Can Absolute Velocity be Measured?

[QuantumCookie]

Just a silly idea i have, please debunk it since i cannot figure it out:

The Thought Experiment:

Provided to us is a spacecraft equipped with a particle acceleration and detection systems.

The craft is set on a course where it would not be interrupted by any interstellar objects and cuts its propulsion systems for the entire duration of the experiment.

A stream of electrons are accelerated to a certain speed (let’s say 0.8c) in the particle accelerator. The mass of an electron in the stream is then measured when the electron stream passes, say, in the direction the spacecraft is traveling.

This measurement of electron mass when the electron stream passes in the direction the spacecraft is traveling is repeated for various speeds (0.82c, 0.84c, 0.86c, 0.88c and 0.9c). We then plot these values on a graph (e.g Figure 1). This graph represents the increase in mass of the electron at various velocities relative to the spacecraft .

(Figure 1)

The graph we obtained is then compared to the graph of:

(Figure 2a)

*9.10938215x10-31 kg is the rest mass of an electron

to see where it fits in. This can be achieved by comparing the change in gradients of both graphs. We then superimpose the graph we got (Figure 1) onto the graph of the equation of Figure 2a

(Figure 3)

We are now able to determine our velocity through spacetime in the direction the electron was traveling when it was measured by taking a point on the graph we obtained and subtracting the relative speed of the electron from its actual speed as reflected from the superimposed graph.

Example: In Figure 1, we measure the relative speed of the electron, W, to be say 0.8c. When we superimpose the graph, the point which contains W now reads off the new graph as X (lets say 0.83c), so we deduce that we are moving through spacetime at a velocity of 0.03c in the direction the electrons were traveling when their mass was measured.

This experiment is repeated where the mass of the electron is measured as it is traveling in various other directions to determine our absolute velocity through spacetime. The direction which yields the largest velocity will give us the absolute velocity of the spacecraft through spacetime.

Further Applications

The absolute velocity of the spacecraft through spacetime can be compared with the velocity of the spacecraft relative to the cosmic microwave background radiation (CMBR) reference frame.

If both velocities are the same, we can assume that the CMBR reference frame (and the black hole/object that gave birth to big bang) is/was moving at an absolute velocity of 0 m/s through space.

However, if both velocities were different, we can deduce that the CMBR reference frame (and the black hole/object that gave birth to big bang) is/was moving through space at a certain velocity.

This comparison would shed some light on the physical nature of the big bang itself, allowing us to eliminate a few of the seemingly infinite number of theories that surrounds the beginning of the universe we know today.

Thanks for reading!

 

[russ_watters]

Welcome to PF!

Without reading your post, I can tell you this: the question posed in the title is misguided in that it assumes that there exists such a thing as absolute velocity. Current theory holds that there is no such thing, so it is meaningless/wrong to ask if it can be measured.

 

[ZikZak]

The experimental graph will always yield the graph of γm and nothing else, leading you to conclude a speed of zero every time, regardless of the spacecraft 's velocity relative to other bodies.

 

[Mentz114]

The absolute velocity of the spacecraft through spacetime can be compared with the velocity of the spacecraft relative to the cosmic microwave background radiation (CMBR) reference frame.

Yawn. There is no absolute velocity through spacetime.

However, if both velocities were different, we can deduce that the CMBR reference frame (and the black hole/object that gave birth to big bang) is/was moving through space at a certain velocity.

You misunderstand the nature of the big-bang. The BB was the actual unfolding of space embedded in a 4-dimensional spacetime. At time t-zero, there was no space.

Imagine a balloon, whose surface is a 2-D space embedded in 3-D. Now try to imagine that the surface is a 3-D space embedded in 4-D. Just as the surface area of a balloon of zero radius is zero, so the volume of the embedded 3-D space was zero at t-zero.

 

[QuantumCookie]

Thank you for your time and comments.

1. There is no such thing as absolute velocity? Maybe I am using the wrong term, but what would you define the velocity at which *God (an observer external to this universe)* measures you traveling at?

If you accelerate (in your point of view) in one direction, time may pass slower for you but if you accelerate in another direction, time may pass faster for you as you are 'slowing down' in space. The point at which time dilation does not apply to you (except for gravitational time dilatation) can be considered a 'special' reference frame where absolute velocity and mass can be measured?

2. I know the most accepted theory holds that space and time exploded from the big bang, but I dun think that there is anything absolutely against space and time existing before the big bang.

Spacetime could have been seen to have exploded out from the singularity as the singularity had sucked so much into begin with. In any case, comparing 'absolute velocity' to the CMB reference frame would mean something either way. Dun want to argue too much about this.

3. ZikZak, would the experimental graph be the same? IMHO, One of the parameters in the equation is the rest mass of the electron measured relative to you. If you were travelling, you would weigh the electron rest mass to be more. The experimental graph you obtain should be of the same shape but translated away from the original graph.

Thank again and have to GOOD Friday.

 

[PAllen]

Thank you for your time and comments.

1. There is no such thing as absolute velocity? Maybe I am using the wrong term, but what would you define the velocity at which *God (an observer external to this universe)* measures you traveling at?

An observer outside this universe is not covered by science. You can posit anything you want - it is untestable, unknowable, and not subject to discussion in a scientific forum.

According to all known physics there is simply no such thing as absolute velocity - period; no qualifications or dancing around it. Any speculations otherwise should be pursued in a forum outside physicsforums (e.g. a religious forum).

If you accelerate (in your point of view) in one direction, time may pass slower for you but if you accelerate in another direction, time may pass faster for you as you are 'slowing down' in space.

This is complete nonsense, at odds with known observations. No matter what your state of motion or acceleration, time flows normally for you. It is true that for two clocks following different histories, one that never accelerates will show more time elapsed compared to one that accelerates away and back such that it meets the first clock. However, the direction this occurs in does not matter, nor does any supposed 'absolute velocity' of the non-accelerating clock matter.

The point at which time dilation does not apply to you (except for gravitational time dilatation) can be considered a 'special' reference frame where absolute velocity and mass can be measured?

No, completely false. Time dilation is relative. If Katy and Robyn are moving apart from each other at 90% of light speen, each concludes the other has a slower clock. If Justin is sees both moving away from him at the same speed, then he concludes both clocks are running slow. Further Katy determines Justin's clock is running slow, and Robyn's clock running even slower. Meanwhile, Robyn concludes Justin's clock is running slow, and Katy's even slower.

2. I know the most accepted theory holds that space and time exploded from the big bang, but I dun think that there is anything absolutely against space and time existing before the big bang.

Your personal beliefs, not subject to publication of a research paper, based on data or analysis, in a reputable journal, are not a proper subject for discussion in physicsforums. Note, we are not called 'idle speculation forums without understanding or knowledge'.

Spacetime could have been seen to have exploded out from the singularity as the singularity had sucked so much into begin with. In any case, comparing 'absolute velocity' to the CMB reference frame would mean something either way. Dun want to argue too much about this.

Forgetting absolute velocity as the nonsense that it is, you can measure velocity relative to CMB. Either the frequency of CMB is isotropic or it is not. If it is not, the degree of anisotropy measures your velocity relative to CMB radiation. This is cosmologically interesting, but it is not an absolute velocity. It is similar to noting that velocity relative to galactic center is interesting.

3. ZikZak, would the experimental graph be the same? IMHO, One of the parameters in the equation is the rest mass of the electron measured relative to you. If you were travelling, you would weigh the electron rest mass to be more. The experimental graph you obtain should be of the same shape but translated away from the original graph.

