A question about the relationship between momentum and forceby Noriskkk Tags: force, momentum, relationship 

#1
Nov2312, 07:09 AM

P: 5

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
Nov2312, 07:12 AM

P: 5

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!!! 



#3
Nov2312, 07:17 AM

Sci Advisor
P: 2,470

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).




#4
Nov2312, 07:29 AM

P: 5

A question about the relationship between momentum and forceThank 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? 



#5
Nov2312, 07:50 AM

Sci Advisor
P: 2,470

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.




#6
Nov2312, 07:58 AM

P: 1,903

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 (d^{2}p/dt^{2}) (dt)^{2} +... Now you know that dp/dt=f(t) If you take the derivative of the above, (d^{2}p/dt^{2}) =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). 



#7
Nov2312, 08:20 AM

P: 5

Yes, I got your idea. Thanks K^2, your replies are of great help! Good luck with you! 



#8
Nov2312, 08:24 AM

P: 5

Thanks again for your help!!! Good luck with you! 



#9
Nov2312, 08:25 AM

Sci Advisor
P: 2,470





#10
Nov2312, 08:36 AM

P: 1,903

The book only discusses second order so this what I was referring to. 


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