Author Topic: A 'simple' Physics postulation...  (Read 6587 times)

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Offline GlennSpriggTopic starter

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A 'simple' Physics postulation...
« on: July 09, 2021, 10:58:32 am »
In reference to JUST the local surface of the Earth where you stand, (NOT the solar system, galaxy etc!!)...
If you were able to throw an object 'exactly' vertical, such that it came back to you due to gravity, did the
object ever 'Stop' at it's apex??  Obviously, it can't have, even for 1/1000000000-th of a second, as that would
defy the laws of physics. So it is only ever going up, or coming down. Without stopping!!?   :phew:
Diagonal of 1x1 square = Root-2. Ok.
Diagonal of 1x1x1 cube = Root-3 !!!  Beautiful !!
 

Offline Brumby

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Re: A 'simple' Physics postulation...
« Reply #1 on: July 09, 2021, 11:09:34 am »
Define "Stop".
 

Offline retiredfeline

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Re: A 'simple' Physics postulation...
« Reply #2 on: July 09, 2021, 11:42:56 am »
Maybe you've discovered oneZ's Arrow paradox.  :-DD
 

Offline Ian.M

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Re: A 'simple' Physics postulation...
« Reply #3 on: July 09, 2021, 01:50:57 pm »
You'd need to go to the Earth's North or South pole to conduct your experiment, otherwise, as you throw it upwards it will drift off to the West, due to the Earth's rotation.  (As the tangential velocity required to remain 'stationary' with respect to a rotating frame of reference is proportional to the rate of rotation and the radius from the axis of rotation, it doesn't have enough tangential velocity to maintain its angular position at a greater radius from the center of the Earth.)   Therefore if you  throw an object so that it lands exactly where thrown from anywhere other than the poles on the Earth's surface, to do so you *CAN'T* thow it vertically, you have to tilt your throw ever so slightly East to give it the extra angular velocity required.  Its trajectory will therefore be a tall extremely skinny* loop, west-bound at the top, and, in flight, it will *NEVER* have zero velocity with respect to the point it was thrown from, and lands at.

At the poles, for an infinitesimal instant at the apex of its vertical trajectory, it has zero velocity.  Its speed with respect to the surface is a continuous function so how else is it going to get from positive (distance increasing = ascending) to negative (distance decreasing = descending) without going through zero speed?

* The tangential velocity difference over 100m height difference at the equator is (if I've done the maths right) only 45um/sec, so is normally absolutely imperceptible.

Define "Stop".
*EXACTLY* *THIS*

Velocity and acceleration both zero for a non-zero time interval (i.e. Physics: 'At rest') would probably be a reasonable definition.   
« Last Edit: July 09, 2021, 04:22:59 pm by Ian.M »
 
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Offline Brumby

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Re: A 'simple' Physics postulation...
« Reply #4 on: July 09, 2021, 02:00:57 pm »
Aside from the polar conversation (which is valid) it is this which contains the key elements of the answer:

...... for an infinitesimal instant at the apex of its vertical trajectory, it has zero velocity.
The word "instant" is key here.  The OP is considering a period of time - which cannot deliver a zero value....

Unless you take that period to be equally before and after reaching the apex in which time the net velocity will be zero.  (But, I don't think the OP will be happy with that.)


However - this statement not only allows, but requires a zero velocity at a point in time:
Quote
Its speed with respect to the surface is a continuous function so how else is it going to get from positive (distance increasing = ascending) to negative (distance decreasing = descending) without going through zero speed?
 
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Online TimFox

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Re: A 'simple' Physics postulation...
« Reply #5 on: July 09, 2021, 02:22:28 pm »
This problem is straightforward in calculus, since velocity is the time derivative of position.  For the parabolic curve of position vs. time, the derivative is zero at the extremum (maximum) position, but for an infinitesimal time. 
I usually think that nothing continuous can go to zero unless it can go negative:  in this case, the if the initial (up) velocity is positive, it reverses at the maximum and changes to a negative (down) velocity.
 

