# Electrons exposed to time-dependent force

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1. Dec 23, 2015

### M-Speezy

I have begun studying Ashcroft + Mermin on my own, and am trying to follow the math in the text. They suggest that an electron in a metal with some momentum p(t) and exposed to a force f(t) will at some time later (t+dt) have a contribution to the momentum on the order of f(t)dt plus another term on the order of dt*dt. My question is where does this dt^2 term enter? My instinct is to say that F=dp/dt, and that the change in momentum can then be given (very simplistically?) by f(t)dt. Obviously, a form of f(t) and an integral is in order, but I cannot see the logic of what is stated in the text.

Any and all help or guidance on the matter would be greatly appreciated!

2. Dec 23, 2015

### Orodruin

Staff Emeritus
The term of order dt^2 comes from the fact that f(t) is not necessarily constant in time. It is related to the derivative of f(t).

3. Dec 23, 2015

### M-Speezy

Why does this matter, though? I would think Newton's 2nd law would be used, and then a change in the momentum would simply be given by f(t)dt. I'm not sure what else should be done to lead to anything else.

4. Dec 23, 2015

### Orodruin

Staff Emeritus
It is just using Newton's second law, but Newton's second law is just the limit when dt goes to zero and so agrees with your result.

5. Dec 23, 2015

### M-Speezy

I figured it out I think. If the force at t+dt is instead expressed using a first-order approximated taylor series, then the extra dt comes out.