A question about the relationship between momentum and force

In summary: I only used the theorem of momentum to understand here but forgot to do the taylor expansion to the p(t+dt).
  • #1
Noriskkk
5
0
Hi guys, I got a question in the solid state physics book written 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!
 
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  • #2
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  • #3
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
K^2 said:
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).


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?
 
  • #5
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
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).
 
  • #7
K^2 said:
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.


Yes, I got your idea. Thanks K^2, your replies are of great help! Good luck with you!
 
  • #8
nasu said:
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).

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!
 
  • #9
nasu said:
The second order is related to the force not being constant in time.
So are the higher orders.
 
  • #10
K^2 said:
So are the higher orders.

Did I imply otherwise?
The book only discusses second order so this what I was referring to.
 

What is the relationship between momentum and force?

The relationship between momentum and force is described by Newton's second law of motion, which states that the net force acting on an object is directly proportional to the rate of change of its momentum. In simpler terms, the greater the force applied to an object, the greater its change in momentum will be.

How does momentum affect an object's motion?

Momentum is a measure of an object's motion and is defined as the product of its mass and velocity. The greater an object's momentum, the harder it is to stop or change its direction of motion. This is why objects with larger masses or higher velocities are more difficult to stop or redirect.

Can momentum and force be conserved?

Yes, both momentum and force can be conserved within a closed system. This means that the total momentum and total force within a system remains constant, even when individual objects within the system are experiencing changes in momentum and force. This conservation is a fundamental principle in physics and is often used to analyze collisions and other interactions between objects.

What is the unit of measurement for momentum and force?

Momentum is typically measured in kilogram-meters per second (kg·m/s), while force is measured in Newtons (N). These units can also be expressed using different measurements, such as pounds (lb) for force or miles per hour (mph) for momentum, but they can always be converted to their standard units for more accurate calculations.

How does the direction of force affect an object's momentum?

The direction of force applied to an object will determine the direction of its change in momentum. If a force is applied in the same direction as an object's motion, its momentum will increase. If a force is applied in the opposite direction, its momentum will decrease. In cases where forces are applied in different directions, the net force and resulting change in momentum will depend on the vector sum of all the forces acting on the object.

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