Offline Ed.Kloonk

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Re: A 'simple' Physics postulation...
« Reply #6 on: July 09, 2021, 02:26:29 pm »
In reference to JUST the local surface of the Earth where you stand, (NOT the solar system, galaxy etc!!)...
If you were able to throw an object 'exactly' vertical, such that it came back to you due to gravity, did the
object ever 'Stop' at it's apex??  Obviously, it can't have, even for 1/1000000000-th of a second, as that would
defy the laws of physics. So it is only ever going up, or coming down. Without stopping!!?   :phew:

When ever I pondered such things as a young man, my father would find and confiscate my stash.
iratus parum formica
 
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Offline esepecesito

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Re: A 'simple' Physics postulation...
« Reply #7 on: July 09, 2021, 02:30:19 pm »
In reference to JUST the local surface of the Earth where you stand, (NOT the solar system, galaxy etc!!)...
If you were able to throw an object 'exactly' vertical, such that it came back to you due to gravity, did the
object ever 'Stop' at it's apex??  Obviously, it can't have, even for 1/1000000000-th of a second, as that would
defy the laws of physics. So it is only ever going up, or coming down. Without stopping!!?   :phew:

I cannot understand why you say stopping would violate a physics law... it doesnt. It does stop.
 

Offline T3sl4co1l

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Re: A 'simple' Physics postulation...
« Reply #8 on: July 09, 2021, 03:47:23 pm »
Yes, without a fully defined scenario, there's no reason, value or meaning in saying something might "stop".

If you mean in terms of a nonrotating, pointlike, classical object, in pure vertical motion with no lateral velocity, following a Newtonian trajectory, without reference to a rotating or accelerating (other than gravity) frame, and no other outside forces (like wind resistance) -- and please take note of how many conditions one must apply to be rigorous here -- then one can say it "stops", as in velocity = 0, at one point during its path: the apex of the parabolic trajectory.  This does not occur for a defined duration of time, but is an instantaneous event only.  We use calculus to describe such conditions.  If you prefer nonzero bounds, then we can define the limit as there being a duration of time dt such that the velocity is v +/- dv, for some dv and dt that exist.  In the limit, we take both dt and dv to zero, but any will do.

Note also, there is nothing special about the point where velocity goes to zero.  We can make just as remarkable a claim about the two points adjacent to zero, or where it's going 1 m/s, or 10, or whatever; any velocity that happens to appear during the trajectory.  (Or including velocities that don't, if you allow extrapolating to earlier and later times while ignoring possible collision, or allowing that the number of points may be zero, one or two, or many, for any velocity we choose to ask about.)

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Online TimFox

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Re: A 'simple' Physics postulation...
« Reply #9 on: July 09, 2021, 03:51:24 pm »
Your discussion of nonzero bounds is handled in mathematics as an "epsilon-delta" proof of a limit.  Stolen from the web:
"The epsilon-delta definition of limits says that the limit of f(x) at x=c is L if for any ε>0 there's a δ>0 such that if the distance of x from c is less than δ, then the distance of f(x) from L is less than ε. This is a formulation of the intuitive notion that we can get as close as we want to L."
This is the modern explanation of Zeno's paradox.
 
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Offline SiliconWizard

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Re: A 'simple' Physics postulation...
« Reply #10 on: July 09, 2021, 05:08:10 pm »
Well, I guess it all comes down to the thought that velocity is mathematically zero when the object reaches its maximum altitude. Then the OP assumes "velocity = 0" means "stop".

Of course, as others have pointed out, first thing would be to define what "stopping" means. Even if it's a term we use in everyday's language without giving it much thought, I'll just dare a simple definition that looks good enough: an object stops iif its velocity reaches 0 *and* the velocity function at this point is non-derivable.

In the OP's example of parabolic motion, velocity indeed reaches 0 at the apex, but is still derivable. So, the object never "stops". Just my 2 cents.
 
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Online TimFox

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Re: A 'simple' Physics postulation...
« Reply #11 on: July 09, 2021, 05:26:30 pm »
It stops, then reverses, with the velocity variable being continuous in time.
 

