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The shortest way is the following one. Define $$f(\eta):= \exp{(\pm\frac{1}{2} \eta \sigma^3)}\:,\quad g(\eta):=\cosh(\frac{1}{2}\eta) \mathbb{I}_{2x2}\pm \sinh(\frac{1}{2}\eta)\sigma^3.$$ Now notice that $$f'(\eta) = \pm\frac{1}{2} f(\eta)\sigma_3$$ and $$g'(\eta) = \frac{1}{2} \sinh(\frac{1}{2}\eta) \mathbb{I}_{2x2}\pm\cosh(\frac{1}{2}\eta)\sigma^3 = \pm\... 1 You just need to use the series expansion and observe that$$(\sigma^3)^{2n} = \mathbb I_{2}$$and$$(\sigma^3)^{2n+1} = \sigma^3$$1 The first step is to expand the exponential as a Taylor series:$$ \exp \left( \pm \frac{1}{2} \eta \sigma \right) = I \pm \left( \eta \sigma/2 \right) + \frac{1}{2} \left( \eta \sigma / 2 \right)^2 \pm \frac{1}{3!} \left( \eta \sigma / 2 \right)^3 \dots$$Noting that \sigma^2 = I, this series can now be split into even and odd terms: the even terms sum to ... -1 It is not momentum conservation that forbids a zero mass particle to decay to two particles. It is the definition of "four vector", it's algebra, and it's "length", the invariant mass, the definition of inertial frames. Two particles will have an invariant mass whereas the incoming particle has mass zero. The "length" of the four vectors is invariant ... 0 Yes, there is a direct mechanical analogue in what engineers do with their fingers when they strain a cube of material by parallel-pipeding it. This is also the same transformation the "cross polarization" of a gravitational wave (GW) does. Suppose the edges of the cube are the x,y,z axes. You look down on the xy plane with z axis sticking in your eye. ... 1 I just want to compile here a curated list of all the things I've seen said about this question here on this site, both right and wrong. Question: Can a massless particle spontaneously split in two? Correct Answer 1:(if the particle is a photon) No, because this would violate conservation of angular momentum. The photon must have helicity \pm 1 (for ... 0 From a purely classical point of view: "Does this [classical mass invariance] stem from other basic principles of classical mechanics or is this an independent experimental fact?" No, this fact doesn't stem from other classical principles. Classically, this is an independent experimental observation. It boils down to Newton's 2nd Law for a ... 3 It's not strictly forbidden for a massless particle to split into two, but the "volume of available phase space" is zero. Remember that the rates for processes are computed by finding a so-called matrix element, and then integrating it over phase space. The allowed phase space here is of measure zero, because you have the additional constraint that the two ... 3 The idea from early treatments of special relativity that mass increases with velocity was superseded in general relativity and is better not used. It is a fundamental principle that the laws of physics are covariant - they are formulated using tensor (& vector & scalar invariant) quantities so as to be the same for all observers. Proper mass, or ... -2 This is a known misconception . There are 2 kinds of masses 1) inertial mass : Inertial mass is found in Newton's law of motion F=ma. This m expresses the resistance of an object to change its velocity. 2)gravitational mass: Gravitational mass is found in Newton's law of gravity. g = G*m/r^2.This m expresses the result of an object to gravity. At low ... 1 Here are two related abstracts that might be of interest to you. (I have not read the papers to comment.) Geometrical interpretation of optical absorption J. J. Monzón, A. G. Barriuso, L. L. Sánchez-Soto, and J. M. Montesinos-Amilibia Phys. Rev. A 84, 023830 – Published 17 August 2011 https://doi.org/10.1103/PhysRevA.84.023830 "We reinterpret the transfer ... 1 Very good question, which has on-going very active research. Short answer is YES, indeed quantum entanglement makes cause and effect indistinguishable, at least in a usual sense that we get used to in everyday. And this is already proven experimental fact,- Physicists demonstrate new way to violate local causality. So we can talk now only about does causal ... 0 I would say that you want a mathematical proof. You can see in Reign's answer a particular case, but here I will post a more general one. As you may know, a Lorentz transformation can be written as a 4-dimensional matrix, \Lambda, so that x' = \Lambda x. This matrix must fulfill this relation, \Lambda^T \eta \Lambda = \eta \end{... 2 [Disclaimer: Throughout this post I will be bracketing the ideas of general relativity because I think they needlessly complicate the story.] The invariance of the speed of the light is more of a statement about the geometry of the universe than it is a statement about light. Suppose I am in an inertial frame S with coordinates (t,x,y,z). If one event ... 0 There seem to be some fundamental mis-understandings. Put yourself inside the rocket. However fast you go with respect to any chosen other body in the universe, your rocket would work exactly the same way, irrespective of your (relative) volicity. What you will notice: the faster you go wrt this other body, the longer your rocket will need to use fuel (... 6 Yes it fires out normally from the ships perspective, because from ships perspective, the ship is at rest. No it does not go faster than light from the point of view of another observer, because velocities are not added the same way as in classical mechanics, but are added by relativistic velocity-addition formula. In classical mechanics, when ship fires ... 1 Let \{ \gamma_0, \gamma_1, \gamma_2, \gamma_3 \} be an orthonormal basis for Minkowski spacetime, with inner product$$\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu}$$and timelike basis vector \gamma_0. Every lightlike vector can be expressed as a Lorentz transformation of \gamma_0 + \gamma_3, to which only two spacelike vectors are orthogonal: \... 2 He then goes on to state that there are three mutually orthogonal space-like vectors. Sure, the proof is by example:$$\epsilon_1^\mu = (0, 1, 0, 0), \quad \epsilon_2^\mu = (0, 0, 1, 0), \quad \epsilon_3^\mu = (0, 0, 0, 1)$$where the time component is on the left. At one point he makes the statement that for given light-like vector, there are only two ... 