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Drift velocity: Why not 1/2 *t*(eE/m)

  1. Apr 2, 2015 #1
    Why is the drift velocity of an electron in a wire defined as: tau*(E*e/m) and not 1/2 *tau* (eE/m) as it is the AVERAGE VELOCITY!!!!?????? because the above defintion of drift velocity is the velocity that is attained before the collision so rather the maximum velocity of the elctron, isn't it???
     
  2. jcsd
  3. Apr 2, 2015 #2

    CWatters

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    Perhaps work through the calculation...

    https://books.google.co.uk/books?id=bihuAgAAQBAJ&lpg=PA225&ots=C_dhBLnKtJ&dq=drift velocity of an electron e/m&pg=PA224#v=onepage&q=drift velocity of an electron e/m&f=false [Broken]

    Edit: That should link to page 224.
     
    Last edited by a moderator: May 7, 2017
  4. Apr 2, 2015 #3
    Thanks a lot :) but unfortunately once again doesn't explain why one multiplies with tau, and therefore uses the maximun/final velocity of the elctron before each collision instead of its average velocity.
     
  5. Apr 2, 2015 #4
    The average thermal velocity drops out of the equations, and the velocity addition from accelerating in the field before the next collision remains.

    What the electron is doing is akin to running down a slope (the field) into a head wind (the collisions). It is the slope adding a running speed, rather than the previous walking speed while looking at the birds and smelling the flowers (random thermal velocity over a population), that decides the final speed of the balance slope/head wind.

    If it is the similar looking relation between the field and the average thermal velocity that trips you up, I suggest you study how the latter comes about. (Quite frankly, I have forgotten. But it looks intuitively correct, the field would heat the electrons.) But it is unrelated to the physics at hand.
     
  6. Apr 2, 2015 #5
    First of all thank you for the quick reply, but, sorry, I'm not a native speaker and when you start talking about flowers and the like as well as the slope comparison, I don'T have a clue what you are trying to say...
     
  7. Apr 2, 2015 #6

    CWatters

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  8. Apr 2, 2015 #7
  9. Apr 2, 2015 #8
    Really? Where the extra factor 2 comes from is a lot simpler than that.

    If you look at an eletron that has just collided, than it will have a mean time of τ before it's next collison, but you now overrepresent short time intervals in the average time.
    You should choose a random electron, and that one will also have an average time of τ before the next collision, but also an average time of τ since the last collsion. (If the times follow the exponential distribution) The average time since the last collision will give you the momentum of τEe/m
     
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