Offline Zero999

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Re: A 'simple' Physics postulation...
« Reply #12 on: July 09, 2021, 07:46:49 pm »
It sounds similar to a child on a swing, who stops, before swinging back the other way, or the voltage and current in the resonant tank of an LC oscillator.
 
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Online TimFox

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Re: A 'simple' Physics postulation...
« Reply #13 on: July 09, 2021, 08:33:47 pm »
Exactly.  All of these time-dependent variables (as functions of time) have extrema (maxima and minima) where the time derivative goes to zero.  (When I learned math, we said that the derivative "vanishes", which usage bothered me as a smart-ass student.)
 

Offline dietert1

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Re: A 'simple' Physics postulation...
« Reply #14 on: July 09, 2021, 09:49:54 pm »
This reminds me of a professor at the university, when i was student of physics. I remember the name, but i won't tell. One day in a lecture he started to explain: Since velocity is a vector, it can never become negative. The vector will just change its direction. Isn't that crazy?
The question in this thread was posed such as to focus on the vertical component of velocity. And that becomes zero before getting negative. Voting about physics questions, that's modern times!

Regards, Dieter
 

Online TimFox

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Re: A 'simple' Physics postulation...
« Reply #15 on: July 09, 2021, 09:59:53 pm »
One can express the vector velocity v in terms of its three scalar components vxx + vyy + vzz
where v is the velocity, and x, y, and z are "unit vectors" in the three Euclidean directions.
The components vx, etc., can be positive, negative, or zero, but the magnitude of the vector v cannot be negative.  If not moving, the magnitude of v in that reference frame can be zero.
 

Offline TerraHertz

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Re: A 'simple' Physics postulation...
« Reply #16 on: July 09, 2021, 11:30:19 pm »
At the poles, for an infinitesimal instant at the apex of its vertical trajectory, it has zero velocity.  Its speed with respect to the surface is a continuous function so how else is it going to get from positive (distance increasing = ascending) to negative (distance decreasing = descending) without going through zero speed?

You're only considering the Earth's rotation. But that is a meaningless exclusion. There are other imposed motions: The Earth's orbit around the Sun (which is a complex spiral due to motion of the solar system) and influences of other objects such as the Moon and other planets.

Glen's question only applies to a hypothetical system of two masses in an infinite void. Initially set in motion moving apart, but not fast enough for gravitational escape. Yes, they both stop and reverse. But then you get into questions of quantum physics, and what is the smallest units of time, space and velocity you can theoretically measure. Even for atomic particles there are limits, and for macroscopic objects with non-zero thermal motion of the component parts, good luck.
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Offline Alex Eisenhut

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Re: A 'simple' Physics postulation...
« Reply #17 on: July 09, 2021, 11:45:35 pm »
When Captain Picard orders a "full stop" of the Enterprise 1701-D, does it stop?

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Offline Rick Law

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Re: A 'simple' Physics postulation...
« Reply #18 on: July 10, 2021, 12:00:46 am »
In reference to JUST the local surface of the Earth where you stand, (NOT the solar system, galaxy etc!!)...
If you were able to throw an object 'exactly' vertical, such that it came back to you due to gravity, did the
object ever 'Stop' at it's apex??  Obviously, it can't have, even for 1/1000000000-th of a second, as that would
defy the laws of physics. So it is only ever going up, or coming down. Without stopping!!?   :phew:

Think of this problem as looking at a video of the event.  You can measure where the object is on each frame, but in between frames, you don't know for sure.  That is how nature works.  In the case of a video, you can increase the frame rate and get more resolution.  In the case of Nature, there is a point where you cannot speed up the frame rate anymore.  There is a limit to the resolution one can get out of Nature.

When one uses Newtonian mechanics to describe an object's travel, one cannot take that solution to apply it to the very fast (near speed of light), very small (quantum scale, or the even smaller Planck length), or very quick  (Planck time).

