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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 ...

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You're not missing anything at all -- it's simply sloppy notation, and the people who do it just don't want to bother putting in the spacing correctly. However, you are missing something about the case of symmetric tensors. In this case, there is no ambiguity: an upper index transforms by contracting against the lower index of $\Lambda^{\mu'}_{\ \ \nu}$, ...

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We then defined $E$ via $E/c=P^0$ and claimed that $E$ is a conserved energy quantity based on the first 2 terms of its Taylor expansion. Taylor expanding $E$ around $v=0$ gives that $E = mc^2 + \frac{1}{2}mv^2 + \mathcal O(v^4)$, which means that at low velocities (ignoring the constant term, because constant shifts in energy are irrelevant in Newtonian ...

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The correct equation is $E=\frac{1}{\sqrt{1-v^2/c^2}}m_0c^2$, or, if you like ($\vec{p}$ is the three-momentum)- $$E^2=|\vec{p}|^2c^2+m_0^2c^4$$. This shows you how the velocity(momentum) of the object comes into the picture when calculating energy. More the $p$, more the $E$, as you'd intuitively expect. The kinetic energy is $$K.E=E-m_0c^2$$ and clearly $... 3 The idea from early treatments of special relativity that mass increases with velocity was superseded in general relativity and is better not used. It is a fundamental principle that the laws of physics are covariant - they are formulated using tensor (& vector & scalar invariant) quantities so as to be the same for all observers. Proper mass, or ... 2 Very good question, which has on-going very active research. Short answer is YES, indeed quantum entanglement makes cause and effect indistinguishable, at least in a usual sense that we get used to in everyday. And this is already proven experimental fact,- Physicists demonstrate new way to violate local causality. So we can talk now only about does causal ... 2 Yes, this is tricky. But it's basic. The Lorentz transformations apply to events,$(x,t)$pairs, which are fundamental. Lengths and times - your$L$and$T- are secondary constructs from events. You measure the length of an object by aligning the ends against a ruler, reading the distances off against the scale, and subtracting them. If the ruler and ... 2 The notation or formalism used to express a physical law doesn’t affect whether that law is covariant under some group of transformations. Maxwell’s equations, for example, are covariant under Lorentz transformations whether written in terms of\mathbf{E}$and$\mathbf{B}$or in terms of$F^{\mu\nu}$. But using Lorentz four-vectors and four-tensors makes a ... 2 Probably the simplest example of this is the first vector law you ever learned, $$\mathbf{F} = m \mathbf{a}.$$ If you define these quantities in the usual way, $$\mathbf{F} = \frac{d\mathbf{p}}{dt}, \quad \mathbf{a} = \frac{d \mathbf{v}}{dt}$$ then this equation is simply not true in special relativity. It's not even true if you artificially allow the ... 2 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 ... 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 ... 2 You are better to take conservation of energy and momentum as a fundamental empirical principle that underpins everything else in physics. You can prove them from Newton's laws, but that only covers classical mechanics. Actually conservation of energy and momentum is a fundamental principle contained in Einstein's equation in general relativity, and they can ... 2 The Lagrangian has to be constructed, demanding Lorentz invariance, symmetry properties, field content, and locality. The last, for example, tells you you must only keep a finite number of derivatives. The field content is what one means by the representation of the Poincare group the field lives in. This information is required to construct invariants that ... 2 In their own reference frames, both observers age at the same rate, on second per second. But because they have different notions of simultaneity, they think that the other ages more slowly. This is illustrated in this diagram 2 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. The only answer I find truly satisfactory is "Correct Answer 3", as given by knzhou above. Question: Can a massless particle spontaneously split in two? Correct Answer 1:(if the particle is a photon) No, because this ... 2 If the proper time interval between two events is the time as measured by a clock following that line, then surely that measurement is independent of path? I think the problem is that you are not understanding the intended use of the words. The path here is the path in 4-D spacetime, and is exactly the same thing as the world line. The difference in world ... 2 The mistake is that, given the time you measure is$t_{0}$, the time the moving observer measures is actually$t=t_{0}/\gamma$with$\gamma=1/\sqrt{1-\frac{v^2}{c^2}}$. You might have mixed up the frames in your definition. Together with$L=L_{0}/\gamma$you have $$L/t = L_{0}/t_{0}$$ 2 Your first mistake is to use length contraction and time dilation formulas, ignoring simultaneity (you need to use the full Lorentz Transform to solve these problems). I call this "gamma slinging", because it trivializes the whole idea of Special Relativity. You cannot just plug gamma in to non-relativistic equations and expect things to work. This is the ... 2 This has nothing to do with length contraction or time dilation . Just the ball in the second case has a 2nd velocity ( its horizontal velocity ) and this makes the path of the ball to be bigger , not time dilation , not length contraction. 1 1) Suppose the events$E(x_1, t_1)$and$E(x_2,t_2)$occur at the same$x_1=x_2$in some Lorentz frame. Then, the interval$\tau=t_2-t_1$in THAT frame is the proper time; it is the time elapsed between events measured at the same spatial coordinate. To make this more obvious, recall$ds^2$vs$d\tau^2$. 2) More properly, it is the time interval measured ... 1 When you extend Newtonian mechanics to special relativity you use the principle that the equations of SR must reduce to the Newtonian in the non-relativistic limit. By this principle you can find suitable new definitions of relativistic energy and momentum, as you describe. So, why are relativistic enery and momentum conserved? There are two slightly ... 1 They are not completely meaningless, but they can be meaningless as is shown in the particular case of quantum entanglement. I must correct one thing you say. Events are not moving at superluminal, or indeed any other, velocity. An event is, by definition, a point in spacetime$(t,\mathbf x)$. As it has particular time, it cannot be moving. Strictly ... 1 The mass energy relation states that mass and energy are equivalent and interconvertible. The law which governs the conversion of mass into energy is$E=mc^2$as you stated. But it seems that you have misunderstood the context of this equation. Think it in this way. The energy$E$is the maximum possible energy the mass can give due to mass energy ... 1 For low speed you could use Newton's relation between kinetic energy$E_k$and momentum$p$: $$E_k=\frac{p^2}{2m}$$ But for high speed this is no longer valid. You need to use the relativistic energy-momentum relation instead: $$(mc^2+E_k)^2=(pc)^2+(mc^2)^2$$ Solving for momentum$p$you get $$p=\sqrt{2mE_k+\left(\frac{E_k}{c}\right)^2}.$$ Applying energy ... 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 ...

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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. ...

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