1 Speed of light is constant wrt to reference inertial frame You always need two things to define physical quantities like distance and speed. Like in your example you have three objects, the sun, the earth and the rocket. If you delete the Sun you still have two observers for these physical quantities to make sense aka measured. If you delete earth the ... 2 SPOILER ALERT: The following is probably not the answer you wanted, but as a recovering ex-engineer I can't help myself: In practical terms, the bridge will not collapse, because the residence time of the relativistic car on the bridge will be insufficient for the applied stresses (the weight of the car) to propagate throughout the bridge structure (this ... 0 Here is the explanation I received on presenting this question to a maths and physics professor. I think it clarifies the 'soul' of the question (ie the point about the conceptual status of speed rather than proving the speed of light which is what most of the other answers seem to be doing): All speeds are relative and without reference to another object ... 2 He can't see a Euclidean metric, except by restricting to space. I think the writer did not mean what he appears to say, as quoted in the comment. I think he only intended to give two examples of diagonal metrics, not to give two examples of metrics which an observer can actually see. 3 If speed is not an entity in itself, but only dependent on other constant factors, how can the speed of anything (let alone light) be a constant? Am I completely missing something here? What you are missing is that distance and time themselves are not constant. Both distance and time depend in part on the velocity of the observer. You are on Earth, and ... 8 how speed can be a constant It's not, it depends on selected reference frame. The only exception is light speed in vacuum, which in any reference frame is c. But what if the Earth itself now starts moving? Now the distance is still changing, but how the individual speeds be calculated? In a pre-relativity times there was a Galilean speed addition rule ... 1 In special relativity speed of light is constant with respect to any inertial frame of reference. That is an axiom of special relativity. Having well defined frame of reference is all you need. It makes no difference if in your particular frame of reference there are other objects like Earth, Sun or some other thing. when a rocket leaves Earth and heads ... 15 My question is: If speed is not an entity in itself, but only dependent on other constant factors, how can the speed of anything (let alone light) be a constant? Am I completely missing something here? The question you are really asking is which is the more fundamental, speed or distance? Think about the distances in space. How are we to measure them ... 2 One has to distinguish between, on the one hand, the physical reality that exists outside of our minds and independently of how we think about it, and, on the other hand, human-made ways to describe this physical reality. Specifically, how speed is dependent on other quantities, how we measure it, etc. are reflections of how we think of it and describe it. ... 41 Speed of light is actually a pretty special case compared to how we typically think of speed (as far as I understand it). Movement is always relative to some frame of reference. In the case of a single isolated object, it's hard to really think about how you could have any frame of reference without at least a second object to measure the speed relative to.... 13 It is counter intuitive, this question or variants of it get asked a lot. Surely if you're travelling towards a beam of light it will appear to be travelling faster? The answer is it won't, every inertial observer measures the speed of light to be exactly c, regardless of their velocity. It is for this reason that the Galilean transformations break down ... 2 Maxwell's equations describe the electric and magnetic fields of a charge moving at any velocity, so long as you account for the fact that a moving charge acts like a current, and that changing electric & magnetic fields also can act like a source of magnetic and electric fields (through the \partial \bf{B}/\partial t and \partial \bf{E}/\partial t ... 0 Maxwell's Equations in their 19th C form are Lorentz covariant, which means they are valid after being boosted into a new frame of any boost-able velocity when transformed by a Lorentz transformation. (Note: v=c is not a boostable speed). It does take some effort to show they are unchanged, unless you put them in their "manifestly covariant" form. So for ... 0 That question doesn't make sense because only photons ( quanta of light ) or gluons ( quanta of strong interaction field ) can only move at lightspeed. And just to be clear Coulomb's law is not valid to describe chemistry because it is a classical theory which doesn't take in account the quantum effects.QED describes chemistry along with some other stuff. ... 0 The time dilation that you have described is how each observe the others time, so if they could "somehow" recede from each other at faster than light, they cannot observe each others time. So there is no meaningful answer. 5 Simple explanation: Its the only type of metric that conserves the speed of light. Mathematical explanation: Let us assume an inertial observer, O, measures the speed of light as dx/dt = c. Let us assume another inertial observer, O', measuring the speed of light in his coordinates, dx'/dt' = c. Naturally at this point one can ask, What is the ... 0 It's true that the specific set of coordinates that we choose is not Lorentz invariant, but when you integrate over all space, it doesn't matter. One however has to check that the measure is invariant. It's a nice excercise to show that, under a Lorentz transformation \Lambda the integral transforms are$$ \int \mathrm{d}^4x\, f(x) \mapsto \int \mathrm{d}^...

