# Tag Info

## New answers tagged special-relativity

1

So you have proper time and proper length. Use rapidity, $\eta$. For proper length $x$ and proper time $\tau$- $\eta=\sinh^{-1}{(\frac{1}{c}{\frac{dx}{d\tau}})}=\tanh^{-1}{(\frac{1}{c}{\frac{dx}{dt}})}$ From this you can calculate the velocity $\equiv dx/dt$

1

The other answers saying that true plane waves don't exist and are mathematical idealizations are perfectly true, but you can certainly have waves that are near enough to plane in reality to give rise to the "problem" you allude to. This is where we meet a subtlety to the oft-cited, but somewhat mashed assertion that nothing can travel faster than light. ...

2

Plane waves are fully compatible with special relativity, since they are Lorentz-invariant objects: $$\psi_k(x) = e^{i px/\hbar} = e^{i(Et-\vec{p}\cdot\vec{x})/\hbar}$$ You seem to be concerned with the fact that plane waves are spread through all space. But in fact, they're spread through all spacetime! They are perfectly periodic both in space and in time,...

1

Plane waves are not real, they are just a mathematical device. In quantum mechanics, particles are represented by wave packets, which do not have infinite amplitude and allow a collection of plane waves to group together by interfering constructively within a certain area and destructively outside that area. From Wikipedia Wave Packets In physics, ...

0

Apart from the 'yes' as the first word in @Daniel's answer, I think the rest of his answer is correct. EPR claimed some 'spooky action at a distance', and that it proved QM was not true. Bell's theorem and inequalities, and the experiments carried out, proved that there was no such issue, that QM does not allow a local hidden variable explanation, by ...

0

Supposing that you know well the Bell test experiments with all its devices. The Bell theorem made a QM prediction about the correlations between entangled photons. This prediction is different than the realists one. Bell expected that the realists may not do better than 50% with a full photons detection. In fact, not-canonical-QM theorists know how to ...

1

Special relativity predicts that either clock runs more slowly than the other, as judged from the other clock's system: http://www.people.fas.harvard.edu/~djmorin/chap11.pdf David Morin, Introduction to Classical Mechanics With Problems and Solutions, Chapter 11, p. 14: "Twin A stays on the earth, while twin B flies quickly to a distant star and back. [.....

1

What would occur if an electron at rest was accelerated to the speed of light? Any charged particle can be accelerated its speed increasing as more energy is supplied, but the limit of the speed is the speed of light. At a Lorentz factor ( = particle energy/rest mass = [104.5 GeV/0.511 MeV]) of over 200,000, LEP still holds the particle accelerator ...

1

Electrons can be accelerated to the speed of light (or practically to the speed of light). If you accelerate electrons to merely 5 MeV the velocity of 0.996 c where c is velocity of light, and yes if they are accelerated to that velocity they will emit gamma like radiation. Here I would like to clarify that the term gamma radiation is mostly used for the ...

2

Yes - these experiments have been conducted, most famously by Aspect et al., but also by others, see Wikipedia. They all observed violations of Bell's inequalities - Our world is therefore not local-realistic in the sense of Einstein. An extension of Bell's inequalities by Legett (Legett inequalities) holds for non-local realistic theories. Their violation ...

4

You're both half right. Your teacher is thinking of the invariant mass or rest mass(see footnote [1]), which is thinking in the more modern and much simpler paradigm of naming properties of a particle-like "thing" in a frame of reference at rest relative to that thing and you're thinking of the transformations of the properties to effective values as seen ...

1

First of All, I am confused by your question. Light travels at a constant speed in all reference frames. What is this "stationary observer in the Aether Field"? That doesnt make any sense to you or me. Forget about that. The experiment supports the non existence of a medium through which light moves in because of the null result. If you look back to ...

6

This is what happens when physicists try to do group theory but don't bother introducing the proper mathematical notions. There is no isomorphism between $\mathrm{SO}(1,3)$ and $\mathrm{SU}(2)\times\mathrm{SU}(2)$, the former is non-compact, the latter is compact. More around this apparently confusing topic can be found in this answer. Furthermore, using ...

4

No, in our own frame (comoving frame) we are not moving at all. Even though the CMB radiation is defining a preferred frame, to us it looks like the CMB is moving with respect to us - that's why we see a kinematic dipole in the CMB maps: http://scienceblogs.com/startswithabang/2013/06/28/our-great-cosmic-motion/

0

You are over complicating what appears to be your interest. You are asking in essence if there is an energy change in a particle as it moves in a gravitational field, i.e., a given spacetime metric. The answer is simple: except for singularities (i.e., what happens there, which is unknown, that's where any particle path ends in general relativity), a ...

0

Light emitted from the sun is gravitational time dilated, which means it travels slower than the speed of light as it is measured here on earth. I don't know if the difference is as great as the questioner assumes, but it would be significant. Einstein's relativity theories are PRINCIPLE-BASED theories. They are based upon established scientific ...