What is stop?  One cannot define position smaller than one Planck length (about 1.6x10-35m), nor can one define time duration smaller than one Planck time (time it takes a photon to travel one Planck length).  So is a particle moving when it is wiggling within a cube where each side of the cube is one Planck length?  There is no answer to that question.  Can something happen quicker than one Planck time?  There is no answer to that question because you cannot measure time less than a Planck time.

That is the resolution Nature is willing to give.

So, at the best nature can offer is: the object went UP to within one Plank-length to the top, for a period of one Planck time, we don't know exactly what happened.  By next Planck time, the object is moving down.  That is the theoretical best one can do, if one can indeed get that accurate.
« Last Edit: July 10, 2021, 12:12:33 am by Rick Law »
 
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Offline Brumby

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Re: A 'simple' Physics postulation...
« Reply #19 on: July 10, 2021, 11:34:05 am »
You're only considering the Earth's rotation. But that is a meaningless exclusion. There are other imposed motions: The Earth's orbit around the Sun (which is a complex spiral due to motion of the solar system) and influences of other objects such as the Moon and other planets.
I did think that as well, but considered the polar perspective was far enough.

One cannot define position smaller than one Planck length (about 1.6x10-35m), nor can one define time duration smaller than one Planck time (time it takes a photon to travel one Planck length).
Also crossed my mind, but I constrained my journey down that track with: Define "stop".
 

Offline Ian.M

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Re: A 'simple' Physics postulation...
« Reply #20 on: July 10, 2021, 12:11:23 pm »
Over the at most couple of minutes* that one *COULD* put an object into a free fall trajectory in vacuo at or near sea level, treating the Earth's orbital motion (including its 'wobble' round the Earth - Moon barycenter) as constant velocity motion in a straight line is a reasonable simplification.

* Deepest mineshafts are around 4Km which gives a free fall time from the top of slightly under half a minute.  One could conceivably wait for such a mine to play out, lease the shaft, vacuum seal it and pump it down and conduct the experiment from the shaft bottom.  You'd need a rifle to launch the projectile, and some mighty fancy doppler laser measuring gear at the top of the shaft!

For convenience: https://www.omnicalculator.com/physics/free-fall
 

Online TimFox

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Re: A 'simple' Physics postulation...
« Reply #21 on: July 10, 2021, 02:20:38 pm »
The late Prof U Fano showed me (and the other guys) how to do quantum-mechanical calculations of macroscopic problems.  To be relevant to a large-scale question, one tracks the "expectation value" of the variable in question, which is well-defined in quantum mechanics.
 

Online TimFox

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Re: A 'simple' Physics postulation...
« Reply #22 on: July 10, 2021, 03:04:53 pm »
Going back to the original simple question about the velocity at the top of the trajectory.
Stripping away the practical complications, such as air resistance and the Earth's rotation, the question goes back to the nature of "infinitesimals", which Leibnitz and others discussed ca. 1700, as calculus was being invented.
In modern mathematics, the question was answered by a rigorous theory of limits, applied to the definition of derivative in the calculus.
A different example of limits:  consider the function  f(x) = sin(x)/x , used in the reconstruction of a continuous analog waveform from discrete sampled data.  (x in radians, of course)
When x = 0, the function would be indeterminate (0/0 is not uniquely defined).
However, applying proper limit theory (left as an exercise for the reader), the limit of f as x goes to zero (from either direction) is unity.
 

Offline T3sl4co1l

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Re: A 'simple' Physics postulation...
« Reply #23 on: July 10, 2021, 03:05:58 pm »
One that's always stuck with me is calculating the quantum number of the Earth-Sun system (applying the parameters to the hydrogenic atom).  N of course is uselessly large, the neat part is calculating the wavelength of the particle (i.e., graviton) emitted in a transition from N to N-1.  Can you guess the wavelength? ;D

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Online themadhippy

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Re: A 'simple' Physics postulation...
« Reply #24 on: July 10, 2021, 03:17:08 pm »
Is the answer 42?
 
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