More complete nonsense. Let's clarify. If one particular observer is accelerating the electrons, then some other observer moving relative to the first will detect anisotropies consistent with their motion relative to the first observer. Similarly, if this second observer accelerates electrons, the first observer will detect anisotropies consistent with relative motion. There will be no observation that can distinguish which one has absolute motion, or even which one has motion relative to some third observer.

You are just throwing out purported observations that are contradicted by experiment, and by the most elemantary understanding of relativity.

Thank again and have to GOOD Friday.

 

[mfb]

1. There is no such thing as absolute velocity? Maybe I am using the wrong term, but what would you define the velocity at which *God (an observer external to this universe)* measures you traveling at?

This is undefined. He can measure whatever speed he wants (less than c), depending on the inertial frame he chooses.

If you accelerate (in your point of view) in one direction, time may pass slower for you but if you accelerate in another direction, time may pass faster for you as you are 'slowing down' in space.

No. For you, time passes with 1 second per second on your clock. And if you compare your clock with another one, only the relative velocity is interesting. And this comparison will be symmetric - both see that time passes slower at the other ship.

In any case, comparing 'absolute velocity' to the CMB reference frame would mean something either way.

You can find an inertial system where the CMB has no dipole moment. But in terms of the laws of physics, there is nothing special about this system.

3. ZikZak, would the experimental graph be the same?

It would.
Note that different observers can measure different particle energys (and therefore different relativistic masses (I don't like this word)).Welcome to the concept of "relativity" ;).

 

[russ_watters]

Thank you for your time and comments.

1. There is no such thing as absolute velocity? Maybe I am using the wrong term, but what would you define the velocity at which *God (an observer external to this universe)* measures you traveling at?

Sorry - yes, you are defining it properly, but no, it doesn't exist.

 

[spikenigma]

There is , of course, no such thing as absolute velocity.

An interesting thought, however, is a spinning cylinder in space and an observer flying past, parallel to it's spin with no acceleration. Everyone in the universe (even in an empty universe) will agree that the cylinder is spinning at x velocity.

The observer can say "I am 5x away from it's moving point at an exact 180 degree angle...I am 4x away from it's moving point at an exact 180 degree angle...I am 3x...etc...

Thus since it is moving with me and everybody agrees that it is moving, I am moving also."

 

[HallsofIvy]

There is , of course, no such thing as absolute velocity.

An interesting thought, however, is a spinning cylinder in space and an observer flying past, parallel to it's spin with no acceleration. Everyone in the universe (even in an empty universe) will agree that the cylinder is spinning at x velocity.

The observer can say "I am 5x away from it's moving point at an exact 180 degree angle...I am 4x away from it's moving point at an exact 180 degree angle...I am 3x...etc...

Thus since it is moving with me and everybody agrees that it is moving, I am moving also."

That's because rotational motion is an acceleration.

 

[spikenigma]

That's because rotational motion is an acceleration.

Yes, but philosophically the non-accelerating traveler can say.

" Everybody agrees that this accelerated point on the frame is moving, I am moving along the x-axis with it, thus I am moving. "

 

[phinds]

Note, we are not called 'idle speculation forums without understanding or knowledge'.

Nicely put ! It's too bad some of the newbies don't get that before posting.

 

[Whovian]

Yes, but philosophically the non-accelerating traveler can say.

" Everybody agrees that this accelerated point on the frame is moving, I am moving along the x-axis with it, thus I am moving. "

No, not everyone agrees that the accelerating point on the frame is moving. As far as I can tell, what you're saying is that if something is accelerating, then it's moving.

This is completely incorrect. Throw a ball straight up. At the peak of its toss, it is not moving. However, it is still subject to gravity, so it's still accelerating. Now, nothing can stay umoving for finite time when it's subject to acceleration.

 

[Naty1]

The absolute velocity of the spacecraft through spacetime can be compared with the velocity of the spacecraft relative to the cosmic microwave background radiation (CMBR) reference frame.

The CMBR is simple one convenient reference frame for comparing measurements. Nothing unique, nor special, nor 'one a kind' frame. Everyone who makes measurements, unless at the same point in time and space, generally obtains different results because space and time vary from point to point in spacetime. We can convert measurements, that is 'transform', from one spacetime point and set of coodinates to another, but yours will be as good as mine as good as every one of the other infinite number of available spacetime points. No frames are special.

 

[Austin0]

No, not everyone agrees that the accelerating point on the frame is moving. As far as I can tell, what you're saying is that if something is accelerating, then it's moving.

This is completely incorrect. Throw a ball straight up. At the peak of its toss, it is not moving. However, it is still subject to gravity, so it's still accelerating. Now, nothing can stay umoving for finite time when it's subject to acceleration.

I would think that as soon as it leaves your hand it is in inertial motion ; freefall. As such there is obviously no point of no motion until it returns to your hand. That is an illusion. Like Zenos arrow.
Based on an abstract dimensionless slice of time that does not apply to real world physics. Zeno's point in creating the paradox.
On the other hand sitting motionless on the ground is a state of acceleration so you're right, not everyone agrees acceleration necessarily implies motion.

 

[Bjarne]

There is no absolute velocity, because c (and maybe everything else) always looks the same for all observers (so fare I have understood).

But this is not the same that there is no absolute motion direction.

We can for example be moving the same or the opposite direction as the Milky Way, and at a larger perspective, the same or the opposite way as the local cluster, and if we one day will reach a larger perspective, for example multi universes moving relative to each other, - we can also be moving the same or opposite way relative to our Universe….etc..

I either cannot see the “problem” or any limit to that absolute motion not should exist.

That out perspective is limited , cannot be the same that an absolute motion perspective not exist.

 

[Naty1]

unsure just what you last post says...

but with Newtonian mechanics, the relative velocity is independent of the chosen inertial reference frame. Time and space are the same everywhere. All frames are the same.

This is not the case anymore with special relativity in which velocities depend on the choice of reference frame. Time and distance vary in different partsof spacetime; any frame is as good as any other; different frames give different results.

 

[mfb]

We can for example be moving the same or the opposite direction as the Milky Way

No, that is not possible, unless you define an arbitrary inertial system in which you want to make this statement. The only physical thing is our velocity relative to the Milky Way, like "we are moving towards the center with 500km/s".

for example multi universes moving relative to each other

And that is even "more undefined", as different universes (with common meanings of the word) are not connected in space. You cannot even quote a distance between universes, as there is no space to measure the distance in it.

 

[Bjarne]

No, that is not possible, unless you define an arbitrary inertial system in which you want to make this statement. The only physical thing is our velocity relative to the Milky Way, like "we are moving towards the center with 500km/s".
.

It can also be relative to the center of a cluster of galaxies

 

[yuiop]

I would think that as soon as it leaves your hand it is in inertial motion ; freefall. As such there is obviously no point of no motion until it returns to your hand. That is an illusion. Like Zenos arrow.
Based on an abstract dimensionless slice of time that does not apply to real world physics. Zeno's point in creating the paradox.

If we take the velocity of the ball 1 second before the apex and the velocity of the ball one second after the apex then the average velocity over that 2 seconds is zero. We can reduce that 2 seconds to as small an interval as we like and still obtain an average velocity of zero. This is part of the slight of hand of how calculus works. Certainly we can set dr/dt = 0 in the equations of motion of a tossed ball and obtain a sensible answer for when the ball arrives at apogee without obtaining any infinities or undefined answers. Mathematically there is a point of no motion for an accelerating ball in free fall. However, you are right that Zeno's arrow paradox argues against this and basically he argues that there is no such thing as a "instant of time". More recently Peter Lynds made a similar point in a published paper. Personally I believe we will end up with a theory of physics that is based on granular quantum time but currently we have no such theory and time is infinitely divisible.