0

I'd like to add to Innisfree's answer, which depends on the orthochronous property of certain Lorentz transformations, so as to clear up the follow-up question to his/her answer:: Why proper, orthochronous Lorentz transformations can't change the sign of $x^0$ of a time-like vector? First let's clear up some terminology. An orthochronous Lorentz ...

1

Well, I'm not going to tell you my opinion, because that would be irrelevant to the actual science. But, what I can do is assess your premises and conclusions. What you should note first, and I'm unsure if you know this or not already, is that the four dimensions of spacetime are the three spatial dimensions and time. We can define a velocity through the ...

-2

I think inertia doesn't depend on speed, it depends on rate of change in speed, i.e. acceleration. The higher you accelerate the more will be the inertia. It can be understood by taking an example of a motorcycle, in which lower gear gives more traction than the higher one. The higher the acceleration you want the more traction is required due to inertia. ...

4

Update 1: 1) Note added in proof: The photon stress-energy densities obtained below more or less heuristically are identical to those obtained in more rigorous approaches from the electromagnetic stress-energy density tensor. 2) The physical reason why the stress-energy argument retrieves the detailed balance result in the OP, but is inequivalent to simply ...

-3

This problem is equivalent to Einstein's original thought experiment with the center of mass of a tubular spacecraft in an inertial reference frame emitting a photon from one end and reabsorbing it at the opposite end. This is the same thought experiment convinced an audience of Newtonian physicists that E=mc^2 by arguments related to center of mass. ...

