# Tag Info

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One shoul think as c as a kind of space-time conversion constat, massless energy travel at this speed. Light and gravity are kinds of massless energy. The idea of E=mc^2 is that mass converts to energy like this.

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by putting $p=\gamma mv$ and then get a value for $m$ (which will be 0 for a photon) and therefore rendering the equation to $E=0$ First, let's write this out in full (in 1D) $$p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}}$$ Then, solve for $m$ $$m = p\frac{\sqrt{1 - \frac{v^2}{c^2}}}{v}$$ Now, holding $p$ constant, see that the limit of $m$ as $v ... 5 You can understand the expression by attempting the limit for$v\rightarrow c$and$m\rightarrow0$. Notice that$\gamma\rightarrow\infty$when$v\rightarrow c$. Therefore$m v \gamma$is an undetermination of the form$0\cdot\infty$. From your expressions, you cannot say that$E=0$. The limit does not exist, and this implies that this expression is not valid ... 6 That's because the relation$p=\gamma mv$doesn't hold universally. As you just showed yourself, using this relation for a photon would lead to a contradiction because the energy of a photon isn't zero. A heuristic way of seeing why this relation won't hold for a photon is by recognizing that $$p=\gamma mv =m\frac{d x}{d\tau}$$ but a photon doesn't ... 0 No, it cannot. The momentum of the particle must be finite in any reference frame where the momentum (for a massive particle) is given by $$\vec p = \frac{m\vec v}{\sqrt{1 - \frac{v^2}{c^2}}}$$ Note that the momentum is undefined for$v = c$or, to put it another way, the momentum goes to infinity as$v \rightarrow c$. So, just as it would be impossible ... 2 The speed of light in a vacuum is invariant: it is the same no matter what point you pick as "stationary". So if I'm on a train, and you're on the ground, and we both measure$c$, we'll get exactly the same number. The speed of light does not depend on the wavelength. Gamma rays travel at the same speed$c$as radio waves. The frequency$f$and wavelength ... 6 The speed of light is there for much more than to look cool, and in fact there are a number of derivations of mass-energy equivalence that shows why$c$is present; I will say that one basic reason is that the units of mass and energy are different, so we require at least some sort of constant factor to make the units work. I'll also say that we often use ... 0 "Now imagine the observer that is looking at the light which is traveling away from him in a direction he is looking." "What's wrong with this reasoning?" Well, this observer will see nothing - not even a point. It's just physically impossible to see light moving away from you. -3 Black holes have light not escape them Beacuse. A black hole has so much mass and gravity can't escape it, not even light! The fastest thing in the universe! 1 In the center of mass frame, if two particle beams have the same energy, using energy-momentum 4-vectors we get:$s =(P_1 + P_2)\cdot(P_1 + P_2) = (E + E, \vec{p}-\vec{p})\cdot(E + E, \vec{p}-\vec{p}) = (2E, 0)\cdot(2E,0) = 4E^2$Therefore the$E_{CM} = \sqrt{s} = 2E$For a fixed target ($E_b$= Energy of the beam and$m_t$= mass of the target):$s ...

1

I'm not particularly confident with experimental Physics, nevertheless I will try to answer to your interesting question. Not every scattering experiment in Particle Physics needs the acceleration of particles in opposite directions, there are a lot of experiment (for example Rutherford scattering) in which a fixed target is used. However in doing so the ...

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As noted within my other post at Time dilation in special relativity, here I have basically the same situation yet the box becomes a 300,000 km long spaceship. The basics are explained. However, the only absolute measure that exists within it, is that all objects are constantly traveling at the speed of light within the 4 dimensional environment known as ...

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Light from beyond the Hubble sphere (the place where recession velocity equals the speed of light) reaches us daily. I'm not good enough a physicist to come up with a nice layman's explanation for this fact, but it might help to think in comoving coordinates: This is a special coordinate system where the coordinate grid expands with space, ie even though ...

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I do not know if the following answer can explain each and every observation, but here goes : The expansion or moving away of galaxies is dependent on the distance between them, if something is moving away at some rate then previously since it must have been close, it must have moved away at a slower pace. While making astronomical observations, we are ...

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We know that some galaxies are moving away from us faster than the speed of light and we know it by measuring the redshift, but how's that possible? The following papers give good explanations: http://users.etown.edu/s/stuckeym/AJP1992a.pdf http://arxiv.org/pdf/astro-ph/0011070v2.pdf In summary, Hubble Law: $v = H(t)D$, where $v$ is recession ...

