Can Absolute Velocity be Measured?

[GeorgeDishman]

Yes , previous to this discussion I was unaware of the convention to consider velocity synonymous with instantaneous velocity.

I suspect that clears it all up but I'll reply to the key points you raise.

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}

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??

Right, so the process of approaching the limit is always approximate while the limit itself is exact. Now look at the definition of f' above. The process involves the ratio of small but finite values while the derivative is defined as the limit which, as you note next, is exact.

[I'll skip the bit about Zeno, I think the rest of my reply makes it redundant.]

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.

Yes, and velocity is defined as a derivative so it is exact, not an approximation.

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

Yes, one process of finding the derivative is defined in terms of infinitesimals but the derivative is the exact limit.

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.

No, see above, the definition of the value is the limit, the process involves the ratio.

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?

No, at the exact instant identified by t=1s, the rate at which the ball is falling is the exact value 20m/s.

The distance it has fallen is 10m so the ratio is the average speed over that 1s period and is only 10m/s.

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

No, the meaning is that, at that instant, the rate at which the distance is changing is a slope of 20m/s but since the acceleration is constant, it has that value only at that instant. That is what is called velocity in physics.

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 ?

"dx/dt" is a notation that indicates use of the derivative applied to the variable 'x'. The derivative is the limit at a particular instant as you have quoted several times above and it is exact, not an approximation. The ratio of infinitesimals is an approximation but that is only a process for finding the limit. The definitions you quote above make that clear so I'm unclear as to where you think there is a problem.

 

[Austin0]

I would really like to try and clear up some of the confusion that has developed in this discussion.

post #32

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.

Back at post 32 I clearly expressed that the question was not about calculus per se, and stated that an instantaneous velocity was exact.

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..

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.

Here you have questioned my logic. Fair enough , it may be flawed.
So let's examine it:
Simply parsing the engish it seems clear that I am referring to an (singular) instantaneous velocity at that (singular , specific) point. I make no statement explicit or implied about the velocity function itself or values derived from it at other points and am clearly referring to a singular specific quantitative value derived from that function.

Now looking at your response:

if the (general) instantaneous velocity were exact only at that point

My statements have been interpreted to become;

Only the instantaneous velocity derived at a single point on the path is exact.
The velocity function is only exact at a single point on the path of an accelerating particle.

1) Since these statements are so obviously boneheaded it was not difficult to refute them.
But as there was no foundation whatever for these interpretations in my actual words. that refutation is itself, a form of invalid argument we are all familiar with.

2) This argument also implicitely suggests that I might be such a bonehead, which may well be true but is not sufficiently proven in this instance.

3) I think if you examine this closely you will find that what you said here

[" 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)" ]

actually supports my point.

That being;
if you want an exact value for a point outside the stated dimensionless point, it is necessary to input a different value of x or t into the function and derive a new and different quantitative value.

Would you disagree with this?

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.

Obviously this value (dx/dt) is incredibly useful as an input value for a variable in other equations

Yes we can calculate an instantaneous velocity and this is a useful abstraction; eg. ICMIF's

The usefulness and validity has never been in question.
But I do have a question. If there are a pair of real world measurements (time-location)
would the derivative of this value differ from the average velocity directly indicated from the measurement?

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->8. However, I would emphasize again, that this limit is not a derivative and has nothing to do with the instantaneous velocity discussion.

Are you changing your mind about the value being c?
I am unclear as to why it is not a derivative?
Is it because the function is a constant , a simple derivative??

I think it is germane to the underlying question.
The relationship of abstract values to the reality they describe.
Calculus is a fantastic tool but is still simply a complex set of algorithms. It produces values but does not interpret them. That requires intelligence and understanding.
As this exsample ; taken literally the value c does not describe the motion or the predicted outcome of real world measurements or conform to the physics of that world as we now understand it. That value must be interpreted as you said.
All I am saying is that an explicit velocity of an accelerating particle , no matter how useful or exact it may be abstractly , as a desription of the motion in the real world it cannot simply be taken literally but must be interpreted.

I hope all this may clear up some of the confusion.
thanks for your input. I know semantics is not our favorite subject either.

