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Additional possibility: when light travels through a medium such as water, it slows down substantially. If light leaves water and enters air, it does in fact speed up when it enters the air. I have no idea how to calculate the time required for this "speed up" to happen, but if the time interval can be calculated or measured, you can calculate an ...


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The speed of light is constant in a vacuum. However, it can change direction in the presence of gravity so in a sense it does accelerate.


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Some background. You want to detect the image of an object. First, either 1) you illuminate it with some light source [a lamp, the sun] or 2) the object itself radiates light out [a star, a fluorescent sample]. Imagine to divide the object in many small parts (voxels): to reconstruct the image of the object, you need to detect independently the light coming ...


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Using a camera that can capture "Motion at a Trillion Frames Per Second", this can be done at the laboratory scale. The technique used has been called femto-photography. (Image credit to Ramesh Raskar, Associate Professor, MIT Media Lab) Of course a camera that literally takes one trillion full frames per second is totally impossible with today's ...


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Let's say you build a ping pong ball counter. It increments the count every time a ping pong ball hits the sensor. You throw a ball, and it hits the sensor: Detected! You throw a ball across the sensor from left to right... no detection, because you didn't hit the sensor. Your eyeball is a light sensor, which creates pictures from the light that hits ...


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If some of the light is reflected off the dust at such an angle that it is diverted to reach the observer, the observer will see that light. However, those specific photons reaching the observer will not reach B (unless they are reflected there by the observer). Similarly, unless the observer is at point B (which is not the case in the question as asked), ...


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My masters project was on something like this (though with hydrogen alpha emission lines for gas clouds between galaxies rather than dust between stars) and the answer in that case (and almost certainly in this one too) is that you can't see it because it's just too dim, though using hundreds of telescope hours can get you close (maybe). Again, this is not ...


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Why are electrons alike but photons not? Because it takes a given amount of an energy to make an electron: 511keV. That's the energy of an electron at rest. A fast-moving electron comprises more energy than an electron just sitting there in front of you, but if you were to stop it by removing the kinetic energy, its rest-energy is 511keV, and its mass ...


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It's a good question, and one that puzzled me for a while as well. However the answer is very simple. For a massive particle like an electron the total energy is given by: $$ E^2 = p^2c^2 + m^2 c^4 $$ where $p$ is the momentum and $m$ is the rest mass of the electron. Electrons can obviously have any momentum you want, so the total energy can be any value ...


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this is a description of an interaction between the electron and photons, which would collapse the wavefunction (right?). No this isn't right. As long as the system stays isolated, the interaction simply means that there are cross terms in the relevant Hamiltonian and that you have a two-particle quantum system, whose state space is the tensor product ...


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It would be possible to see the progress of photons through space if the light pulse were exceedingly intense, and if the dust cloud from which they reflect were positioned and shaped to reflect the light toward us. Rather than shooting a beam from Point A to Point B, it would be better if the light source were between us and the dust cloud, as light ...


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Let's look at where the electromagnetic interaction comes from in hydrogen. At first quantization you have a multiparticle system so the wavefunction is defined as $\psi=\psi(x_1,y_1,z_1,x_2,y_2,z_2,t)$ and the point is to write the Hamiltonian. And the Hamiltonian comes from the Lagrangian. For a single particle of charge $q$ in an external ...


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Sometimes we do, and the phenomenon is called a light echo. What you're looking at there is NOT moving gas. It's an "echo" exactly as you describe. The problem is that you need a pulse of light. If you have a constant stream of light, the "light echos" will be exactly like what you see in fog on earth.


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The following explanations are for electrons, but the are true in some kind for protons, positrons and antiprotons too. Every electron has the same electric field. It rotational symmetric. Additional every electron has a magnetic dipole moment (a tiny magnet) and an intrinsic spin. This spin is really a rotation of the electron around the axis of the ...


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Light changes wavelength in the presence of gravity I'm sorry Alex, but actually, it doesn't. You know how if you're moving away from a light source you see a redshift? The photon energy appears to reduce and the wavelength appears to increase? Well the photons coming from that light source haven't changed one iota. Instead you've changed. You also ...


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Actually, you have it backwards. The magnetic field isn't made of photons. Photons are made of magnetic (rather, electromagnetic) fields. To be specific, photons are ripples in the electromagnetic field. So, a magnet is surrounded by a magnetic field. If the magnet is not moving, then the field is stationary, and there are no photons. Wiggle the magnet, and ...


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I tried answering this by going to the XCOM database where you can get a calculation of the stopping power of elements and compounds. First - pick a few likely candidates. I found a table of elements with density which is a good place to start. The highest density elements are also among the highest Z ones: proton density ...


