# Tag Info

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Or I have to say object is really measured to shrink? That's the right answer. Imagine a thought experiment. Let's say we are shooting another sequel of the Speed movie, 5 (I lost track :) ). So there is a bomb on the board of a spaceship, and a terrorist says you cannot slow down below the speed of 0.5c, because if a mechanism that involves a ...

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A particle moving at the speed of of light does not experience time, as it has no rest frame. Furthermore, a particle cannot continuously accelerate and eventually reach the speed of light, since massless particles can only move as fast as light. They either move at the speed of light or do not exist at all.

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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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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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The point at rest in $k$ moves as $(x(t),y(t),z(t)) = (x_0+vt,y,z)$ in the frame in which $k$ itself moves with $v$ in positive $x$-direction. Thus, we have to set $x'(t)=x(t)-vt = x_0 +vt-vt$ to obtain a stationary $x'(t) = x_0$ coordinate for a tupel $(x',y,z)$.

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