0

I build up on Brian Trundy's answer, giving a more explicit derivation. Indeed, when you do a Lorentz boost your orientation changes and you cannot think of $dV$ transforming as a simple length. You need to think of it as the component of a four-vector $dV^\mu=u^\mu dV$, where $u^\mu$ is the four-velocity. In a static reference frame, the four-velocity is $u^... -1 There is a very quick and clean way of doing this, which is presented in Building an Orthonormal Basis from a 3D Unit Vector Without Normalization. JR Frisvad. J. Graphics Tools 16 no. 3, 151 (2012). Suppose you have a normalized vector$\vec n=(n_x,n_y,n_z)^T$, and you want a rotation matrix that will take the$z$axis into$\vec n$. (Here it's ... 1 What you are asking is just another version of the Twin Paradox. A twin leaves on a near lightspeed journey to space. When he comes back his twin is now much older than he is. In a Lorentz clock each bounce is a second. Since the speed of light remains the same, the distance goes larger between bounces. The distance of the light appear as longer to the ... 0 Anything moving at speed of light loses its reference system. This is the reason why your calculation yields strange results. Everything is multiplied by zero so that not only photons (which indeed have no length) but also long distances are reduced to zero. 2 Yes, there is a principle related to this, but we need to be precise in stating it. Suppose we take two points in spacetime$A$and$B$. For example in the twin paradox point$A$could be when the twins part and point$B$could be the point when the twins return. There are an infinite number of possible paths linking the two points. For example the twin who ... 1 It is shown in movie "The contact" where it passes just a second or so on earth, but during the same time, the astronaut records many hours of static. What people on earth saw was that the space ship crashed before even taking off, but astronaut experienced having made a journey to other words. As far as I remember, the concept is not clearly described in ... 1 I am only a layman, so don't take this answer seriously. This length contraction formula, and the whole Special Relativity in its original form, is for macro-sized, non-quantummechanical objects. Thus, the formulas work if you want to calculate the size of a spaceship nearing the speed of the light. And not if you want to calculate the size of photons with ... 0 While I'm sure it's a translation issue, your claim that the Moon (an astronomical object) is more than a light-year away is false. As a matter of fact most of our visible Solar System is only light-minutes or light-hours distant. By the way, while is may be a cultural issue, it is never correct to claim "we all know X". The wide variety of human experience ... 1 No, this is not a thing that can be directly observed -- for several reasons. First, in order to observe a difference, some concrete event would have to happen out at Alpha Centauri, which can be observed here with sufficient accuracy that the two observers can even form concrete impressions about when it happened. That's not a commonplace occurrence over ... 1 Consider two clocks$A$and$B$at the two ends of your pipe, and two identical clocks$A'$and$B'$right next to$A$and$B$. An observer stationary with respect to the pipe then synchronizes$A$with$A'$, then (all at once) synchronizes$A$with$B$using light and$A'$with$B'$using sound. Now, according to that observer, all four clocks are ... 1 The conclusion would be the same, but quantitative results would be harder to derive. Why the conclusion would be the same: The following relies on the assumption that the speed of sound in the train frame is known to be lower than the speed of light. Now imagine the train observer, in the middle of the train, sending out both light and sound waves ... 1 1) Usually special relativity is taught at the end of the semester, after the class got through rotations, which they, on average, don't understand; torques, which are pseudo vectors and for this reason blow their minds up... By the time they get to relativity they are done! If you teach the same kind of course, it is unavoidable, that they will get confused.... 8 Yes, this is indeed a missing piece of the argument. The 'real' argument is to just derive the Lorentz transformations from first principles, then note that$y' = y$for a boost along the$x$direction. But this isn't pedagogically useful, because usually people want to see proofs of time dilation and length contraction individually first, so they can ... 8 Even if nothing propagated at the speed$c$, it would still be a universal speed limit, and we could still measure it. In fact, it's not impossible that light has a (very tiny) mass in reality. If it does, that wouldn't change anything about special relativity. It would make teaching it even more of a nightmare than it already is, because we'd have to deal ... -10 Articles published in Science and Nature say the speed of light is not constant: http://science.sciencemag.org/content/347/6224/857 "Spatially structured photons that travel in free space slower than the speed of light" Science 20 Feb 2015: Vol. 347, Issue 6224, pp. 857-860 http://www.nature.com/nature/journal/v406/n6793/full/406277a0.html Nature 406, ... 6 Above all, speed of light is the speed of propagation of fields through space. While light may be slowed down when crossing matter, fields (electromagnetic fields, gravity) are always propagated at c. One of the consequences is the "speed limit for causality" mentioned by DavidZ and the speed limit for transmission of information. 0 The numerical value of$c$does not have any fundamental significance. Rather it is the number we get based on the experimental fact (according to the number & unit system employed) . If some alien civilization ended with some different value of$c$compared to us. Even that is not a problem. They will reach the conclusion that this is upper bound of the ... 46 It's the second one: the reason the speed$299792458\ \mathrm{m/s} = c$is special is because it's the universal speed limit. Light always travels at the speed$c$, whatever that limit may be. The reason there is a "universal speed limit" at all has to do with the structure of spacetime. Even in a universe without light, that speed limit would still be ... 1 Why both the observers should see the speed of light be exactly the same? To cut the long story short 'It is an experimental fact.'! The principle of relativity suggests that the laws of Physics must be the same in all the frames. Experiments suggest Maxwell's equations to be the correct laws of Physics. So a theorist can argue that everything that is ... 1 To carry forwards with John Rennie's response let us divide the metric by$dt^2$$$\left(\frac{ds}{dt}\right)^2~=~1~-~a(t)\left(\frac{dr}{dt}\right)^2$$ and with the generalized Lorentz gamma factor$\Gamma~=~\left(\frac{dt}{ds}\right)^2$means we have $$\Gamma~=~\frac{1}{\sqrt{1~-~a(t)\left(\frac{dr}{dt}\right)^2}}$$ This factor explodes at$t~=~0$and ... 3 This is essentially the same as the narrowing of the light cones that happens as you approach the event horizon of a Schwarzschild black hole, and it occurs for the same reason i.e. the coordinate velocity of light tends to zero as you approach the horizon. There is nothing physically interesting in this. It is a result of the coordinates we are using. If ... 0 No, they can't. In fact sound waves would lead you to Galilean relativity, and in all the cases where it differs from Einsteinian relativity they would lead you to the wrong result. This is a catastrophically bad model for Einsteinian relativity. As a simple example of where things go most horribly wrong consider a supersonic aircraft, and that the ... 1 No. In "STR" the speed of light is the same value no matter how the measurer is moving relative to anything else. For sound waves, the speed of the wave is measured relative to the medium through which the wave travels. Consider a sound source and receiver at rest relative to each other. The source and receiver have coordinated watches and have agreed on ... 1 When we say scalar, spinor, vector, and so on, field, we mean which representation of the frame bundle the field belongs to. Or in index notation, which spacetime indices the field has: none, spinor, vector, and so on. We can combine this with internal symmetries which are$G$-bundles for some gauge group$G$, for example$SU(2)$. In indices this is some ... 0 I would like to get some help concerning the definition of information [in the context of sending information] Suppose Alice and Bob are playing a game. The communication game. The communication game goes as follows: A referee flips a coin, and tells Alice the result. (Alice and Bob apply some strategy X.) Elsewhere, Bob tells a second referee what he ... 3 Let us consider an example and take the Weinberg-Salam Lagrangian: $$\mathscr{L} = i\bar{\psi}\gamma\cdot\partial\psi - m\bar{\psi}\psi$$ and let us adapt it to the case describing electrons and neutrinos as$$\mathscr{L} = i\bar{\textrm{e}}_R\gamma\cdot\partial\textrm{e}_R + i\bar{\textrm{e}}_L\gamma\cdot\partial\textrm{e}_L + i\bar{\nu}_L\gamma\cdot\... 6 I can't give an answer using fiber bundles, but I don't think it is important as the confusion is at a much simpler level. A field can be in different representations for different symmetry group. The Higgs field is in the trivial representation of the Poincarre group, that is, under Lorentz transformations,$\phi(x)\to \phi(\Lambda x)$, but in non-trivial ... 5 Whether or not you're allowed to include non-particle-number-conserving terms in your Hamiltonian has nothing to do (at the mathematical level) with whether your system is relativistic or not - it has to do with what Hilbert space you're using. If your Hilbert space takes the form$\mathcal{H} = \otimes_{i=1}^n \mathcal{H}_1$, where$H_1\$ is a single-body ...

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