How do you change a force? For instance, since Force is just mass times acceleration, doesn't this mean we need to have a change in acceleration? Where does this third derivative of position with respect to time actually come from to cause a force to change?

Question

theeric · Answer

Hi! I'm nowhere near your level of math, but acceleration is second derivative of position with respect to time! :)

I often think of the force equation as a relationship, but not to be read as cause and effect! I think like that just because it's never intuitive to me. So I would say that the change in the velocity implies that a force has been enacted. That's just how inertia is defined!

So that's how I view it!

And force is the change of momentum, which is a nice way to think about it, too. We can just think of $F=\dfrac{\text dp}{\text dt}$ as $F=ma$ by noting that momentum is $mv$ so
$F=\dfrac{\text d(mv)}{\text dt}$

Assuming constant mass, which is often the case,
$F=m\dfrac{\text dv}{\text dt}$
because the derivative of a constant that is multiplied by a function is equal to that constant multiplied by the derivative of that function. Mathematically,
$\dfrac{\text d(c\dot\ f(x))}{\text dx}=c\dfrac{\text d(f(x))}{\text dx}$

And since $\dfrac{\text dv}{\text dt}=a$ by definition....
We get to
$F=ma$

So force is mass times acceleration assuming constant mass.

I'm sure you knew this, @Kainui , but I think others might like to see it :)

kainui · Answer

Yeah, I appreciate your willingness to help, and unfortunately (or fortunately?) I already know this pretty well. The problem I have is with the creation of the force itself.

So suppose you're standing still. Then someone pushes you. You went from no force to a force moving you. This would be represented by $$\frac{dF}{dt}=\frac{d(ma)}{dt}=m\frac{da}{dt}=m\frac{d^3x}{dt^3}$$ So this is what I mean when I say third derivative of position with respect to time. According to wikipedia the third derivative is "jerk" and dF/dt is called "yank". 

But how would I go about thinking of this quantity? What would it look like if the net yank acting on an object was zero as some sort of analog to net force being zero? How do I jerk something? I suppose magnets or electric force exerts a jerk. Even the gravitational force does. $$F=k_e \frac{q_1 q_2}{r^2}$$ After all, the position of an object relative to the one forcing it will be a distance r which is a function of time right? But these aren't really the questions I'm looking to answer you see? This is just one level higher than acceleration.

I'm really more interested in how to think of all higher order derivatives on an object, for example, what's the tenth derivative of position with respect to time look like? At that point, it feels like things similar to "instantaneous velocity" or "instantaneous acceleration" start to break down. What would an instantaneous change of a change of ... to the tenth derivative of position really look like at an instant?

This might seem really obscure, but the reason I'm thinking this is because I was considering that in the creation of a taylor series of a function, like sin(x), we're able to say that if we know all the derivatives at one point we're able to say what happens to it at some further point in time. This sort of gives me an interesting idea that maybe we can impart a particle in free space with a velocity, acceleration, jerk, etc to higher derivatives and if we are able to control all of these precisely by hitting a particle at only a single moment in time we would be able to not only know exactly where it would be in the future but also hit objects so that they travel in sinusoidal paths or something weird.

Ok, don't let that overwhelm you, half of that or more is just pure speculation or garbage, so don't worry about it and discuss what you think or where the reasoning seems to fall short or whatever. What do you think?

theeric · Answer

Haha, I'm a little slow to grasp things, so I'm definitely not the best person to bounce ideas off of in they're new to me! I see what you meant going into the beginning and middleish! It's easier to think of things we can identify with, like position, velocity, acceleration, and even jerk. Another example of jerk is when you accelerate a car forward. Changing acceleration means changing force meaning that what you feel from the seat on your back. If you accelerate quickly, you feel a stronger push. Slower acceleration, weaker push. But then, how to think about things that you can't personally identify with? Things that are difficult to comprehend. That's where I fall on math, which really isn't my strong point anyway. Which, I suppose, is a good segway to: I don't understand the rest completely. If it goes into the uncertainty principle, I'll be lost anyway! And, maybe I'm just tired... I am tired, but maybe it's just my slowed thinking... But are you talking about knowing the values of the derivatives at one point in time, and using it to get other values at other times? In taking the Taylor series, you would need that high order function for force given time. Or the acceleration function. I suppose a function of position would be complicated, but I don't know. But we'd have to know some function. $F(t)=\sin t+t^6+t^5+t^4+t^3+t^2+t^1+1$ http://www.wolframalpha.com/input/?i=plot+sin+t+%2B+t^6%2Bt^5%2Bt^4%2Bt^3%2Bt^2%2Bt^1%2B1 $\rightarrow F(t)=\sin t+t_0^6+t_0^5+t_0^4+t_0^3+t_0^2+t_0^1+1\ \qquad+(\cos t_0+6t_0^5+5t_0^4+4t_0^3+3t_0^2+2t_0^1+1)(t-t_0)\div2\ \qquad+(-\sin t_0+30t_0^4+20t_0^3+12t_0^2+6t_0^1+2)(t-t_0)^2\div3!\ \qquad+(-\cos t_0+120t_0^3+60t_0^2+24t_0^1+6)(t-t_0)^3\div 4!\ \qquad+\dots$ It seems like we can use the Taylor series of a function about a point to get other points when we have a function like $\sin(t)$. And I would think that the particle's behavior would really depend on the force field. The particle can go through changing forces with many non-zero derivatives, and/or the field can be changing. Thus, as time continues, the force as a function of time could have many non-zero derivatives. Given the conditions at some point in time, and considering the force field, we could say how it would behave. As long as it's just classical mechanics. That's what I got.... Hehe. Hopefully one of the physics pros can get in on this.