Recent content by voila

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    Why Does Diamond/Silicon Have 8 Bands in Its Band Structure?

    Just to update with the answer: In the case of diamond, there's a two atom basis. In this atom basis, we can think of the orbitals as the bonding (filled) and anti bonding (empty) orbitals, which are the ones to give rise to their respective bands. In the case of hydrogen, there's just one atom...
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    Why Does Diamond/Silicon Have 8 Bands in Its Band Structure?

    Just found around that each sigma sp3 bond splits into a bonding and an anti bonding state, the former being full and viceversa. The bonding states correspond to the therefore valence bands and the anti bonding to the conductions band. I've got no problem with that, but it seems I have a...
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    Why Does Diamond/Silicon Have 8 Bands in Its Band Structure?

    Homework Statement I am asked to discuss the band structure of diamond. I saw the band structure of diamond has 4 filled valence bands and then 4 conduction bands. Silicon, the same. Homework Equations --- The Attempt at a Solution I'm feeling really silly because I don't understand why it is...
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    Partial derivatives and thermodynamics

    @Charles Link Let me ask it to the point: is ##\left(\frac{\partial u}{\partial u}\right)_T=0## (assuming ##u=cRT##)?? I think that could kind of make sense, though I had never considered it. Maybe knowing that I may avoid future problems (but I can't promise it!).
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    Partial derivatives and thermodynamics

    I know keeping T, v constants forbid me from changing the energy, but still I applied the definition correctly, right? The thing is, I know you are right: "the variation of pressure with respect to energy, keeping v and T constants" makes no sense because it is physically impossible. However, it...
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    Partial derivatives and thermodynamics

    Then how is it that ##\left(\frac{\partial P}{\partial u}\right)_{T,v}## can give a different result depending on how I pick it?
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    Partial derivatives and thermodynamics

    The problem I seem to have is the following. Say I want the variation of P with respect to u keeping the v and T constants. Starting from the ideal gas law: $$\left(\frac{\partial P}{\partial u}\right)_{T,v}=0$$ since T and v re constants. However, if I start from ##P=\frac{u}{cv}##, I get...
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    Partial derivatives and thermodynamics

    Hi all. Suppose I have the ideal gas law $$P=\frac{RT}{v}$$If I'm asked about the partial derivative of P with respect to molar energy ##u##, I may think "derivative of P keeping other quantities (whatever those are) constant", so from the formula above I get $$\frac{\partial P}{\partial...
    • voila
    • Thread
    • Derivatives Partial Partial derivative Partial derivatives Thermodyamics Thermodynamics
    • Replies: 9
    • Forum: Thermodynamics
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    I Values of Lagrange multipliers when adding new constraints

    Say we have a Lagrange function with one multiplier a times a constrain. I minimize and solve the system to find a. I now add another constrain to the same system multiplied by the constant b. Is the value of a the same or can it change?
    • voila
    • Thread
    • Constraints Lagrange Lagrange multipliers
    • Replies: 1
    • Forum: Calculus
  10. V

    I Why does low surface energy imply hydrophobicity?

    @hilbert2 Say you have a metal surface. How are the bondings between the metal and the liquid?
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    I Why does low surface energy imply hydrophobicity?

    I don't understand why materials with low surface energy are hydrophobic and viceversa. All I can find are quick phenomenological explanations that don't quite deal with the physical (microscopic) process going on. Could anyone provide a good microscopic picture of why it is that way? What's...
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    I Is the Fermion number operator squared equal to itself?

    @Demystifier yes I did, @stevendaryl pointed that out. Thanks everyone for the answers, I understood it.
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    I Is the Fermion number operator squared equal to itself?

    How come?
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    I Is the Fermion number operator squared equal to itself?

    Yup, I meant number operator for a fermionic state (number of particles equals either 0 or 1). So ##\exp({\hat{n}})=\sum{\frac{\hat{n}^k}{k!}}=\sum{\frac{\hat{n}}{k!}}=\hat{n}\sum{\frac{1}{k!}}=\hat{n} e ##?
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    I Is the Fermion number operator squared equal to itself?

    What the title says. Acting on a fermionic state with the number operator to a power is like acting with the fermionic operator itself. Does this allow us to define ## \hat{n}^k=\hat{n} ##? Or is there any picky mathematical reason not to do so?
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