As far as the CMBR, I hope the OP is aware that two observers separated by billions of light years could have a relative velocity of say 0.9c with respect to each other and yet both be at rest with their local CMBR in the expanding universe so there is no absolute sense of who is moving or who is stationary even when the CMBR is taken into account.

 

[Whovian]

I would think that as soon as it leaves your hand it is in inertial motion ; freefall. As such there is obviously no point of no motion until it returns to your hand. That is an illusion. Like Zenos arrow.
Based on an abstract dimensionless slice of time that does not apply to real world physics. Zeno's point in creating the paradox.
On the other hand sitting motionless on the ground is a state of acceleration so you're right, not everyone agrees acceleration necessarily implies motion.

Wait. I thought the derivative got rid of Zeno's Paradox.

The post I was replying to basically assumed, from what I gathered, that in all inertial reference frames, an accelerating object is moving at all points in time. I was trying to find a certain reference frame with a certain time that this didn't hold. (Note that gravity complicates things a bit, but we can move back to the reference frame of the ball and see the velocity of the observer, which is 0 at a certain point.)

Wait. One small thing. Sitting on the ground isn't an inertial reference frame.

Note, again, that having acceleration but no velocity cannot persist for any extended period of time.

 

[Austin0]

This is completely incorrect. Throw a ball straight up. At the peak of its toss, it is not moving. However, it is still subject to gravity, so it's still accelerating. Now, nothing can stay unmoving for finite time when it's subject to acceleration.

The ball is subject to acceleration through the whole course from the hand and back.
So according to your next statement; it can not stay unmoving for finite time through that whole time while it's subject to acceleration.
Finite time is the key. As long as the time duration is non-zero (finite) there will be movement. It is only when you consider an instant (interval with no duration) that there is no motion. Like a snapshot with a camera with an infinitely fast shutter speed. But then anything in motion is motionless for that instant. That is the basis of Zeno's arrow paradox.

I would think that as soon as it leaves your hand it is in inertial motion ; freefall. As such there is obviously no point of no motion until it returns to your hand. That is an illusion. Like Zenos arrow.
Based on an abstract dimensionless slice of time that does not apply to real world physics. Zeno's point in creating the paradox.
On the other hand sitting motionless on the ground is a state of acceleration so you're right, not everyone agrees acceleration necessarily implies motion.

Wait. I thought the derivative got rid of Zeno's Paradox.

The post I was replying to basically assumed, from what I gathered, that in all inertial reference frames, an accelerating object is moving at all points in time. I was trying to find a certain reference frame with a certain time that this didn't hold. (Note that gravity complicates things a bit, but we can move back to the reference frame of the ball and see the velocity of the observer, which is 0 at a certain point.)

Wait. One small thing. Sitting on the ground isn't an inertial reference frame.

Note, again, that having acceleration but no velocity cannot persist for any extended period of time.

I think you must be referring to Achilles and the tortoise. A reductio ad absurdum argument against thinking the mathematical concept of infinite divisibility of a line, applied to the real world.
The irony is that, for thousands of years after, many mathematicians failed to get the joke [see the point] and seriously tried to solve the puzzle and mathematically justify the real world result. I.e Achilles does catch up with the tortoise.

Another example is the abstract concept of a dimensionless mathematical point.
When applied in the real world in the form of a point charge in electrostatics, it produced Zenoesque calculation results (infinities) and required renormalization. In effect assigning it a finite dimension.

One small thing Note above where I said sitting on the ground is an accelerating frame.
No disagreement.

 

[Austin0]

If we take the velocity of the ball 1 second before the apex and the velocity of the ball one second after the apex then the average velocity over that 2 seconds is zero.
We can reduce that 2 seconds to as small an interval as we like and still obtain an average velocity of zero. This is part of the slight of hand of how calculus works.

1) AN average velocity of 0 over a course does not necessarily imply any interval of 0 velocity at any point in that course

2) More importantly it definitely doesn't imply no motion.

Would you agree that many of the tricks of calculus ,like limits , provide exceedingly useful and accurate approximations?
SO yes; as dt ---->0, dr ----->0 and v ---->0 but it is only at dt=0 (dimensionless instant) that v=0

Certainly we can set dr/dt = 0 in the equations of motion of a tossed ball and obtain a sensible answer for when the ball arrives at apogee without obtaining any infinities or undefined answers.
Mathematically there is a point of no motion for an accelerating ball in free fall.

Yes you can find an accurate figure for when the ball reaches apogee dt_a.
And an equally accurate figure for when the ball will then return to the initial point dt_i if you add them together and subtract the sum from the time for the whole course dt_c - (dt_a+dt_i) = what do you get?
0 time actually at apogee ,right?
For the ball to actually be motionless at this point in space would require a finite time interval spent at this exact location. Which would require an external acceleration/force, anti-gravity or magic.

As far as the CMBR, I hope the OP is aware that two observers separated by billions of light years could have a relative velocity of say 0.9c with respect to each other and yet both be at rest with their local CMBR in the expanding universe so there is no absolute sense of who is moving or who is stationary even when the CMBR is taken into account.

 

[yuiop]

...
The irony is that, for thousands of years after, many mathematicians failed to get the joke [see the point] and seriously tried to solve the puzzle and mathematically justify the real world result. I.e Achilles does catch up with the tortoise.

I agree with this 100%. Mathematicians have failed to get the point of Zeno's paradoxes which for the most part demonstrate that if time and space are infinitely divisible, as mathematicians assume, then the world simply would not work the way we observe it does. Slowly and quietly the revolution is taking place and Zeno type effects are being taken seriously in particle physics at the quantum scale involving virtual photons etc. I believe if we eventually arrive at a quantum theory of gravity that supercedes GR, then mathematicians might finally get the point and Zeno will be vindicated. Until then, calculus says the velocity of a particle at apogee is zero.

 

[GeorgeDishman]

I think you must be referring to Achilles and the tortoise. A reductio ad absurdum argument against thinking the mathematical concept of infinite divisibility of a line, applied to the real world.

The irony is that, for thousands of years after, many mathematicians failed to get the joke [see the point] and seriously tried to solve the puzzle and mathematically justify the real world result. I.e Achilles does catch up with the tortoise.

The error in Zeno's Paradox is that he implicitly assumes that the sum of an infinite series must also be infinite, Achilles cannot catch the tortoise in any finite number of steps, i.e. on a graph plotted with equal x-axis increments representing a halving of the separation, the point at which they meet cannot be plotted. In a sense it has a parallel in GR and cosmology in that some coordinate schemes do not cover the whole of a manifold.

 

[GeorgeDishman]

As far as the CMBR, I hope the OP is aware that two observers separated by billions of light years could have a relative velocity of say 0.9c with respect to each other and yet both be at rest with their local CMBR in the expanding universe so there is no absolute sense of who is moving or who is stationary even when the CMBR is taken into account.

True but it is worse than that, the observers can easily be moving apart at greater than c because the intervening space is expanding by more than one light year per year, and it then becomes difficult to even define velocity.

Galaxies at a red shift more than roughly z=1 were moving faster than c when they emitted the light we receive. The material that emitted the CMBR was expanding away from us at ~50c at that time and the galaxies it turned into are still "moving" at ~3c (IIRC).

 

[Austin0]

1) AN average velocity of 0 over a course does not necessarily imply any interval of 0 velocity at any point in that course

2) More importantly it definitely doesn't imply no motion.

