Everything is quantum mechanical. The dynamics of any physical system will always follows the laws of quantum mechanics. It is only when you zoom out sufficiently far that the laws of quantum ...

You are comparing the spectral radiance in frequency $B(f,T)$ with the spectral radiance in wavelength $B(\lambda,T)$. Because these are differential distributions, you cannot simply replace $f=c/\... View answer Accepted answer 4 votes You're forgetting that for fixed$m$,$u_0$is a function of$L$.$J(u,L)$is function of two variables, not one, so your differential equation for "$J'(u)$" isn't so simple. If we take$m$... View answer Accepted answer 4 votes First off, a system composed of two independent spin-$1/2$particles, with definite spin-projections, would be represented by the tensor product state$|s_1,m_1\rangle\otimes|s_2,m_2\rangle$, not ... View answer 4 votes It was realized long ago that the idea of a "cross product"$(A_x,A_y,A_z)\times(B_x,B_y,B_z)=(A_yB_z-A_zB_y,A_zB_x-A_xB_z,A_xB_y-A_yB_x)$had surprising relevance in the real world (e.g. with torque ... View answer Accepted answer 4 votes The phrase "combined mass" is misleading here. It means "mass of everything except the stones". So the final mass of the boat (and everything else still on it) is indeed$M$. What's important is that$...

The article you linked merely provided counterevidence against one single direct dark matter detection done by the DAMA/LIBRA collaboration in Italy nearly 20 years ago. Nobody has ever found a ...

This is not true. See section 3 of this article where the full analytic expression is derived. For finite mass, we can write the $x^2\rightarrow 0^+$ fermionic propagator (the $x^2\rightarrow 0^-$ ...

As a quick extension to the above answers, let me repeat that none of quantum mechanics is "derived" from any preceding theories. Yes, there are many correspondences that are quite striking - ...

As a partial answer, I took the advice of @JonCuster and pulled the following equations from "Quantum Mechanics of One- & Two- Electron Atoms" by Bethe and Salpeter. All results are in C.G.S units....

The Hamiltonian of a rigid rotor electric dipole is given by $$H=\underbrace{\frac{p^2}{2m}}_{\text{translation}}+\underbrace{\frac{L^2_{\theta}}{2mR^2}+\frac{L^2_{\phi}}{2mR^2\sin^2\theta}}_{\text{... View answer Accepted answer 3 votes The area of this wedge, swept out by a Keplerian body in a time small time dt, is given by$$\begin{align*} dA&=\frac{1}{2}(r)(r\sin(d\phi))\\ &\approx\frac{1}{2}(r)(r d\phi)\\ &=\frac{...

Many simplifying assumptions of our reality must be made to arrive at the results you're looking for (and that the other answers gave). I'll try to give my own 2 cents. I hope that anything incorrect ...

I believe you are confusing the work done by the electric field with the work done on the particle. By the electric field: The radial force of the electric field is always pointing outwards, and the ...

The sum over states that you've written is supposed to be literally the sum over all states, with the restriction that the total spatial momentum of each state should be $\vec p=0$. You are in ...

To answer the question in your title, it's because it really is much colder up there. It's colder up there because atmospheric pressure decreases as you increase altitude, which is a result of gravity....

Current is amount of charge passing through a cross-sectional area per unit time. Consider a cross-sectional area $A$. Suppose in a time period $\Delta t$ you measure a total amount of charge \Delta ... View answer 2 votes Polarization is, generically, a normalized vector in spin-space, i.e. the spin-state of a particle (When I say spin, I actually mean total angular momentum, spin+orbital angular momentum). Usually we ... View answer Accepted answer 2 votes With the help of @Slereah, I found the PDG article regarding the mesonic decay constant. Leptonic Decays of Charged Pseudoscalar Mesons In that document, you can find a summary of all the known ... View answer Accepted answer 2 votes I believe they are making a pedagogical point here that for you may be intuition/obvious. The point of this section is that there may be more distinct terms in a Lagrangian than there are free ... View answer 2 votes Maxwell's (local) equations in a linear, homogeneous, and isotropic medium read \begin{align*} \nabla\cdot\mathbf{E}&=0\\ \nabla\cdot\mathbf{B}&=0\\ \nabla\times\mathbf{E}&=-\frac{\... View answer 2 votes First off, just to set entropy straight, classical (Boltzmann) entropy is an emergent phenomenon in the study of multi-component systems. We may know the basic dynamical principles of each individual ... View answer 2 votes Velocity is defined as dx'/dt', which in your case comes out to be:v'=\frac{dx'}{dt'}=c\tanh\left(\frac{g\tau}{c}\right)$where$g$is the proper acceleration. Notice that as$\tau\rightarrow\...

The energy (per particle) is defined up to a global constant/reference. If we scale the energy per particle (of all particles!) by a constant $\mu_{\text{ext}}$, then if we add another particle to our ...

First let me clarify something. I think what you mean by "Coulomb's law" is the solution to the electrostatic Poisson equation with the assumption (boundary condition) that it vanishes at spatial ...

Entropy is a measure of our ignorance of a thermodynamic system, or in its precise mathematical form (Boltzmann's entropy formula), $S=k\log \Omega$, where $\Omega$ is the number of microstates of the ...

As others have stated, it really depends on why you want to learn quantum mechanics, and how deeply you want to learn it. (1) If you want to learn it as badly as you want to watch a movie at the ...

This is an answer to the point I brought up in the comments. To begin, I'd like to point out that the quoted formula for the gravitational binding energy of a spherically symmetric body isn't quite ...