# Tag Info

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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} - ... -2 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. ... -1 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 ... 1 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 ... 1 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 ... 1 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 ... 0 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 ... 1 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 ... 5 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 ... 0 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 ... 3 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 ... 0 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 ... 0 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 ... 4 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 ... 2 When a wave travels through a rope, the rope goes up and down, the position of all the 'rope-particles' changes, they oscillate and this makes up the wave. With light, it is indeed the electromagnetic field oscillating, but you shouldn't think of the arrows that represent that field in your first picture of light as 'extending into the rest of the space'. ... 0 It is very important to keep in mind that the double slit interference is a quantum mechanical phenomenon. The ideal would be to solve the quantum mechanical problem and get the wavefunction , i.e. photon , two slit field pattern, screen , detector and impose the boundary conditions to get the system wave function and see the probability of the photon ... 1 The key point here is the notion of distinguishability. If the two paths are made distinguishable, by associating different polarization to each path, then there is no way for the two amplitudes to interfere when they meet again because the "which-path" information is present. It doesn't matter whether or not the observer decide to actually look at the ... 3 You haven't said precisely what instrument you are using, but what you are seeing is normal for grating monochromators. Suppose we have light of one and only wavelength \lambda incident normally on a grating. Scattered light will be seen at the following angles, \theta:$$ d \sin \theta = m \lambda \qquad \text{for } m=0, \pm 1, \pm 2, \pm 3, \ldots ...

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Although it has been said in other comments and answers, it bears repeating succinctly: photons (as far as any experiment can tell) are massless and therefore always move at the universal, invariant speed of light. There is NO non-relativistic description of the photon. Even the "classical" description of light - Maxwell's equations - can be interpreted as ...

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People are addressing the speed question, but just to be clear: a photon can be very low energy. For instance, radio waves are much lower energy than gamma rays, even though both are made of photons (and, in vacuum, both travel at the speed $c$). What determines the energy of a photon is the frequency of the excitation (frequency of the corresponding light ...

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Refractive Index is when light travels more slowly in a medium. Here is an example of light being slowed down to 38 miles per hour. The speed of a photon does not affect its energy. It has zero mass, therefore zero kinetic energy. The energy it has is due to its frequency (color), and nothing else. (However, it does have momentum!)

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Intensity is an objectively measurable attribute of light. It is the rate at which energy is delivered to a surface. Intensity is energy delivered per unit time per unit area. The intensity of light is a measurement of photon irradiance, which is the number of photons delivered per square meter per second. You can measure intensity with a photoelement, ...

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