# Tag Info

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Basically, the egg-white (just another protein structure) is made of long chains of amino acids. They're held in shape by weak bonds. When you heat the structure, the energy you put in, is enough to break those bonds, thereby destroy the structure. This process has a name for it. Once the temperature is high enough, new covalent bonds can form between the ...

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A very simple diagram shown below is from the wikipedia article on Permittivity. "Electronic" encompasses level/band transitions, atomic represents such phenomena as vibrations (as shown) and rotations (which is how microwave ovens work). Here is a more complicated spectrum for semiconductors/solid state matter taken from here.

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TylerHG: Yes it is easy to calculate the density of states. But what I'm really asking here is "why." Note that a thin circular ring in $\mathbf{k}$-space of thickness $dk$ has area $dA=2\pi k\,dk$ (by elementary geometry). In $E$-space, since $E\propto k^2$, that ring corresponds to a patch of width $dE=2k\,dk$. Thus $$\frac{dA}{dE}=\pi.$$ But ...

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My quick read of a few articles on Q, the reaction quotient, suggest that it's generally rather small. Thus $ln(Q)$ is negative, so $- T * ln(Q)$ is positive and the potential difference increases with temperature.

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It is easy to show that the total number of electrons in a 3D fermi sphere is : $$N(e)=\frac{V}{3\pi^2}*k_F^3$$ Where $k_F$ is the Fermi wave vector and $V$ is the real space volume of your sphere. Now if you rearrange for $k_F$ in terms of the total number of electrons you'll get a particular equation. It is know that ...

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To address the follow up question, which requires more characters than a comment allows: In a simple idealized view, the Fermi level is the top energy level in the solid occupied by electrons. In silicon with no doping it sits at mid-gap: the valance band is full, the conduction band empty. In a thought experiment, if you had two separate chunks of ...

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But why when I connect a volt-meter across the whole diode will I not see this built in potential? Because at equilibrium (V=0) the Fermi-level throughout the whole device is flat. A voltage will only exist when there is a gradient in Fermi-level over the component (i.e. a voltage drop). Edit Maybe a more visual explanation would help? Think of the ...

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The classical situation with no symmetry breaking is the case of the, so-called, isostructural transitions. The word "isostructural" is misleading, since what is meant is "isosymetric". However, historically the term emerged. There is a number of examples of such transiotions. One is the alpha-alpha' transitions in the hydrogen-metal systems, another is ...

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Let's first consider a finite crystal with $M$ cells. By Bloch theorem we have $$\psi_n(\vec k)=e^{i\vec k\vec x}u_{n,\vec k}(\vec x),$$ where $u$ is periodic with crystal period. Now, for fixed $n$ we have exactly $M$ points in Brillouin zone, each corresponding to some $\vec k$. In some sense, $M$ can be understood as proportional to volume of Brillouin ...

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Here we go : Lets considere the lagrangian density of our free membrane. For convenience, we will use a cartesian set of coordinates, so that : $$\mathcal{L}=\frac{\rho}{2}\dot{h}-\frac{\kappa}{2}\left[{\partial_x}^2 h+{\partial_y}^2 h\right]^2$$ Apparently, this lagrangian depends on a quite large number of variables : ...

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Quantised energy levels/bands arise from electrons moving in a potential well. In say a white dwarf, the degenerate electrons can have kinetic energies of order 100-keV to 1MeV and can usually be assumed to behave like (almost) free particles. In which case there are a set number of available momentum eigenstates per unit volume $$g(p)\,dp = \frac{8 \pi ... 0 Why do you say for N particles you have N! distinguishable state? Lets just make sure you got it right. Lets say you have N particles on a lattice having M sites. In the ideal case, no excluded volume therefore each particles can access M sites. The total number of states for distinguishable particles in M^N. Now, if you considere that the particles are ... 1 With$$u(n)=u_0e^{-i(\omega t+k n a)}$$we have$$p=\sum_{n=1}^Nm\frac{d}{dt}u(n)=i\omega m u_0e^{-i\omega t}\left(\frac{1-e^{-iakN}}{1-e^{iak}}\right).$$With cyclic boundary conditions we have$$u(N+1)=u(1)\Rightarrow k=\frac{2\pi j}{Na}\text{ for }j\in\mathbb{Z}$$and inserting k into p gives$$p=i\omega m u_0e^{-i\omega t}\left(\frac{1-e^{-2 i \pi ...

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Many places, and especially in everyday life, metals are defined by examples and descriptions of their properties. In physics(solid state physics) metals are defined by the Fermi level within an electronic band. I do not see anything that makes copper special from other metals in this sense.

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It's all about the electronic structure. "Electronic structure" is the term for the available energy states and transitions of electrons in the crystal. Largely, the nuclei in a crystal are much of a muchness; big positive things that don't move a lot. The electrons, however, can move around to different degrees depending on how many of them there are, the ...

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The formula you use to calculate the skin depth is just an approximation and not necessarily valid at THz frequencies. How do you model the optical properties of gold in your simulation? If you know the refractive index $n$ at a given frequency $\omega$, you can easily calculate the exact skin depth ($1/e$ amplitude decay length): $c/(\omega\text{Im}(n))$

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I am not sure I understand your question either but will try something else. When you write a probability distribution $\mathbb{\pi}: \: \mathbb{R}^3 \rightarrow \mathbb{R}_+$ for the cartesian variables $x,y,z$, you can also look at the induced probability measure for the function $r^2 = x^2 + y^2 + z^2$. To figure out what is the corresponding probability ...