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The likelihood of a photon scattering off a particle in its path is proportional mostly to the amount of mass per unit area in the path of the photon beam. This means that, though higher atomic mass is important, it is really density that makes a good gamma shield. Wikipedia cites lead as being only 20-30% better at shielding than a similar shield made of ...


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However it did pass within Δx of the electron. The Δx is not the difference in space with the electron, as the electron is bound to a nucleus with a potential simulated by "an infinite potential well" . The Δx is related to the whole system, from the center of its mass as a possible location to start with. So the problem is : "photon + atom" as a ...


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The equation $E = m^2c^2 + p^2 c^2$ is restricted to Special Relativity. However, in classical physics we have $$ \vec{F} = m \vec{a}, $$ and $$\vec{F} = m \vec{g},$$ whence $$ m \vec{a} = m \vec{g}. $$ This can be written as $$ m \big( \vec{a} - \vec{g}\big) = \vec{0}. $$ From a mathematical point of view we have $$ \big( m = 0 \big) \vee \big( \vec{a} - ...


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It's quite wrong to assert that relativistic mass is obsolete. Gary Oas of Standford university did a study of a sample of the SR and GR textbooks. Of the ones published between 1995 and 2005 there were 60% of them which use relativistic mass. In this video, Alan H. Guth of MIT describes why the mass of a photon is not zero. ...


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Look at the paragraph "gravity and the photon" in this link: In the relativistic framework, i.e. large velocities, any energy is also a relativistic mass: For the photon this means the following equation: m is the relativistic mass of a photon with energy h*nu. Gravity attracts relativistic mass, and the photon has one. Read the link further to ...


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You misunderstand. All objects have some escape velocity, which is the velocity needed for anything (photons or matter) to escape from that object's gravitational field. And that's not the velocity it needs to maintain under some sort of constant thrust, but the initial velocity it needs to, shall we say, coast away from the object. For a black hole that ...


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The possible room for the nasty alternative scenario mentioned in WetSavannaAnimal's answer has been worked out in detail in this article. When the OPERA experiment due to a flawed cable connection appeared to have seen faster than light neutrinos, it was possible to calculate that this was impossible, because the room there exists for Lorentz invariance ...


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I'd like to continue from Jims Bond's already comprehensive answer. Suppose that we had conclusive proof of our observation of a $V>c$ and that we've ruled out alternative (5) of Jim's answer. I agree with Jim that the next most likely of Jim's alternatives is: 2) We would be forced to conclude that $c$ is not the limiting speed of information ...


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I'm going to try to teach you the right way to think about this, but possibly that will be very difficult to visualize. So I wanted to give you a starter course on what you're getting wrong. What you're getting wrong Your colors are indeed orthogonal states that can be measured differently. On your second screen you'll see green and orange light hit the ...


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In Minkowski spacetime, the spacetime interval of lightlike movements is zero. That means, from the (hypothetical) point of view of a massless particle such as a photon, it does not even exist one Planck time. At a proper time zero, any wavelength becomes meaningless, even if the physical process is the same that we observe. For the answer you have to take ...


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We don't really have a good perspective on what a photon "feels" or, indeed, anything about what its universe would look like. We're massive objects; even the idea of "we must travel at the speed of light because we're massless" makes little sense to us. But we can talk, if you like, about what the world looks like as you travel faster and faster: it's just ...


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After some investigation it turns out that my question is a bit of a convoluted way to ask for the momentum, which contains the information as follows. The momentum-energy relation $$ E(t)^2/c^2 - \vec{p}(t)^2 = m_0^2$$ with $\vec{p}$ the momentum, $E$ the engery and $m_0$ the invariant rest mass, simplifies to $$ \vec{p}(t)^2 = E(t)^2/c^2 $$ for ...


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First, webcams do use CCD or CMOS sensors, usually whichever chip is cheapest at the time. You can catch photons, but not reliably. In other words, for every photon that you catch, you will miss several. There will also be a noise signal, typically equivalent to many photons, Consider a CCD sensor. When a photon arrives, it may successfully excite an ...


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What Maxwell derived was from radio waves. Radio waves are modulated photon radiation. Electrons in an antenna rod get accelerated at once and emit photons. The density of this photons distributes in space. Detecting this radiation with a receiver one get a sinus wave form. The frequency of this wave has to do with the frequency of the antenna generator and ...


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It turns out that one photon states of the electromagnetic field can be written in a way such that the state "propagates" fulfilling Maxwell's equations. This is an exact model as I discuss this in more detail in my answer here. So we begin with a one-photon Fock state of the quantized electromagnetic field. Let's keep our discussion to one mode, so one ...


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No, it is sufficient for the photon energy to exceed the band gap. Any excess energy is transformed into kinetic energy for the electron in the new band. You get exactly the same effect when ionizing an atom - the excess energy simply powers the electron into a faster continuum state. You should also take into account that photon energies are never exactly ...



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