# Tag Info

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All inertial observers (including A) will see light move at exactly $c$. So both A and B see the light move at $c$. But from A's frame, it appears that the light moves with a speed relative to B of $(c - 0.8c) = 0.2c$. Likewise the light in the other direction would be observed by A to be moving relative to B at $(c + 0.8c) = 1.8c$ As long as you are ...

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The speed of light can be derived from simple classical mechanics using the equations of Maxwell. This page offers such a derivation, utilising simple vector calculus. The speed of light in vacuum is seen to be $$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$ Where $\mu_0$ and $\epsilon_0$ are the permeability and permittivity of vacuum. The answer to the second ...

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We don't know why; it's a physical fact that's been established by experiment. One could suppose after all, any finite speed was possible; and this is what Newtonian Mechanics allows.

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To make a long story short, when we write $c = 299792458$ meter/second, then the number 299792458, while in principle just an arbitrary number that defines the meter relative to the second, can be interpreted as a physical measure of the human body just like the number 25 appears in the expression 25 kg/meter^2 for the threshold BMI that separates overweight ...

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First, we do not know why the speed of light is what it is. Of course its numerical value depends on the particular unit system, but the basic fact remains that light has some particular speed, and we cannot explain that any further. This is similar to other physical constants such as $\hbar$ and $G$. Dimensionless constants such as $\alpha$, the fine ...

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Basically yes. Firstly instead of your dispersion relation you must now use that $E = \hbar \omega$ and that $p = \hbar k$, then your new dispersion relation (valid for energetic electrons) is $$\omega = k c$$ and the result follows.

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No, there is nothing that is "keeping light from going faster". The local velocity of light in vacuum can not be different than the standard $c=3 \times 10^8{ m\over{sec}}$. There are two parts to my answer. 1) When light passes near you in vacuum, you will always measure the standard $c=3 \times 10^8{ m\over{sec}}$ using your local meter stick and ...

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I wonder why light does not have infinite speed. It does not have infinite speed because the experiment of Michelson and Morley has proven that it has the same constant speed in any reference frame. Moreover, any experiment on electromagnetic waves shows the presence of retarded potentials, that is, events need a certain time to propagate in space once ...

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If the speed of light were infinite then the laws of physics would be non-local. The way things work in our universe is that the state of a system in the future only depends on the present state of itself an its local neighborhood. E.g., the future state of the Earth one year ahead depends only on the present state inside a bubble of one light year ...

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The speed of light is determined (in terms of other fundamental constants) by Maxwell's equations. In particular, the speed of light $c$ must satisfy $c^2=1/(\mu_0\epsilon_0)$, where $\mu_0$ and $\epsilon_0$ are the permeability and permittivity of the vacuum. Because neither $\mu_0$ nor $\epsilon_0$ is equal to zero, the speed of light cannot be infinite, ...

1

To an extent, you can exceed the speed of light exactly as you described. Because of time dilation and the Lorentz contraction, if you measure an object at 100 light years away, then you accelerate to some speed really close to light speed, it will take you less than 100 perceived years to get there. While you're traveling, the object will seem to be closer ...

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The answer to your question depends on a lot of information you haven't given us. First, as Bill N. points out in a comment, if you start out with infinitely many balls, then you've got an infinite amount of mass, so throwing a ball will not cause you to recoil. So the first piece of missing information is: Where do you keep getting new balls? I'm ...

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So once i get to the speed of light. If i throw another ball why do i not accelerate faster then the speed of light? But there's the rub: you may get arbitrarily close to c, but you'll never reach it. As to why this occurs, I'm afraid you'll need to actually study some special relativity. One way of looking at it is that, as you go faster and faster ...

