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To measure time, a duration, you need two moments – when you press "start" and "stop" button on the stopwatch, respectively. But because the two objects are moving relatively to each other, it isn't possible for them to "meet" at both moments. If their locations coincide at the "start" moment, for example, so that their clocks may be compared at this "start" ...


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There are plenty, plenty and plenty of equations with constants under the different degrees. The main reason for this: the $c^2$, $\hbar^2$ and other doesn't have any direct physical meaning; contrary the $c$, $\hbar$, $e$, various masses and other.


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It's a mixture of $c_\infty = c_0 = c$ and "the question doesn't make sense". So, first, how it does not make sense: What's the "speed" of a quantum object? It has, in general, no well-defined position, so $v = \frac{\mathrm{d}x}{\mathrm{d}t}$ is rather ill-defined. Instead, we should probably look at the mass of the photon, since all massless objects ...


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There is no contradiction. From your quote: The proper length of an object is the length of the object in the frame in which the object is at rest. and The proper time between two events - such as the event of light being emitted on the vehicle and the event of light being received on the vehicle - is the time between the two events in a frame ...


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From the reference frame of the space ship, your body is stationary. When you swing your arm back and forth, it has a non-zero speed in this frame, and thus its mass, or rather momentum, increases. The space ship's velocity of $v_\mathrm{ship} = 0.866c$ doesn't add to this. That means that the increase in momentum is exactly the same as if you swing your arm ...


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The energy of a photon is given by the equation E = hf where h is Planck's constant and f is frequency. The energy would decrease, making the frequency decrease (since h is constant). So, if the photon was blue light, then it would get redder and redder as time when on. There is a point, however, when your system eventually stops working. This is because the ...


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It is a standard exercise in most quantum field theory books that the 2-point function does not vanish outside the light-cone of a particle. Less technically; the probability that some particle $r$ away from a source feels the effect of the source quicker than light to travel would $r$ is non-zero. That is a good definition of superluminal motion. See for ...


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After some searching, I found a very simple way to do so. The blue disks shows show light traversing up a rod. If the rod were stationary then the light would reach the top in ct where t starts at t' = 0. However, with the horizontal motion of the rods, the light takes a diagonal path, indicated by the red line. By the Pythagorean Theorem, we get $$d^2 = ...


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Yes, I do believe there is and that is the approach I outline in my answer to the Physics SE equation "What's so special about the speed of light" as well as my answer here. I like to think of SR as simply Galileo's basic idea but with the assumption of absolute time relaxed (not to devalue Einstein's bold step in making this relaxation). One begins with ...


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Complex mass means gravitational mass + i.lambda.higgs mass. The weak coupling constants are all proportional to (higgs) mass with Higgs vacuum value (=246 GeV) as the proportionality constant. Thus mass necessarily becomes a complex number. The real part produces attractive gravity forces and the imaginary part produces repulsive "weak" forces. The factor ...


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einstein did not give e=mcc at first rather he had given l=mvv where l is energy and v is velocity o f light. e =mcc is written in this way to show the unique dependence of energy to the speed of light.


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Where is the flaw in my thinking? In your concept of time. It's little more than a cumulative measure of local motion, see A World without Time: The Forgotten Legacy of Godel and Einstein]. Your macroscopic motion relative to some other guy results in you measuring his local motion to be slow, whilst he measures your local motion to be slow. This sounds ...


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$\newcommand{ket}[1]{\left| #1 \right>}$ $\newcommand{bra}[1]{\left< #1 \right|}$ $\newcommand{bk}[2]{\left< #1 \big| #2 \right> }$ In the level of QM you really don't have kinetic energy and potential energy. Some likes to call the expectation value of the operator $\frac{\hat{p}^2}{2m}$ on the state $\ket{\Psi}$ the kinetic energy and the ...



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