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Conductivity of metals

  1. Mar 27, 2008 #1
    [SOLVED] conductivity of metals

    1. The problem statement, all variables and given/known data
    http://www1.union.edu/newmanj/lasers/LaserTypes/conductionbands.gif

    I do not understand this picture. Why does the fact that the valence band for the metal (conductor) is half-filled imply that it has high conductivity?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 27, 2008 #2
    There is no gap. The upper band is a conduction band. The bigger the gap between the conduction and valence band (the filled band), the lower the conductivity.
     
  4. Mar 27, 2008 #3

    olgranpappy

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    in the first pic there is a big gap between the valence and conduction. electrons can't be excited without absorbing energy on the order of the gap.

    in the second pic there is a little tiny gap.

    in the third pic the band is only half full and so electrons can be moved around to give a non-zero current without absorbing a large amount of energy. e.g. an applied dc electric field E can shift the fermi sphere by an amount [itex]e\tau \vec E[/itex] and give a non-zero current... which you can calculate by assuming the the equilibrium distribution function f_0(\epsilon_k) gets shifted to
    [tex]
    f(\epsilon_k)=f_0(\epsilon_k+e\tau\vec E\cdot\vec v)\approx f_0(\epsilon_k)+e\tau\vec E\cdot\vec v \frac{df_0}{d\epsilon_k}
    [/tex]
    and so there is a nonzero current
    [tex]
    \vec j \approx -e \int \frac{d^3 k}{{(2\pi)}^3}\vec v e\tau\vec E\cdot\vec v \frac{df_0}{d\epsilon_k}
    \approx e^2\tau \int \frac{d^3 k}{{(2\pi)}^3} \vec v \vec E\cdot\vec v \delta(\epsilon_k - \epsilon_{\rm fermi})
    [/tex]
     
  5. Mar 27, 2008 #4
    hmm...what is f the distribution function of? when you say the electrons can "move around," you mean move around in k-space, right?
     
  6. Mar 27, 2008 #5

    olgranpappy

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    electrons
    Yeah, the electrons get redistributed in k-space.
     
  7. Mar 27, 2008 #6
    How did you get this? Why is E dotted into v? Is v the drift velocity?
     
  8. Mar 27, 2008 #7

    olgranpappy

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    Well, the argument I'll give isn't rigorous, but can be made a little bit more rigorous using the Boltzmann equation... but still, can never be too rigorous, so whatever. Anyways...

    The distribution function for a bunch of free electrons (or the relevant electrons in a good free-electron-like metal) is given by
    [tex]
    f_0 = \theta(\epsilon_{\rm fermi}-\frac{p^2}{2m})
    [/tex]
    which is a sphere in momentum space (fermi sphere).

    Now, here's the "non-rigorous" part:

    Suppose an electron field acts on the electrons and that it can accelerate the electrons (change their momentum) but not indefinitely due to the finite mean free path (or time). We have to assume this because otherwise the electrons will accelerate indefinitely in the dc field and we won't find a current proportional to the field. So, we assume the force due to the field shifts the momentum of the electrons like
    [tex]
    p \to p +\tau F
    [/tex]
    where \tau is the MFT.

    you can picture this as shifting the distribution function (the fermi sphere) in momentum space. The new distribution function (f) is thus related to the old distribution function (f_0) by
    [tex]
    f(\vec p)=f_0(\vec p-\vec F\tau)
    [/tex]
    but since the distribtion function is only a funciton of the energy \epsilon, this is the same as letting
    [tex]
    \epsilon \to \epsilon - \frac{\vec p \cdot \vec F \tau}{m}
    [/tex]
    to first order in \tau (and note that F=-eE, and that v=p/m)

    then you can expand f in a taylor series to get what I got.
     
  9. Mar 27, 2008 #8
    What is \theta?
     
  10. Mar 27, 2008 #9

    olgranpappy

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    "theta" function ("step function") defined as
    [tex]
    \theta(x)=1
    [/tex]
    if x>0.

    [tex]
    \theta(x)=0
    [/tex]
    if x<0
     
  11. Mar 28, 2008 #10
    OK. Thanks for your really detailed explanation but I don't really have time to try and understand all of the approximations and the mathematics. Is there a more qualitative way to explain why that third picture represents a conductor?

    I kind of understand what you said about how the partially filled band means that there is room for the Fermi sphere to move around in k-space, but something is just not "clicking" for me.

    The definition of a conductor is something that has a high conductivity i.e. a large rate of current when you apply an electric field. So, is there a qualitative reason why having a partially filled energy band means that the electric field causes lots of current? I guess what I do not understand is that olgranpappy said "in the third pic the band is only half full and so electrons can be moved around to give a non-zero current without absorbing a large amount of energy." But the electrons ARE ALREADY MOVING AROUND APPARENTLY BECAUSE THE BANDS ARE IN K-SPACE.

    So, I guess I don't see what a current is really. My book says that you get a current occurs when an electron goes from the valence band to the conduction band. But what if the conduction band for one material IS the valence band for another material i.e. they are in the same place in k-space. Does that mean that the second material ALWAYS has a current even when there is no electric field?

    Ahh, I am so confused...
     
  12. Mar 28, 2008 #11

    Gokul43201

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    Yes, you are terribly confused! There are no two materials involved.

    Let me try a different, hand waving approach:

    What can you say about the mean momentum of electrons in a metal with no applied external field? Is this consistent with a Fermi sphere centered on the origin (k=0)?

    Next, what happens to the mean energy and momentum of an electron when a small field is applied? For a small field to make a small increase in the electron's energy (i.e., excite it to a low-lying excited state), there must be unoccupied low-lying states available. In which of the 3 pictures do you see electrons occupying states just below unoccupied, but allowed states? The increase in energy comes with an increase in momentum along the direction of the applied field (what does this do to the Fermi sphere?), and that's basically what a current is.

    So, in only one of the pictures, is it possible to produce a current by applying a small voltage, and that is the band picture for a conductor.

    When you understand things better, go back to posts #3 and #7 and make sure you understand them too.
     
  13. Mar 28, 2008 #12
    No, I am not that confused. I was saying hypothetically consider two materials such that material 1 has a FILLED valence band at the same height as the CONDUCTION band in material 2. Why does the material 1 not have a steady current when it is just sitting on a table?

    I think you kind of answer this by saying that it is only the center of the Fermi sphere that determines the current. Whenever there is no applied electric field the Fermi sphere is always centered at k = 0. Can you give me some sense of how the the third picture will change when you apply an electric field? Will the red part of the upper box just move up? Do you really need a 4D diagram to understand this?
     
  14. Mar 28, 2008 #13

    olgranpappy

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    your phrase "at the same height" in the above quote does make much sense to me. what do you mean? the drawings are schematic, the "height" doesn't really mean anything. the only thing that matters is if the band is full or not.

    next, a FILLED valence band will not produce a steady current when it is just sitting on the table because, as I will show, the current due to a filled band is zero.

    The current due to a filled band is
    [tex]
    \vec j = -2e\int_{\rm B.Z.} \frac{d^3 k}{{(2\pi)}^3}\vec v
    =-2e\int_{\rm B.Z.} \frac{d^3 k}{{(2\pi)}^3}\frac{\partial \epsilon}{\partial \vec k}
    [/tex]
    which is zero because the integral of the gradient of a periodic function (the energy is periodic wrt the BZ) over the whole BZ is zero. see Ashcroft and Mermin Appendix I
     
    Last edited: Mar 28, 2008
  15. Mar 28, 2008 #14
    That makes sense.
     
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