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Notice that the movement of light must always be treated as a local phenomenon, which in your thought experiment means the light is travelling only within the moving box. It is not travelling from the moving box to a stationary observer. Therefore, throughout the whole experiment the box is stationary for the light and no wall is moving toward or from it. ...

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1 - Actually, John is not correct. When each of the two beams return to the beamsplitter, half of each (for a "standard" 50/50 beamsplitter) will reach the detector. The other half of each beam will return to the laser. 2 - And yes, you do get interference. Depending on the exact relative lengths of the two paths, the interference will be either ...

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Before the escape velocity reach the light speed, photons, which are produced on the star can escape from there and therefore star can shine (it does not have to be a visible light, though). As you have pointed out, while the mass of the star increases the escape velocity rires too. At some point of time, this velocity, as calculated, is higher then the ...

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The key point that you are missing is that the speed of light is constant for all inertial frames of reference. If you are going $0.99\ c$ and you are holding a flashlight and you turn it on, the photons emitted from the flashlight will appear to you to be leaving the flashlight at exactly $c$ (the speed of light). The key point of special relativity is ...

3

Your understanding is spot on, as is PhotonicBoom's Answer. Something that might give you a bit more insight along the lines that you are thinking is if I answer your question backwards: the property we call "mass" (or "rest mass") is acquired by a particle with a rest mass of nought when that particle is confined in some way. If you look at my thought ...

2

Any body as you say with rest mass cannot fully reach the speed of light, as you would need to supply an infinite amount of energy to accelerate it to that exact speed. We do know thought that all massless particles do travel at the speed of light. Are they pure energy? They are, but then, everything is as we know from Einstein's relation $E = mc^2$. I ...

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What you're asking is essentially whether anything can rotate faster than the speed of light. Just like how it would take infinite energy to accelerate an object to the speed of light in a straight line, it would also take an infinite amount of energy to rotationally accelerate an object to the speed of light. In any practical sense, this tower would be ...

0

Then how do we know that photons are moving at $c$ [...] the question isn't exactly "why the light moves with the speed of light?". So the question seems to be more precisely: "How do we know that the signal front of any signal that's been exchanged between (suitable collections of) electro-magnetic charges is attributed to the exchange of quanta of ...

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Consider a very distant supernova; for example, suppose that the photons of the explosion have to travel a billion lightyears to reach us. If these photons had different velocities, then these differences would cause an accumulating difference in their travel time. Even if their velocities would differ by as little as a billionth, then the fastest, most ...

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Assumptions we start with an observer and a star at rest with respect to each other and 1000 lightyears apart. Both observer can agree on these facts. The observer then sets out toward the star in a reasonable fast starship, arriving after 10,000 years as measured in their original frame, recording the light from the star as he goes. The traveller ...

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There is no way to be 100% sure, but we can put upper limits on the mass. Massless particles don't have a rest frame, so it doesn't make sense to talk about time dilation in the photon's frame. A massive photon would have a rest frame, so you could eventually catch up to it and move alongside it. List of experimental limits on photon mass more ...

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In special relativity the energy is related to mass and momentum by $E^2 = (pc)^2 + (mc^2)^2$, where $p$ is the momentum. $m$ here is the rest mass of the particle, so for the photons case there is only energy from the momentum. The $E = mc^2$ you are likely familiar with ignores the momentum term, and hence only involves the rest mass. Photons are the ...