 

[Mentz114]

I am unclear as to why it is not a derivative

A derivative is a specific limit. The derivative of x² can be found by working out
<br /> \frac{(x+\delta x)^2-x^2}{\delta x} = \frac{x^2 + 2x\delta x + (\delta x)^2 - x^2}{\delta x}=2 x + \delta x<br />
and if we let δx = 0 the answer is exactly 2x. You might find it amusing to work out the derivative of xⁿ with this method.

In general a limit is just the value of any function when one of its arguments goes to some other value.

 

[Austin0]

A derivative is a specific limit. The derivative of x² can be found by working out
<br /> \frac{(x+\delta x)^2-x^2}{\delta x} = \frac{x^2 + 2x\delta x + (\delta x)^2 - x^2}{\delta x}=2 x + \delta x<br />
and if we let δx = 0 the answer is exactly 2x. You might find it amusing to work out the derivative of xⁿ with this method.

In general a limit is just the value of any function when one of its arguments goes to some other value.

Hi Mentz just looking at the 3rd power the handwriting on the wall seems to indicate nx^n-1 ?

I am still not sure why the equation for constant acceleration taken to the limit \Deltat ---->\infty is not a derivative? because of the infinity?

In the velocity equation v=\lim_{\Delta t\to 0} \, \frac{ x(t+\Delta t )- x(t)}{\Delta t}=dx/dt
am I correct in thinking the final dx/dt to the right of the equals sign is the actual result??
In a case with actual inputs this represents the quantitative output value??

Thanks

thinking about it more, would the explicit answer in terms of x be: I don't know the notation but something like the series x(n-k)(n-k)... where k = the incrementing integers from 0 ,1,2...(n-1) ?

 

[Mentz114]

Hi Austin, you're right about nx^n-1.

I am still not sure why the equation for constant acceleration taken to the limit Δt ---->∞ is not a derivative? because of the infinity

I'm don't know which formula you're referring to, but a limit as Δt -> ∞ cannot be a derivative, as you conjecture. In the definition of the derivative the δ term is an infinitesimal which is taken to zero.

In the velocity equation
...
am I correct in thinking the final dx/dt to the right of the equals sign is the actual result??

uh, yes, that's the definition of dx/dt

In a case with actual inputs this represents the quantitative output value??

Not sure what you mean. An example. Suppose x(t) = x₀ + ht, where h is a constant, then the velocity is

dx(t)/dt = h.

I have to say that this is probably not the place for coaching in calculus. I was hoping you could see how straightforward it is and get a textbook ...

 

[GeorgeDishman]

In the velocity equation v=\lim_{\Delta t\to 0} \, \frac{ x(t+\Delta t )- x(t)}{\Delta t}=dx/dt
am I correct in thinking the final dx/dt to the right of the equals sign is the actual result??
In a case with actual inputs this represents the quantitative output value??

uh, yes, that's the definition of dx/dt

I may be over-analysing but it may avoid some confusion if I check something here.

Austin0, if you are asking if the final "dx/dt" represents the division of a value dx by another value dt, the answer is no. The bit that gives you the value of v is

v=\lim_{\Delta t\to 0} \, \ ...

The notation "d/dt" is a way of writing that limit so as Menzt114 says, that is the definition of dx/dt.

 

[Dale]

Sorry I have been delayed. I was traveling for business very heavily. Some of your questions may have already been resolved, so I apologize for any redundancy.

Here you have questioned my logic. Fair enough , it may be flawed.
So let's examine it:
Simply parsing the engish it seems clear that I am referring to an (singular) instantaneous velocity at that (singular , specific) point. I make no statement explicit or implied about the velocity function itself or values derived from it at other points and am clearly referring to a singular specific quantitative value derived from that function.

Now looking at your response:

if the (general) instantaneous velocity were exact only at that point

My statements have been interpreted to become;

Only the instantaneous velocity derived at a single point on the path is exact.
The velocity function is only exact at a single point on the path of an accelerating particle.

Yes, upon reading your statements that is exactly how I had understood them. Even now, I have a hard time interpreting them any other way. But I am glad you have clarified.

1) Since these statements are so obviously boneheaded it was not difficult to refute them.
But as there was no foundation whatever for these interpretations in my actual words. that refutation is itself, a form of invalid argument we are all familiar with.

2) This argument also implicitely suggests that I might be such a bonehead, which may well be true but is not sufficiently proven in this instance.