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I am not quite sure I understand your question. In your second equation, $E$ is treated as a variable and not as function anymore, so why do you want to find $E$ function? The density of states is basically a function counting the number of state in $x\in\mathbb{X}$ that give the same energy: $\Omega(E_0)=|\{x:E(x)=E_0\mbox{ and } x\in\mathbb{X}\}|$ ...

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This is expected. For simplicity, consider a one dimensional lattice, with two atoms per unit cell, let's call them A and B. This misses some details of the three-dimensional Si structure, but it gets to the kernel of the argument. First look at what is happening at the $\Gamma$ point. For the LO branch, each "molecule" in the lattice is vibrating in ...

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Fermi pockets (or Fermi surfaces) are contours of Fermi energy in the Brillouin zone. Depending on the effective mass $m^*$ of quasi-particles, the Fermi pockets can be divided into electron pockets (if $m^*>0$) and hole pockets (if $m^*<0$). For weakly interacting Fermion systems, according to the Fermi liquid theory, all the low-energy physics ...

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Not possible, because if you want a very sharp blade, it should be very thin. On the other hand, cut material into two parts need to break the chemical bond of the molecules. However the blade is too thin that even if it cut off the bond(rather than saying get through), the bond will form again.

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Because in the bulk, the energy between two dislocations is proportional to the distance between them. This means the dislocations are confined in the bulk. So they can not appear in side the crystal.

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The surface of the film is a [001] plane through the silicon (or diamond) lattice, space group Fd-3m (#227),

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Such a blade (zero pressure required to cut a solid material) is not achievable. Even if you can produce an indefinitely thin edge, the body of the blade still has to force the sides of the cut apart. Think of cutting a steel block which is, say, one inch thick. Let's say that the blade is 1/2 inch deep with a back edge 1/8 of an inch wide. When you've cut ...

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A blade must be near-atomically thin to generate immense local pressures to cleave "anything." The edge must also be immensely stiff to avoid contact deformation. The brittle edge could be mounted in a tough matrix to avoid brittle failure, or erode as it cut to maintain its edge (e.g., pattern-welded Samurai/Damascus steels, Talonite; uranium long rod ...

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Pressure is a measure of the force applied to a given area. Blades are sharp because they have a small cross sectional area, allowing you to create very high pressure whilst applying only a modest force. This force generates so much of shear stress on the object getting cut that it crushes through the molecular bonds in that object. Cutting through something ...

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I think that no blade made of atomic matter (not degenrated matter) would be able to cut through everything else (like neutron star for example). As for a blade made from degenrate matter I also think that it would not be possible, because it takes huge amounts of force to keep matter stable in such form.

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One can notice that: $$(n_{i\uparrow}-1/2)(n_{i\downarrow}-1/2) = n_{i\uparrow}n_{i\downarrow} -\frac{1}{2}(n_{i\uparrow}+n_{i\downarrow}) +\frac{1}{4}$$ To show the equivalence you can absorb the $(n_{i\uparrow}+n_{i\downarrow})$ term in the chemical potential. We don't care about the kinetic term, and have:  ...

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The dependence comes through the magnetic flux, $\Phi$, through the sample. The number of states per Landau level is proportional to $\frac{\Phi}{\Phi_0}$, where $\Phi_0=\frac{hc}e$ is the fundamental quantum of flux. For a given sample, the total number of charge carriers if fixed. Thus, as $B$ and hence $\Phi$ increase, the number of filled Landau levels ...

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These two models are exactly the same. The first model has the additional advantage that particle-hole symmetry takes place at $\mu = 0$. The second model takes place at $\mu = U/2$.

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A common structure for LEDs the ppin-junction, [p-layer][p electron blocking layer][i layer][n layer] The p-layer, i-layer and n-layer is just your standard pin-junction structure. The p and n layers provide an electric field, which under forward bias will drive electron and hole towards i-layer where they can recombine radiatively. Normally the p and n ...

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Let me answer your first question: Phase transition do not necessarily imply a symmetry breaking. This is clear in the example your are mentioning : The liquid-gas transition is characterized by a first order phase transition but there is no symmetry breaking. Indeed, liquid and gas share the same symmetry (translation and rotation invariance) and may be ...

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Notice that the Fermi-Dirac integral comes from the use of the Fermi-Dirac distribution. Another way to define a system degenerate or non degenerate is to compare the Fermi temperature $T_{F}$ to the actual temperature $T$ of the fermionic system under the study.

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The reason is the following. If you write down the formula for Fermi-Dirac distribution, you can see that when x is negative, expanding the Fermi-Dirac distribution function using Maclaurin series, you will get the Boltzmann distribution which holds for a classical gas. So it means that your gas is nondegenerate.

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Your value is within the range of literature values. Hydrocarbon Lithography on Graphene Membranes states "the Fermi wavelength of the electrons in graphene of 0.74 nm". Many references cite this value. Another reference says ~0.14nm.

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The energies of electrons in tightly bound states (some of which might be referred to as "core states", depending on circumstances) are very nearly the same as the energies of the free molecule. These electrons do not interact very much with their neighbors, so their properties are nearly as if they are isolated. In such cases, with little or no overlap ...

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