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A particle is said to be on-shell if it satisfies the relativistic dispersion relation, $$E^2 = p^2 +m^2$$ in units wherein $c=\hbar=1$. If you graph it, you obtain a parabolic surface for massive particles, and a cone for massless particles, like a photon. This is known as the mass shell, it is quite literally a shell or surface. The momentum of a real ...


4

Such an ordering arises from the fact that are arranged chronologically, i.e., according to the dates they were "discovered". The principle quantum number $n$ entered the picture with Bohr's theory of the Hydrogen atom in 1913.Bohr introduced $n$ in his quantization of angular momentum postulate where $n$ is the allowed orbit. Mathematically, $L = n{h ...


4

From "The Transistor, A Semi-Conductor Triode", by J. Bardeen and W. H. Brattain, Phys Rev. 74(2), 230-231 (1948): "The device consists of three electrodes placed on a block of germanium as shown schematically in Fig. 1. Two, called the emitter and collector, are of the point-contact rectifier type and are placed in close proximity (separation ~0.005 to ...


4

Remember, operators are nothing but maps. Expectation value of an operator is pretty much defined (I guess) in general operator theory. It just turns out that in QM (Hermitian) operators correspond to dynamical variables. In general you can also calculate expectation values of operators like $L_+$ and $a^{\dagger}$ etc., which don't have any dynamical ...


4

The term "shell" originally derives from the non-relativistic version of the answer by @JamalS. In a non-relativistic theory, a free particle satisfies the following dispersion relation $$ E = \frac{ {\bf p}^2 }{ 2m } $$ For a fixed energy a particle satisfies $$ {\bf p}_x^2 + {\bf p}_y^2 + {\bf p}_z^2 = 2 m E $$ In momentum space, this is precisely the ...


3

Lets define some variables first. Lets say that the length of the column of air that you trap in the tube between the water level and your thumb is $h_0$ and it is initially at the same pressure as the surrounding air which we will call $P_0$. Lets also define the cross-sectional area of the tube to be $A$. As you draw the tube out of the water, the water ...


3

Because as far as we understand general relativity, it's not doing "the opposite of what gravity does." Gravity can be locally attractive or repulsive, depending on whether the stress-energy content satisfies or violates the strong energy condition. For ordinary matter, the stress-energy is dominated by the mass, the SEC holds, and its gravity is attractive. ...


2

I depends on the pressure ratio between the chamber and ambient . As long as the ratio is critical, the ambient air will flow through the orifice at the speed of sound (with respect to the state in the chamber). For undercritical pressure ratio the flow can basically be found by the Bernoulli Equation.


2

The full statement seems to be: $T_{dS}\sim\frac{1}{R}\sim \sqrt\Lambda \implies non-SUSY$ In a de Sitter universe, that the temperature (at the horizon) is inversely proportional to radius (distance to horizon) and proportional to the square root of the cosmological constant implies breaking of super symmetry. See for example Temperature at horizon in de ...


2

In the situation you gave, it's immediately clear what is meant, and there's no possibility for misinterpretation, so yes, it's perfectly acceptable. (Remember that torque is mathematically defined as a vector for convenience, but the direction of that vector isn't really physical.) The only issue I can see with that is that as you leave the simple ...


1

Indeed, the two effects are very much related! I don't know how your background is, so let me start by defining the four-vector $x^\mu=(t,x,y,z)=(t,\vec{x})$ such that $x^0=t$ and $x_i=x,y,z$ for $i=1,2,3$. (Note that it is convention that greek indices run from $0$ to $3$ (space-time) while latin indices run from $1$ to $3$ (space only). Summation over ...


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I think you are attaching excessive significance to your order of magnitude estimate. If I say something is $10^5$ to within an order of magnitude I mean that the $\log_{10}$ of that quantity is nearer $5$ than $4$ or $6$. If some quantity has a $\log_{10}$ of about $5.5$ does it really matter whether our order of magnitude estimate is $10^5$ or $10^6$? ...


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According to commentator "bn" at The Chalkboards of Quantum Mechanics Professors (via Wired), this one is mostly about holography, how much information can be contained on a 2-D screen of a certain size. The diamonds are referring to domains of causal dependence. The square brackets are commutator brackets (Canonical commutation relation - Wikipedia), and ...


1

Is it therefore, incorrect to talk about "expectation value of an operator"? Yes, because when you write those integrals you're asking for the average of a dynamical variable $A$ whose associated operator is $\hat A$. The point is that you're asking for a number , e.g. the average position of the electron in a gaussian wavepaket. Then, when you want to ...



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