Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Yahya on 07/01/2017 17:58:04

when I throw a stone upward exactly vertical ( 90 degrees angle )with a particular initial speed, what is the duration at which its velocity is exactly zero? suppose it stops at a particular height, how long it will take while it is at this height ?

de4pends on gravity. atmosphere, mass, speed, etc. You want to specify it? Maybe the idea is one of there being a 'equivalence' adapting those factors?

when I throw a stone upward exactly vertical ( 90 degrees angle )with a particular initial speed, what is the duration at which its velocity is exactly zero? suppose it stops at a particular height, how long it will take while it is at this height ?
Your question is deep. It depends on whether the flow of time is smooth and indivisible. Or whether time is made of discrete particles such as "chronotrons"
If time is made of chronotrons, then your upthrown stone would stay still  at zerovelocity  for precisely 1 chronotron. Then it would reverse direction, and fall back down again.

Trying to answer this very precisely would require more precise definitions. For instance, all of the atoms within the stone are moving randomly at incredibly high frequencies, so are we talking about the center of gravity of the stone, or the average velocity of all of the constituents of the stone? Also, whose perspective are we using as the frame of reference for determining the velocity of the stone?
I think it would be very hard to specify an amount of time that the stone actually had zero velocity. We could calculate some incredibly short window of time for which the momentum of the stone would be within the Heisenberg Uncertainty limit of zero given what is known about the location of the stone....

I think the answer is : it spend at that point zero time , it spends nothing , a moving particle spend nothing when it is at a particular point apart from its velocity , when we want to measure the time elapsed for this point , we set the time to be zero at the start point, because in fact the time is zero.

I think the answer is : it spend at that point zero time
This is the traditional answer that Newton would have given.
 The acceleration of the stone is (almost) constant.
 the velocity starts off negative (upwards) and increases in an (almost) straight line until it reaches the same value, but positive (downwards)
 the amount of time spent crossing the v=0 axis is 0 seconds
 this answer doesn't really change if you add in the inverse square law of gravity, or air resistance

This is a really nice question that highlights a common misconception. Many people, especially younger learners, get the idea that a ball or stone launched skywards "pauses" for a while at the top of its travel before it falls again.
Actually that pause is only instantaneous. The object is constantly slowing down as gravity accelerates it downwards. At some point the velocity is transiently zero before it begins to accelerate groundwards again.
The reason the object appears to pause is that as it approaches standstill it is moving progressively more slowly. As it begins to accelerate down again it is also initially travelling only very slowly.
The net result is that it looks like it's stopped in mid air for a bit.

"everything should be as simple as it can be, but not simpler" Albert Einstein

This is a really nice question that highlights a common misconception. Many people, especially younger learners, get the idea that a ball or stone launched skywards "pauses" for a while at the top of its travel before it falls again.
Actually that pause is only instantaneous. The object is constantly slowing down as gravity accelerates it downwards. At some point the velocity is transiently zero before it begins to accelerate groundwards again.
The reason the object appears to pause is that as it approaches standstill it is moving progressively more slowly. As it begins to accelerate down again it is also initially travelling only very slowly.
The net result is that it looks like it's stopped in mid air for a bit.
Surely the object doesn't just "look like it's stopped". It must actually have stopped going upwards, before it can start going downwards.
I mean the object can't be moving in two directions  upwards and downwards  at the same time. Surely there must be a "instant" in time at which the upward movement stops? Before the downward movement can start.
And mightn't this "instant" be defined, in the particlebased physics which is absolutely required at present  as 1 chronotron.
Couldn't the chronotron explain Time, just like the Higg's Boson explains Mass. The Higg's Boson was (supposedly) detected by the LHC.
Would it possible for us to build a"Chronotron" detector?

Surely the object doesn't just "look like it's stopped". It must actually have stopped going upwards, before it can start going downwards.
The standard explanation is that it has stopped instantaneously before reversing directions. But I believe what chris was referring to is the illusion that an object hangs for a finite moment before reversing course. For instance, if one counts g as 9.8 m/s^{2}, an object would spend about 200 microseconds (a far cry from instantaneous) with a speed less than 1 mm/s (a far cry from 0, but maybe not distinguishable by a casual observer)
I mean the object can't be moving in two directions  upwards and downwards  at the same time. Surely there must be a "instant" in time at which the upward movement stops? Before the downward movement can start.
This depends on your definitions of "moving," "object," and "same time." :)

It's Zeno's arrow all over again  after 2,000 years!

