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TL:DR; There is no "classical" explanation... The speed of light is given by: $$c = {1 \over {\sqrt {\mu_0\epsilon_0}}}$$ $\mu_0 =$ permeability of free space, $\epsilon_0 =$ permettivity of free space So this simple equation shows that the speed of light depends on the ability of free space (i.e., the vaccuum) to support electric and magnetic ...

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Speed of any wave is property of the medium through which it travels. So, it is property of empty space that electromagnetic waves travel at a certain speed (no more, no less). The vacuum is not a medium. With a medium the propagation speed is related to the bulk and/or Young's modulus depending on the wave type. That's why it's a property of the ...

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Mathematically, yes. Physically, no. Tachyons are a sign of an unstable theory and need to be dealt with. tachyons are these weird particles which move faster than the speed of light. Special relativity tells us that mass tends to infinity as an object's velocity tends towards light speed i.e. $$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}},$$ which as ...

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A photon doesn't see anything. You cannot attach a frame of reference to the photon and describe what the photon would see. If you were to go through with it, there is not really a time for the photon (infinite time dilation) which makes it difficult to describe collisions. If physicists describe the energy of a collision, they mostly choose the centre of ...

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The fact that speed is frame-invariant for photons doesn't prevent other parameters related to energy to be frame-varying ! For photons, energy relative to a frame would corresponds to the appearant wavelength. ( $E = h\nu$ )