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Action is constructed as Lorentz invariant dencity integrated over all Minkowski space. $\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$ is Lorentz invariant dencity. $d^4x$ is Lorentz invariant measure of integration in rectangular coordinate system. If you wanna to choose some another coordinate system (curvilinear coordinate system), in general you need ...

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Time dilation is based on clock time. Clock time is the time we measure with when doing experiments. It is also the time according to our eyes. In clock time, the speed of light is constant. Most of Physics study is done using clock time. Things age, and happen more slowly when the local rate of Clock Time is slower that the rate of Clock Time it is ...

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How would this observer explain the tensile stress force he observes on the crystal which someone standing next to the crystal will not detect? Stress is the space-space components of the stress energy tensor. For the stationary observer the stress energy tensor is $$\left( \begin{array}{cccc} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 ... 0 The moving observer won't see any stress. This is essentially Bell's spaceship paradox. http://math.ucr.edu/home/baez/physics/Relativity/SR/BellSpaceships/spaceship_puzzle.html From the point of view of the moving observer it is the cube that is moving. In detail, at the atomic level, from the point of view of the observer, if you solve for the required ... 2 Let us simplify the example without loosing the essential elements. Suppose instead that the gates shoot a laser towards the center once the door is closed, and that at the center you have a bomb with a coincidence detector so that if both lasers hit simultaneously the bomb explodes destroying everything inside the gates (and the gates too, why not?). In ... 1 By little group scaling, the three-particle MHV amplitude for three massless spin-1 particles is given by$$ \mathcal{M}^{--+} =\frac{\langle 12 \rangle^3}{\langle 13 \rangle\langle 32 \rangle}~.$$This is not permutation invariant in the sense that if I exchange legs one and two I pick up a minus sign. The equivalent graviton amplitude doesn't pick up a ... 1 Scattering amplitudes should obey little group scaling, i.e. a transformation on the momentum that leaves it invariant, which for four-momentum is p_i^\mu = \lambda^\mu_{~~\nu} p_i^\nu. In spinor helicity notation, and for massless particles, this transformation acts on the spinors as \lambda_i \rightarrow e^{i\theta}\lambda_i. The scattering ... 1 When doing a complex shift in order to enact the BCFW transformation, we take, as you pointed out,$$ \hat{p}_1 = p_1 - zq~. $$Choosing to do a \langle 12] shift means we make a clever choice for q and work with spinors, which gives$$\hat{p}_1 = |1\rangle|1] - z|2\rangle|1] = (|1\rangle - z|2\rangle)|1] = |\hat{1}\rangle|1].$$The choice of q is ... 2 You are doing it correctly. In reference to your comment, to find the minkowski metric in cylindrical coordinates, it is easiest to transform the line element so:$$ ds^2=g_{\mu\nu}dx^\alpha dx^\beta \tag{1}, $$Where usually the minkowski metric is denoted by \eta_{\mu\nu} Which is$$ ds^2 = -c^2dt^2+dx^2+dy^2+dz^2 \tag{2} $$In Cartesian. If you ... 1 A little over a hundred years ago this was a red-hot topic. The Michelson-Morley experiment sought to detect any variation in the speed of light from different directions and found none. Subsequent refinements have produced the same result. 1 The result that light's speed in vacuum doesn't change due to relative motion among the light source, the light receiver, or a hypothetical background medium was a major discovery at the beginning of the twentieth century. You are correct that it's a non-trivial problem, but you're incorrect that a non-electromagnetic observation is needed. Most books on ... 2 Yes, the proper length, the spacial distance between two points measured in the rest frame. 1 Assume you have \mathbf p = p^{\alpha'} \mathbf e_\alpha. Then$$ \mathbf p\cdot\mathbf e_\beta = p^{\alpha'} \mathbf e_\alpha \cdot\mathbf e_\beta = p^{\alpha'}\eta_{\alpha\beta}$$so$$ p^{\alpha'} = \eta^{\alpha\beta}\mathbf p\cdot\mathbf e_\beta$$In particular$$ p^{0'} = \eta^{00} \mathbf p\cdot\mathbf e_0 = -\mathbf p\cdot\mathbf e_0$$And yes, you ... 1 It is correct that moving with the charge, there will no longer be an observed magnetic field. In fact, it hints on something much deeper. Instead of a point charge, consider an infinite line of current. From Ampere's law, we can calculate the magnetic field$$ \vec{B} = \frac{\mu_0 I}{2 \pi r}\hat{\phi}  If we then stop the line current, we have an ...

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