What is the position of this moving particle over time?

[Mike_bb]

I think up following example.

Let we have kinematic equation of motion for moving particle: $f(t)=t^2$.

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

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

1)My first question is: is it true that moving particle has position f'(t)=0 $0$ from $0$ to $dt$ time?

2)My second question is: if 1) is true then how is it possible that moving particle has position at $dt$: $f(dt)=dt^2$ but not $0$?

What is position of moving particle accurate?

Thanks.

 

[BvU]

1) No. $f'(t)=0$ only at $t=0$
2) is then a non-starter

$\$

 

[Mike_bb]

1) No. $f'(t)=0$ only at $t=0$
2) is then a non-starter

$\$

I edited first message : Is it true that moving particle has position $0$ from $0$ to $dt$ time?

 

[berkeman]

I edited first message :

Just a Mentor Note -- it is best to just post a new reply with the correction, since going back and changing things that have been pointed out as incorrect can really confuse future readers of the thread. If you do make a correction to an earlier post, it is best to at least use "strikethrough" to line out the incorrect item and then post the correction with a note about the change. Thanks.

 

[PeroK]

I edited first message : Is it true that moving particle has position $0$ from $0$ to $dt$ time?

First, $dt$ is not a finite unit of time. $dt$ is a differential. A finite unit of time is usually written as $\Delta t$. There is often confusion between $\Delta t$ and $dt$, which causes problems.

Second, consider the function $f(t) = t^2$ as a mathematical object. You can draw its graph. This function is only zero for $t = 0$ and is greater than zero for $t > 0$. This shouldn't be problematic.

Note that:
$$f(0) = 0, f'(0) = 0, f''(0) = 2$$So, although the derivative of $f$ at 0 is zero, the second derivative is not zero. You can extend this by considering $f(t) = t^3, t^4 \dots$. In each case, more and more derivatives of the function are zero at $t=0$, but eventually some nth derivative is not zero.

If ALL derivatives of a function were zero at $0$, then the function would be identically zero. But, not if only the first derivative is zero.

 

[Mike_bb]

First, $dt$ is not a finite unit of time. $dt$ is a differential. A finite unit of time is usually written as $\Delta t$. There is often confusion between $\Delta t$ and $dt$, which causes problems.

Second, consider the function $f(t) = t^2$ as a mathematical object. You can draw its graph. This function is only zero for $t = 0$ and is greater than zero for $t > 0$. This shouldn't be problematic.

Note that:
$$f(0) = 0, f'(0) = 0, f''(0) = 2$$So, although the derivative of $f$ at 0 is zero, the second derivative is not zero. You can extend this by considering $f(t) = t^3, t^4 \dots$. In each case, more and more derivatives of the function are zero at $t=0$, but eventually some nth derivative is not zero.

If ALL derivatives of a function were zero at $0$, then the function would be identically zero. But, not if only the first derivative is zero.

Perok, I work in Non-standard analysis framework. See tag.

 

[PeroK]

Perok, I work in Non-standard analysis framework. See tag.

That's non-physical. Physics works on a real, not a hyperreal, continuum.

 

[Mike_bb]

That's non-physical. Physics works on a real, not a hyperreal, continuum.

Real line is incomplete because it doesn't include infinitesimals.

 

[PeroK]

Real line is incomplete because it doesn't include infinitesimals.

The Real line is complete. That's the Archimedean principle.

If you want to study physics using hyperreals, then it's up to you to sort out the physical issues. Nature is under no obligation to obey your hyperreal axioms.