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Is there any explanation/description of actual/physical mechanism of how the curving takes place? Yes. And it's similar to the same story for electromagnetism. Let's do that first. In electromagnetism you start by admitting that electromagnetic fields exist, even in vacuum, and that they evolve in time according to equations like $$\frac{\partial \vec ... 0 I can sense from your question that you are looking for a simple and basic explanation without jargons. I will give it an honest shot and will keep it really simple and classical. I am a classical thinker, so, I do not even have any more complex explanation. Hope more qualified and accreditted users will not frown upon the answer. Let me break the question ... 2 One of the premises of special relativity is that observations made in different non-accelerating (or "inertial") frames are equally valid, and that there are certain quantities that all such observers agree on. One such quantity is the proper acceleration \vec{a}^2-a_0^2. All inertial observers agree on whether an object is accelerating or not. This ... 2 I agree with Симон Тыран. "If you get accelerated forces act upon you. Therefore you can tell exactly that it is you who accelerates and not the whole universe" To answer the follow-up comments - Even if we can not distinguish between who has actually accelerated, the body that is accelerated, does feel the acceleration. And clock speed is influenced by ... 3 If you get accelerated forces act upon you. Therefore you can tell exactly that it is you who accelerates and not the whole universe. 2 Why do we have a universal speed limit? Is there a more fundamental law that tells us why this is? The more fundamental laws are causality and locality. Causality expresses the fact (or assumption) that effects cannot precede causes, and locality expresses the fact (or assumption) that fundamental causal relations are described by differential equations. ... 0 So, just to clarify your approach: You take events A and C occurring at the same location x'_a = x'_c but different times t'_a and t'_c in the primed frame. You also take event B occurring at the same time as A in the primed frame, t'_b = t'_a, but at the same location as C in the unprimed frame, x_b = x_c. For events B and C you then apply the ... 1 You seem to be asking two things, could a black hole's magnetic fields cool nearby matter, and could this cooling produce cold fusion. But maybe we should first ask whether "cold fusion" is a real thing. Nuclei contain protons and neutrons held together by pions. Fusion is when two nuclei become one. The barrier to this happening, is the positive electric ... 0 Note that \int^{\infty}_{-\infty}f(t)dt=\int^{\infty}_{-\infty}f(-t)dt and also \int^{\infty}_{-\infty}\frac{df(t)}{dt}dt=\int^{\infty}_{-\infty}\frac{df(-t)}{-dt}dt 0 It seems Wikepedias article "Two-body problem in general relativity" adresses the subject in detail. Obviously things are very much more complicated than I imagened as it involves GR and Einsteins field equations to which there are no closed form general solutions. To sign off on the matter I have found the following supposedly correct answers to my 7 ... 1 This an attempt to give a more detailed explanation since the question really is quite fundamental and has mostly been explained by referring to the impossibility of a co-moving observer detecting any effects of the non accelerated linear motion whatever the speed might be. Its the same as saying you must just trust Einstein without explaining the mechanism ... 2 To understand the matter-energy conversion you need to understand how quantum field theory describes matter. Quantum field theory postulates that for every type of particle there is a corresponding quantum field that fills all of spacetime. Particles are described as excitations of these fields. If you add a quantum of energy to a field the energy appears ... 0 If you are in a flat space(time), i.e. without any source of gravitation, in a spaceship, and you emit a ray of light across the spaceship, both the spaceship and the light will be in the same frame of reference. The frame will be inertial - not accelerated - and therefore the light will follow a straight path. Yet if a boost is applied to the spaceship, it ... 2 In physics, frames moving at the speed of light are not valid. What "list verse" means, more accurately, is that as you approach the speed of light, your time, as seen by a "stationary" observer, ticks slower. This is a well-documented effect that needs to be accounted for in all manner of applications, ranging from particle accelerators to GPS satellites ... 0 Usually the energetic process results in a very localized and extremely high energy density; e.g., a group of electrons are ejected from a target by means of a very short, intense laser pulse; some of the electrons will return due to the strong electrical attraction of their negative charges with the equally positive target; when the returning electrons ... 1 The proper time \tau for the journey for the traveler is 4 years. The time that passes for someone on earth will be \gamma \tau, \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}. For the earth observer, the distance to the star divided by the speed of the rocket is the time the earth observer experiences and equals \gamma \tau. So ... 0 From the perspective of Earth or Proxima Centauri, the spaceship will take$$t = \frac{d}{v}$$time to make the trip, where d is the distance traveled and v is the velocity of the spaceship. The passengers on the spaceship will experience time dilation, so they will experience$$t' = \frac{1}{\gamma}\left(\frac{d}{v}\right),$$where \gamma is the ... 0 When he fires he will see the white spot where it was the light time away so it will have moved into the right position when it is hit at instant later. By the way I dont understand fedino's remark about light speed having a vertical component. Are we talking about the photons momentum vector, the Poyinting vector or what? 0 This is the whole reason why we call it 'relativity'! There is no global unambiguous way to define an absolute velocity, so only the relative velocities matter. Everything else is just from a point of view. When one says 'Bob moves at 10km/h and Alice is stationary' they are implicitly defining a reference frame. Usuallay in day to day life, we define our ... 1 I have just watched one of Brian Greene's videos which gave this Blocktime impression as well, yet it is misleading. https://www.youtube.com/watch?v=VYZQxMowBsw Perhaps this may help you. Say we have a very long train that is 600,000 km long. Clocks are located at the opposite ends of the train, and there is also one clock located in the middle of the ... 1 You have\frac E{m_0}, the energy divided by the rest mass is \gamma=\sqrt{1-\frac{v^2}{c^2}}. The proper time is lab time divided by \gamma. Since you have a fixed E, as m_0 \to 0, \gamma \to \infty and the proper time goes to 0. For the last part, you are supposed to assume that an 11 MeV neutrino arrived 7 seconds before a 7 MeV ... 0 1) The S-matrix must be Lorentz Covariant, rather than Lorentz Invariant. That is, if \alpha and \beta the in and out states, they must BOTH transform as the corresponding free-particle states (free particle state \ne in/out state). S_{\alpha,\beta} = \langle \beta | \alpha \rangle = \sum c(\alpha,\alpha') c(\beta,\beta') \ S_{\alpha',\beta'}  (1). ... 3 Just one comment to the higgsss answer. Formally from the Wigner theorem we have that if there exist time shift symmetry, for which the scalar product of quantum mechanical rays is conserved,$$ \tag 1 |\langle \psi (t)|\kappa (t)\rangle| = |\langle \psi{'}(t+\tau)| \kappa{'}(t+\tau) \rangle|, $$then the symmetry transformation acts on |\psi\rangle as ... 