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2

Particles do not constantly appear out of nothing and disappear shortly after that. This is simply a picture that emerged from taking Feynman diagrams literally. Calculating the energy of the ground state of the field, i.e. the vacuum, involves calculating its so-called vacuum expectation value. In perturbation theory, you achieve this by adding up Feynman ...


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Technically, little vacuume can carry huge wight even with little air flow. For example typical vacuum cleaner can lift column of water 2 m high other dimensions will go together with the vacuum surface. This means that it can carry some 80 cm column high of concrete or 25 cm column high of steel, assuming it is perfectly sealed. If not perfictly sealed such ...


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The commonly accepted interpretation of Special Relativity is that it's impossible to determine an (inertial) frame of references absolute motion (by performing experiments within that frame). So the default answer (given what we know) would be no, you can't determine your absolute motion relative to the background sea of virtual particles. However this ...


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Your argument against is based on the fact that in one frame there is no magnetic field in one frame, but there is a magnetic field in a different frame. So there must be magnetic virtual particles in some frames but not in others. Hence magnetic virtual particles can't exist. However there aren't separate magnetic and electric virtual particles. There are ...


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Virtual particles refer to actual, nonzero features in the quantum fields of real objects, but they are features that are not particles in many ways so you should not expect anything from their being named "particle". Basically, the idea of virtual particles was invented as a device for when you want to hold on to the particle picture while doing quantum ...


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Virtual particles are not observable by definition. They represent "internal lines" in Feynman diagrams. For example, this diagram: Here two electrons move toward each other, interact, then move away from each other. The external lines represent "real" electrons which we can measure/observe. The internal line here is an excitation of the electromagnetic ...


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I believe you are all using things you know little about to make your claim seem reliable. For example, from this simple strain of comments I have heard anything from black holes to the expansion of the universe. These things, although interesting, have nothing to do with whether or not a stone will halt after being thrown into space; the forces you have ...


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Virtual particles aren't real. It's in the name. The reason people say there are "vacuum fluctuations" in form of virtual particle-pairs forming and annihilating again is because they misunderstood Feynman diagrams. In Feynman diagrams, internal lines are called virtual particles, since the external lines (i.e. the open-ended lines) correspond to real ...


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I'm not a fluid mechanics expert, but my mechanical systems knowledge suggests it might be simply a natural oscillatory behavior, which is always present but in this case is more noticeable due to the aggressive initial response (i.e fast influx of air) your chamber experiences. So what is causing this inexplicable pressure drop? Once the chamber has ...


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There are various general ways to see the point. Let me give you a simple one that is still general enough (see also last comment at the end). Imagine you have a lagrangian density $L$ that is invariant under the continuous group transformation that acts linearly on the fields as $$ \phi\rightarrow \phi^\prime=U\phi=e^{i\omega^a T^a}\phi=(1+i\omega^a ...


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It is not needed for the vacuum to fluctuate all the time, but one can say the the probability of having a vacuum fluctuation at this point $x,t$ is non-zero


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Since the conceptual problem is not really coming from field theory itself, let's look at the 0D case. This case is too naive (no SSB here), but the discussion is still correct qualitatively. The partition function is $$Z(j,j^*)=\int dz dz^* e^{-V(|z|^2)+ z^*j + j^* z}, $$ where $dzdz^*$ really means $d\Re z\,d\Im z$ and $V(x)=r\, x+x^2$ where $r$ can be ...


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It's true that the vacuum ought to be an eigenstate of the full interacting Hamiltonian. But as seen from the perspective of the Hamiltonian of the free theory (all interactions being treated as perturbations around this free theory) the actual ground state is "dressed" by many vacuum fluctuations on top of the free ground state.


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No the denser sphere pulls away from the lighter sphere immediately. Just think about the instant they're both dropped and they fall some tiny amount of distance. They both run into the same amount of air molecules in the short distance but the net effect of those air molecules is less on the heavier of the two objects because it has more mass and therefor ...


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$a^\dagger a\! \mid \! n \rangle = n\mid \! n \rangle$. For vacuum you can write $a^\dagger a\! \mid \! 0 \rangle = 0\mid \! 0 \rangle$, if you like. No need to write 0 solely.


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It means it's "the end of the line". The vacuum state is, as you correctly say, not the zero state. It has energy content, and physical meaning - it's the state with no particles. Annihilating the vacuum leaves...nothing. Trying to take a particle out of it is not possible - it gives you the zero vector, which does not represent a physical state, since it is ...


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Actually what I got is equivalent to the form given in Ryder, we have $$P\Delta_F(x)=-\delta^4(x)$$, then $$\Delta_F(x)=-P^{-1}\delta^4(x)$$, from which we also have $$\Delta_F(x-y)=-P^{-1}\delta^4(x-y)$$, where $P^{-1}$ will not be affected since it's a differentiation operator with respect to $x$. Consequently, we can extract $P^{-1}$ from the above ...


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Your relationship (solution) $\bar{\phi}=P^{-1}J$ is an integral relationship in fact where $P^{-1}$ is a Green's function or the Feynman propagator. You have to write down the arguments properly and you will get the right result.


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From your brief description, it sounds like you are operating a vacuum chuck. What kind of pumps are these? A very basic diaphragm or rotary vane pump will easily reach an ultimate vacuum of 100mbar or much lower. Since you are using the pump to hold an item, you probably do not need high vacuum or high pumping speed, because the atmospheric pressure outside ...



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