# Tag Info

1

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 ...

2

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 ...

-1

Looking at natural aspects of the world we can see first that a lack of pressure in no way means water won't and some point coalesce amongst themselves given the ability (eg clouds). So does; by creating a lack of pressure do we also decrease the relative amount of water molecules? NO, but pulling out the molecules via vacuum pump will decrease the overall ...

3

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 ...

0

Though there is 'nothing' in the space, it have some ability to 'hinder' that means, if you could put the 2 leads of a Ohm-meter, you would get a very high resistance reading (very low current). Just think of a Cathode ray tube. So much strong voltage is required to create the current through that low-pressure gas (though not a perfect vacuum) Whereas, if ...

2

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 ...

0

I'm not a physicist but based on studies I did : all types of elementary particles and forcrs also have fields in the whole universe. and fields always are there so even if we don't see any particle in a place it does't mean that there is "nothing" in that place because fields are always there. based on the uncertainty principle Since we can't accurately ...

0

Concerning point 2: Operators do not always come through each other cleanly, but there are some very basic rules that always apply, which can be turned into less tedious rules that apply in special cases. Often the latter are taught first, causing mass confusion. General rule: Operators can be expressed as (Sum over a in the set of eigenvectors ) |a > ...

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 ...

3

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 $\langle ... -1 On a practical note, there was a space walk a few years back when they replaced a failed ammonia pump that was the size of a refrigerator. The astronaut simply gave it a swift push away from the station knowing that it would soon deorbit fast enough that it wouldn't be a collision hazard. PS- If you have Kerbal Space Program, this would be a fun thing to ... 37 You don't need to throw the ball! At the altitude of the ISS the atmosphere is thick enough that it loses 50-100m of altitude every day due to the drag. At that rate over your ten year timescale the ISS would lose 180 to 360km. When you take into account the increased drag at lower altitudes ten years is enough to bring the ISS crashing to a fiery end. So ... 3 According to the rules of qft there are virtual photons in the vacuüm. No, according to QFT the vacuum is static, in the sense that$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$. This means that the picture of particles constantly appearing and disappearing ... 0 I don't really know anything about Hawking radiation but one thing bothers me in this question. In flat spacetime, generators of boosts are Killing vectors (generators of isometries: the symmetries of spacetime itself). In curved spacetime there are no such Killing fields. You cannot boost your spacetime and get the same thing. It is not true that all frames ... 1 There are many ways to think about the entropy of a vacuum (assuming there is no radiation and thus T=0), but all give the same result, the entropy is zero. One easy way is to notice that the walls are made of something (it doesn't matter what) that cannot change its state, so the number of microstates,$\Omega$, is equal to 1. Then$S=k\ln\Omega=0\$.

0

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 ...

11

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

@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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