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Another set of experiments which support $E=mc^2$ are Compton scattering experiments. The mass-energy of the electron is an important quantity in analyzing these events, and the results are consistent across a wide range of energies for the primary photon and scattering angles. The energy of the secondary photon is given by $$E_{\gamma '}= ... 6 Emilio's answer was also the first that came to my mind, but I was not quick enough to post. However, even more precise experiments come from particle accelerators, and similar devices. https://en.wikipedia.org/wiki/Tests_of_relativistic_energy_and_momentum The power of the magnets in the LHC is determined by the relativistic mass of the particles going ... 5 There is a huge mass of experimental evidence that confirms the mass-energy equivalence. The clearest example that it happens is, of course, nuclear power, both in its explosive and civilian forms. If you want a detailed breakdown of all the experiments that corroborate it, I would recommend the entire archive of Physical Review C or a similar ... 4 Typical ballpark figure for the diameter of a pulsar is 10km. Therefore circumference is Pi multiplied by the diameter, which is about 30km. If it is rotating at (say) 1000 revolutions per second (for a millisecond pulsar) we get a velocity of 30,000 km/s or about 10% the speed of light 0 Photons are emitted and absorbed. Between emission and absorption there is no proper time because the spacetime interval of lightlike movements is zero. Light is generated by the absorption process, not by the travel of light. So there is no reason to fear that it should be dark. 1 You've mixed up which time dilates for which observer, and written yourself into a paradoxical corner: if going closer to the speed of light slows time for the object going that speed, and if time slowing down means going slower, then the conclusion is that "the faster you go, the slower you go." Which obviously doesn't make sense as you've pointed out. ... -2 If a photon carried a wristwatch, it would not tick. But us subluminals still see the photon move because for us, time has not stopped. -1 Time being a measurable quantity according to the beat of cesium atomic clock, it cannot be stopped. If the light ray hitting your eye is taken as a measure of time. Then you can absolutely stop the time by travelling at the speed of light. 2 Being a good physics question answer we'll ignore a few things, what the ball is made of, why it isn't tearing it's self apart and where we get the energy to spin it up seem like good choices since we're interested in what it looks like not how we build one. The trick to understanding how this looks is to realise that there are two different frames of ... 0 I remember wondering this exact same thing in high school. I did the math on it and figured that an infinitely thin bubble would take the shape of a peanut. The Lorentz Factor for 99,5% of C would shrink the equator to be 10cm in diameter, but if you look at the ball half way between the pole and the equator it will be traveling at 70% of the speed of light. ... 4 The real problem here is what is meant by a rigid body in relativity. In Newtonian mechanics a rigid body is described as a system of mass points that maintain their relative distances to each other. You can count the degrees of freedom by adding the mass points one by one. For the first mass point you have no constraints, so you get 3 degrees of freedom ... 1 You can't get the speed of the individual particles above speed of light. Once they get close, the amount of energy required to accelerate just a little is increasing very fast. So I guess that, if the ball surface is rotating already very close to the speed of light, pushing sideways the ball will hardly accelerate in that direction, as if it is very ... 1 It seems that none of the answers address the most interesting part of your question, that is: what happens to the side of the ball which is spinning in the "forward" direction? Obviously it cannot reach 99.5+99.5=199% of the speed of light. While I am not an expert on relativity, we know that from a stationary observer's point of view, time on objects ... 8 From an SR point of view, I suspect that two things will happen. If the ball is moving across the observer's field of view, rather than directly towards or away, it will appear to be a prolate ellipsoid of revolution due to apparent length contraction. Additionally, the rotation will appear to gain a phase shift caused by differential time of flight of light ... 1 How ever you will accelerate something, you will end up with photons as the mediator. This is the real reason why it is not possible to accelerate a body to the light velocity. What happens to a rotating ball? Suppose, you are looking in the direction of the light beam and the ball is rotating in a way, the left side is moving in your direction and the right ... 5 Intuitively, if you paint longitude/latitude lines on the ball, the linear movement will, if I remember the remnants of my special relativity class, result in a rotation of the ball. So far that's pretty tame. What can the fast rotation do? It will distort the coordinate system on the ball. What will happen to the colors? One side of the ball will be blue, ... 6 What happens if a super fast rotating ball accelerates near speed of light? The ball breaks up even before you accelerate it linearly. This is a bit of a problem with flywheels. In practice you just can't spin the ball at 99.5% of the speed of light. But never fear, we can still make progress. Assume we have a ball with diameter 1 meter and mass 1 ... 1 One difference that might help you is the following. A moving charge certainly has fields that move along with it, so I understand why you might think it would produce waves. But these fields (the electric field, at least) decay as 1/r^2, just like the field of a static charge. They also "propagate" at the speed of the charge, because they're following it. ... 2 For any charged particle in uniform motion there is an inertial frame in which that particle is at rest, and vice versa. So if the particle shed energy as EM waves due to uniform motion you would have the odd situation that a motionless particle would also have to shed energy as EM waves. Likewise if a motionless particle doesn't create EM waves then neither ... 2 Suppose you're on Earth watching the spaceship. The spaceship departs at time zero and returns at some time t = T. In between its departure and return the spaceship's velocity is given by some function v(t). The time measured by the astronauts on the spaceship, \tau, is given by:$$ \tau = \int_0^{T} \sqrt{1 - \frac{v^2(t) }{c^2}} dt \tag{1} $$... 2 In SR, it is not acceleration which causes time dilation, so whether you accelerate or decelerate makes no difference. What matters is relative velocity, and the calculation for the difference factor,$$\gamma = \sqrt{1 -\frac{v^2}{c^2}} produces a result which is independent of sign. So the answer is no. And actually, I'm not at all sure what ...