Austin0, I never said nor implied anything about you being a bonehead. This site is for people who are trying to learn and people trying to learn make honest mistakes. It isn't boneheaded for someone to make an incorrect conclusion about a brand new concept and I understand that. So please don't get offended and imagine unintended insults.

My whole involvement in this thread was to correct the misunderstanding that the velocity is an approximation i.e. inexact. Your statements indicated to me that you still felt that it was inexace because it was a derivative. So I explained. That is all, no insult was provided nor implied.

3) I think if you examine this closely you will find that what you said here

[" 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)" ]

actually supports my point.

That being;
if you want an exact value for a point outside the stated dimensionless point, it is necessary to input a different value of x or t into the function and derive a new and different quantitative value.

Would you disagree with this?

I agree. Mathematically f'(x)≠f'(x+Δx) in general, for Δx≠0. But both f'(x) and f'(x+Δx) are exact if f is smooth.

But I do have a question. If there are a pair of real world measurements (time-location)
would the derivative of this value differ from the average velocity directly indicated from the measurement?

In general the velocity does not equal the average velocity.

Are you changing your mind about the value being c?

No. The limit is exactly c.

I am unclear as to why it is not a derivative?
Is it because the function is a constant , a simple derivative??

No, it is not a derivative because it doesn't fit the form of a derivative. Recall the definition of a derivative:

f&#039;(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+x)-f(x)}{\Delta x}

Now, look at the limit you are trying to evaluate:

\lim_{t \to \infty}\,v(t)

There is no Δt which is approaching 0, so the limit cannot be expressed in the form of the above limit. Therefore it is not a derivative. Now, v(t) itself is a derivative v(t)=x'(t) so we can write the limit as

\lim_{t \to \infty}\, \left( \lim_{\Delta t\to 0} \, \frac{x(\Delta t+t)-x(t)}{\Delta t} \right)

This makes it clear that the overall expression is not a derivative, but a limit of a derivative. One of the key things that makes it not a derivative is the fact that the limit is taken to infinity. In a derivative the limit is always taken to 0.

This is important, because a limit to infinity is fundamentally a different thing than a limit to some finite value. A limit to some finite value means "as I get arbitrarily close to X ..." but you cannot get arbitrarily close to infinity. For instance, the real number .0001 is within .001 of 0. Can you give me a real number which is within .001 of infinity?

So a limit to infinity is a fundamentally different thing, it means what does f(x) tend to asymptotically as x increases without bound. In this case, that number is c. So the limit is c meaning that as t increases without bound v asymptotically approaches c.

I think it is germane to the underlying question.
The relationship of abstract values to the reality they describe.
Calculus is a fantastic tool but is still simply a complex set of algorithms. It produces values but does not interpret them. That requires intelligence and understanding.

I agree. I think you have the required intelligence, and hopefully you now have the required understanding also.

 

[Austin0]

Hi GeorgeDishman Thank you for your patience.There has been much confussion in this discussion and I have no doubt a certain frustration on both sides.

In the velocity equation v=\lim_{\Delta t\to 0} \, \frac{ x(t+\Delta t )- x(t)}{\Delta t}=dx/dt
the final dx/dt is the result?

I may be over-analysing but it may avoid some confusion if I check something here.

Austin0, if you are asking if the final "dx/dt" represents the division of a value dx by another value dt, the answer is no. The bit that gives you the value of v is


[\QUOTE]

I thought the final dx/dt represented the answer .A ratio like 20m/s not an operation of division to obtain the answer. Is this incorrect?

"dx/dt" is a notation that indicates use of the derivative applied to the variable 'x'. The derivative is the limit at a particular instant as you have quoted several times above and it is exact, not an approximation. .

There seems to be a question regarding the meaning and use of delta.
The term delta itself seems to have another intrinsic meaning , that being; interval.

Wrt dynamic systems it is found by subtracting one value of some variable at a point from the value at some other point to get a difference , an interval ..Correct?

Now delta may have a more specific meaning as an operator in the context of derivation but it is certainy widely used outside that context as meaning interval with no implied reference to the operation of derivation,limit etc.

Example

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.

DO you agree that this is in common usage??
Is there some distinction between small d delta and capitol D delta you are referring to?


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?

Are you saying this definition is invalid or has no relevance to a velocity as a description of motion in the world.??

Isn't this essential meaning validly expressed with symbols using the fundamental definition of delta as v=dx/dt ?

