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I think I have understood the problem. Since I want to compute the length in the moving reference frame $R'$, I must assume that $t_2' = t_1'$ but not $t_1=t_2$, because I want to measure the position of the two ends of the bar at the same time in the moving frame. The calculation is as follows: $$ x_2-x_1 = \gamma(x_2'+ut_2')-\gamma(x_1+ut_1) = ...


4

Yes it is - well, sort of. The coordinate invariant form of velocity is the four velocity, and the magnitude of any four velocity is always $c$ (or $1$). So even if you are stationary in space in your chosen coordinate system the magnitude of your four velocity is still $c$. Whether moving at the speed of light on the time axis is a good way to state this ...


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With 4-position $$ x^\mu = \left(ct,{\bf r}\right) $$ the 4-velocity is defined as ($\tau$ is proper time) $$ v^\mu = \frac{d x^\mu}{d \tau} = \gamma \frac{d x^\mu}{dt} = \gamma \left(c,{\bf \dot{r}}\right) $$ Taking the dot product of $v$ with itself gives: $$ v^\mu v_\mu = \gamma^2 \left(c^2 - {\bf \dot{r}}^2\right) = c^2 = \left(c ...


3

I realize that in essence there is no object which can be considered as "not moving in space". No object at all is moving in space if you are taking the point of view of its reference frame! The law of conservation of energy is requiring that the energy of its mass (e = mc2) is "transported through time", or in other words, that time is passing for ...


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If I understand correctly you are asking how observer dependent is electromagnetic radiation. The first thing is that non uniformly accelerated charges are described in a inertial frame by Larmor's formula and Abrahm-Lorentz force which take into account the radiated field and the recoil on the particle. Now in Newtonian mechanics and special relativity ...


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Since this is also true for sound propagating in air No, actually, it isn't. There's something crucial that you've left out. Look more closely at the 2nd postulate: light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body. Now, this holds for the emitter of the light ...


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There are apparently some extensions of the Standard Model that allow for Lorentz symmetry to be violated, although from what I understand the symmetry is broken by spontaneous symmetry breaking which means the theory would have been symmetric in the era when the forces were unified, and the symmetry was broken by a random decay to different vacuum state. So ...


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No. My answer is negative, even if I confirm the statements of other answers: "The first thing is almost completely arbitrary, especially in full general relativity. The second thing is an unambiguous result of an experiment."(Jerry Schirmer) "In Einsteinian relativity all observers can still agree on a number of facts, they are just ...


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As stated in the comments: The Doppler shift is a change in observed frequency due to relative speed difference. However, the speed with which the signal propagates is the speed of light.


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(Edit) Notice one thing, Geremia. When two particles finally meet, or when a material body (spaceship in my second example below) and a photon meet (or even when two material bodies meet), none of them separately have exceeded the speed of light. Each of them travel at either c or $v<c$, and they simply meet on their way toward each other. So the closing ...


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If one thinks in terms of spacetime and world lines, the answer is plain to see. While each particle has a world line which must be within the light cone of any event along the world line the distance between the two particles is not a physical entity and does not have a world line. Thus, it is not directly constrained by the 2nd postulate. Put another ...



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