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## Hot answers tagged terminology

8

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 ...

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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 ...

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Refs. 1 and 2 define a canonical transformation (CT) $$\tag{1} (q^i,p_i)~\longrightarrow~ (Q^i,P_i)$$ [together with choices of Hamiltonian $H(q,p,t)$ and Kamiltonian $K(Q,P,t)$] as satisfying $$\tag{2} (p_i\mathrm{d}q^i-H\mathrm{d}t)-(P_i\mathrm{d}Q^i -K\mathrm{d}t) ~=~\mathrm{d}F$$ for some generating function $F$. On the other hand, Wikipedia (March ...

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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, $L = n{h ... 3 The short answer to (1).$F^\mu{}_\nu$and$F^{\mu\nu}$are related by$F^{\mu\nu} = g^{\nu\rho}F^\mu{}_\rho$where$g^{\mu\nu}$is the metric ($g^{\mu\nu} = \operatorname{diag}(1, -1,-1,-1)$in Minkowski spacetime). Since the metric is invertible, either of$F^\mu{}_\nu$and$F^{\mu\nu}$uniquely determines the other. You can pick whatever version you ... 3 Without a doubt, it is the zeroth law of thermodynamics, as it defines an equivalence relation. It states that If two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. 3 Because as far as we understand general relativity, it's not doing "the opposite of what gravity does." Gravity can be locally attractive or repulsive, depending on whether the stress-energy content satisfies or violates the strong energy condition. For ordinary matter, the stress-energy is dominated by the mass, the SEC holds, and its gravity is attractive. ... 2 In the situation you gave, it's immediately clear what is meant, and there's no possibility for misinterpretation, so yes, it's perfectly acceptable. (Remember that torque is mathematically defined as a vector for convenience, but the direction of that vector isn't really physical.) The only issue I can see with that is that as you leave the simple ... 1 Before the kink it had momentum density$\frac{\vec{p}}{V} = \frac{\rho V \vec{v} }{V} = \rho \vec{v_1}$; afterward$\rho \vec{v_2}$. The change in momentum density is$\rho (\vec{v_2} - \vec{v_1})$. Multiplying by$A v t$(the volume that moves past the kink in time$t$), we get the change in momentum that must be supplied to maintain the kink (i.e. the ... 1 In the relativistic theory, you suppose a metric tensor to be$g^{\mu\nu}=g_{\mu\nu}$such that$\left(ds\right)^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$is the infinitesimal length element. For special relativity, you define$g_{11}=g_{22}=g_{33}=-1$and$g_{00}=1$. The$0$-component stands for the time and the$1,2,3$-components stand for the space (other ... 1 The notation is short-hand for an expression utilizing the Backer Campbell Haussdorf formula. Let$X$and$Y$be operators, then $$e^{x}Ye^{-X} = Y + [X,Y] + \frac{1}{2!}[X,[X,Y]] + \frac{1}{3!}[X,[X,[X,Y]]] + ...$$ I assume$[X,Y]_{(n)}$refers to the$n\$th term in this expansion; it roughly counts how many times the commutators are nested in each other. ...

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This is not standard notation, and one would typically expect any text that uses it to define it at its first occurrence. Since you understandably cannot provide us with a reference, your best bet is hunting for all occurrences of that notation, starting from there and going up through the text, until it explains what it means. Trust me, it will be there.

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