2

One of the tricky things with general relativity is that different observers may use different coordinate systems and measure very different things. The exterior geometry of any static spherically symmetric object is described by the Schwarzschild metric: $$ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{r_s}{r}\right)} + r^2 ... 1 I assume you don't mean the speed of light, but you are essentially asking: Will light escape that strong gravitational pull? If this is your question then first: Direct quote from wikipedia -> "An object whose radius is smaller than its Schwarzschild radius (r_s = \frac{2GM}{c^2}) is called a black hole. The surface at the Schwarzschild radius acts as ... 1 The speed of light is a constant regardless of where or when you measure it. The speed of the light as it leaves the star will be c=299792458\frac{m}{s}. The speed of the same light far from the star will also be c. Instead of slowing down like newtonian objects, the light will instead lose energy as it attempts to leave the star. This will correspond ... 0 if you consider a rigid rod and apply a torque at one end, by definition of being rigid, the whole body must start rotating at the same instant. But [...] I'm wondering if one can even define a rigid body in a relativistically correct manner In the strict sense indicated: one can not. An important clue in the argument has been stated by J. L. ... 2 Firstly, if we travel at 90% the speed of light, the relativistic effects will be too magnified to ignore. If you are travelling at a speed of 0.9c towards the star, according to your frame of reference, the star is coming towards you at 0.9c too. You'll observe that the speed of rotation of the star slows down by a factor of ... 5 We certainly would, or at least we would if we had telescopes powerful enough. However, better still, we could choose to watch its history unfold at an arbitrarily high fast-forward rate! Suppose our universe were classical/Newtonian/Galilean (or whatever you want to call it) but with a finite speed of light propagation (and let's just say that we're still ... 0 No. Speed of light in Vacuum isn't dependent on Solar System's motion. It's a constant. It'd be same even if the motion wasn't there. Due to your question type problems, we've even calibrated our scales to create Relativistic Physics. 4 No the speed of light in vaccuum is an absolute constant c = 299 792 458 m/s The way to add up relativistic speeds is: u' = \frac{u-v}{1-\frac{uv}{c^2}} to account for the constancy of the speed of light You cannot simply add them up. Edit: This also applies to normal everyday speeds. The reason we don't use this formula is because the speeds we are ... 1 Indeed, what we infer about stars from the light we see at the Earth is "old news". However for almost all practical purposes in stellar astrophysics this doesn't matter. The phases of a star's life last millions if not billions of years and most of the individual stars that are studied are within say 30 thousand light years of the Earth. The example you ... 0 Perhaps your question is whether the speed of approach of the two particles is 2c, is this so? Yes, it is 2c and this does not violate the principles of relativity, because such speed is not the speed of a particle, but it is just a derived value. On the other hand, the speed of a photon is c regardless of the inertial frame, and is calculated by the ... 0 This is quite a subtle issue. The more charges medium has in unit volume, the more it produces secondary EM waves. Common belief backed by the success of dispersion theory is that the relation \mathbf j = c\mathbf E is valid, where c is a constant dependent on the frequency of the wave, \mathbf j is current density and \mathbf E is total macroscopic ... 4 The simplest picture is that light always travels at the speed of light. But in a material it travels at the speed of light until it hits an atom. It is then absorbed and re-emitted in the same direction, which takes a small amount of time. The more this happens, the slower the effective average speed. The denser the material, the more atoms there are in the ... 0 by definition, any object is stationary relative to itself, regardless of any other frame. from the perspective of that object, no dilation exists, and time is experienced at it's full normal rate rather than a fractional amount as relative to another objects perspective. the rate of time doesn't increase to infinity, but rather can decrease by a fractional ... 2 Here is a diagram that describes what happens when a photon hits your eye. It is in two dimensions, one is time the other is space. The photon interacts with the electromagnetic interaction with an atom in the retina of the eye, where the electron which is bound in the atom goes to a higher energy level and the photon is absorbed giving its energy to ... 5 There is some discussion about this in the question How does a photon experience space and time?. You'll commonly hear it said that photons don't experience time, but this is somewhat misleading. Observers moving at different velocities have different coordinate systems, and these systems are related by the Lorentz transformation. If you apply the ... 0 The SR postulate is talking about an "observer". Your problem is that your question is failing to describe an observation process. The "photon of your look" will never be reflected by the photon you want to observe. By consequence, the way you described will never permit you to observe a photon. 0 You're messing up with simple time dilation. Time intervals are relative quantities. Two observers may not be agree with measured time intervals of an event. You see other moving observer's time dilated. Also, you see other observer's time dilated if she is deep in Gravity well than you are. Meaning, you find other observer's measured time interval more than ... 1 There is no absolute stationary object, an object may only be stationary with regard to an observer. If e.g. the relative velocity of an object is zero in our reference frame, we observe an object which is not moving with regard to our own reference frame. In this case Lorentz factor is 1, that means that there is no time dilation at all. -2 Lorentz factor, the speed-dependent dilation function, is$$\lim_{v \rightarrow 0} \sqrt{1-v^2/c^2} = \sqrt{1-0^2/c^2} = \sqrt{1} = 1. We see a considerable time dilation at v=0.

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The result of a research team in China is faked and wrong In Ref.[1], it is pointed out that Tang and coworkers’ result cannot be obtained from normal process and their result is faked. In Ref.[2] it is concluded that their result seems to be drawn based on the misinterpreting of the known formula and the less careful data analysis. Reference [1] Y. Zhu, ...

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