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No, all observers do not agree on the speed of a beam of light in a medium. However, that fact doesn't break special relativity's edict that the laws of physics are the same in all inertial frames of reference. The presence of the medium might make it more convenient to express the laws of physics in an inertial frame of reference in which the medium is at ...

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In some sense it is not the only possibility. And that's ignoring the obvious typos like missing signs or inconsistent use of superscripts and subscripts, and the fact that your equations require velocities to be dimensionless to even be dimensionally correct. In particular the first principle tells you next to nothing if you haven't specified which ...

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Light always travels at a (local) velocity of $c$, but light in a medium is not just light, and that's why its velocity can be lower than $c$. Light is an oscillating electromagnetic field, and when it passes though anything that contains charged particles (i.e. any matter made from electrons and protons) the electric field of the light interacts with those ...

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Let us Consider a Rocket of Mass $M$. As Rocket going up, its Fuel Will Decrease so its mass also Decrease, from Mass Equation. $$M=\frac{M_o}{\sqrt{1-\frac{v^2}{c^2}}}$$ If $V^2=c^2$ $$M=\frac{m_o}{0}$$ $M=\pm\infty$ $\text{So rocket's mass should be much higher which is not possible}$ $\text{Conclusion: It is not Possible for a particle to travel in ... 2 Honestly, nobody has the answer. I spent hours every day to search for answers after the question was posted. I never find a beginning of independent explanation to "why there is a speed limit ?". The constancy of speed of light was postulated after Michelson-Morley experiments. The theory is consistent after a century of observations. When comes the time ... 0 The speed of light in a vacuum (or the constant c) is what is the same to all observers in a reference frame. It is possible to move faster than the speed of light within a medium (see: Cherenkov radiation). 0 I've noticed that the closer one gets to fundamental physical theories, the ones that describe the most basic interactions in our universe, the more the equations all start to look like coordinate transformations. Sometimes these coordinates are in abstract spaces--the groups of the Standard Model of particle physics and the Hilbert spaces of quantum ... 2 The laser pen will point in the same direction for the moving observer, but the light ray will emerge at an angle. So the direction of the light ray won't be the same as the direction of the laser tube. To see why this is imaging looking at a single pulse of light as it travels along the laser pen towards the aperture: From your perspective, with the ... 2 Prologue This is a work in progress - in particular the third section on the arrow of time could be improved. All suggestions for improvements are welcome. This a community wiki answer so anyone should feel free to edit it. However if you want to make large changes, e.g. completely rewrite a section, please post the revised text as a separate answer and I ... 0 In GR, the speed of light is not constant, it varies with the curvature of space--time. So the constancy of this universal speed depends on space--time's having constant curvature. Which it doesn't, but this is locally a useful approximation, and in order to address the OP's intention, we will from now on assume that the Universe is a space of constant ... 0 In GR, the speed of light is not constant, it varies with the curvature of space--time. So the constancy of this universal speed depends on space--time's having constant curvature. Which it doesn't, but this is locally a useful approximation, and in order to address the OP's intention, we will from now on assume that the Universe is a space of constant ... 0 I must object by saying, I could go about questioning everything, but one must rather understand how? and why? things are the way they are. One must understand the Maxwell Equations, the problems and issues that come with it, special relativity, how it solves the problems, how to derive the Lorentz factor and how it breaks down at the speed of light. There ... 0 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 ... 0 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 ... 4 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 ... -1 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 ... 0 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. ($E = h\nu$) 0 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 ...

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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.

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