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A question about the relationship between momentum and force

  1. Nov 23, 2012 #1
    Hi guys, I got a question in the solid state physics book writen by ashcroft and mermin. This question is about the relationship between momentum and force.
    Suppose we have an electron with momentum p(t) at time t. If there is a force f(t) acted on this electron in the ongoing infinitesimal time dt, its momentum will change to p(t+dt) at time t+dt.

    The book says that, p(t+dt)=p(t)+f(t)dt+O(dt)~2 where O(dt)~2 denotes a term of the order of (dt)~2.

    I don't know why there is a term O(dt)~2 ?


    According to dp/dt=f(t) we have dp=f(t)dt+a(dt) where a(dt) means a higher order term of dt. But why is this high order term is the term of the order of (dt)~2 instead of 3, 4...


    Can you help me about this question, I will show you the snapshot of the book in the next floor, thank you!
     
  2. jcsd
  3. Nov 23, 2012 #2
    Snapshots from the book,

    I can't not insert an image. Pleas go to the second floor of this link

    http://tieba.baidu.com/p/2004868633

    I promise it is clean. Thanks a lot!!!
     
    Last edited: Nov 23, 2012
  4. Nov 23, 2012 #3

    K^2

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    It actually means order 2 or higher. It includes 3, 4, etc. Point is, if you take dp/dt, only first order survives. Higher orders remain infinitesimal, so you get dp/dt = F(t).
     
  5. Nov 23, 2012 #4

    Thank you very much for your reply!! Would you please check the snapshots of the book in the second floor of http://tieba.baidu.com/p/2004868633 for me? I can't insert an image or upload an attatchment. Sorry for the caused inconvenience.

    In the book, it only says that the term behind the f(t)dt is of the order of (dt ) square . It does not mention the higher order like 3, 4 ...
    So I think there might be something wrong in the textbook, am I right?
     
  6. Nov 23, 2012 #5

    K^2

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    The book just isn't being terribly clear. When used in context of expansions, +O(x²) means second order or higher. Higher orders simply aren't going to matter in expansion, though, as expansion error is completely determined by highest order unaccounted for. Hence the notation mentions only that second order.
     
  7. Nov 23, 2012 #6
    No, in this specific case he means only second order terms in dt.
    The second order is related to the force not being constant in time.
    p(t+dt)=p(t)+(dp/dt) dt + 1/2 (d2p/dt2) (dt)2 +...
    Now you know that dp/dt=f(t)
    If you take the derivative of the above,
    (d2p/dt2) =df/dt

    So if the force is constant, second order is zero. If the force is time dependent you have second order (but can be neglected as being, well, second order).
     
  8. Nov 23, 2012 #7

    Yes, I got your idea. Thanks K^2, your replies are of great help! Good luck with you!
     
  9. Nov 23, 2012 #8
    Thanks for your reply. Yes, you are right. And for the author's purpose the second order is enough so he omitted the three and higher order terms. I only used the theorem of momentum to understand here but forgot to do the taylor expansion to the p(t+dt).
    Thanks again for your help!!! Good luck with you!
     
  10. Nov 23, 2012 #9

    K^2

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    So are the higher orders.
     
  11. Nov 23, 2012 #10
    Did I imply otherwise?
    The book only discusses second order so this what I was referring to.
     
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