicists reading this, I apologize in advance but in my ...
2
The "color" of a photon can be ultraviolet. Visible light is just a small part of the electromagnetic spectrum. Ultraviolet is the part of the spectrum with slightly shorter wavelengths than blue and purple. Many materials have a threshold wavelength in the ultraviolet. And for any material with a threshold wavelength in the visible, ultraviolet light will ...
2
I think it's deeply related to the fact that photons are bosons, ergo they follow the Bose-Einstein Statistics, or in this case they make a Bose-Einstein Condensate.
If you are not familiar with this exciting concept, I suggest you make a look at this Wikipedia article or any other statistical mechanics textbook you have around. Anyway, the two photons ...
1
Take a billiard ball sitting on a table. Glance off it another billiard ball. Part of the energy of the moving ball will be taken up by the scattered, and the resulting energy and momentum conservation will be subject to measurement errors, rotational possibilities, friction on the table and the balls etc. These are the normal measurement errors.
Now if you ...
1
The TL;DR version: even if we could form a synthetic event horizon, it wouldn't help us learn about the black hole interior.
The long version:
The phenomena you describe where light essentially orbits a black hole is called the "photon sphere" and it doesn't happen at the event horizon. The radius of a black hole, $R$ is where the event horizon is and the ...
1
One visible photon has a ridiculous amount of energy to harm us: about $2\times10^{-19}$ Joules. That's about 50 000 000 000 000 000 times smaller than the energy of a raindrop falling on your head (0.01 Joules). But in one second, a sunlight beam of the size of a raindrop sends $10^{17}$ photons which makes it about as powerful as a raindrop.
A sunlight ...
1
Xrays are photons, so as suggested in the comments above, the photoelectric effect could be exploited by using an appropriate material.
For example, SAXS (Small Angle XRay scattering) machines use many kinds of Si based detectors.
1
So I understand the electromagnetic spectrum -- electromagnetic
radiation is mediated by photons
Briefly, electromagnetic radiation is due to real (observable) photons; electric and magnetic force are due to virtual photon exchange.
The macroscopic electromagnetic wave phenomena we observe are due to an almost unimaginable number of photons, ...
1
The short answer is that when an electron in the valance band absorbs energy from a photon to become a conduction band (mobile) electron, both that electron as well as the hole "left behind" in the valence band can participate in an electric current. Thus, it is said, an electron-hole pair is created.
Where there was no mobile charge carrier, there are now ...
1
Each photon leaves its energy in the molecules of the screen. Destructive interference observed at the line x=1mm for example , means that the probability of finding a photon at x=1 is close to zero. Instead, the photon has very high probabiliy of depositing its energy at the construcive interference fringe.
1
Your teacher is right that photons have no rest mass.
You are right that photons have momentum.
The equation $p = m v$ is only valid for objects moving much more slowly than the speed of light, and therefore it is not valid for photons, which move at the speed of light. Therefore (1) and (2) do not conflict
The momentum of a photon can be found from its ...
1
The metal's threshold wavelength is a wavelength of light. So yes, you would use a chart converting wavelengths of light to the color to identify it. For some metals, the threshold wavelength is not visible light; it might be ultraviolet. But whatever chart you're using would identify the wavelength you have as either ultraviolet or visible, and which color ...
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