So any explicit velocity value , independent of the method of arriving at it , is expressed as a ratio of intervals of change eg. 20m/s

This value has an explicit literal meaning under the fundamental definition of velocity which is: the ratio of a specific interval/change of position with respect to a specific interval/change of time. Eg. 20m/s

I am not talking about the interpretation here but the literal meaning.
Would you agree?

No, the meaning is that, at that instant, the rate at which the distance is changing is a slope of 20m/s but since the acceleration is constant, it has that value only at that instant. That is what is called velocity in physics. [\QUOTE]
I was of , course, aware of this as it is the point of this discussion but:

Here you are talking about interpretation which is fine.. So what is your interpretation of this value as a description of the real world motion of the particle??

Thanks

 

[Austin0]

Yes, upon reading your statements that is exactly how I had understood them. Even now, I have a hard time interpreting them any other way. But I am glad you have clarified..

Hi . Why is it hard to interpret them as written? Is my english not clear?? Actually this very thing seems to have repeatedly occurred throughout this discussion, not just with you but generally.
Everything i have said has been reinterpreted into a statement or argument directed to the proposition that a derivative or instantaneous velocity it not exact. This is in spite of the fact that right from the beginning and repeatedly after, I have clearly stated this was not the question , that it was given that a velocity was exact at an instant.
The result is that my communication has not been understood and just as in this case , my arguments and questions have not ever been addressed.

Austin0, I never said nor implied anything about you being a bonehead. It isn't boneheaded for someone to make an incorrect conclusion about a brand new concept and I understand that. So please don't get offended and imagine unintended insults..

I thought that the inclusion of a cartoon face would make it clear enough I was not serious.
But on the other hand your response here almost has me wondering if it was I and not you that made the incorrect conclusion about a brand new concept ;-)
But seriously, you have been nothing but helpful as long as I have been in the forum and I consider you ,with DrGreg, as being among the most considered and considerate members, so I never assumed any negative intent on your part whatever.

post 32

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...

Your statements indicated to me that you still felt that it was inexace because it was a derivative. ..

The post above is #32 we are now up to #57 and it is apparently still unclear that I never felt or said the value of an instantaneous velocity was inexact , I have been trying to talk about the meaning and interpretation of the derived value itself, independent of its method of determination. What that value meant as a description of the motion of a particle in the real world. But somehow this is not possible and it always, circularly, ends up back focused on derivatives.

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??..

SO it is in this sense that I feel that no matter how exact a velocity may be abstractly , as a description of the motion in the real world it is an approximation confined to a small region of application.

This is also a rational interpretation of the statement ;" the tangent at a point is a linear approximation of the values of the curve near to that point of exact value."
As the curve represents the motion in the real world this means within a short distance on the actual path.

If you disagree with this interpretation of the tangent could you explain your alternative?

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.

.

But I do have a question. If there are a pair of real world measurements (time-location) would the derivative of this value differ from the average velocity directly indicated from the measurement?

In general the velocity does not equal the average velocity.

Since you didn't answer the question I will take a stab at it.
Given two measurements, the accuracy of the derived value is inversely dependent on the spatial separation between the two events. Increasing as the interval decreases. In this case derivation is no more accurate at interpolation than the simple dx/dt
and would return the same value . Both are approximations in this circumstance. Is this correct??
This is irrelevant to the initial question but there is another question.
Given an exact value at a point (ideally close measurements) what would you say about this value as a predictor of measurements of position or time at other points?

So what is your interpretation of an instantaneous velocity as a description of the motion of an accelerating particle.?

Thanks for your patience and help
.[/QUOTE]

 

[GeorgeDishman]

Hi GeorgeDishman Thank you for your patience.There has been much confussion in this discussion and I have no doubt a certain frustration on both sides.

No problem, sometimes it takes a few mails to get to the core of a problem.

I thought the final dx/dt represented the answer .A ratio like 20m/s not an operation of division to obtain the answer. Is this incorrect?

Yes, that is incorrect.

There seems to be a question regarding the meaning and use of delta.

The problem is that you are reading "dx/dt" as the ratio of two deltas. Instead you need to read it as an operation "d/dt" applied to a variable "x" which is a function of "t".

The definition of that operation is the derivative.

The term delta itself seems to have another intrinsic meaning, that being; interval.

There is a dual meaning, a "delta" is also used to mean a small finite step but we are not interested in that meaning here, that is a large part of the confusion.