Would you agree that many of the tricks of calculus ,like limits , provide exceedingly useful and accurate approximations?
SO yes; as dt ---->0, dr ----->0 and v ---->0 but it is only at dt=0 (dimensionless instant) that v=0


Yes you can find an accurate figure for when the ball reaches apogee dt_a.
And an equally accurate figure for when the ball will then return to the initial point dt_i if you add them together and subtract the sum from the time for the whole course dt_c - (dt_a+dt_i) = what do you get?
0 time actually at apogee ,right?
For the ball to actually be motionless at this point in space would require a finite time interval spent at this exact location. Which would require an external acceleration/force, anti-gravity or magic.

I agree with this 100%. Mathematicians have failed to get the point of Zeno's paradoxes which for the most part demonstrate that if time and space are infinitely divisible, as mathematicians assume, then the world simply would not work the way we observe it does. Slowly and quietly the revolution is taking place and Zeno type effects are being taken seriously in particle physics at the quantum scale involving virtual photons etc. I believe if we eventually arrive at a quantum theory of gravity that supercedes GR, then mathematicians might finally get the point and Zeno will be vindicated. Until then, calculus says the velocity of a particle at apogee is zero.

I have noticed you did not respond to any of my points above, but instead seem to have turned the question into a matter to be decided by relative authority. Zeno vs Calculus

Since I never brought up Zeno ,other than as an interesting aside, this might appear to be an ingenious combo, Straw Authority argument on you part

In any case: Did you miss the above post? Agree?Disagree?

 

[yuiop]

1) AN average velocity of 0 over a course does not necessarily imply any interval of 0 velocity at any point in that course

2) More importantly it definitely doesn't imply no motion.

Forget about the average velocity thing. Let's look at the standard equations of motion. http://en.wikipedia.org/wiki/Equations_of_motion#Constant_linear_acceleration

E.g. the first one v = at +u.

Lets say the initial velocity (u) is -20 m/s and the acceleration (a) is 10 m/s^2. The velocity after 2 seconds is exactly zero. This is the velocity at the exact instant labelled "2 seconds" and is not an average of one millisecond before and after or anything like that. Calculus does not require velocity to be defined in terms of a finite distance over a finite non zero time interval.

Would you agree that many of the tricks of calculus ,like limits , provide exceedingly useful and accurate approximations?

Mathematicians would say they are not approximations.

SO yes; as dt ---->0, dr ----->0 and v ---->0 but it is only at dt=0 (dimensionless instant) that v=0

Yep, but mathematicians and physicists find that acceptable.

Yes you can find an accurate figure for when the ball reaches apogee dt_a.
And an equally accurate figure for when the ball will then return to the initial point dt_i if you add them together and subtract the sum from the time for the whole course dt_c - (dt_a+dt_i) = what do you get?
0 time actually at apogee ,right?

We can do the same for any other point or velocity in the trajectory. If the initial velocity is 100 m/s upwards and we find the time to slow down to 70 m/s and the time to get to the apogee and return to -100 m/s we find that the time spent at 70 m/s is exactly zero. We cannot find any velocity that the object spends a finite time traveling at. The object does not spend a finite time being at any given velocity. I am sure that Zeno would have found that interesting ;)

For the ball to actually be motionless at this point in space would require a finite time interval spent at this exact location. Which would require an external acceleration/force, anti-gravity or magic.

Lets look at that equation of motion again: v = at +u

Lets say we throw the ball up at an initial velocity of u = 100 m/s and the acceleration of gravity is a = -10 m/s^2. At what time is the velocity exactly zero?

v = at + u

0 = -10*t +100

t = 10 seconds.

So after 10 seconds the velocity is exactly zero (at the apogee).

Put it another way. If an object had a velocity of 7 m/s and after accelerating it had a velocity of 9 m/s would you agree that at some point its velocity must have been 8 m/s? If your answer to that is yes, then you would also have to agree that if an object had a velocity of +1 m/s and later had a velocity of -1 m/s that its velocity at some point in between must have been zero.

 

[space_buster]

If you state there is absolute velocity , the statement violates the General Relativity concept as well as applicability of Physics Laws to frame of references .
Existence of absolute velocity would mean existence of absolute frame of reference and all the other frame of references are relative to it and that would mean existence of absolute SPACE-TIME which is not possible .
I hope this might help you .

 

[Austin0]

" Everybody agrees that this accelerated point on the frame is moving, I am moving along the x-axis with it, thus I am moving. "

Throw a ball straight up. At the peak of its toss, it is not moving. Now, nothing can stay umoving for finite time when it's subject to acceleration.

I would think that as soon as it leaves your hand it is in inertial motion ; freefall. As such there is obviously no point of no motion until it returns to your hand. That is an illusion. Like Zenos arrow.
.

As long as the time duration is non-zero (finite) there will be movrmrnt. It is only when you consider an instant (interval with no duration) that there is no motion. .

when the ball arrives at apogee Mathematically there is a point of no motion for an accelerating ball in free fall. time is infinitely divisible.

.

As you can see from the original parameters of this thread the question was motion not velocity
You changed the conditions to velocity which is of course a different concept with a different definition.

1)]Would you agree that many of the tricks of calculus ,like limits , provide exceedingly useful and accurate approximations

Mathematicians would say they are not approximations.

Are you really sure about this assertion? You did leave out the exceedingly accurate qualification in my question.

1)]
0 time actually at apogee ,right?
For the ball to actually be motionless at this point in space would require a finite time interval spent at this exact location. Which would require an external acceleration/force, anti-gravity or magic..

We can do the same for any other point or velocity in the trajectory. If the initial velocity is 100 m/s upwards and we find the time to slow down to 70 m/s and the time to get to the apogee and return to -100 m/s we find that the time spent at 70 m/s is exactly zero. We cannot find any velocity that the object spends a finite time traveling at. The object does not spend a finite time being at any given velocity. I am sure that Zeno would have found that interesting ;)

I never suggested that the point of apogee was at all special ,in fact that was my point so this is no surprise ,,,I don't know about Zeno?? ;-)

Forget about the average velocity thing. Let's look at the standard equations of motion. http://en.wikipedia.org/wiki/Equations_of_motion#Constant_linear_[B]acceleration[/B]

E.g. the first one v = at +u.

Lets say the initial velocity (u) is -20 m/s and the acceleration (a) is 10 m/s^2. The velocity after 2 seconds is exactly zero. This is the velocity at the exact instant labelled "2 seconds" and is not an average of one millisecond before and after or anything like that. Calculus does not require velocity to be defined in terms of a finite distance over a finite non zero time interval. Lets look at that equation of motion again: v = at +u

Lets say we throw the ball up at an initial velocity of u = 100 m/s and the acceleration of gravity is a = -10 m/s^2. At what time is the velocity exactly zero?

v = at + u

0 = -10*t +100

t = 10 seconds.

So after 10 seconds the velocity is exactly zero (at the apogee).

Put it another way. If an object had a velocity of 7 m/s and after accelerating it had a velocity of 9 m/s 1)would you agree that at some point its velocity must have been 8 m/s? If your answer to that is yes, then you would 2)also have to agree that if an object had a velocity of +1 m/s and later had a velocity of -1 m/s that its velocity at some point in between must have been zero.

Yes I am somewhat familiar with the equations of motion and their application in this case, that was never a point of contention ,just the interpretation of the results of that math.
Correct me if I am in error but:

1) I would say that strictly speaking the term velocity and its math only applies to inertial particles. With a constant motion , unchanging vector in both direction and magnitude..

