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8

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 ...


6

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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@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., an electron and a positron) are created from the energy in the very strong gravitational field near the horizon of the black hole ...


2

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 ...


2

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.


2

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 ...


1

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 ...


1

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 ...


1

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 ...


1

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 ...


1

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 ...


1

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 ...


1

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


1

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 ...


1

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 ...


1

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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