New answers tagged

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Yes, Einstein's postulates entail symmetrical (reciprocal) time dilation, but in 1905 Einstein deduced, invalidly (in the sense that this does not follow from the postulates), asymmetrical time dilation - when the moving clock passes the stationary one, the former lags behind the latter. Nowadays you can often hear the same incorrect conclusion: "Moving ...


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It took me three days and many pages of calculations, but I think I've solved it.


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Proper time is the reference time all observers can agree on. Makes for calculating things much much easier. Since tau is attached to the particle the space differentials are zero Imagine we used some other reference time (which you are more than welcome to do). Then the spacetime line element would have dx, dy, and dz not equal to zero. Furthermore ...


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In relativity time is no longer a universal concept, it is a quantity specific to a frame of reference. It isn't meaningful to compare the "age" of two objects in two different frames of reference using a single "frame time." "Frame time" denotes time as measured in a specific frame of reference. This frame of reference could be that of either object for ...


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4-vectors have invariant length as defined by $$\vec{v}\cdot \vec{v} = g_{ab}v^a v^b.$$ Coordinate velocity $\frac{\mathrm{d}\vec{x}}{\mathrm{d}t}$ does not have this property. Proper velocity $\frac{\mathrm{d}\vec{x}}{\mathrm{d}\tau}$ does. Coordinate velocity is defined. It's not a 4-vector, so it's not that useful.


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Given a path $x(s)$ on a manifold, the velocity with respect to that path is defined taking derivatives with respect to the invariant parameter the path is described with, on the manifold (the arc length). In general relativity such parameter is the proper length $s$ (or proper time $\tau = \gamma s$).


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Yes and no. Coordinate acceleration doesn't need to be relative, but proper acceleration is always invariant.


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The thing about the relativity principle is just that: all non-accelerating frames are equal, in the sense that no inertial frame is more "real" or "accurate" than an other inertial frame. If two frames are moving in a Minkowskian manifold with a constant velocity relative to each other, it doesn't matter which one you choose. No matter what, all the ...


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The boost generators are First of all, note that You've chosen specific representation of Lie algebra generators of Poincare group, which is vector-like matrix representation. There are many representations in general (below I'll write about them). In Your question, You've chosen the matrix representation of Poincare group algebra generators in pseudo-...


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This definition doesn't depend on the metric signature convention. Note that in definition of $\gamma$ the metric doesn't appear anywhere. It is defined purely in terms of "3-vectors" and "3-scalars" measured by particular observer. So it is impossible for metric to appear here explicitly.


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The question has not been addressed fully so far. The question is better re-framed as, "Can the Uncertainty Principle be written in co-variant form?" and the answer is yes. For example I can consider the four vectors (x,y,z,ict) and (px,py,pz,iE/c) as conjugate and write the uncertainty relation between them.


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Yes and No. Yes because it can be "postulated" and No because, apart from all experimental subtleties of the notions involved, there are also inequalities outlining the region of its applicability. Point-likeness, classicality, etc. - all are approximations with the corresponding inequalities. Inequalities are an essential part of any branch of theoretical ...


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This turns out to have a really boring answer. We can find the two circular orbits by finding the maxima and minima of the effective potential, and we get: $$ r = \frac{L^2}{2M}\left(1 \pm \sqrt{1 - \frac{12M^2}{L^2}}\right) \tag{1} $$ where the $+$ gives the outer stable orbit and the $-$ gives the inner unstable orbit. Note that both orbits exist only ...


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The Earth is approximately a magnetic dipole, and your electromagnet is also a magnetic dipole. So you need to consider the force between two dipoles. This turns out to be quite complicated. If you're interested Emilio gives a thorough description of the physics involved in his answer to Force between two point dipoles. The bottom line is that the force can ...


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The constant speed of light is confirmed by the Michelson-Morrey experiment . But I don't know why it is constant. It's a measured quantity, just as G is.


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I'll address your issues with definition (1): $E$ is a function of $\vec p$ because $\lvert \lambda_{\vec p}\rangle\sim\lvert \lambda_0\rangle$ where by $\sim$ I mean that they are related by a Lorentz boost. That is, to "construct" these states, you actually first sort out all the states $\lvert \lambda_0\rangle,\lambda_0\in \Lambda$ ($\Lambda$ now denotes ...


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The speed limit c for massless particles is a characteristic of space. In accordance with Maxwell's equation, speed of light which is the speed of massless particles is reflecting the characteristics of vacuum vacuum permittivity $\mu_0$ and vacuum permeability $\epsilon_0$. $$ c = \frac {1}{\sqrt {\mu_0 \epsilon_0}}$$ Such a speed limit of Maxwell'...


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$\Lambda_{\mu\nu} = {\Lambda_\mu}^\sigma\eta_{\sigma\nu}$. It doesn't "do" anything. $\delta_{\mu\nu}$ and $\delta^{\mu\nu}$ are not tensors, as I explain at length in this answer of mine. The matrix elements of the identity are $\delta_\mu^\nu$, which you could have determined by thinking about the fact that the identity must send vectors $v^\mu$ to other ...