2) Acceleration is a constant state of changing motion with a dynamic vector, varying in either; direction, magnitude or both.
It is a continuum of states, moving not only in space but along this continuum. Any finite interval of time must result in a change of location on the scale I.e. a point of different instantaneous velocity. To pick ANY durationless point on this continuum and declare there is no motion along the continuum is exactly equivalent to picking a durationless point on the path of an inertial particle and saying it is motionless in space.

3) Yes we can calculate an instantaneous velocity and this is a useful abstraction; eg. ICMIF's.
But it is not an actual velocity, which I think is demonstrated if you take the derived value and apply the math for velocity... dx=v x dt for some finite dt.
This will not produce an exact prediction for the position of the particle. For that you need to use the applicable math for acceleration.

Calculus does not require velocity to be defined in terms of a finite distance over a finite non zero time interval.

4) As far as I know that is exactly the definition of velocity. You are using the variable v from the equation for acceleration and presenting it as a general definition of velocity. But it is only an actual velocity if it is final. If acceleration ceases at that point.
Otherwise it is an instantaneous velocity and has a different meaning.

Motion of a particle has a very simple definition. A change in any spatial component between two events.
How do you propose to define it in terms of a single event??

5) The question of a state of motion or motionlessness is a boolean proposition and as physics is an exact and precise science this eliminates approximations in this case. So yes time is infinitely divisible** but no matter how infinitesimal you make the time interval there will be an infinitesimal change in both position and velocity.

** I never questioned this and it is irrelevant to the arrow paradox.
1) and 2) I didn't directly answer these questions but I trust you can infer my answers from what followed.

 

[pervect]

Yes I am somewhat familiar with the equations of motion and their application in this case, that was never a point of contention ,just the interpretation of the results of that math.
Correct me if I am in error but:

1) I would say that strictly speaking the term velocity and its math only applies to inertial particles. With a constant motion , unchanging vector in both direction and magnitude..

I haven't been following the rest of this thread, but since you asked for corrections...

I don't see why you say this. Accelerating particles have velocities. There's nothing wrong or ill-defined about an instantaneous velocity, just as there is nothing wrong with taking the derivative of a function, or finding a vector that's tangent to a curve at a point. In fact, all of these ideas are closely related.

 

[Austin0]

Yes I am somewhat familiar with the equations of motion and their application in this case, that was never a point of contention ,just the interpretation of the results of that math.
Correct me if I am in error but:

1) I would say that strictly speaking the term velocity and its math only applies to inertial particles. With a constant motion , unchanging vector in both direction and magnitude..

I haven't been following the rest of this thread, but since you asked for corrections...

I don't see why you say this. Accelerating particles have velocities. There's nothing wrong or ill-defined about an instantaneous velocity, just as there is nothing wrong with taking the derivative of a function, or finding a vector that's tangent to a curve at a point. In fact, all of these ideas are closely related.

Hi I never meant to suggest there was anything wrong with instantaneous velocity, or derivatives or integration, I am well aware of their power. Only that instantaneous velocity
is only well defined or exact for an instant (zero duration) of time. So you can calculate an accurate momentum with this value which will be exact if there happens to be a collision at that same instant but it is a transitory value.
That there is a difference between velocity and instantaneous velocity.
You may disagree and be correct or think that I am splitting hairs [or infinitesimals ;-)) but if you look at the question at hand that is what it comes down to.

Thanks for your input. I don't enjoy being mistaken but I am always happy to be put straight if that is the case.

 

[GeorgeDishman]

1) Would you agree that many of the tricks of calculus ,like limits , provide exceedingly useful and accurate approximations

Mathematicians would say they are not approximations.

Are you really sure about this assertion?

The statement is correct, a derivative is not an approximation.

1) I would say that strictly speaking the term velocity and its math only applies to inertial particles. With a constant motion , unchanging vector in both direction and magnitude.

Strictly speaking, velocity is a derivative and is exact at a particular instant.

2) Acceleration is a constant state of changing motion with a dynamic vector, varying in either; direction, magnitude or both.

To pick ANY durationless point on this continuum and declare there is no motion along the continuum is exactly equivalent to picking a durationless point on the path of an inertial particle and saying it is motionless in space.

I don't think you said what you meant. An inertial particle is motionless in it's rest frame at all times by definition.

3) Yes we can calculate an instantaneous velocity and this is a useful abstraction; eg. ICMIF's.
But it is not an actual velocity, which I think is demonstrated if you take the derived value and apply the math for velocity... dx=v x dt for some finite dt.

That is not the definition of velocity. v=dx/dt is a derivative, not a ratio. The value of the derivative is not dx divided by dt, it it the single unique limit value which that ratio approaches as dt -> 0.

This will not produce an exact prediction for the position of the particle. For that you need to use the applicable math for acceleration.

To predict the position, you cannot use simple multiplication in general, you have to integrate the velocity (the exception of course is when the velocity is constant) and if acceleration isn't constant then you need to integrate the jerk, etc..

Calculus does not require velocity to be defined in terms of a finite distance over a finite non zero time interval.

4) As far as I know that is exactly the definition of velocity.

"yuiop" is correct, velocity is the slope of the tangent to the line, not the ratio of finite deltas.

 

[russ_watters]

I'd take that one step further: only the instantaneous velocity as obtained by a derivative is exact. A velocity taken using change in position over a delta-t is an average over that delta-t and therefore only an approximation if used as a proxy for an instantaneous velocity.

Austin0 has the issue backwards.

And IMO, Xeno's paradox is little more than a silly riddle for non-scientists today even if it was a profound problem 2000 years ago. It isn't a difficult to figure out and doesn't even require much math, much less a discussion of the quantization of space/time.

 

[harrylin]

Welcome to PF!

Without reading your post, I can tell you this: the question posed in the title is misguided in that it assumes that there exists such a thing as absolute velocity. Current theory holds that there is no such thing, so it is meaningless/wrong to ask if it can be measured.

Also without reading that post, and while obviously it cannot be measured according to current theory, clearly your answer is equally misguided: current theory treats questions about things that in theory cannot be measured as belonging to philosophy; modern theories of physics make no statements about the existence or non-existence of such things.
Nevertheless, a modern interpretation of relativity as affected by quantum theory (based on Bell's theorem) does hold such a thing as a plausible model of reality - but that's another topic!

 

[harrylin]

[..] According to all known physics there is simply no such thing as absolute velocity - period; no qualifications or dancing around it. Any speculations otherwise should be pursued in a forum outside physicsforums (e.g. a religious forum).

Actually, such models (as well as block universe, many worlds interpretations etc.) are discussed in the QM forum as well as in this forum; physical models and possible explanations have always belonged to physics even though they are philosophical in the sense that they cannot be directly verified.

 

[harrylin]

Just a silly idea i have, please debunk it since i cannot figure it out: [..]

I have other things to do today and there is an infinite number of paradoxes that one can (and does!) invent (and some of which I did solve) - so, my excuses for not reading it. As I came late to this thread, has the error been pointed out to your satisfaction? and how good is your knowledge of relativity?

 

[Austin0]

Would you agree that many of the tricks of calculus ,like limits , provide exceedingly useful and accurate approximations?

Mathematicians would say they are not approximations.

The statement is correct, a derivative is not an approximation.

Wiki et al.

The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value.

so that "f(x) becomes arbitrarily close to L" means that f(x) eventually lies in the interval (L - ε, L + ε),

\lim_{x \to c}f(x) = L
the derivative of the position of a moving object with respect to time is the object's instantaneous velocity.
means that f(x) can be made to be as close to L as desired by making x sufficiently close to c.

Are these not correct?