The object will stay exactly as newtonian mechanics would predict an infinitesimal amount of time.
In newtonian mechanics you can throw an object perfectly vertical and then at some point it stops and stays there for an infinitesimal amount of time.
If I interpret this using my theory, there is no such a thing as an object at rest. This also violates time asymmetry. However you can theoretical assign a centre to a structure with a certain degree of precision (you can't have perfect precision not even theoretically, sounds a bit like the uncertainty principle but this is something else) and that center will at some point stop and stay for an instant like in newtonian mechanics.
The best way to answer your question in the real context is by the motion or a beam of light in a gravitational field. If you imagine sending a beam of light at 1 degree instead of 90deg, it will reach a certain height then it will go down following a parabolic trajectory . You can see that it never actually stop,
but it will stay at the top for an instant of time.

time being infinitely small violates velocity definition , if there is distance elapsed then there is time taken and vice versa, if there is not any distance elapsed( supposing this is the definition of this stone at zero velocity because if it take any tiny distance downward its velocity will start to increase ) then there is not any time taken, this for any object moving, if its distance is zero and it takes time then its instantaneous velocity is 0/t =0 , for a moving object its instantaneous velocity can not be zero.
we can apply this rule for this stone because when its velocity is zero , that mean it is in an exact point, think of a number line , to the right are positive numbers and to the left are negative numbers, where is the zero ? it is at the exact point between the two.
you can think of the stone as a particle moving through the number line from the right to the left.

v=dx/dt But dx and dt are not 0.

they are. 0/0 can equal any number according to the situation , according to the smallest number the denominator and numerator were before they became zero, if the ratio equals a specific number, and the nominator decrease and the denominator decreases as well but keeping the same ratio , then they will end up in 0/0 but still the ratio the same in this case 0/0 equals the instantaneous velocity (instantaneous velocity is constant) , remember also that : 0/0=any number, because 0=0*anynumber. and in the real world they are.

You can never divide by pure zero.

You can never divide by pure zero.
it's just my personal explanation.

You need to learn calculus to understand the mathematical concepts. However we don't know how infinity and infinitesimal quantities occur in reality.
An example of an infinitesimal number can be 10.(9) 0.(9)is 0.999... basically you add an infinite amount of 9 decimals.
For the speed of light c=3*10^8 m/s = x/t=dx/dt
x=3*10^8
t=1
dx=x/n=3*10^8/n
dt=t/n=1/n
where n is a very large number.
Even if n tends to infinity dx/dt remains the same 3*10^8m/s
If we take a number r0=1 we can't really say what is the next real number because there is no next number.
If n tends to infinity then the next number after r can be thought as r1=r0+1/n then r2=r0+1/n+1/n=r0+2/n
All of these are only mathematical solutions.

v=dx/dt But dx and dt are not 0.
In calculus, v is normally expressed as "the limit as dt approaches 0. "
For a wellbehaved function (like an object lobbed in a gravitational field), v is wellbehaved as dt approaches 0, and when dt = 0. See L'Hopital's rule.

v=dx/dt But dx and dt are not 0.
... and when dt = 0. See L'Hopital's rule.
dt can never be pure zero. That is the purpose of "d"

dt can never be pure zero. That is the purpose of "d"
an object at stationary , it elapses zero distance and the time it takes zero at an instant ( because when start to count time will first be zero ) how do you determine its velocity , there is not limits here , zero distance and zero time ? similar problem for the stone. but we all know this velocity is zero , and the stone velocity has the instantaneous velocity , how 0/0=0 ? 0/0=anynumber
but fortunately we know its velocity is zero , because it will still not make displacement and then v=0/t=0 , for the stone its velocity is the instantaneous velocity taken from the slope of the graph.

v dt = dx.

In calculus, v is normally expressed as "the limit as dt approaches 0. "
For a wellbehaved function (like an object lobbed in a gravitational field), v is wellbehaved as dt approaches 0, and when dt = 0. See L'Hopital's rule.
Who was L'Hopital?

In calculus, v is normally expressed as "the limit as dt approaches 0. "
For a wellbehaved function (like an object lobbed in a gravitational field), v is wellbehaved as dt approaches 0, and when dt = 0. See L'Hopital's rule.
Who was L'Hopital?
https://en.m.wikipedia.org/wiki/Guillaume_de_l'H%C3%B4pital (https://en.m.wikipedia.org/wiki/Guillaume_de_l'H%C3%B4pital)

The object will stay exactly as newtonian mechanics would predict an infinitesimal amount of time.
does that mean that I'm the only one who has this idea?

You need to learn calculus to understand the mathematical concepts.
I learned calculus....
v=dx/dt But dx and dt are not 0.
but I learned that if dx/dt=0/0 that does not mean an answer does not exist.

'It depends on whether the flow of time is smooth and indivisible. Or whether time is made of discrete particles such as "chronotrons" '
If time is quantized, it hasn't shown up at the very small timescale events probed by the high energies in the LHC.
A Stone has much lower energy per nucleon than particles in the LHC, so we can effectively treat time as continuous.
It is possible that time is quantised on very small timescales; and it may even have macroscopic sideeffects. Just as quantised energy levels in atoms avoided the immediate collapse of all atoms in the universe, it is possible that quantised time may someday explain some other phenomenon like persistent infinities around black holes.
But for stones falling in Earth's gravity, treat time as continuous.