3 (1) The operator U(\Lambda,a) is a unitary "rotation" in the Hilbert space corresponding to an inhomogeneous Lorentz transformation of the spacetime coordinates. When U(\Lambda,a)=\exp(iH\tau), it is an operator that adjusts the clock forward by \tau. Conceptually this is not a physical time evolution of the system. (2) A unitary rotation U in the ... 2 I'll reduce your question to its simplest expression: "What is mass?" And give you my best, simplest answer:"It is a measurement of how much an entity opposes acceleration or deceleration". I believe that in the end it all comes to that... 0 You can't deduce this, because all metrics with different \mu are invariant under Lorentz transformations. The best you can do is to choose the units in which you measure the invariant interval such that \mu is equal to whatever you like it to be. 0 I think that there's two things which you're leaving out: First, slanty lines. This was adequately covered by @Jaywalker but if you need a further synopsis: suppose in reference frame R_1 both A and B are at rest and A emits a laser pulse at B, describing the trajectory x = 0, y = c \tau. We transform to a reference frame R_2 moving in the ... 0 This is a very good question. For your second example, the light does actually reach B, but the paths it takes to get there are different for each frame of reference. If you are in A and B's reference frame, you could argue to be at rest and the light travels a direct path. Looking from the labs perspective however, the light moves along with the frame AB. ... 0 Always remember this in mind when dealing with special relativity: there is not a "better reference" If they are all frames in constant velocity,say Bob and Alice in your question. You can't distinguish them apart from "any" physical experiment. They are theoretically the same(and indeed they are!! What you are wondering is that "which reference" are you ... 2 The ray will hit the right hand wall at its midpoint, and both the passenger and the outside observer will agree on this. The equivalence principle states that the laws of Physics in a free fall and in an inertial moving elevator are the same, meaning that a passenger cannot expect an experiment to distinguish between the two (but deduce its motion ... 3 An observer outside the elevator would not see the laser beam as being parallel to the floor of the elevator. Imagine you are playing ball with a friend. You stand to the west of him and he runs from south to north as you throw him the ball. You throw the ball as he is directly east of you; in order for him to catch the ball, you must throw it northeast, ... 0 Let the Minkowski spacetime be the set \mathbb{R}^4 parametrized by coordinates \underline x = (t,x_1,x_2,x_3). You can visualize this set with a 4-dimensional cartesian diagram, with axes labeled (t, x_1, x_2, x_3). You can help your intuition just by drawing the 3-dimensional one (suppressing one of the space dimensions). A Lorentz ... 1 let's assume we have 2 rays moving at 180°. If the container ( and the background ) move instead of the photons, the container would have to move in the direction of both rays. But, it's impossible since we assumed 2 antiparallel rays. How can container move forward and backward simultaneously. 3 Your question is the one that Einstein pondered for long time and from which Special Theory of Relativity was born. He wondered what could happen if you travel at the speed of light how would you see a ray light. The problem was that according to Maxwell's Electrodynamics, explained light as oscillating E and B vectors along space and time, so as a ... 1 Elementary particles do not have consciousness, individuality or volition. They follow the rules of the boundary value solutions of the quantum mechanical equations they obey. The relativistic quantum mechanical mathematics have zero mass particles moving at velocity c, and in all valid frames massive particles move at velocities less than c. It is the ... 1 https://en.wikipedia.org/wiki/Geodesic_deviation This is due to tidal forces. The equations of geodesic deviation which are valid for objects that are small compared to the (radius) of curvature are the easiest way to qualitatively see that. If you plug in the coefficients for the Schwarzschild metric in the conventional distant observer form you will get ... 2 The above picture I drew to expand on Kostya's wonderful answer. Basically, imagine people who measure height of buildings in degrees of angle of the buildings' visibility from the certain fixed distance. This is not at all unreasonable if you fix the distance C large enough compared to the building heights'. However, for taller buildings you'd notice that ... 3 The time taken to travel to the planet, as seen by the bystander at point B, is the distance from A to B, is simply the distance divided by your speed. The time experienced by you doing the travelling is the time seen by B divided by your relativistic factor. It's just like moving from A to B in non-relativistic physics. So the faster you go, the shorter the ... 1 We won't be able to observe the light beam because photons don't interact with each other -- there is no way for light to bounce off the other photon so we won't be able to see it! Even if we imagine a particle travelling parallel to our path, both at the speed of light, if a photon were to bounce off of the particle, it could never reach our position ... 0 In physics, you always have to know the meaning behind the formula. The formula itself is meaningless, it's just a way of writing the meaning in a short and universal form. Even if you have two variables that describe the same quantity, but in a different context, equating them will probably lead to nonsense. In this case, E is precisely the REST energy of ... 0 I think it would help for you to first understand what does this E in E=mc^2 means. Yes it means energy. But more specifically, E in this equation is the energy you can generate by converting m amount of rest mass into energy. So the idea this equation describe is that you can convert rest mass into energy, or you can convert energy into rest mass. It is ... 0 The mistake is that bodies at rest don't have zero energy. The relativistic energy is given by$$ E^2=p^2c^2+m^2c^4 $$where p is the body's momentum and m its rest mass. Hence, if the object is at rest, p=0, so$$ E=mc^2 \rightarrow m=\frac{E}{c^2}\neq 0 $$and$$ W=mg=\frac{E}{c^2}g\neq 0 

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There's a new equation which serves as a correction to the previous one, ${ E }^{ 2 }={ { m }_{ 0 }^{ 2 } }{ c }^{ 4 }+{ p }^{ 2 }{ c }^{ 2 }$ Or as previously stated in a comment, objects at rest do not have $E=0$ or $m=0$. Misunderstandings including this have made people reluctant to talk about relativistic mass. If the given system is at rest, ...

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Some constant c with dimensions of a velocity is necessary because boosts do not commute, and therefore boosts must be done by dimensionless (mathematical) radians. The constant c converts velocities to radians. The constant c can not be infinite because that would make boosts commute. Giving an object a velocity (boosting) in the x-direction does not ...

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The spin group $Spin(3,1)\cong SL(2,\mathbb{C})$ is the double cover of the restricted Lorentz group $SO^+(3,1)$, cf. e.g. this Phys.SE post and links therein.

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