1

There is no difference between acceleration and deceleration, except in the difference in direction. This is no different from non-relativistic physics. With regards to your application, what you would do is to perform the calculation on Earth and you would use the twin-paradox and send the person who wants to know the answer on a long journey. He would come ...

1

If two particles are entangled at a distance of light years, does changing one affect the other instantaneously, or does it take years to occur? It isn't like each particle had its own individual state and that you could change one and by changing it change the other. What you had is a joint state of the combined system. For instance if you have ...

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If we want to observe the universe expansion with the most straightforward and direct way we have to measure redshifts of galaxies. The minimum distance where this effect would start to be observable is that at which the speed of recession is larger than the average noise speed, which is around a few hundred $km/s$. Given that the value of the hubble ...

0

This problem can be answered if we treat Light as a Particle. Due to Heisenberg's Uncertainty Principle, $\Delta x\Delta p\geq \dfrac{\hbar}{2}$, saying that we cannot exactly know the particle's Position and Momentum at the same time. In the case of a laser, a photon cannot have 0 momentum in any arbitrary direction so the photons cannot just go in one ...

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Backing up what zeldredge said, what you asked about is known as "relativity without light". According to the intro of this paper (arXiv link) for instance, the original argument was given as early as 1910 by Ignatowski, and has been rediscovered several times. There is a modern version due to David Mermin, in "Relativity without light", Am. J. Phys. 52, ...

0

in my opinion, the concept of Planck Units sorta settles what "constants" might meaningfully vary from those that it doesn't matter because we wouldn't know the difference. express everything in terms of Planck Units. then $c, \hbar, G, \epsilon_0,$ and $k_B$ do not even exist to vary. they are all $1$ except (from convention) $\epsilon_0 = \frac{1}{4 ... 1 You seem to be confusing "transmission rate" with "speed of propagation". In communications engineering terms, the latter is linked to the notion of network latency, i.e. how long your message "disappears" in the network before it shows up at the receiver. But both "information transmission rate" and "speed of propagation" are very well defined and precise ... 0 I am starting to doubt the premise of my question. That premise ,I think is that we are somehow "entitled" to have knowledge of an event (and that the universe is connected by these communications ). A universe without this possibility of knowledge of other things seemed "impossible" to me . But I am now starting to consider that there are events that we ... 1 Why don't electromagnetic waves need a medium to propagate? That's a "Why question". It's dangerous to ask "Why" in physics, because the answer is simply "they do". "How" is more interesting, and in this case it is very complicated. Just know that for a very long time, people really thought that electromagnetic waves needed a medium to propagate, ... 0 That a rigid body can move without medium - in vacuum - is without any doubt. The same is possible for gas or single atoms or electrons ... Beside matter exist energy. Max Planck explored the black body radiation and found, that energy consists of energy packages. Planck not believe, that the energy packages in his formula are real. Einstein stated, that ... 0 When you say that an object$A$has some velocity with respect to another object$B$, you are implying that there is a reference frame where$B$is at rest. This means that additional participants ($J$,$K$, ...$P$,$Q$...) can be thought of (or can even be actually identified) who were and remained at rest with repect to$B$, and with respect to each ... 0 As mentioned by physicist Brian Greene in his book The Elegant Universe, (See the_Elegant_Universe-B.Greene.pdf, pages 26 and 27, titled as "Motion trough Spacetime"), all objects are constantly on the move within space-time, and that they do so at a speed that is identical to the speed of light. For a view of Brian Greene's mathematical representation of ... 0 This is a rigorous a definition as I could find: In information theory, the Shannon–Hartley theorem tells the maximum rate at which information can be transmitted over a communications channel of a specified bandwidth in the presence of noise. It is an application of the noisy-channel coding theorem to the archetypal case of a continuous-time analog ... 7 When you say that an object$A$has some velocity with respect to another object$B$, you are implying that there is a reference frame where$B$is at rest, and$A\$ moves with the aforementioned velocity in that reference frame. It makes no sense to talk about the movement of a photon with respect to another, because photons have no rest frame. See: Why ...

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The light is not attracted by mass (at least by a small mass such as Earth). Hence there is no gravitational pull or acceleration by earth on light. Hence the velocity will be "c", that is all, no need to do a correction to it.

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How about using something which has well-defined speed but not at the speed of light? As an extreme example, suppose you have a conveyor belt, marked at regular intervals, and very well-calibrated speed. Then you know how long it takes for one tick-mark to go from the starting point to where the observer is. Synchronize the output pulse of light to a ...

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