The object will stay exactly as newtonian mechanics would predict an infinitesimal amount of time.
does that mean that I'm the only one who has this idea?
The object really stops at the top and v reaches absolute zero according to newtonian mechanics. However according to my concept this never happens because my concept uses waves propagation to explain motion and all propagations are at a constant speed c and obviously never stop.

does that mean that I'm the only one who has this idea?
No. Newton and his peers knew about this which is one of the reasons they were interested in the development of calculus.
The answers are within this thread see
I think the answer is : it spend at that point zero time
This is the traditional answer that Newton would have given.
 The acceleration of the stone is (almost) constant.
 the velocity starts off negative (upwards) and increases in an (almost) straight line until it reaches the same value, but positive (downwards)
 the amount of time spent crossing the v=0 axis is 0 seconds
 this answer doesn't really change if you add in the inverse square law of gravity, or air resistance
but the confusion can be better explained if you avoid talking about an object only having a vertical component of velocity.
Consider an object thrown with both vertical and horizontal velocity components. The object will reach the apex and have zero velocity at a specific time  the turnaround point. That time is a specific point on the curve and if you zoom in to that point it gets smaller and smaller trending towards 0 in both velocity and time. So the problem is not another Zeno paradox as suggested earlier, but a fractal effect.
If we look at the way Newtonian mechanics calculates the total time the object is in flight we find it is symmetric around the apex ie no time is allowed for any hold or hesitation at the apex  this you will find on many school homework sites.
As @evan_au says, it can only be considered to spend any time at the apex if time is quantised and as yet we have no evidence for that.

According to quantum mechanics neither space nor time are quantized. Only energy has quanta. From the continuums of space and time the magnitude of the energy quanta can be derived

Is the object not just following its trajectory in curved spacetime ?
What would its trajectory look like if it was modeled in this way?
Would it not be similar to an object moving on the surface of the Earth with the difference that the latter is being constantly accelerated out from the centre of the Earth whereas the stone is in freefall immediately following its initial upward acceleration?
I wonder whether this "infinitesimal pause" might be modeled as simply a point on a geodesic in the curved spacetime model and the question not arise.

I wonder whether this "infinitesimal pause" might be modeled as simply a point on a geodesic in the curved spacetime model and the question not arise.
Firstly, it isn’t an infinitesimal pause it is genuinely a dt=0 at v=0. It is only infinitesimal if you choose to zoom out and sacrifice accuracy.
Yes, it can be modelled as a point on the geodesic, but that is exactly the same as modelling the point on the apex of the trajectory curve as described above, just a different perspective.

Firstly, it isn’t an infinitesimal pause it is genuinely a dt=0 at v=0. It is only infinitesimal if you choose to zoom out and sacrifice accuracy.
I've not waded through this thread, so the answer may be in there somewhere.
Are you saying that the stone does not actually stop? Would that mean that the ve > +ve acceleration happens at such low speed that the pause is only an illusion?

I think the answer is : it spend at that point zero time
Just as an aside; what is the difference between being at a point for zero time, and not being there at all?

Are you saying that the stone does not actually stop? Would that mean that the ve > +ve acceleration happens at such low speed that the pause is only an illusion?
The answer is in @evan_au post #5.
Yes the stone does not stop because stopping implies spending time at a location in space. Yes, the pause is an illusion, as you say due to the low speed, also meaning the transition between up and down motion is seamless.
To suggest otherwise would mean there is a flat spot at the apex, and clearly there isn’t.
Just as an aside; what is the difference between being at a point for zero time, and not being there at all?
There is one for the philosophers.
However, it passes through that point so it must have been there ;)
Or does it just kiss the apex and move on ;D
EDIT: PS Here is another thought for you @Bill S  is it possible to say that any moving point (eg the tip of your nose as you walk down the street) spends any time at a particular point in space. If it does it is stationary. So, has your nose been anywhere?

You can never divide by pure zero.
it's just my personal explanation.
Thanks for clarifying your lack of understanding.

So, has your nose been anywhere?
Assuming that walking produces a continuous motion in a constant direction, it would seem that your nose does not stop. Therefore, it spends no time at any specific point. If every point along the way is designated as “a specific point”, it follows that your nose has spent no time at any point. So, if spending no time at a point (a) is the same as not being there (b), your nose has been nowhere. Since we know that your nose has been somewhere, there must be a difference between a and b.
So far, so good; but how do you define the difference? Consider the following: If points A, B and C are contiguous points along your way, there is no space between them. Your nose spends no time at A, B or C. Between A>C there is nowhere for your nose to be, so how long does it take your nose to move from A to C. If you extend this “journey”, say, from A to Z; does that make any difference to the time taken?

Another thought: If it takes your nose no time to travel from A to Z, it must travel at "c". In order to do that, it must be massless.
There's no accounting for hooters.

"Keep a clean nose. Watch the plain clothes. You don't need a weatherman. To know which way the wind blows."
The model is not the same thing as what is observed.It is like Aesop's tortoise trying to catch up with the hare

"Keep a clean nose. Watch the plain clothes. You don't need a weatherman. To know which way the wind blows."
Look out kid
It's somethin' you did
God knows when
But you're doing it again
Hare today, gone tomorrow
Another thought: If it takes your nose no time to travel from A to Z, it must travel at "c".
Or instantaneous?