# Tag Info

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 ...

6

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 ...

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 ...

0

I disagree with @Raskolnikov I think the principle of equivalence is described by the carnot engine: So basically $$Q_H =Q_C + W$$ This is equivalence as I understand, it doesn't have to be a law.

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.

0

This is all just terminology. 'Force' is a term from Classical mechanics really. 'Fundamental Force' is a term for any one of the set of four theories, gravity, and the three Standard Model interactions, Strong nuclear, weak nuclear and electromagnetic. The strong nuclear interactions (plural) for example could be said to be eight 'forces'. It's just that ...

0

Okay, I think I got it after reading a lot of scattering-related papers. I will omit the wavefunctions normalization and the constants like $\hbar$, $m$, $\pi$, etc., where possible since the precise values are not so relevant. Consider a 2D rectangular pipe (for simplicity), infinite in $x$ direction and of finite height $H$ in $y$ direction. Under the ...

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. ...

0

The term "Dark Energy" is just a name that we have given it while we try to determine what exactly it is and what a better name for it would then be. However, calling it an energy is appropriate. Our best model, the cosmological constant, says that dark energy has a constant energy density (that is a constant amount of energy per unit volume) in 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 ...

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 ...

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. ...

3

I think this is the Baker-Campbell-Haussdorff formula, and the notation means to iterate the commutator. That is, $$[L, M]_1 = [L, M]$$ And $$[L,M]_{n+1} = [L, [L,M]_{n}].$$

1

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.

4

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 ...

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 ...

3

"Equilibrium" means thermal equilibrium. The solid has one well defined temperature, and a constant Fermi energy. The Fermi energy is an energy value against which energy levels are compared to determine how fully occupied (or not) an energy level is. Generally when the Fermi level is constant throughout a solid electrons diffuse equally in all ...

0

Terminology doesn't matter much as long as people understand the ideas involved. You should call and idea whatever you want to call it as long as you are clear about the substance of your ideas. Some commentators have stated that the Standard Model or other scientific results are well-grounded. So far, these ideas have not been refuted. If somebody invented ...

Top 50 recent answers are included