Hot answers tagged

6

No because solid is a state of matter. Light cannot be considered matter since it is made up of particles which have no mass and I'm pretty sure occupy no space (i.e. photons have no volume). Edit: Since photons are at the quantum level, we can't actually fathom what it would mean for them to occupy space. But on this thread someone pointed out that there ...


5

In general the wavenumber is a vector. That is, $e^{i(\vec{k}\cdot\vec{x}-\omega t)}$ is a solution to the wave equation in 3 (or any number) dimensions. We say this solution is a plane wave propagating in the $\hat{k}$ direction with wavenumber $|\vec{k}|$ or wavelength $\lambda = 2\pi/|\vec{k}|$. So properly the de Broglie relation is $\vec{p} = \hbar \...


5

Conclusion: The particles are, in-fact, magnetic This is doubtful. There are two possibilities here: If the objects used to grind (let's call them grinders) the dust grains were made of a ferromagnetic material like iron/steel and/or nickel, then it is possible that the grinders deposited some material onto the dust. There is no reason to think that ...


4

No, we know enough of the "bulk properties" of antimatter to rule this out. Antimatter interacts with the electromagnetic field in exactly the same way as regular matter, just with the opposite charge. Therefore, antimatter should be detectable using most of the techniques we use to detect regular matter in astronomy. This works even if the antimatter is a ...


4

The result $\omega^2=\frac{c_1+c_2}{M} \pm \frac{1}{M}\sqrt{c_1^2+c_2^2+2c_1c_2 \cos ka}$ leads to two real solutions for $\omega^2$, since $ -1 \lt \cos ka \lt 1$ and the square root lies between $|c_1 - c_2|$ (for $\cos ka = -1$) and $c_1 + c_2$ (for $\cos ka = +1$), so that $\omega^2$ is always positive. The frequency $\omega$ is taken as a positive ...


2

This is a general property of square potential wells. De broglie wavelength of the ground state is approximatly twice the well width. For this reason wider wells have smaller wave-numbers ($k=\frac{2 \pi}{\lambda}$) of the groundstate. Hence the lower energy of the ground state (and others) (since $E=\frac{\hbar^2 k^2}{2 m}$).


2

Mostly kinetic energy. The kinetic energy of a free particle is not quantized. It becomes so when the particle is closed in a box. But even in this case the energy levels are often so closely spaced that the spectrum is almost continuous. In fact, if you solve the Schroedinger equation for a particle in a 1D infinite square well you will find the following ...


2

The fractional quantum Hall effect (FQHE) is a physical phenomenon in which the Hall conductance of 2D electrons shows precisely quantised plateaus at fractional values of e^2/h . It is a property of a collective state in which electrons bind magnetic flux lines to make new quasiparticles, and excitations have a fractional elementary charge and possibly ...


2

Ordinary silica glass really is amorphous. You can study the short range order of a material by measuring its radial distribution function. That is, take any atom and plot the density as a function of distance from that atom. This is easily measured using X-ray or neutron scattering. In a crystal the RDF is just a function of the crystal structure and is ...


2

I think we can pretty safely say this is not the case. The main reason is that we have a pretty good idea of where dark matter is--to some degree, it can be reconstructed from the gravitational influence it has on surrounding matter. The dark matter appears to be distributed evenly throughout the galaxy. Thing is, a galaxy is pretty "dirty," as far as space ...


2

For wave mechanics there is the phase velocity and group velocity. For the energy $E~=~\hbar\omega$ the phase velocity is $$ v_p~=~\frac{\omega}{k}~=~\frac{\hbar}{2m}(k~+~ck^3). $$ This is the velocity of a wave front, or where the phase of the wave is constant. There is also the group velocity that is $$ v_g~=~\frac{\partial\omega}{\partial k}~=~\frac{\hbar}...


1

Your question is really broad. As correctly pointed out by @John Rennie there can be so many signals that can pass from the solids. It must be noted that even if the signal can pass through solid it will experience certain losses. Mechanical waves such as sound took advantage of elasticity of the material. The sound oscillations are transferred through ...


1

Remember that $\eta^{q}$ are Fermionic operators, therefore $\eta^{2}_{q}=0=\eta^{\dagger 2}_{q}$.


1

One way to do this is to start by solving for the allowed $\mathbf{k}$ values for an $L \ \times \ L \ \times L$ crystal using the usual procedure found in any book on solid state physics (if you're unfamiliar, see A & M pg. 430-432). The allowed $| \mathbf{k} |$ values are integer multiples of $\frac{2 \pi}{L}$. The corresponding density of k-states is ...


1

In Solar Cell Device Physics, the author Stephen Fonash makes a distinction between "electrostatic (electric) fields" and "effective fields", and you have stumbled upon an example that needs the distinction. Electrostatic fields arise from a real electric charge density, which would be caused,for example, by uncompensated donor atoms. In your schematic as ...


1

This depends upon how you want light to solid. If you want light to be solid in the way the "Star Wars" movies have light sabers, I would say no. There are however materials that trap photons so they have zero velocity. Photons are in a sense trapped, and these are sometimes called artificial black holes. The energy of the photons then contribute a tiny ...


1

In general, or as far as I'm aware, the band diagrams and Density of States shown in your question are electronic band structures. You can also see this from the graph's legend which shows that you're dealing with d and p orbitals belonging to W and S. However, looking at the numbers in your question, I can see where the confusion comes from. The direct ...


1

The fluctuation-dissipation theorem is a relation between the symmetrized and retarded correlation functions $$ G_s(\omega,k)=\coth\left(\frac{\omega}{2T}\right) \,{\rm Im}\, G_R(\omega,k). $$ I would not say that it is due to the "$i\epsilon$" terms. The $i\epsilon$ simply reflects the different analyticity properties of $G_{R,A}$ etc. Physically, the ...


1

Luttinger liquids are a case of bosonization. Two fermions $\psi$ and $\psi'$ are written according to a boson field $\phi$ $$ \psi^\dagger\psi' = exp(i\phi), \psi'^\dagger\psi = exp(-i\phi^\dagger), $$ where the "field" $\phi$ serves as a phase and is real valued. For the fermion operator $c_j$ the number operator $n_j = c_j^\dagger c_j$ permits us to ...



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