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There is no way to explain this without explaining relativity first. In Galilean universe (the "classical" physics, which is what most people intuitively assume and think about), light speed cannot be explained. Indeed, the Maxwell equations which is describe how light works, were the first clue that our understanding of space-time was flawed. So the real ...


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In the special relativity, as explained by Einstein, there are two possibilities. Either the speed of any particle is limited at all or not. Now if the speed is not limited, the Galilean theory pops out. But as we already know, that does not properly explain the transformation of velocities from one to another reference frame accurately when dealing with ...


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That light moves with a fixed speed in vacuum, in all reference systems is an experimental fact. Maxwell's equations fit so well all macroscopic electromagnetic data that the speed of light is fixed is not under question. It is inherent in the construction of the classical theory. Light is made up by a zillion of photons. Photons are elementary particles in ...


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I'm assuming you mean Bernard Schutz's book A first course in general relativity, and you're looking at section 1.6 Invariance of the interval. Schutz's book is an excellent introduction to relativity for readers who are mainly just curious and not intending to do research in GR, but Schutz does skip over things he thinks are obvious. In this case he ...


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Since I finally got the first one right, I might as well answer my own question. \begin{align*} [\gamma^{\mu}, S^{\rho\sigma}] &= \dfrac{i}{4}[\gamma^{\mu}, [\gamma^{\rho}, \gamma^{\sigma}]] \\ &= \dfrac{i}{4}\left( \gamma^{\mu}\gamma^{\rho}\gamma^{\sigma} - \gamma^{\mu}\gamma^{\sigma}\gamma^{\rho} - \gamma^{\rho}\gamma^...


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Distance contraction is a case of length contraction. The length of a rod in its rest frame is contracted for observers moving near light speed. Distance contraction is working in an analog way: The length of the rod corresponds to the distance between a mass object at the point of departure and a mass object at an end point. For this purpose we must ...


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First: You've chosen to express "the distance from me to the star" as some formula involving a quantity called $L_0$ --- and then you've asked us to tell you what $L_0$ means. But you're the person who wrote down this expression, so only you can know what you meant by it. Second: It's very hard to tell what you're actually asking, but I am pretty sure ...


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There is no frame in which "space" is at rest. Using the length contraction formula $$ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} $$ for the distance to the star is slightly misleading. The distance $L$ in your frame corresponds to two points, $x_0 = 0$ for you and $x_1 = L$ for the star, that you are observing simultaneously, say at time $t_0$. In the star's frame ...


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Suppose, as you said, we have a thin solenoid pointing upward, going through the middle of a loop one light-year wide. Then the three equations $$\nabla \times E = - \frac{\partial B}{\partial t}, \quad \mathcal{E} = -\frac{d\Phi_B}{dt}, \quad B = \mu_0 n I$$ together imply violation of causality, as shown in your question. Maxwell's equations are already ...


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Nothing is problematic with it. As FraSchelle says above—and is also true in the development of many other physics theories, in that they are, over time, purified of the scaffolding that helped construct them*—the original motivation doesn't affect the content of the developed theory. *cf. the top of p. 90 (PDF p. 91) of Stefano Bordoni's When ...


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Yes, Einstein knew about the Michelson-Morley experiment. Toward the end of his life, he saw it as increasingly important for relativity theory. Here's an excerpt from Clark, Ronald. 1971. Einstein: life and times. New York: World Pub. Co. p. 78: As Einstein said years later, talking to Sir Herbert Samuel in the grounds of Government House, Jerusalem:...


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Aside from simply looking at the Lorentz transformation, seeing a divergence and concluding "meh, it doesn't work", another way to gain insight into the divergence is through the statement: no finite sequence of finite boosts will get you to a speed $c$ relative to your beginning inertial frame. Imagine yourself in a spaceship with orientation controls ...


3

To see what's going on, it's enough to do this in two dimensions, with the Lorentz form $\pmatrix{-1&0\cr 0&1\cr}$. (I've set $c=1$.) The Lorentz group is the group that preserves this form. A typical element is $$\pmatrix{\pm\sec\theta&\tan\theta\cr \tan\theta&\pm\sec\theta\cr}$$ where $\theta$ runs through the open interval from $-\pi/...


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Lorentz transformations apply to objects with nonzero mass. For an object with mass, it would require an infinite amount of energy to reach light speed.


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The difference arises because you are confusing the three spatial components of the 4-momentum, which are in fact $p^j = \gamma\,m\,v^j$ with the three components of the Newtonian momentum $m\,v^j$. If you write: $$\frac{p^2}{m^2\,\gamma^2} = v^2$$ and solve for $v$ in terms of $p$, then substitute back into the Watkin's lecture notes, you'll get your ...


1

The expansion is not that hard. We have $$ K~=~(\gamma~-~1)mc^2~=~\left(\frac{1}{\sqrt{1~-~v^2/c^2}}~-~1\right)mc^2 $$ where we do the binomial expansion on the Lorentz factor $$ K~\simeq~\left(1~+~\frac{v^2}{c^2}~+~\frac{3}{8}\frac{v^4}{c^4}~-~1\right)mc^2 $$ $$ =~\frac{1}{2}mv^2~+~\frac{3}{8}m\frac{v^4}{c^2}. $$ I do not know how you get the $m^3$ in the ...