Isn't the fundamental definition of velocity; rate of change of position as a function of time??

Yes, exactly. In your previous example some posts back, you had a velocity of 20m/s which was the rate of change of position at a specific instant.

The difference in position with respect to the difference in time?

No, in your example the position had only changed by 10m in 1s.

Are you saying this definition is invalid or has no relevance to a velocity as a description of motion in the world.??

We are all telling you that the first definition is correct for "velocity" while the second is "average velocity" and they are different things. Both are valid as you can see from the example but they are not the same thing (the first had the value 20m/s while the second was 10m/s).

Isn't this essential meaning validly expressed with symbols using the fundamental definition of delta as v=dx/dt ?

The fundamental definition of velocity is "rate of change of position" which is a derivative, not a ratio.

So any explicit velocity value , independent of the method of arriving at it , is expressed as a ratio of intervals of change eg. 20m/s

No, it is explicity the derivative.

This value has an explicit literal meaning under the fundamental definition of velocity which is: the ratio of a specific interval/change of position with respect to a specific interval/change of time. Eg. 20m/s

The change of position in 1s was 10m, the velocity after 1s was 20m/s downwards.

I am not talking about the interpretation here but the literal meaning.
Would you agree?

I already answered that:

No, the meaning is that, at that instant, the rate at which the distance is changing is a slope of 20m/s but since the acceleration is constant, it has that value only at that instant. That is what is called velocity in physics.

I was of , course, aware of this as it is the point of this discussion but:

So what is your interpretation of this value as a description of the real world motion of the particle??

The interpretation is that velocity is the instantaneous value of the rate at which the particle's position is changing, the velocity vector is the derivative of the position vector.

 

[Dale]

Hi . Why is it hard to interpret them as written? Is my english not clear?? ...
Since you didn't answer the question I will take a stab at it.

I believe that not only did I answer the question I did so in as unambiguous a fashion as possible. Since we are not communicating and since we are using english, then apparently your english is not clear and apparently mine is not either.

I am sorry, but I don't think I can help. I have given you the mathematical definition of the derivative and explained it in english to the best of my ability. The physical position can be represented as a vector-valued function of time, x(t). The velocity is defined as the derivative of the position with respect to time, x'(t) = dx/dt. It seems to me that this is a complete description of the topic, so I don't see what else can be added and I don't understand where you think there is some ambiguity in physical interpretation.

 

[Whovian]

Back at post 32 I clearly expressed that the question was not about calculus per se, and stated that an instantaneous velocity was exact.

And derivatives are exact ...

 

[Dale]

This is also a rational interpretation of the statement ;" the tangent at a point is a linear approximation of the values of the curve near to that point of exact value."
As the curve represents the motion in the real world this means within a short distance on the actual path.

If you disagree with this interpretation of the tangent could you explain your alternative?

I think this is the third or fourth time that I have mentioned that the tangent line is the first order Taylor series expansion of a function and it is explicitly an approximation. That does not in any way imply that velocity is approximate. They are different things.

 

[Austin0]

I think this is the third or fourth time that I have mentioned that the tangent line is the first order Taylor series expansion of a function and it is explicitly an approximation. That does not in any way imply that velocity is approximate. They are different things.

Yes you are right. I apologize for not responding sooner , I planned but they slipped away in the confusion.
I did look up the Taylor expansion and it is exacty as you said but is a different thing than the sources I quoted. Every source and also GeorgeDishman and Russ_Waters seem to agree that the tangent comes directly from the derivation to the limit and is exact.
I may be mistaken but in a way it doesn't matter as I came into this with the understanding I had gotten from Minkowski diagrams of accelerating systems that the tangent was exact so that was never a question.
So my interpretation of the statement that "the tangent is an approximation of the values of the curve near the exact point" , is with the understanding that the line is exact and the point of congruency is exact. I explicitely stated this in an earlier post.
And that the tangent was also congruent with the ICMIF at that point.
Thanks

 

[russ_watters]

Where did I say that?!

Perhaps the issue here is you are confusing "a line" with "the slope of a line"?

 

[Dale]

Every source and also GeorgeDishman and Russ_Waters seem to agree that the tangent comes directly from the derivation to the limit and is exact.