SO I guess the first question is the meanings of exact and approximate.

ap·prox·i·ma·tion
n.
2. Mathematics An inexact result adequate for a given purpose.


approximation [əˌprɒksɪˈmeɪʃən]
n 4. (Mathematics) Maths
a. an estimate of the value of some quantity to a desired degree of accuracy
approximate
adj [əˈprɒksɪmɪt]
1. almost accurate or exact

Considering a particle under constant acceleration : to the limit \Deltat ---> \infty would you say the derived value was exactly c or approximately c?

1) I would say that strictly speaking the term velocity and its math only applies to inertial particles. With a constant motion , unchanging vector in both direction and magnitude.

Strictly speaking, velocity is a derivative and is exact at a particular instant

.

To pick ANY durationless point on this continuum and declare there is no motion along the continuum is exactly equivelant to picking a durationless point on the path of an inertial particle and saying it is motionless in space.

I don't think you said what you meant. An inertial particle is motionless in it's rest frame at all times by definition.

There was no reference to the rest frame of the particle.It was an inertial test particle and an accelerating particle relative to one observation frame
If we pick a point of the inertial path and say: this point is an instant and having no duration, there can be no change of location in this instant. So therefore the particle is motionless in space at this point. I.e. Has no velocity.

DO you think this is a valid or meaningful logical conclusion or description of reality?

I certainly don't.
It is trivially true but meaningless. An abstraction divorced from reality and falsified by that reality, wherein the particle is in continuous motion throughout.
I was comparing that with picking a durationless point on an accelerating path and saying it was motionless (not changing velocity) at that point.

This will not produce an exact prediction for the position of the particle. For that you need the applicable math for acceleration.

To predict the position, you cannot use simple multiplication in general, you have to integrate the velocity (the exception of course is when the velocity is constant) and if acceleration isn't constant then you need to integrate the jerk, etc .

If you will read what I wrote I think you will find that what you are saying here is exactly what I said. That I was comparing the case of constant Velocity (inertial motion) with Instantaneous Velocity (accelerated motion)
In the first case you simply apply the value dx/dt to predict position,, in the second case you could not and needed to apply other maths,integration or in this case the equation for constant acceleration.

3) Yes we can calculate an instantaneous velocity and this is a useful abstraction; eg. ICMIF's. But it is not an actual velocity which I think is demonstrated if you take the derived value and apply the math for velocity dx = v(dt) for some finite dt.

That is not the definition of velocity. v=dx/dt is a derivative, not a ratio. The value of the derivative is not dx divided by dt, it it the single unique limit value which that ratio approaches as dt -> 0.

There seems to be a miscommunication. You are defining velocity in terms of a process for its determination.
I was talking about the definition of the term as applied to the result of that process.The meaning of the actual value thus derived and called velocity.
Which in all cases is in fact a ratio...correct? some dx/dt

Am I communicating this distinction I am talking about clearly??

Isn't the fundamental definition of velocity; rate of change of position as a function of time??
The difference in position with respect to the difference in time?

Isn't this validly expressed using the fundamental definition of delta as v=dx/dt
In the case of constant linnear motion , this function v=dx/dt. produces a value (dx/dt) which completely and exactly describes the motion of the particle over time without exception. correct??
Would you agree it is strictly applicable without restriction or qualification to a inertial particle??

Calculus does not require velocity to be defined in terms of a finite distance over a finite non zero time interval.

4) As far as I know that is exactly the definition of velocity. You are using the value v from the equation for acceleration and presenting it as a general definition. But it is only an actual velocity if it is final. If acceleration ceases at that point. Otherwise it is an instantaneous velocity and has a different meaning

"yuiop" is correct, velocity is the slope of the tangent to the line, not the ratio of finite deltas.

As you can see I was referring to the meaning and definition of the derived value

If the velocity is the slope of the tangent, then what is that but a simple ratio?
Expressed as a ratio of finite deltas? Eg. 2mm/sec

The tangent as a line, is indefinitely extended in space and time.
Isn't it true that it only exactly describes the motion of a world line or segment of a world line that is congruent or parallel ??
The instantaneous co-moving inertial frame , yes?.
That it is only congruent with the accelerated world line at a mathematical point. I.e. dimensionless in space and time.

Obviously this value (dx/dt) is incredibly useful as an input value for a variable in other equations but what is it's kinematic meaning as describing or predicting the motion of this accelerating particle in the real world other than an approximation??

I hope you now understand the mix up in the subject of our definitions and would like to know if you still think I was incorrect on any point.
Thanks

 

[Austin0]

Hi yuiop I don't know if you remember, but this debate actually started a long time ago in another thread.
We didn't conclude it then because it was off topic.
This discussion seems to have run into semantic snarls and I have let it sit for awhile. But I hope that those questions where you feel I am incorrect can be clarified so we might continue the discussion.

Put it another way. If an object had a velocity of 7 m/s and after accelerating it had a velocity of 9 m/s would you agree that at some point its velocity must have been 8 m/s? If your answer to that is yes, then you would also have to agree that if an object had a velocity of +1 m/s and later had a velocity of -1 m/s that its velocity at some point in between must have been zero.

Yes I would agree that it had to pass through a point where the calcuated velocity would be zero for an abstract dimensionless point of time .
Theb question is dos this imply motionlessnss??

Would I be wrong to say that the equation for determining instantaneous acceleration as applied to the apogee would be
a= limit as dt --->0 of dv/dt =0 also ?

Would you say this was a meaningful description of the acceleration of the particle at this point?

Or would it be more correct to say it was undergoing constant (non-zero) acceleration the whole time?


Would you agree that time, motion (change of position) and acceleration (change of velocity) are continua , not series of discrete steps??

 

[Austin0]

I'd take that one step further: only the instantaneous velocity as obtained by a derivative is exact. A velocity taken using change in position over a delta-t is an average over that delta-t and therefore only an approximation if used as a proxy for an instantaneous velocity.

Austin0 has the issue backwards..

A velocity taken using change in position over a delta-t with regard to an inertial particle is quite exact and that is the context that I was talking about.Not only is the value produced exact, but that value itself provides an eact description of the motion of the particle over time.
In contrast to an instantaneous value which when applied to an accelerating particle provides only an approximate description of the motion over a very limited range.Which was my point.

And IMO, Xeno's paradox is little more than a silly riddle for non-scientists today even if it was a profound problem 2000 years ago. It isn't a difficult to figure out and doesn't even require much math, much less a discussion of the quantization of space/time.

I essentialy agree with you. I only directly mentioned it in this discussion in one two word sentence;
"Zeno's Arrow"
I entered this not as an argument but rather as an illustration of a fallacy (IMO) which I assumed was generally known and understood. In this I may have been in error ..
Thanks for your response

 

[Dale]

Wiki et al.

The derivative is exact. The Wiki quote is referring to a Taylor series expansion. The series expansion uses a derivative to approximate a function. The series expansion is an approximation, not the derivative.

 

[Austin0]

The derivative is exact. The Wiki quote is referring to a Taylor series expansion. The series expansion uses a derivative to approximate a function. The series expansion is an approximation, not the derivative.

All the sources I found seemed to agree , not just Wiki. If a function is taken to a limit isn't it a derivative? If the result is within a range (L+infinitesimal,L-infinitesimal) isn't this an exceedingly close approximation?

SO does the Taylor expansion use the same symbolic representation?

So maybe I should ask what you mean by exact?

Considering a particle under constant acceleration : to the limit Δt --->\infty

v= a(dt) = ? ...would you say the derived value v was exactly c or approximately c?

You didn't answer this question.

