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You ask: Do atoms of a surface excite to reflect the light? but it's actually the other way round. Reflection occurs because the oscillating electric field of the light produces oscillating dipoles in the electrons in the substrate. These oscillating dipoles in turn radiate light isotropically and the reradiated light interferes constructively only in ...


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Short answer: Light is absorbed and re-emitted all the time. Colors and reflectivity all depend on the electrons in the material, and what the material is. Sometimes photons are converted to heat in a material. Other times photons pass through a material or are re-emitted (so it may look like they just "passed through"). Depending on what photons are ...


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Light of different color is understood as a wave of different frequency of oscillation. Matter in surfaces has characteristic behaviour for each of these frequencies, based on its chemical composition: electrons in the surface oscillate and absorb a lot of light when its frequency is around one of resonance frequencies of the matter; and the electrons ...


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In my answer to this question: (What is the sun's spectral series?), I give a very detailed answer about why mixing colours of light produces other colours and how it is purely a result of biology and evolution. I also delve a bit into the structure of the human eye and why, in fact, only three colours are necessary to reproduce all of the colours we can ...


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A "normal, healthy" human eye has two types of light-sensitive cells in the retina: rods ("color blind", but capable of sensing low light levels) and cones: cells that are sensitive to different bands. See this figure for their relative sensitivity (from http://hyperphysics.phy-astr.gsu.edu/hbase/vision/colcon.html) When you look at a spectrum of light, ...


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The cells in our retina that detect by frequency (read: colour) detect most strongly in three slightly different bands we know as Red, Green and Blue. To make a slight correction I would say an incadescent bulb is quite far from white, so I would rather proceed talking about sunlight on a clear day. The reason why sunlight appears as white as say a white ...


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It is possible that the assumption about a photon "it does not stop" may not be totally true. Light certainly interacts with the environment such as in an eye or solar generator. It may not stop in the traditional sense but may be captured both in an eye and in a black hole. For more information on the possible properties of light http://youtu.be/B1DCP4C4MnY ...


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So the key point to understanding this problem is to understand that it is the modes that contain information about the physical parameters of your photons (such as the momentum or angular momentum), and quantization is just a description of excitation of those modes. For instance the canonical quantization of the plane-wave expansion which you've ...


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by putting $p=\gamma mv$ and then get a value for $m$ (which will be 0 for a photon) and therefore rendering the equation to $E=0$ First, let's write this out in full (in 1D) $$p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}} $$ Then, solve for $m$ $$m = p\frac{\sqrt{1 - \frac{v^2}{c^2}}}{v}$$ Now, holding $p$ constant, see that the limit of $m$ as $v ...


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You can understand the expression by attempting the limit for $v\rightarrow c$ and $m\rightarrow0$. Notice that $\gamma\rightarrow\infty$ when $v\rightarrow c$. Therefore $m v \gamma$ is an undetermination of the form $0\cdot\infty$. From your expressions, you cannot say that $E=0$. The limit does not exist, and this implies that this expression is not valid ...


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That's because the relation $p=\gamma mv$ doesn't hold universally. As you just showed yourself, using this relation for a photon would lead to a contradiction because the energy of a photon isn't zero. A heuristic way of seeing why this relation won't hold for a photon is by recognizing that $$p=\gamma mv =m\frac{d x}{d\tau}$$ but a photon doesn't ...


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Is the photon physically oscillating through space as it travels? I wouldnt imagine so. Which periodic occurrence is referred to when one talks about the frequency of a particle? No the photon is not oscillating through space. It is an elementary particle of the standard model which is the quantum mechanical description of most of our experimental ...


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The easiest way to see frequencies is in interference. Imagine you have waves coming towards a wall. Imagine too that the frequency of the waves is way higher that what you can see. You cannot directly observe the waves, but you will see that the wall is wet a few centimetres over the surface. Now, instead of one wave, you have two coming from different ...


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Just another hint for Enos Oye: 2 u know that the photon emission in a resonant cavity can be greatly reduced by changing the size of the cavity? This suggests that the photon already knows when/where it'll meet a receiver.. its receiver


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Yes. Actually photons exert pressure on any surfaces exposed to them. For example, photons emitted by the Sun exert pressure of $9.08 \mu N/m^2$ on the Earth. Reference: https://en.wikipedia.org/wiki/Radiation_pressure


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Yes - you can even propel spaceships with it - Solar Sail Although Solar radiation pressure at the Earth is around 9E-6N/m2 while the thrust from a Saturn V rocket is 34 MN, so you would need a solar sail something like 2000Km on a side to get the same acceleration


3

Ah, all that talk about curved space-time. Well, there is a simpler argument. The fundamental axiom of general theory of relativity, "principle of equivalence", says: The effect of a homogeneous gravitational field is equivalent to that of a reference frame in uniform acceleration in the direction opposite to that of the gravitational field. All that ...


