Hot answers tagged

2

There is a well-known isomorphism between the linear space ${\mathcal M}_{m, n}$ of $m\times n$ matrices and typical (vectorial) linear spaces ${\mathcal L}_{m\times n}$ of dimension $\text{dim}({\mathcal L}_{m\times n}) = m\times n$. Everything that is valid in ${\mathcal L}_{m\times n}$ has an equivalent in ${\mathcal M}_{m, n}$ and conversely. For this ...


2

My main concern is: does this blurring/loss of knowledge come from any well-known physical law/principle? For instance, should we link the quantization of the phase space to ΔxΔp≥ℏ2ΔxΔp≥ℏ2, or to some kind of observer effect? The description in the Wikipedia article is misinformed and misguiding. The blurring, or coarse-graining, is merely one possible ...


2

The physical principle being invoked is the finite resolution of any experiment, independent of the value of $\hbar$, together with coupling between observable and microscopic degrees of freedom, i.e. it applies to both classical and quantum systems. Technically energy conservation and Liouville's theorem, or unitarity in QM, are also needed to prevent ...


1

In general, $\frac{\partial L}{\partial \dot{q}}$ is the canonical (or generalized or conjugate*) momentum, and $m\dot x$, for $x$ the actual position, is kinetic momentum. Likewise, the cross product of the former with the generalized coordinate vector $q$ might be called "canonical angular momentum", and the cross product of the latter "kinetic angular ...


1

This is more or less an exercise in chasing definitions. The adiabatic invariant $I$ is defined as $$ I\equiv \oint p \frac{\mathrm{d}q}{2\pi}\tag{49.7}$$ where the integral is taken over the path for given $E$ and $\lambda$. The external parameter $\lambda(t)$ is a slowly varying function of time $t$ in $\S49$, but is assumed to be a constant in $\S50$. ...


1

It is, a priori, completely correct to add both primary and secondary constraints to the Hamiltonian density by Lagrange multipliers. What is not correct is how you determined the equations of motion: There is no "$F^{i0}$" in the Hamiltonian theory! It is called $\pi^i$ there and it is not dependent on $\partial_i A_0$, it is an independent canonical ...


1

The derivation of ideal gas equation from Hamilton's equations will take the same procedure as what you have seen in Wikipedia. Since you said you haven't understood the way in which the equation is derived I will give you a step by step explanation on it. So we have a system of perfect gas molecules. Of course they are non-interacting. We are going to ...



Only top voted, non community-wiki answers of a minimum length are eligible