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37

John Rennie already gave the practical answer considering the atmosphere, noting that without doing anything objects near the ISS will deorbit quickly from drag. But that's letting reality get in the way of a good physics problem. I'll show that while a human can't send a ball crashing into the surface in one orbit, they can come close. The ISS is listed as ...


37

You don't need to throw the ball! At the altitude of the ISS the atmosphere is thick enough that it loses 50-100m of altitude every day due to the drag. At that rate over your ten year timescale the ISS would lose 180 to 360km. When you take into account the increased drag at lower altitudes ten years is enough to bring the ISS crashing to a fiery end. So ...


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According to the rules of qft there are virtual photons in the vacuĆ¼m. No, according to QFT the vacuum is static, in the sense that $P^\mu|\Omega\rangle=0$. Or put it another way, The vacuum at a time $t$ is exactly the same vacuum at a time $t+\Delta t$ for any $\Delta t$. This means that the picture of particles constantly appearing and disappearing ...


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Definition of resistivity: Electrical resistivity (also known as resistivity, specific electrical resistance, or volume resistivity) is an intrinsic property that quantifies how strongly a given material opposes the flow of electric current. A low resistivity indicates a material that readily allows the flow of electric current. So you are using the ...


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The annihilation operator is a linear operator. A linear operator can be applied to ANY state. And yes, it returns zero when applied to the ground state. You can really take this as a definition. The definition of the momentum operator $\hat{p}$ is the operator such that $\langle x|\hat{p}|\psi\rangle=-i\hbar \psi'(x)$. One could write this as $\langle ...


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It depends on what you mean by vacuum. If you mean a field configuration that has $F=0$, then the gauge potential is locally pure gauge (cf. this answer by Qmechanic), so there can only be global obstructions to the gauge equivalence class of the $A$ corresponding to $F=0$ being $A=0$ everywhere. On $\mathbb{R}^{1,3}$, there is no such obstruction. If ...


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Other answers have handled the oxygen issue quite well. In regards to the temperature, space itself has no temperature because it's a vacuum. Objects in space, however, do have a temperature. If a human is exposed, unprotected, to space near the sun (or any other star), the temperature change in their body could very well be terminal. Even near our ...


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It is a bit of a hypothethical scenario where the body can breathe oxygen and does not rupture from the pressure difference (dissolved gases in the intestine and blood bubbling), but at the same time can evaporate and radiate freely. If we assume so, and also assume that there is no sunlight, then there are two major mechanisms for heat loss: radiative ...


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The temperature of a true vacuum would be a measure for the energy distribution of the photon gas in that vacuum. You can derive the occupation of the electromagnetic modes in a volume with Bose-Einstein statistics, which is essentially what Planck did to describe the emission spectrum of a black body. However, you don't need to do understand the details of ...


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Quick answer My question is how does the presence of nonzero $J(x)$ results in a non-trivial spacetime dependent value of $\langle 0|\phi(x)|0\rangle$? The equation $\phi(x)=\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}$ works both for $J=0$ and $J\neq 0$. Therefore, $$ \langle \phi(x)\rangle_J= {}_J\langle 0|\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}|0\rangle_J ...


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Almost everything from the wikipedia page you link is just false, or at best very misleading. IMHO, that page was written by someone that doesn't know anything about quantum mechanics beyond what one could find in TV documentaries. "Not even wrong" came into my mind many times as I was reading the article. In quantum physics, a quantum fluctuation (or ...


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There are many ways to think about the entropy of a vacuum (assuming there is no radiation and thus T=0), but all give the same result, the entropy is zero. One easy way is to notice that the walls are made of something (it doesn't matter what) that cannot change its state, so the number of microstates, $\Omega$, is equal to 1. Then $S=k\ln\Omega=0$.



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