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If (and that's a big if) tomorrow we had a $70\sigma$ detection in a repeatable experiment of a particle that travelled faster than $c$, then one of several things would be true. 1) We would be forced to conclude that $c$ is not, in fact, the limiting speed of information transfer; everything based on this assumption would have to be scrapped (pretty much ...


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But what if tomorrow we happen to observe a particle X that travels with a speed V>c? We would have made the first observation of a tachyon. In special relativity, a faster-than-light particle would have space-like four-momentum, in contrast to ordinary particles that have time-like four-momentum. It would also have imaginary mass. Being ...


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Newton's formula is an approximation of how "gravity really works". We actually still don't know how gravity really works, but we have vastly refined our understanding of it thanks to Einstein's general theory of relativity. Gravity is simply a measure of curvature of a 4-dimensional manifold we earthlings call space-time. Local concentrations of mass or ...


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You have misunderstood the nature of a singularity. The singularity at the centre of a black hole is not a point in space. Instead it is a place where spacetime becomes infinity curved, and it isn't possible to describe what happens there. Well, it's not possible using General Relativity, but we hope some future theory of quantum gravity will explain what ...


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The problem here is that you need to think of it from the point of view of an observer. If we, on Earth, (or anywhere else for that matter) try to watch a photon (which travels along a radial null geodesic) approach a black hole, we will never see it 'enter' the back hole. It takes an infinite amount of time for the photon to reach the black hole. So indeed ...


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The equation you are mentioning is the gravitation force derived by Newton. This force doesn't apply to particles such as photons for two reasons: Photons are too small, and you can't use Newtonian physics to describe their properties. Photons travel too fast (their velocity is the speed of light) and at such a velocity Newtonian mechanics cannot be ...


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you are correct in your thinking, because you are using newton's laws. However, newton's laws were not completely accurate. For circumstances like ones with photons and extreme gravity like black holes, you must use Einstein's general theory of relativity. These laws say that all particles follow the shortest path along spacetime, including photons. But ...


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In Newtons theory of gravity photons are not affected by gravity (created by masses). So your conclusion is correct. But in General Relativity the curves of free objects like test particles or photons (geodesics) are determined by the space-time geometry. The geometry is described by the metric which is given by the energy and mass distribution of the ...


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Your logic is ultimately wrong because that equation doesn't reveal the true nature of gravity. According to general relativity, objects themselves bend space-time. Imagine space like a rubber sheet. If you stretch it and place a mass in the middle and roll a ping-pong ball past the mass, it will curve towards the mass. Similarly when space-time is being ...


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Energy conservation may be better stated as, The total energy, i.e. energy in the form of mass + all other forms of energy is conserved for an isolated system. This would mean that annihilation is simply an example of interconversion of energy from 'mass-energy' to 'light(electromagnetic) energy'.


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Photons aren't pure energy - they are a particle like all other particles. Admittedly photons are massless but then so are gluons, and indeed above the electroweak phase transition temperature so are all particles. So pair production from photons and annihilation into photons is just a scattering process like any other particle interaction. However if is ...


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Yes, it must happen at once. From the point of view of the EM field, you have created one large entity - so it can only react as an entity. You confuse the issue by assuming that the energy required to get to the next energy level for this object is 100 times that of one particle. The arrangement could be anything, and indeed, larger objects usually have ...


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I have another perspective on it. Interference is a wave phenomenon which nobody knows why and how it happens. I am just putting my thoughts and see if it answer yours. When a photon reaches slit it needs to decide whether to continue as a photon or convert to a wave. Conditions which are favorable for conversion is wavelength and phase correlation of nearby ...


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Yes, but it happens all the time. If other observers have absorbed all the photons which would otherwise have reached you, then they have blocked your view.


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Let say we have a candle. There are photons emitted from the candle in all directions, a finite number of photons per unit of time. If an observer is far enough away from the candle, the observer will only occasionally see (detect) a photon from the candle. So as far as I can understand the question, this is the type of situation referred to. At a far ...


0

Then how do we know that photons are moving at $c$ [...] the question isn't exactly "why the light moves with the speed of light?". So the question seems to be more precisely: "How do we know that the signal front of any signal that's been exchanged between (suitable collections of) electro-magnetic charges is attributed to the exchange of quanta of ...


1

Consider a very distant supernova; for example, suppose that the photons of the explosion have to travel a billion lightyears to reach us. If these photons had different velocities, then these differences would cause an accumulating difference in their travel time. Even if their velocities would differ by as little as a billionth, then the fastest, most ...


