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Consider this previous exchange on the Cosmic Microwave Background The crucial bit is this: However, the crucial assumption of Einstein's theory is not that there are no special frames, but that there are no special frames where the laws of physics are different. There clearly is a frame where the CMB is at rest, and so this is, in some sense, the ...

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You can look at the problem like this: Whether or not one person gets to see all the other persons as connected for some period of time depends on 1) the disk size, say radius R 2) the duration $\tau$ of the connection in the disk frame 3) the speed $v$ of individual components on the rim relative to the disk. Let's denote $T$ the astronauts' period of ...

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When the people see the light they fire thrusters to get up to speed and move in a circle and then after that they fire their thrusters just to maintain uniform circular motion... OK, we have a disk, and light, and a rotating ring of people. No problem. So there are frames where all of the one events happen before all of the other events. And ...

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Since all the velocities are constant working out $\frac{dx^{1}}{dx^{0}}$ is simple division (no calculus required). Note that the entries in your vector in $S$ are the values you're interested in: $$(dx^0, dx^1, 0, 0)=(\gamma(d {x^0}' - v d {x^1}'), \gamma (d {x^1}' - v d {x^0}'), 0, 0)$$. So we have: \frac{dx^{1}}{dx^{0}} = \frac{d ...

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If you have two vectors you can project one onto the other. It you orthogonally project a spacetime vector onto your unit tangent then the length of that projection is how much time you observe separating the two events. That might be enough to answer your question right there. So to get an average velocity between two events first compute the vector A ...

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The answer is YES. It's true that the coordinate system (u′,v′,w′) is related to (x′,y′,z′) also by the orthonormal matrix A, at least under the Lorentz Transformations used in the following. But please, let use other symbols (for example it's custom to use $\;\upsilon\;$ for the algebraic magnitude of the velocity $\:\mathbf{v}=\upsilon\mathbf{n}\:$). ...

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This is not exactly the twin paradox, but it's close. First, let's make the problem more precise. Let's assume the train is one light year away from a planet, traveling near light-speed, and at the very beginning of the journey, a person on the planet views the clocks as synchronized. Then a person on the train does not view the clocks as synchronized. This ...

