# Tag Info

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I don't really know anything about Hawking radiation but one thing bothers me in this question. In flat spacetime, generators of boosts are Killing vectors (generators of isometries: the symmetries of spacetime itself). In curved spacetime there are no such Killing fields. You cannot boost your spacetime and get the same thing. It is not true that all frames ...

0

If we setup the camera to record like above but NEVER EVER EVER look at the result of what was recorded. Does the wave function still collapse? The answer is that we just don't know. We can tell that the wave function has collapsed (in Copenhagen terms) only when we humans look at the system -- in the canonical experiment that means looking at the ...

3

Any observer outside the Schwarzschild radius sees the same thing: matter approaching the Schwarzschild radius at slower and slower (asymptotically zero) speed, forming a thin shell around the event horizon. The matter takes an apparently infinite time to collapse, and infinity is infinitely larger than a large finite the same way it's infinitely larger than ...

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Inside the Schrodinger's cat's box, the moment the radiation is detected by the counter, doesn't this mean the system already has a fixed eigenstate (a collapsed wave function, or is decoherent, whatever you like to call it)? At the end of your question, you give a list of things that might have happened to the system, and treat them as equivalent, but ...

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The answer whether the cat is dear or alive is not certain before the measurement. It is all measurement which determines the probability density function and the collapse of the entanglement

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I think you are probably misinterpreting the context here. If you read the previous line carefully it says "there is always an undetermined interaction between observer and observed; there is nothing we can do to avoid the interaction or to allow for it ahead of time. And later he just says due to the fact that photon can be scattered within the 2θ' angle ...

2

This might or might not be responsive to the question you intended to ask: Suppose you've got a meter stick. Over a period of time, I apply identical forces to the front and back ends of that meter stick, causing them to accelerate in the same direction. Therefore the entire stick, being a rigid body, accelerates in that direction. After a while, the ...

1

Despite their superficial similarity Lorentz contraction and time dilation are different things and this is why there isn't a distance version of the twin paradox. To see the difference you need to understand that a clock is a form of odometer. Suppose you start at the origin and travel 100 metres, then the odometer you carry will show the total distance ...

1

According to both observers, if nothing else happen, the 2 ships are going to collide in exactly 10 seconds. For concreteness, stipulate that both spaceship's clocks will read 0 at the instant of the collision (assuming the bomb does not explode first). Now consider the following question regarding the spaceship without a bomb: when the spaceship's ...

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The following will try to briefly address your specific issues: Questions 1-2: The discussion leading to the paragraph in your link starts in Sec.VII of that ref. (see first 3 paragraphs therein), with the assumption of the postulate of the speed of light. Hence Einstein is endeavoring to show that accepting the light postulate necessarily implies ...

1

I think that, rather than trying to pick apart the logic of the thought experiment which may or may not be well-worded (yes, I realise I just criticised Einstein), it is easier to just work one out of your own and see what is going on. Experimental setup To that end, consider two frames of reference moving relative to one another: one is going to be his ...

1

This is not really a relativistic effect. This is simply a consequence of information delay due to the speed of light being finite. $L/c$ is the time it takes for light to travel the distance $L$. Hence the information transmitted by light over the distance $L$ will be delayed by $L/c$. If the two clocks were exactly synchronized (in the stationary reference ...

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Black holes don't form. And you should keep your hand on your wallet when the word clearly is involved. If you have a spherically symmetric star and a spherically symmetric shell around it and you are inside the shell but outside the star then you notice the mass of the star but not the shell. When the shell contracts and gets to where you are it changes ...

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