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There is no way to be 100% sure, but we can put upper limits on the mass. Massless particles don't have a rest frame, so it doesn't make sense to talk about time dilation in the photon's frame. A massive photon would have a rest frame, so you could eventually catch up to it and move alongside it. List of experimental limits on photon mass more ...


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In special relativity the energy is related to mass and momentum by $E^2 = (pc)^2 + (mc^2)^2$, where $p$ is the momentum. $m$ here is the rest mass of the particle, so for the photons case there is only energy from the momentum. The $E = mc^2$ you are likely familiar with ignores the momentum term, and hence only involves the rest mass. Photons are the ...


2

One of the tricky things with general relativity is that different observers may use different coordinate systems and measure very different things. The exterior geometry of any static spherically symmetric object is described by the Schwarzschild metric: $$ ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{r_s}{r}\right)} + r^2 ...


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Nothing in the universe is continuous. Matter and light are made up of particles, and physicists have strong reason to believe that even space is quantised. As for the mirror, photons travel from the lightsource to the object. There they are either absorbed or reflected, depending on their wavelength and the properties of the object at the point of ...


0

The approximate number of photons which strike a mirror and enter your eye over the period of a couple seconds is a massive quantity (in Mathematica): << PhysicalConstants` Convert[(1 Milli Joule 600 Nano Meter)/(PlanckConstant SpeedOfLight), 1] Output: 3.02047*10^15 Since this is in the quadrillions, for all practical intents and purposes, ...


1

I assume you don't mean the speed of light, but you are essentially asking: Will light escape that strong gravitational pull? If this is your question then first: Direct quote from wikipedia -> "An object whose radius is smaller than its Schwarzschild radius ($r_s = \frac{2GM}{c^2}$) is called a black hole. The surface at the Schwarzschild radius acts as ...


1

The speed of light is a constant regardless of where or when you measure it. The speed of the light as it leaves the star will be $c=299792458\frac{m}{s}$. The speed of the same light far from the star will also be $c$. Instead of slowing down like newtonian objects, the light will instead lose energy as it attempts to leave the star. This will correspond ...


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As Ross pointed out, two scenarios are possible: free electron / electron as part of an atom. They're treated in two totally different ways. Free electron: free electrons can't really "absorb" photons. They can collide with them, and some things can happen (this, for instance). Those types of collisions are described by QED and there are a bunch of ...


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Note that electron in isolation can never absorb or emit a photon. It is only a system of 2 particles (*) than can. P.S. 2 or more; Theoretical consideration of electron in a static field requires something to create said field.


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When an electron absorbs a photon, it remains an electron and the photon disappears. The electron energy and momentum are altered to account for the energy and momentum the photon was carrying. For a free electron, it will not be possible to balance energy and momentum simultaneously. There will have to be another interaction to make that work. If the ...


0

Quantum scientists came to the realization that they had to forget about the why question. Did you notice that the answers above only talk about what happens, but not about the mechanism by which it happens? There are some calculations that one can do about the energy of photons causing electrons to move off at a particular angle, and the energy of the ...


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the simple answer is that a quantum state has several variables (degrees of freedom), so if you measure only one of them and leave the others unchanged, then you detect the photon , change its state but do not destroy it completely. this is what they say in introduction Second, nondestructive detection can serve as a herald that signals the presence ...


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The problem with your idea is that each time the light reflects off the mirror it transfers some of its energy to the mirror (to increase the mirror's kinetic energy) and is red shifted as a result. So the thrust would fade as the light red shifts away to nothing. For obvious reasons the light can only transfer as much energy to the mirrors (in the form of ...


1

In undergraduate physics courses (c.f. MIT 8.03 course on waves), electromagnetic radiation is introduced as a pair of time-dependent oscillating magnetic and electric fields which satisfy the electromagnetic wave equation, as derived by Maxwell: $$\frac{1}{c^2}\frac{\partial^2 E(x,t)}{\partial t} - \frac{\partial^2 E(x,t)}{\partial x^2} = 0$$ ...


0

1) Basically, it seems what they implemented a version of the superdense coding theorem. What's that about? Well, the idea is the following: As always, Alice wants to send bits to Bob. Before they start, however, they already establish a connection by sharing a maximally entangled state $|\Phi\rangle=\frac{|00\rangle+|11\rangle}{\sqrt{2}}$. So, in ...


1

There can be a pulse made up of many photons whose temporal length is a femtosecond, or there could be a single photo - one energy quantum, with a single wavelength. So, if you know the spectral bandwidth of your short-pulse signal, compare that with the spectral transmittance of your beam splitter. I'd suggest that unless you're right on the cutoff edge ...



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