# What is position of moving particle accurate? Example from non-standard analysis

Consider the kinematic equation of motion for moving particle: $$f(t)=t^2$$.

$$df(t)/dt=2t$$ and $$df(t)=2tdt$$

For $$t=0$$ we have $$df(t)=0dt$$.

Also we have $$f(0+dt)-f(0)=0$$. It proves that moving particle position is $$0$$ from $$0$$ to $$dt$$. (moving particle rest $$dt$$ time)

My question is: if moving particle rest $$dt$$ time then how is it possible that position of particle at $$dt$$ is equal to $$f(dt)=dt^2$$

How is it possible?

Thanks.

• If $f$ represents the position of the particle then it has position $0$ at $t=0$ only. The quantity $dt$ does not exist in the domain of $f$. Commented Jul 30 at 14:21
• @JohnDouma it's Non-standard analysis. The quantities like $dx$ exist. Commented Jul 30 at 14:23
• @MikhailKatz what is average velocity over infinitesimal interval $0$ and $dt$?What is it equal to? Commented Jul 31 at 8:00
• Not in this case, if the context is nonstandard analysis and classical logic. Commented Jul 31 at 8:07

You are confusing instantaneous velocity at t=0 and average velocity over the infinitesimal interval between 0 and dt. The latter is infinitesimal. The former, being the standard part of the latter, is 0. The average velocity is the change in position divided by the length of the interval; namely $$\frac{dt^2}{dt}$$.

• Ok. I want ask you about following thing. Abezhiko wrote that $f(\mathrm{d}t)-f(0) = \int_0^{\mathrm{d}t} 2t\mathrm{d}t = \mathrm{d}t^2$. We can write $f(\mathrm{4})-f(0)$ = $\int_0^{4} 2t\mathrm{d}t$. Also $f(\mathrm{d}t)-f(0) = dt^2$ than $f(2dt)-f(dt)= 3dt^2$ and $f(3dt)-f(2dt) = 5dt^2$. Let's summarize these infinitesimals: $f(4)-f(0)$=$dt^2+3dt^2+5dt^2 + ...$. But this infinite sum isn't equal to $f(4)-f(0)=16$. How is it possible? @Mikhail Katz Commented Jul 31 at 10:07
• What makes you think this infinite sum isn't equal to f(4)-f(0)=16 :-) Commented Jul 31 at 10:11
• I did it. I try to sum these infinitesimals again. Perhaps, I am wrong. @Mikhail Katz Commented Jul 31 at 10:12

One may start from the Taylor series of $$f$$ in order to write $$f(t+\mathrm{d}t) = \sum_{n=0}^\infty \frac{f^{(n)}(t)}{n!}\mathrm{d}t^n,$$ with the derivatives being defined from $$f'(t) = \operatorname{st}\left[\frac{f(t+\mathrm{d}t)-f(t)}{\mathrm{d}t}\right],$$ where the standard part $$\operatorname{st}[\,\cdot\,]$$ "kills" all infinitesimal terms by rounding to the closest real number below its argument. As a consequence, the expression $$f'(t)\mathrm{d}t$$ is not equivalent to $$f(t+\mathrm{d}t) - f(t)$$ anymore; indeed, one has $$f'(t)\mathrm{d}t = \operatorname{st}\left[\frac{f(t+\mathrm{d}t)-f(t)}{\mathrm{d}t}\right]\mathrm{d}t$$ but $$f(t+\mathrm{d}t)-f(t) = \sum_{n=\color{red}{1}}^\infty \frac{f^{(n)}(t)}{n!}\mathrm{d}t^n.$$ Loosely speaking, you can see the left-hand side as a "regularized" derivative and the right-hand side as a kind of "raw" derivative, which cannot be the same unless $$f$$ is (locally) affine $$-$$ or under an assumption such as $$\mathrm{d}t^2 = 0$$, but it doesn't correspond to the standard setup of nonstandard analysis (no pun intended).
With your concrete example, namely $$f(t) = t^2$$, the LHS is equal to $$2t\mathrm{d}t$$, which vanishes at $$t = 0$$, while the RHS is given by $$(t+\mathrm{d}t)^2 - t^2 = 2t\mathrm{d}t + \mathrm{d}t^2$$, which is reduced to $$\mathrm{d}t^2 \neq 0$$ when $$t = 0$$.
I’m assuming the function $$f(t)=t^2$$ represents the position of the particle over time. If $$dt=0$$, then there is no change between $$f(0)$$ and $$f(dt)$$. If $$dt>0$$, then the change in position is $$f(dt)-f(0)$$, or simply $$dt^2$$. If $$dt<0$$, then the change in position is $$f(0)-f(dt)$$, or simply $$-dt^2$$.