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On a practical note, there was a space walk a few years back when they replaced a failed ammonia pump that was the size of a refrigerator. The astronaut simply gave it a swift push away from the station knowing that it would soon deorbit fast enough that it wouldn't be a collision hazard. PS- If you have Kerbal Space Program, this would be a fun thing to ...


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I think the source of your confusion can be traced back to the statement that "Vacuum is SUCTION POWER". There exists a household appliance called a vacuum which induces a pressure gradient that causes "suction" (actually it's the atmosphere pushing, but that's neither here nor there) but this is not what "vacuum" means in the context of physics. In physics, ...


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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 vacuüm. 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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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 ...


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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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Don't let the size and length of the tube confuse you. What you described can be viewed as a big nozzle with a diameter of 20 meters with a water pressure of 500 atm. entering the back side of the nozzle. The formulas that can be used are: V = sq. root of 2gh where: V = velocity in ft/s. g = acceleration const = 32.2 ft/s^2 ...


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To answer this we need to talk a bit about how particles are described in quantum field theory. For every type of particle there is an associated quantum field. So for the electron there is an electron field, for the photon there is a photon field, and so on. These quantum fields occupy all of spacetime i.e. they exist everywhere in space and everywhere in ...


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@JohnDuffield: I can give you both a correct answer in simple terms and the fairy tale, together with references to an explanation how the fairy tale is related to the real thing! The dry facts are that two real particles (e.g., two photons, or an electron and a positron) are created from the energy in the very strong gravitational field near the horizon of ...


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I think the question was already answered here, look at either my answer of the one by Adam. The punch line is that that the expectation value of a field (such as the field $\phi$ at the bottom of the mexican hat potential) is fixed by the way the source $j$ that couples to $\phi$ in the path-integral for the functional generator is sent to zero. As there ...


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Yes, one could build very large structures in the upper atmosphere, but the economic advantages of such structures on Earth are not clear. The problem if launching into space is not that of providing "lift", the problem is to provide enough velocity. An air launch from a high flying plane only saves approx. 10% in total fuel due to atmospheric drag and ...


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The vacuum Einstein equation is just $$ G^{\mu\nu} = 0 \,.$$ Of course, that does not help much, if one does not specify this tensor. That is given by $$ G^{\mu\nu} = R^{\mu\nu} - \frac 12 \mathcal Rg^{\mu\nu} \,.$$ Then we need to specify the Ricci tensor and the Ricci scalar. Those are $$ \mathcal R = g^{\nu\beta} R_{\nu\beta} \qquad\text{and}\qquad ...


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Many answers discuss the transition from the unstable state to the stable one. Let me thus discuss the issue of choosing the ground state itself. I will suppose a two dimensional Mexican hat potential. As Numrock realised, it has degeneracy. There is nothing which lift this degeneracy in principle. Then you can change the ground state without energy. This ...


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In quantum field theory, the vacuum is the state containing exactly zero particles anywhere in space and at all times. Since it is an eigenstate of the number operator, there is no uncertainty at all about this. On the other hand, empty space between matter (i.e., what is informally called a vacuum) is never completely empty; it is still filled with the ...


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In relativistic QFT, this cannot be a process in time. The unstable initial state does not exist at all: An unstable ground state is impossible in relativistic QFT at temperature T=0 (i.e., the textbook theory in which scattering calculations are done) since it would be a tachyonic state with imaginary mass, while the Kallen-Lehmann formulas require $m^2\ge ...


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The thermodynamic definition of entropy of a system is where T is the absolute temperature of the system, dividing an incremental reversible transfer of heat into that system (δQrev). (If heat is transferred out the sign would be reversed giving a decrease in entropy of the system.) You may consider a vacuum as a system under study, but in this ...


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A explain this practically, you need to place a spherical mass of 0.5 kilometre diameter at the distance of 10 metre from LIGO sensor to detect its gravitational wave which may read the maximum of 1mm in the reading. Simple answer would be Yes you can create a gravitational wave with quadruple movement of atom or subatomic particles Theoretically. But our ...


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Yes, even tiny objects produce gravitational waves as they move. It's just that their gravitational waves will be way too tiny to measure. Just consider that the recent gravitational wave detection was caused by 2 black holes weighing 36 and 29 times the mass of our sun. Even those enormous black holes only caused a tiny change a thousand times smaller than ...


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For an ideal balloon-pop (no interactions with the walls of the balloon) $dU = 0$. In the case of an ideal gas, $U = U(T) \propto NkT$. So the temperature, strictly speaking, does not change when the balloon expands. You end up with a gas at the same temperature but with a lower density.


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The true mediators of forces are the quantum fields. A thorough discussion of virtual particles and their properties (and possible way of existence) is given in the following two recent essays of mine: The Physics of Virtual Particles Misconceptions about Virtual Particles From the introduction to the second essay: ''virtual particles are defined ...


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In fact in some sense the full state of the quantum system shouldn't break any symmetry. It's just that the overlap of the different vacuum states (in field theory) vanishes. So in other words the full quantum state is a superposition of all the different vacua, but when we make observations we "collapse the wave function" (feel free to insert your ...


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Physically, you look for the ground state of your theory in order to make a correct predictive calculus around the ground state. The "random" choice of the ground state depends on the dimension of the theory, in fact, you can obtain a "hat" shape or a simple 2-dimensional shape with just 2 possibilities for the ground state (look the scalar quantum ...


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The quality of the vacuum at LHC is pretty good but particles are constantly accelerated/bent by strong electromagnetic fields along the circumference of the accelerator. Thus the situation is not the one felt by a free particle. So conceptually, in your diagram, you can simply replace the photon coming from the nucleus by a photon from the electromagnetic ...


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only if tenperatur is not 0K, any moleculus will move randomly ,it is so called thermal motion. when all moleculus is in one close capacity, they move to herr and there, so what you see is they are always uniform, since you open the capacity, it is obvious that they will move out


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See a vacuum chamber has no air inside it. So that is a low pressure region. But the capillary tube contains some pressure exerted by the air molecules on the walls of the tube. The fluid flows from a high pressure region to a low pressure region. So the air molecules spread out into the vacuum chamber filling the empty spaces. This continues as long until ...


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Really Are you Trying to move Sound in a Medium less Region . 1. Sound Require Medium to Travel So No chance If they Are Hearing Each Other . 2. IF there is Small Contact Then There will be Small Passing Of vibration and result in A little whisper to Both of them if they Shout Or whatever they Do. 3.as i said Transfer Of vibrations will be Small and Hence ...



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