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4

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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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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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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The states inside the Fermi sphere are occupied, so $c^{+}_{k,\sigma}$ applied to any of those states gives zero. Not a state with zero electrons, but an identically zero state function. Not helpful in this context. On the other hand, if one applies $c_{-k, -\sigma}$ to states inside the Fermi sphere (hence occupied), an electron having properties ($-k, ...


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They are shown at the $\Gamma$ point in special diagrams called the reduced zone scheme in which a band will be shown folded back on itself. This way of showing the band structure is convenient for a few reasons, one of which is that it saves space on the page. If you look at that band gap at $\Gamma$ and follow the lower band down to lower energies, you ...


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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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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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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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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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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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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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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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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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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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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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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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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 ...


1

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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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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I may be misinterpreting the situation (as you didn't mention which book you're getting the expression from), but assuming that $\mathbf{n}=(n_x,n_y,n_z)$ is actually the location of the center of the cell in question, by assuming an infinite spatial extent you naturally have the identity $$\begin{align} \Phi_{\mathbf{n}ai}^{\mathbf{m}bj} = ...


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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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The Fermi Sphere Imagine a filled Fermi Sphere, every state with $k<k_f$ is occupied and the states with $k>k_f$ are empty. Here the particles we are talking about are electrons, hence a state is occupied by an electron with a momentum $\mathbf k$ and spin $\sigma$. $$ |F\rangle=\prod_{|\mathbf k| <k_f,\sigma}c_{\mathbf k,\sigma}^{\dagger} ...



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