Thanks

 

[yuiop]

Yes I would agree that it had to pass through a point where the calcuated velocity would be zero for an abstract dimensionless point of time.

I would agree that it was an abstract notion.

Theb question is dos this imply motionlessnss??

Similarly there are other points where the velocity is non-zero for a similarly abstract dimensionless point of time and I am sure you would use those abstract points to show that there is motion. A lot depends on how "motionlessness" is defined. If we define it as a change in location over a finite non zero time interval, the zero result for dr/dt at the apogee does not prove by itself, that the particle is "motionless at the apogee. However, since you made it clear that you prefer to work with non abstract intervals of non zero time intervals, then it easy to show that the average velocity of the particle averaged from just before the apogee to just after the apogee, is zero. You might then claim that is only an average, but its instantaneous velocity at any instant is non zero during that averaging period and we would be back to discussing abstract dimensionless points of time again. (No win situation )

Would I be wrong to say that the equation for determining instantaneous acceleration as applied to the apogee would be
a= limit as dt --->0 of dv/dt =0 also ?

No, we have defined the accleration as constant in this case and so the acceleration remains non zero even at the apogee. dr/dt=0 does not imply dv/dt=0. Again, if dismiss abstract instantaneous points of time (which are very handy mathematically) then we can work in terms of average velocity over a finite time interval that is as small as you like and we can then demonstrate that the average velocity over that tiny time interval is zero and the acceleration is non zero, if we zoom into a tinier (but non zero) time scale.

Would you say this was a meaningful description of the acceleration of the particle at this point?

Or would it be more correct to say it was undergoing constant (non-zero) acceleration the whole time?

The latter.

Would you agree that time, motion (change of position) and acceleration (change of velocity) are continua , not series of discrete steps??

Well I won't comit myself to that one. I have Zenoist sympathies so I am uncommitted on what happens physically at the quantum scale, but my hunch is that any ultimate TOE or theory of quantum gravity will involve discrete quantities rather than continua. Certainly Planck demonstrated that the only way to resolve the "ultraviolet catastrophe" is to assume the energy comes in discrete quantities and Einstein also demonstrated that in his Nobel prize winning analysis of the photoelectric effect and quantum theory by its very nature involves discrete quantities.

 

[Dale]

All the sources I found seemed to agree , not just Wiki.

You misunderstood Wiki, you probably misunderstood the other sources also. We can go through them one by one if you like.

If a function is taken to a limit isn't it a derivative?

Not necessarily, a derivative is a limit, but not all limits are derivatives.

SO does the Taylor expansion use the same symbolic representation?

Usually a Taylor series expansion is written explicitly as an approximation using "big O" notation. E.g.
f(x)=f(0)+f'(0)x+O(x^2)
Where the O term represents an error of the order of x^2 or smaller.

So maybe I should ask what you mean by exact?

I mean
f'(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+x)-f(x)}{\Delta x}
This quantity is the definition of the derivative, and it is exact. There is no ≈ nor any O. If the limit exists then the derivative is exactly that value, by definition.

Considering a particle under constant acceleration : to the limit Δt --->\infty

v= a(dt) = ? ...would you say the derived value v was exactly c or approximately c?

You didn't answer this question.

That is not a derivative, it is a limit, but it is equal to c.

 

[and9]

You want to measure absolute velocity. The problem is all of your measurements are determining velocity RELATIVE to another object in motion (CMB, spaceship, etc). What do you consider the frame for which your velocity is absolute?

I like the time you put into the OP but the premise is flawed since you are attempting to disprove special relativity within the framework of special relativity.

 

[GeorgeDishman]

the derivative of the position of a moving object with respect to time is the object's instantaneous velocity.

Are these not correct?

They are correct but I think you already said you accept that:

Hi I never meant to suggest there was anything wrong with instantaneous velocity, or derivatives or integration, I am well aware of their power. Only that instantaneous velocity is only well defined or exact for an instant (zero duration) of time.

That is correct, where you go wrong (and it may only be terminology) is when you say:

That there is a difference between velocity and instantaneous velocity.

In physics, the unqualified "velocity" means the instantaneous value while the value over a finite period might be called the "average velocity", thus you might see a statement such as "If a car travels at a velocity of 10k/hr east for 3 hours then north at the same speed for 4 hours, the average velocity is 7.1km/hr on a heading of 37 degrees."

You may disagree and be correct or think that I am splitting hairs [or infinitesimals ;-))

IMHO, the definition in the form of a function is more useful and may remove your reservation:

http://en.wikipedia.org/wiki/Derivative#The_derivative_as_a_function

but if you look at the question at hand that is what it comes down to.

I looked at the original question but can't see the connection, the answer to your question is that the graph of the passing particles' "relativistic mass" has a minimum when they are at rest relative to the craft.

Considering a particle under constant acceleration : to the limit \Deltat ---> \infty would you say the derived value was exactly c or approximately c?

I would say the velocity is asymptotic to the exact value c.

To pick ANY durationless point on this continuum and declare there is no motion along the continuum is exactly equivelant to picking a durationless point on the path of an inertial particle and saying it is motionless in space.

I don't think you said what you meant. An inertial particle is motionless in it's rest frame at all times by definition.

There was no reference to the rest frame of the particle.It was an inertial test particle and an accelerating particle relative to one observation frame
If we pick a point of the inertial path and say: this point is an instant and having no duration, there can be no change of location in this instant. So therefore the particle is motionless in space at this point. I.e. Has no velocity.

Look at your first statement, I've highlighted the key terms. If the particle is "inertial, its velocity is unchanging. If it is "motionless" at anyone instant then it follows that it is "motionless" at all times. By definition, the frame in which a particle is "motionless at all times" is its rest frame. Do you see what I was saying? I think you were trying to make some other point, such as considering an accelerating particle rather than an "inertial particle". You can't equate that to an accelerating particle because it is not inertial so I'm not sure what you were trying to say.

DO you think this is a valid or meaningful logical conclusion or description of reality?

I certainly don't.
It is trivially true but meaningless. An abstraction divorced from reality and falsified by that reality, wherein the particle is in continuous motion throughout.

The velocity has an exact value at any instant but that value varies with time.

A 'particle in continuous motion' has an exact value for its location at any instant but that value varies with time (ignoring QM for the moment).

The same applies to acceleration if the jerk is non-zero and so on. I see no difference in those descriptions.

If you will read what I wrote I think you will find that what you are saying here is exactly what I said. That I was comparing the case of constant Velocity (inertial motion) with Instantaneous Velocity (accelerated motion)
In the first case you simply apply the value dx/dt to predict position,, in the second case you could not and needed to apply other maths,integration or in this case the equation for constant acceleration.

You can apply the operation d/dt to either the function x(t) to get the function v(t) or to v(t) to get a(t), there is no difference. If x(t) is proportional to t, v(t) is a constant but it is only a special case of the more general function.

That is not the definition of velocity. v=dx/dt is a derivative, not a ratio. The value of the derivative is not dx divided by dt, it it the single unique limit value which that ratio approaches as dt -> 0.

There seems to be a miscommunication.

Yes.

You are defining velocity in terms of a process for its determination.
I was talking about the definition of the term as applied to the result of that process.

It is the other way round. The definition of the result is the exact value of the limit found by applying the operation known as differentiation and shown as "d/dt". You are confusing that with the approximate value obtained by process of dividing infinitesimals as a means of finding the exact limit.

The meaning of the actual value thus derived and called velocity.
Which in all cases is in fact a ratio...correct? some dx/dt

It is not a ratio, it is the exact value of the limit towards which the ratio is an asymptotic approximation.

Am I communicating this distinction I am talking about clearly??