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Lets put the observer at A and the particle at B. The position kinematics are: $\vec{r}_A(t) = (a,0,0)$ $\vec{r}_B(t) = (r \cos \phi,r \sin \phi,0)$ The velocity kinematics are $\vec{v}_A(t) = (0,0)$ $\vec{v}_B(t) = (-r \omega \sin \phi,r \omega \cos \phi)$ The acceleration kinematics are $\vec{a}_A(t) = (0,0)$ $\vec{a}_B(t) = (-r \omega^2 \cos ... 0 A Lorentz transformation is a transformation that leaves$dt^2-dx^2-dy^2-dz^2$unchanged. So a rotation (which leaves$dx^2+dy^2+dz^2$unchanged and doesn't change$t$) is a special kind of Lorentz Transformation, one that has$t'=t.$So can you rotate by L then rotate by A? Sure. Can you rotate by A then rotate by L? Sure. Do you get the same answer ... 1 The bad news is that your understanding of the coordinates x, t, x', t' is badly jumbled . The good news is that your last sentence asks a good question, so let's start with it: Q: "how can I be certain which object in a question is the 'moving' one, and which one is the 'stationary' object?" A: You just "name" them that way! You can even switch the ... 0 No, your understanding of the variables$t',t,x',x$is incorrect. Lorentz transformation relates two coordinates that are in relative motion. One way to state the principle of relativity which serves as an axiom of SR is that there's no absolute state of rest, so given two observers who are in relative motion and whose coordinates are given by$(x,y,z,t)$... 0 Statement than one reference frame is stationary means that we choose its point of view for the problem at hand. It doesn't mean that it is special in any way - only that it is convenient for problem at hand. So "stationary", " static" and other statements like that are always some sort of shortcut. It can be misleading at first. Remember that we could have ... 3 Further to udrv's answer, and, the technicalities he raises aside, there are two ways to argue the reciprocity relationship that the boost from observer$A$to$B$is the boost from$B$to$A$but with$v\mapsto-v$. By the detailed arguments in the afterword, we find that the transformations between inertial frames form a group and that group acts linearly ... 0 I think udrv's answer hits the nail on the head, but I'll expand a little bit on the intuitive way to look at it. In your example, you have a ship leaving earth flying towards Neptune at relativistic speed. While you might think that the ship is moving fast and the Earth is essentially still (though in truth, the Earth is moving around the sun and around ... 9 Your question seems to be: if the observer on Earth sees the spaceship moving at velocity v, how do we know that the observer on the spaceship will see Earth moving at velocity -v? This is known as the "reciprocity principle" problem and it is a good one, in the sense that it raises the following issue: "Does the reciprocity principle follow from the basic ... 1 What is missing: this must be recognized as a relativistic process and be treated accordingly. Why is it a relativistic process: energy transforms into mass or conversely. This is something that cannot be accounted for using only classical mechanics. In older times I would've added "because it needs to account also for relativistic mass", but since using ... 1 There is no absolute inertial frame because all motion is relative. The earth is a good approximation, and the frame of reference we use for most situations. However the earth is accelerating around the sun (i.e. towards the sun, that acceleration keeps the earth in orbit). The closest thing to an absolute inertial frame might be a space ship traveling ... 0 There is nothing called an absolute inertial frame however in an inertial frame obeying the laws of motion your calculations will hold true. 2 version1 The laws of physics are the same in all inertial reference frames. This is a correct principle, but it is useless unless you also state which things are the laws of physics. If they the laws include Newton's laws as usually written then you get Galilean Relativity, if the laws include Maxwell's equation then you get special relativity. ... 0 Special Relativity has different postulates correspond to several kinds of symmetries. Symmetry means everything will be the same if you do certain operations. Two of the symmetries are the two postulates you stated, that are written at beginning of every textbook that teaches special relativity. The postulate one says a symmetry of uniform velocity in a ... 5 That motion was relative was realized by Galileo, so there was a theory of relativity -- Galilean relativity -- long before Einstein. That the speed of light should be the same according to all observers is indeed inconsistent with the Galilean relativity. This is because in Galilean relativity time is absolute. But it is not mathematically inconsistent to ... 0 Say in the Lab frame the total 4-momentum of the system is p = (E/c, p), where E is the total energy and p is the total 3-momentum. Also let M be the total mass of the system. Like in non-relativistic mechanics, in Special Relativity the COM frame is the frame where the total 3-momentum is null,${\it p_C} = (E_C/c, {\bf 0})$,${\bf p}_C = {\bf 0}$. With ... 0 Particle 1 and particle 2 each have relativistic momentum 4-vectors: $$\pmatrix{\frac{E_1}{c}\\p_{1x}\\p_{1y}\\p_{1z}}\text{ and } \pmatrix{\frac{E_2}{c}\\ p_{2x}\\p_{2y}\\p_{2z}}$$, so the total momentum 4-vector is $$\pmatrix{\frac{E_1+E_2}{c}\\p_{1x}+p_{2x}\\p_{1y}+p_{2y}\\p_{1z}+p_{2z}},$$ where$E_1=m_1c^2+K_1$and$E_2=m_2c^2+K_2$. These have squared ... 0 Lorentz Transformation$\:O x y z \longrightarrow O_1 x_1 y_1 z_1\:$and its inverse$\:O_1 x_1 y_1 z_1 \longrightarrow O x y z \:$,$\:\left(y_1 = y, z_1 = z\right)\:$\begin{bmatrix} x_1\\ \\ \\ c t_1 \end{bmatrix} = \gamma_1 \begin{bmatrix} 1 & -\dfrac{v_1}{c}\\ & \\ -\dfrac{v_1}{c} & 1 \end{bmatrix} ... 0 Use the Lorentz Boost parameter ($\lambda$) for each shell, where $$v/c=\tanh(\lambda)$$ $$\gamma = \cosh(\lambda)$$ $$\beta \gamma = \sinh(\lambda)$$ For boosts in the same direction the boost parameters just add or subtract. So$\lambda_1 -\lambda_2$is the boost with which one shell is approaching the other. Consider the trigonometric identity ... 2 Let the equations of motion be expressed in a frame with coordinates$q$. We now want to switch over to another (arbitrarily moving) frame, whose corresponding coordinates are$Q$, given by: $$Q = f(q, t)$$ For example, if the frame itself is moving with position$x(t)$, we will have: $$Q = q - x(t)$$ (where$x\$ is not dynamic, but is completely specified in ...

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