The tangent line is:
f(x_0)+f&#039;(x_0)(x-x_0)\ne f&#039;(x)

Since you are now saying that you believe that the tangent line is also "exact", I really have no idea what you think is inexact. Perhaps you can clarify, and also perhaps clarify what you mean by "exact" (eg exactly what).

 

[Whovian]

The tangent line is:
f(x_0)+f&#039;(x_0)(x-x_0)\ne f&#039;(x)

Since you are now saying that you believe that the tangent line is also "exact", I really have no idea what you think is inexact. Perhaps you can clarify, and also perhaps clarify what you mean by "exact" (eg exactly what).

The slope of the tangent line is exact. We can also find an exact tangent line. It can be used, through first-order Taylor approximation, to approximate the values of the function in question.

What I'm saying is there's no uncertainty in what the tangent line is, but it can be used to approximate stuff with some uncertainty.

 

[GeorgeDishman]

Yes you are right. I apologize for not responding sooner , I planned but they slipped away in the confusion.
I did look up the Taylor expansion and it is exacty as you said but is a different thing than the sources I quoted.

The definition of a Taylor Series is quite simple and all sources should give you the same meaning even if differently written:

http://en.wikipedia.org/wiki/Taylor_series

In a post some time back that I can't find at the moment, there was a comment to the effect that if you wanted to approximate a curve with a straight line, the best value for the slope of that line was to make it equal to the value of the derivative at the desired point. There was no implication in the cited source that a derivative was an approximation but possibly some readers might have mistakenly taken it that way.

Every source and also GeorgeDishman and Russ_Waters seem to agree that the tangent comes directly from the derivation to the limit and is exact.

I have never used the phrase "derivation to the limit" nor would I. What everyone has told you is that the mathematical operation known as a derivative produces a result which has an exact value for a given function. A derivative has nothing to do with any form of derivation.

I may be mistaken but in a way it doesn't matter as I came into this with the understanding I had gotten from Minkowski diagrams of accelerating systems that the tangent was exact so that was never a question.

That is not the impression you gave in post #30 which is probably the root of much of the confusion:

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?

Are we to take it that you now agree with Yuiop's statement in post#28?

 

[Dale]

What I'm saying is there's no uncertainty in what the tangent line is, but it can be used to approximate stuff with some uncertainty.

Yes.

I think that part of the problem here is the ambiguous use of the words exact and approximate. Austin0 should clarify.

 

[vin300]

When the tangent line is used to plot a point on a curved line in the neighbourhood of a given point, this point becomes an approximation because the tangent line is an approximation of the curvature in the vicinity of the given point.
The velocity at this point lies on the tangent, is not an approximation because we use the limit delta t tends to zero; in the domain of this limit, the infinitesimal displacement is almost an exact straight line(no room for curvature), hence the velocity is said to be exact.

 

[Austin0]

Where did I say that?!

Perhaps the issue here is you are confusing "a line" with "the slope of a line"?

I am confused but not on that point. I assumed that if the point of congruence was exactly determined and the slope was exact that this by itself was enough to exactly define the tangent line. Graphically: pick an arbitrary \Delta x and the consequent \Delta y at that point to define a second point. The line intersecting that point and the point on the curve would be the exact tangent line.
Apparently I was again mistaken as you seem to be saying it is necessary to resort to other processes to define the line.
Is this the case??

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

f&#039;(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).

a being the point of exact value.

SO is the tangent line referred to here not a direct product of the derivative??
Is it not to be taken, as I did, as being exact??


So is the approximation that is mentioned here referring to the line itself or is it referring to the relationship of the line to points on the curve. It seems to me to clearly be the latter , am I wrong??
Thanks

 

[Austin0]

I have never used the phrase "derivation to the limit" nor would I. What everyone has told you is that the mathematical operation known as a derivative produces a result which has an exact value for a given function. A derivative has nothing to do with any form of derivation.

yes I am still learning the terminology. I mistakenly assumed the process of determining a derivative would be derivation. Oops.
Also I use the word ratio interchangebly with rate . I am beginning to suspect this may be semantically wrong in this context also?

That is not the impression you gave in post #30 which is probably the root of much of the confusion:

When I made post thirty I only had a geometric understanding of instantaneous velocity as the tangent to an accelerating world line gained after hearing the concept , by looking at such a worldline where it was clear that the tangent at various points would indeed be equivalent to an inertial worldline intersecting that point. I assumed it was exact and never researched the actual method of determining the tangent and so had no conception of what a derivative was outside a vague idea that it was the return of a function of some kind. I would certainly never make any statement about derivatives whatever.

