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The fact that different observers in relative motion can measure the same light ray to move at a speed of c has to do with the fact that each observer defines the "speed" in terms of distance/time on rulers and clocks at rest relative to themselves. It's crucial to understand that different observers use different rulers and clocks to measure speed, because ...

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The problem is that your analysis is all done from the perspective of the frame that measures the box to be moving at 0.1c -- in this frame, it's true that the time for the light to get from the source to the wall is different from the time for the light to get from the wall back to the source. But if these same events are measured by someone inside the box ...

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It appears from your edit that you added the velocity of the ship and the speed of light classically. However, the velocity of the ship and the speed of light need to be added relativistically, using a Lorentz Transformation. Here's a quick way to know something is amiss with your moving ship: You state that the stern observer sees a wave propagation speed ...

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The speed of a photon, and all massless particles, is essentially always $c = 299792458\space m/s$. The Doppler effect is entirely contingent on the fact that the velocity is the same throughout. We have a source, $S$, emitting light at a certain speed: ...

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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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At what speed does our universe expand? This question doesn't make sense in the form in which it was posed. To see why, let's start by thinking about how we know the universe is expanding. The expansion of the universe was originally discovered by Lemaître and Hubble, who found that the redshifts of galaxies were proportional to their distances from ...

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I haven't done this in a long time, but my understanding from Feynman's QED is that the speed of a photon is unknown (Heisenberg) - photons traveling in a vacuum are around the speed of light +-, but at any instant the speed differs due to uncertainty. The photons going faster and slower than light speed cancel in the same manner that photons bouncing off a ...

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Comment to the question (v4): It seems relevant to mention that there in principle could be a difference between the universal speed limit constant $c$ (which is usually casually referred to as the speed of light in vacuum), and the actual speed of light in vacuum if the photon has a rest mass, at least from an experimental point of view. Of course, no ...

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In quantum mechanics a particle can be treated as a wave and a wave can be treated as a particle. This is the notorious wave particle duality. I won't go into this any further here because it's been discussed to death in lots of previous questions. Search this site for wave particle duality if you're interested in finding out more. Anyhow, assuming I ...

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If rest mass does not change with v then why is infinite energy required to accelerate an object to the speed of light? The momentum of a material particle, a conserved quantity, is theoretically and experimentally a non-linear function of velocity given by $$\vec p = m \frac{\vec v}{\sqrt{1 - \frac{v^2}{c^2}}}$$ which goes to infinity as $v ... 1 In relativity the rest mass is the mass of an object measured from a reference frame in which it is at rest. But this is not the mass involved in acceleration or inertial mass. Inertial mass, or the opposition of the body to the change of movement (directional or in magnitude), will grow with the speed of the body: $$m = \frac{m_o}{\sqrt{1-v^2/c^2}}$$ ... 3 The beginning of the chapter, the author does use$cfor the Lorentz transformations (cf., Equation (13.1)). \begin{align} A_0&=\gamma(A_0'+(v/c)A_1') \\ A_1&=\gamma(A_1'+(v/c)A_0') \\ A_2&=A_2'\\ A_3&=A_3' \end{align} Shortly after Equation (13.1), the author lists several enumerated remarks. In particular is #4: Lest we get tired ... 1 Here is a slightly different take on this using the boundary conditions for electromagnetic fields at an interface. A key boundary condition, that is derived from Faraday's law, is that the component of the E-field tangential to the boundary must be continuous. So take an EM wave travelling at normal incidence with the electric field solely in a direction ... 0 There is another effect to take into account as you accelerate your ship. It is the Unruh Effect 0 Firstly, the immediate answer is no, the intensity of radiation is higher when you fly into it, like the rain situation (but the maths is different). There are three interrelated effects that you need to know about regarding your question, which are frequently omitted in popular or introductory SR: aberration of light, the full angle-dependent doppler ... 1 Not only does your speed affect the amount of radiation that you receive, but this actually happens to the Earth and has been measured experimentally. You say: So basically in space, there is bound to be stray radiation, whether from the stars, or cosmic background, floating around right. and the most obvious example of this is the cosmic microwave ... 0 the would-be superluminal neutrino speed [...] The relevant comparison is between the arrival of neutrinos at a suitable detector, e.g. regarding neutrinos which had been travelling from CERN to LNGS; and the (first possible) detection of the corresponding "signal front", e.g. regarding signals due to any one neutrino bunch having been released at ... 1 The common answer is that nothing can be faster then light. If you look deep enough you will find out that the interaction between the rocket (or whatever else) and the force that is pulling or phushing it is always based on electromagnetism. Go from fuel to gas to molecules and you came at the end to the interaction between the electrons. And sometimes with ... 3 In our frame, the light beam approaches the rear end of the train at speedc+v$, because both are moving and in opposite directions. Note that this does not mean anything is traveling faster than light. Similarly, the light beam approaches the front of the train at speed$c-v\$. We want to choose the position of emission so that they arrive at the same ...

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