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You are quite correct that we don't have to bring the twins together at the same point in space. It's enough to make their relative speed zero i.e. bring them into the same inertial frame. The twin paradox generally involved bringing the twins together just for simplicity. When two observers are at the same point in space they can directly compare their ...


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The question may be about the covariance or otherwise of temperature, in which case have a look here. As well, have a look at the paper "Temperature in special relativity" by J. Lindhard, Physica Volume 38, Issue 4, 5 June 1968, Pages 635-640.


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This is a simple variation on the so called "Twin Paradox", which is not a paradox in the logical sense (i.e. not a logical contradiction). Each cycle of the oscillator's motion is like the journey of the spacefaring twin. One possible cycle on a spacetime diagram is drawn below (source: Wikipedia "Twin Paradox" Article with my own additions). The ...


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Lorentz transformations are transformations between inertial frames of reference, i.e. frames of reference that move with constant velocity relative to each other. An oscillation frame as you described it is not an inertial frame of reference, so Lorentz transformations do not apply here.


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In the kinetic theory of gases, you only really define the temperature for molecules that are in constant, random, and rapid motion. So if you have a container with a gas at temperature $T$ you don't change the internal energy of the gas by uniformly moving the container. Uniformly moving the container gives all the molecules a non-zero average motion, but ...


1

No, many other couplings are possible. For example, in the very simple Lagrangian $$L = \frac{1}{2} mv^2 - mgh$$ we have coupled the particle to the gravitational field $\phi = gh$. This is already in relativistically invariant form, since both $m$ and $\phi$ are scalars. (Of course the real story for coupling to gravity is more complicated, but this works ...


1

I think you can avoid all these troubles if you define the temperature as proportional to the variance of velocity, i.e. $$E[(v-\overline{v})\cdot(v-\overline{v})]=E(v\cdot v)-\overline{v}\cdot\overline{v}$$ Here $E$ means expected value, $v$ ranges over the velocities of the individual particles, and $\overline{v}=E(v)$. Clearly this is frame-...


1

All kinds of weird things happen if you try to define temperature in a moving object. The paradox to me (not a generalized accepted answer) resolves by realizing that temperature should only be defined as measured when the object is stationary. Not only is not a scalar but it is not even well defined for areference frame in relative motion. Is temperature ...


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Temperature is related to kinetic energy in the rest frame of the fluid/gas. In non-relatvistic kinetic theory the distribution function is $$ f(p) \sim \exp\left(-\frac{(\vec{p}-m\vec{u})^2}{2mT}\right) $$ where $\vec{u}$ is the local fluid velocity. The velocity can be found by demanding that the mean momentum in the local rest frame is zero. Then $\vec{u}...


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[June 19,2016: thoroughly revised, giving a more detailed, comparative presentation and better references] General case. In relativistic thermodynamics, inverse temperature $\beta^\mu$ is a vector field, namely the multipliers of the 4-momentum density in the exponent of the density operator specifying the system in terms of statistical mechanics, using the ...


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In general, it's more useful to think of special relativity problems in terms of the spacetime interval $\Delta s^2$ than in terms of the question of "who is moving relative to whom." A lot of those explanations (in terms of things like time dilation, length contraction, etc) are very ad-hoc, and are designed to make the idea of the spacetime interval more ...


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A boost is "the only other" plausible choice for transformations that mix spacetime co-ordinates, aside from rotations. The two notions in this sense are complementary. See for example: Palash B. Pal, "Nothing but Relativity", Eur. J. Phys. 24, pp315-319,2003 Given certain "reasonable" assumptions about our Universe; roughly The first relativity ...


2

Even if you interpret the Dirac equation the original way (not as a quantum field theory), the Lorentz transformation doesn't mix electron and positron. Consider two types of plane wave with positive frequency and negative frequency $$u(p)e^{-i (E t-p x)},\quad v(p)e^{+i (E t-p x)}$$ The idea is the energy $E$ is always positive, but the positive frequency ...


4

To understand this you need to understand what we mean by speed. If I want to measure positions and times I need to set up a coordinate system. For example I can take my stopwatch and my metre rule and construct some Cartesian axes $t$, $x$, $y$ and $z$, then I can describe every point in spacetime by its position in my coordinates $(t,x,y,z)$. Once I have ...


1

The force is given by: $$ F = -\frac{dp}{dt} $$ where $p$ is the momentum of the particles being ejected. The momentum is given by: $$ p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} $$ And the momentum is related to the energy by the relativistic equation for the total energy: $$ E^2 = p^2c^2 + m^2c^4 $$ so: $$ p = \frac{\sqrt{E^2 - m^2c^4}}{c} $$ The ...


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You simply need to remember the first relativity postulate to answer this one; essentially Galileo's postulate that there is no experiment that an inertial observer can do to detect their motion relative to other frames without making use of information from outside that frame. Therefore, the forces that any of the charges feels do not change from what ...



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