I think you did, you just have the role of the limit and the ratio reversed.

I hope you now understand the mix up in the subject of our definitions and would like to know if you still think I was incorrect on any point.
Thanks

I've snipped a bit here as I think it is already covered by the replies above, you should now see why I think you have it the wrong way round.

I was going to dig out another reference from a book by Penrose that goes farther but I can't lay my hands on it at the moment, sorry for the delay.

 

[Austin0]

Only that instantaneous velocity is only well defined or exact for an instant (zero duration) of time..

That is correct, where you go wrong (and it may only be terminology) is when you say:.

That there is a difference between velocity and instantaneous velocity..

In physics, the unqualified "velocity" means the instantaneous value while the value over a finite period might be called the "average velocity", .".

Yes , previous to this discussion I was unaware of the convention to consider velocity synonymous with instantaneous velocity..
While I might think the term more appropriately applied to inertial motion and then average and instantaneous applying to accelerating motion I certainly won't argue with convention.

Believe me I am well aware of the differences and confusion on that point has had nothing to do with the miscommunication.

IMHO, the definition in the form of a function is more useful and may remove your reservation:

http://en.wikipedia.org/wiki/Derivative#The_derivative_as_a_function.

On the samee page right under the fundamental expression of derivative :

f'(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+h)-f(x)}{\Delta h}

was this:
which has the intuitive interpretation (see Figure 1) that the tangent line to f at a gives the best linear approximation to f near a (i.e., for small h).

Considering a particle under constant acceleration : to the limit Δ t ---> \infty would you say the derived value was exactly c or approximately c?.

I would say the velocity is asymptotic to the exact value c..

I agree completely ,which of course was my point. Asymptotic means never reaching the point of exact value. Only exceedingly close = proximate = approximate. Agreed??

To pick ANY durationless point on this continuum and declare there is no motion along the continuum is exactly equivelant to picking a durationless point on the path of an inertial particle and saying it is motionless in space..

I don't think you said what you meant. An inertial particle is motionless in it's rest frame at all times by definition.

There was no reference to the rest frame of the particle.It was an inertial test particle and an accelerating particle relative to one observation frame
If we pick a point of the inertial path and say: this point is an instant and having no duration, there can be no change of location in this instant. So therefore the particle is motionless in space at this point. I.e. Has no velocity..

Look at your first statement, I've highlighted the key terms. If the particle is "inertial, its velocity is unchanging. If it is "motionless" at anyone instant then it follows that it is "motionless" at all times. By definition, the frame in which a particle is "motionless at all times" is its rest frame. Do you see what I was saying? I think you were trying to make some other point, such as considering an accelerating particle rather than an "inertial particle". You can't equate that to an accelerating particle because it is not inertial so I'm not sure what you were trying to say..

I was describing a fallacy (IMO) In fact the initial premise of Zeno's arrow. SO just substitute arrow for inertial particle and it might make more sense.
I was not saying it actually was motionless but rather, that to say it was motionless was nonsense.
If you look at the following sentence below it is clear that the arrow was actually in continuous motion, never actually motionless . The point was the correspondence between this case and picking a similar durationless point on the path of an accelerating object and saying it was not accelerating.
In one case you are saying it is not changing position and in the second you are saying it is not changing velocity.

DO you think this is a valid or meaningful logical conclusion or description of reality?

I certainly don't.
It is trivially true but meaningless. An abstraction divorced from reality and falsified by that reality, wherein the particle is in continuous motion throughout..

Is it clear yet?

You are defining velocity in terms of a process for its determination.
I was talking about the definition of the term as applied to the result of that process..

It is the other way round. The definition of the result is the exact value of the limit found by applying the operation known as differentiation and shown as "d/dt". You are confusing that with the approximate value obtained by process of dividing infinitesimals as a means of finding the exact limit..

Isn't this what we are talking about??

f'(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+x)-f(x)}{\Delta x}
This quantity is the definition of the derivative, and it is exact.

If this is it then all the descriptions I have encountered including the page you referenced above described the process in terms of infinitesimals.

ANother source:

inputThe derivative of a function at a chosen input value describes the best linear approximation of the function near that value.
so that "f(x) becomes arbitrarily close to L" means that f(x) eventually lies in the interval (L - ε, L + ε),

Infinitesimals yes?

The meaning of the actual value thus derived and called velocity.
Which in all cases is in fact a ratio...correct? some dx/dt.

It is not a ratio, it is the exact value of the limit towards which the ratio is an asymptotic approximation..

Once again you are talking about the process and I the value.

We drop a ball. After 2 seconds we can calculate the instantaneous velocity is 20m/sec. This is the value of the velocity. It is a ratio. A dx/dt

dx=20,dt=1s correct?

The meaning of this value is a change of 1 sec in time results in a 20m change in position Right?

How do you propose to explicitely express a velocity value for a real world state of motion in a form which is not a dx/dt ?

Thanks for your help

 

[Austin0]

The derivative is exact.

So maybe I should ask what you mean by exact?

I mean
f'(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+x)-f(x)}{\Delta x}
This quantity is the definition of the derivative, and it is exact. There is no ≈ nor any O. If the limit exists then the derivative is exactly that value, by definition.

Well I can't argue with that perfect self referential symmetry.

So:
Given that an instantaneous velocity of an accelerating particle is exact at that point:
That point being a mathematical point , it would follw that it was only exact within a restricted region of zero dimension, both spatially and temporally.
Therefore any finite interval of time must inevitably fall outside that region and be inexact.
Since any explicit value for that velocity would necessarily have a time term of finite duration
it would necessarily be approximate as applied to the motion of the particle in the real world..
SO would you say it was only abstractly exact?

Considering a particle under constant acceleration : to the LIMIT Δ t ---> \infty would you say the derived value was exactly c or approximately c?.

That is not a derivative, it is a limit, but it is equal to c.

So would this be a case where "Calculus says" something that does not have meaning or truth in the physics of the real world??
Entirely not the fault of calculus of course

Thanks

 

[Dale]

which has the intuitive interpretation (see Figure 1) that the tangent line to f at a gives the best linear approximation to f near a (i.e., for small h)

Note that this quote says that the tangent line is an approximation to f. That is not at all the same as saying that the derivative is an approximation. The tangent line is the first order Taylor series expansion, so this confirms my previous point.

Given that an instantaneous velocity of an accelerating particle is exact at that point:
That point being a mathematical point , it would follw that it was only exact within a restricted region of zero dimension, both spatially and temporally.

That doesn't follow. That would follow if the instantaneous velocity were exact only at that point, but that is not the case. The instantaneous velocity is exact at other points also (i.e. plug in a new value for x, take another limit, and you have the exact velocity at another point). If f is smooth (i.e. no teleportation) then the velocity function is defined at all points in the domain of f.

SO would you say it was only abstractly exact?

Although I don't agree with your logic, I do agree with your conclusion, but not for the reason you stated. The reason that I would agree is that in the abstract you can deal with the exact calculated values, but in reality you deal with measured values. Even classically, measured values always include some error, so they are not exact. However, the exact values of abstract theory are accurate predictors of our best inexact measurements, so the use of an instantaneous derivative is experimentally justified.

So would this be a case where "Calculus says" something that does not have meaning or truth in the physics of the real world??

No, calculus agrees in this case with the physics of the real world, as long as you understand what is meant by a limit as t->∞. However, I would emphasize again, that this limit is not a derivative and has nothing to do with the instantaneous velocity discussion.

 

[russ_watters]

Note that later in that same wiki, it does indeed affirm that the derivative at a point is the exact slope.