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Observer C is situated in line with B but "above" the spaceship. C and B have synchronized their clocks and agreed on a reference frame. This statement is not clear. If you are saying that C is at rest with respect to B then it will also observe the same time as done by B. If you are saying that B is at rest with respect to A then it will measure ...


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If we know the wave function then we also know the time of arrival in a statistical sense. Consider a laser pulse. Suppose that the electric field is a wave package travelling at speed v, say a 3D gaussian. Let's assume the spread to be constant for simplicity. The probability of a transition in the sensor is proportional to E$^2$ by Fermi's golden rule. E ...


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The following is a failed attempt (at best with an extra assumption it can only work with cases where momentum is conserved) and too long for comment. Hopefully it illustrates the difficulty of the problem. Let us solve in a one dimensional universe (but can be further generalised) and let the last possible time the electron can hit the detector be $T$ and ...


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Once you collapse the waveform, the information you see is gone. However, there are some interesting corner cases which have been explored. Weak measurement comes to mind. Instead of measuring the system completely, you do some clever tricks (typically involving entanglement) to only collapse part of the system. This lets you restore the waveform, other ...


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No, this would be like asking if a coin is fair by only flipping it one time, or determining in what way a die is weighted by only tossing it once. If you wanted to determine the wavefunction prior to collapse then you would need many similarly prepared systems measured in the same way. From there you could construct the position probability distribution, ...


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I just finished a thesis on this subject and I'm happy to share. None of the linked papers are my own. The time of arrival in quantum mechanics is actually a subject of ongoing research. It is certainly a question which begs for an answer, as experiments have been able to measure the distribution of arrival times for decades (see for example Fig. 3 of this ...


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I think it will help to first consider a simple scenario involving a straight flight from A to B, at a speed high enough to make the Lorentz factor equal to 2. We suppose A and B are on the ground (of some huge planet), at rest relative to one another. Suppose observers in the aircraft find that their clocks advance by $1$ minute on the journey from A to B, ...


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Congratulations, you have discovered an example that the speed of light is not isotropic by Einstein's theory, despite the claim . Another point is that the light clocks in motion, oriented perpendicular to motion (y- and z-axis, in both positive and negative directions), all run synchronously, and that non-simultaneity only exists in one direction. Usually ...


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Thought experiments are used as catalyst for boosting ideas and development of theory. Outcome of thought experiment doesn't have to be unambiguous and usually can be paradoxical or unintuitive. The only purpose of such activity is to raise questions, lay some "emotional foundations" for discussed theory at hand. And Btw, thought experiments are ...


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The purpose of a thought experiment is usually not to come to a conclusion, but to clarify thinking on an issue. A thought experiment is an extremely simplified imagined situation--far simpler than could be achieved in real life [1]. A simple situation is easy to reason about because we can have complete knowledge. In this way, it's more like math than ...


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You do a thought experiment every time you solve a textbook problem. Thought experiments use existing theory plus logic to make predictions. If the existing theory is right, then the results of the thought experiment and of the (eventual) real experiment must match up. Usually the theory is right. This makes thought experiments useful first because they ...


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Yes, they should observe the same result (although what exactly this means might depend on which interpretation you work in).


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The equation you wrote down for the system in the reversible case corresponds physically to using a continuous sequence of ideal reservoirs at slightly different temperatures (running from Ts to To) rather than a single reservoir at To. This equation correctly determines the entropy change for the system in the irreversible case. The equation you wrote ...


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If it was heated from all sides the temperature gradient in the middle would only vanish if gravity was zero and pressure uniform. Assuming there is gravity and you have a ball filled with air then there is going to be convection of particles inside of it as some gets warmer fast (the ones close to the edges) and some takes longer (the ones in the middle). ...


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The temperature through the ball will have zero slope in the middle. It will not have a sharp peak or sharp minimum. That means it has a gradient of zero in the middle.


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This is primarily a response to one of your comments, but is a tad too long to fit in a comment box. You wrote: I forgot to take into account the fact that a signal has to travel from the horizontal lightclock to the golf ball and that, from the perspective of person B, this signal takes a different amount of time depending on which mirror is hit. No, ...


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Typically we would not consider each mirror to constitute a “tick” but rather a full round trip. However, that is rather minor and is just a small semantic issue. The resolution to this paradox is that the nudging of the golf ball cannot be simultaneous with the “ticking” of your horizontal light clock in all frames. Again, as with your previous paradox (and ...


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So if you placed the ball at the center of the light clock, then, yes, person B would see the ball nudged uniformly, but it would occur before the the leading bounce and after the trailing bounce. He would also see the time between leading bounce to training bounce be shorter than the time from trailing bounce to leading bounce.


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Photons have momentum and energy so by GR there can only be so many photons above a certain energy in a volume before it collapses into a black hole.


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Now this matter which the observer was observing seems to be stopped at the boundary outside the event horizon, thus forming the accretion disk. That's not an accretion disk. Accretion disks are mostly made of matter at a substantial distance from the black hole, which isn't dramatically redshifted and isn't on the verge of crossing the event horizon. I ...


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Because of the intense gravitational pull near the event horizon, very little light reaches us from the parts of the accretion disk close to the horizon (and none, of course, from the infalling part of the accretion disk that has crossed the eveny horizon). In practice, the parts of the accretion disk that are bright enough to see have to be a certain ...


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You're assuming that the nudging of the golf ball and the bouncing of the light are simultaneous with respect to the rest frame of A and the rest frame of B. That isn't possible, just like $x>y$ and $y>x$ isn't possible. You'll have to change your assumptions. It would be logically consistent to assert that they're simultaneous with respect to only one ...


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The size of your clock is important. If the size of the clock is effectively zero, there is no problem. A clock of size zero is one based on local events, such as detecting light at a fixed location (the clock in which the light moves back and forth qualifies as a zero size clock). If your clock ticks in the way you defined it, then it is not based on local ...


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Dale's answer is right. Here is a little more detail: Let's set the clocks right next to each other, so that the bottom of Clock 1 coincides with the left end of Clock 2. Let's conveniently make each clock one light-second long. When the clocks start, light leaves the origin in both the up and rightward directions. In the train frame, the first tick of ...


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Typically we would not consider each mirror to constitute a “tick” but rather a full round trip. However, that is rather minor and is just a small semantic issue. In principle your analysis is correct. The effect you have noted is due to the relativity of simultaneity. Indeed, the time between the rear and the front tick is different than the time between ...


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