Tag Info

Hot answers tagged

5

This can be computed for small changes in the pressure by considering the partial derivative of the temperature w.r.t. pressure at constant entropy. If we suddenly raise the pressure a bit, then this is to a good approximation an isentropic process as no heat is exchanged and it is not a violent process causing large irreversible effects. So, we want to ...


5

Not sure about the condensed matter context, but in general the answer is NO. For instance, quarks have fractional charge but are regular fermions.


4

The thermal expansion coefficient for the frame poles must be larger than the thermal expansion coefficient for the fabric. This is a factor to keep in mind whenever you're considering thermal expansion for two materials at once. For instance you could have a container full of fluid which can be filled to the brim at one temperature, then spill when heated ...


3

Optical conductivity and AC electric conductivity experiments are indeed quite similar. However, they operate in different frequency regimes and measure slightly different quantities. Optical conductivity refers to an experiment using light, such as a reflectivity measurement and then using a Kramers-Kronig transform to deduce the real part of the ...


3

In a metal the Fermi energy is somewhere in an unfilled band. At any temperature above absolute zero (which you can never reach) there are states available for electrons to get to and result in conduction at the Fermi surface. This will occur in any metal. Superconductivity is a separate phenomena that I won't touch on here.


2

At present, there is a belief (though obviously not verifiable) by solid-state physicists that a metal cannot exist at absolute zero. The Fermi surface of the metal will be unstable to order of some sort such as superconductivity, charge density waves, magnetic ordering, etc. With that said, let us concentrate on your scenario though. If there are no ...


2

Tungsten has been known to bait gold bars (historically). There are a few methods we use to determine if something in front of use is gold or if it is alloyed, or if its plate, fill or scrap. You can cut the bar in half...You will then know immediately of you got bunk gold. You can do a specific gravity check of your gold. There are scales designed for ...


2

No, “this conclusion” is based on the topological properties of rotation groups. Namely, for any $n > 2$ $\mathrm{Spin}(n)$ is the universal cover of $\mathrm{SO}(n)$, whereas for $n = 2$ it is not. That’s why in $n > 2$ any thing has to be controlled by a representation of the Spin group, whereas in $n = 2$ it has not.


1

There is an ambiguity. Although I did not understand your analysis of the problem completely, charge carriers certainly can run against the (averaged) electric force due to difference in available bands and other particle statistics effects. The gauge freedom is irrelevant. There are two cases for the “ubiquity”. First, these non-Maxwellian deviations ...


1

In the image above, you can see a series of Bragg planes drawn in the crystal. This is called one "set of planes". Another "set of planes" would be if one would just draw a series of horizontal lines through the atoms. (Of course by lines I mean planes, but they are projected here onto a 2D image). The planes are those formed by the atoms, so in that ...


1

If time-reversal symmetry is broken at the surface of a topological insulator, a gap could open at the Dirac point of the topological surface state. The Dirac point, where forward- and backward-moving electrons have the same energy, is located at a time-reversal invariant momentum point (also called a Kramer's point) in the reciprocal space (the crystal ...


1

There may be a few options for describing the crystalline order. Personally, I love the one used by Anderson in his book 'Basic Notions of Condensed Matter Physics'. Following him, one writes the atomic density as $\rho(\vec{r}) = \sum_{\vec{G}}\rho_{\vec{G}}e^{i\vec{G}\vec{r}}$. The appearance of Crystalline order is signified by a set of finite ...


1

The spin referred in condensed matter is the spin of the electrons least bound to the atoms (usually valance electrons). The atoms reside on the lattice sites. A spin half problem means the atoms have only one valance electron. But there are other possibilities like spin 1, 3/2 and all. As qeb has already mentioned it can also be used for nuclear spins also. ...


1

A spin-half on a lattice site is a theoretical 'particle' with the property of having a spin of one half. It can be either 'up' or 'down' along the measuring-axis (in most textbooks the spin operator for assigning spin up-or down is the $\sigma_z$ operator, the third Pauli matrix. You could also use the $\sigma_x$ or $\sigma_y$. I believe the spin on a site ...


1

The answer is 'no' in condensed matter also. As anyons are neither bosons nor fermions, they can follow some statistics other than BE or FD, but it has nothing to do with fractional charge. just-learning has already given you a perfect example.


1

This is a really interesting question, and I've no precise answer... So let me just tell you a few random things. First, I'd like to say that mesoscopic physics emerges simply because it was technologically available, as most of the emerging field actually. It is indeed a big field, also because of its potential industrial interests (the famous nanoscale = ...


1

Wait a minute. Surrounding the protons are a matrix of electrons. These are the electrons not "cool enough" to exist in the conduction band. I imagine these are what collide with the electrons in the Drude model approach. The closer you get to those electrons, the more they're going to push back by the Coulomb potential. Furthermore, I imagine some Pauli ...


1

The Drude model is fine for thinking about some things. (It is still taught.) The electrons collide with the atoms. (or a bit more precisely, the outer electrons of the atoms.) Since this is a classical picture perhaps it's OK to have a classical picture of the atoms. Imagine they are little sphere's all joined to the other atoms by little springs. ...


1

I can answer that yes, momentum is definitely transferred to the surrounding lattice, though not through a direct scatting of electrons with protons. This is important in increasingly small electronics. When the cross-section of the wire or metal is extremely small, the collisions can be enough to displace the metal, and wreck a circuit. Tungsten is ...


1

Without doing the experiment I think it would be hard to pin down the precise reason. However my guess is that this is because most fabrics become more elastic, i.e. less stiff, in the presence of water. In the heat of the day when you pitch the tent the fabric is dry and relatively stiff, but if it rains or dew forms overnight the fabric becomes less ...


1

Your reasoning is fine and indeed the band gap of silicon is $1.12$ eV, which is $43kT$ at room temperature, so thermal promotion of electrons from the valence to the conduction bands at room temperature should be negligable. The trouble is that it's exceedingly hard to get silicon so pure that there are no gap states, and while the conductivity of ...


1

Have you heard of superconductivity? This is a phenomenon where a material exhibits zero resistivity near absolute zero: it clearly contradicts your assertion that thermal excitation is needed for conductivity near absolute zero. For a semiconductor, it is true that electrons need to be kicked into the conduction band by thermal fluctuations - but for a ...



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