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

This statement was directed to the limits I was familiar with: asymptotic approach to c with constant acceleration, the "Slow clock transport" as velocity approaches 0, etc.

When this was misinterpreted to apply to instantaneous velocity , tangents etc I immediately responded (below) that this was definitely not the case

post 32

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...

You either did not read post 32 (above) or completely disregarded it and responded with

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

So here the subject was now switched to derivatives and that confusion continued in spite of what I had said previously and continued to reiterate.

Are we to take it that you now agree with Yuiop's statement in post#28?

I was clearly mistaken as a general statement about limits. But unless you have changed your mind and now state that constant acceleration to the limit at infinity, is exactly c
I was not completey wrong.

Now I would say "like some limits and the Taylor series expansion" provide exceedingly useful and accurate approximations

Just kidding. I plan to carefully avoid any such provocative statements in the future.

Thanks

 

[Austin0]

I think that part of the problem here is the ambiguous use of the words exact and approximate. Austin0 should clarify.

I think that the confusion is not regarding the meaning of exact or approximate but is connected with the meaning of abstract.
I hope you will bear with me if I state the obvious in an effort to clarify my meaning.
Fundamentally all of mathematics is abstraction. But within that structure there are varying degrees of correspondence with the reality it is designed to describe.

A velocity regarding an inertial particle has complete unqualified correspondence both as a description of the motion of that particle and as a predictor of future positions and times as long as it remains inertial.

AN average velocity is essentially a complete abstraction. Equivalent to an average family having 2.73 children.

A particle with an average velocity between two points may have never actually traveled any distance at all at that velocity.
As a description or predictor it has almost no value regarding short intervals and only an approximate value over intervals equal or longer than the original

AN instantaneous velocity regarding a particle under continuous acceleration is also an abstraction.

The particle in question never actually travels any distance at all at that exact value.

To say this value is an exact description of the motion of the particle is equivalent to saying an exact description of motion over no distance. A self contradictory description with no correspondence to the real world.

The dimensionless mathematical point is itself somewhat problematic.
It works fine for assigning coordinates to a position, because events are static by definition. But to attach a dynamic evaluation to such a point is a different story.

A rate of change with respect to no duration is meaningless.
As an abstract exact value as an input for other calculations [like calculating proper acceleration relative to ICMIFs, or input as u to v=u +at] it is not a problem because those functions use the exact value as a starting point , not as motion over an interval.

But as a description of the motion of a particle, it is meaningless without an interval ,and as soon as you define any interval, (of time or space), the exact value no longer applies exactly to the motion through that interval.

Mathematically f'(x)\neqf'(x+\Deltax) in general, for \Deltax\neq0.

the tangent line to f at a gives the best linear approximation to f near a (i.e., for small h).

I interpret this, in the case of velocity, as f representing the worldline of the particle in question and the tangent approximating the velocity near (a) the point of exact value.

would you agree with this interpretation??

So my conclusion from all of the above is that an instantaneous velocity regarding an accelerating particle is exact as an abstract value but not as a meaningful description of the real world motion of that particle.
But it is a meaningful and accurate approximate description of that motion over a very small interval.

Hope this will clear up some of the confusion

 

[Dale]

I think that the confusion is not regarding the meaning of exact or approximate but is connected with the meaning of abstract.

No, I don't think that has been a source of confusion. As far as I recall the word "abstract" hasn't even come up in our conversation.

I interpret this, in the case of velocity, as f representing the worldline of the particle in question and the tangent approximating the velocity near (a) the point of exact value.

would you agree with this interpretation??

No. The tangent line is approximately equal to the position (to first order). You can take the derivative of the tangent line and get an approximation to the velocity, but it is constant (0th order), not linear (1st order). To get the best linear approximation to the velocity you would need to do a first order Taylor series on the velocity, or equivalently take the derivative of a second order Taylor series on the position.

So my conclusion from all of the above is that an instantaneous velocity regarding an accelerating particle is exact as an abstract value but not as a meaningful description of the real world motion of that particle.
But it is a meaningful and accurate approximate description of that motion over a very small interval.

In the first statement, the velocity is exactly equal to what? But in the second statement, it is approximately equal to what?

Btw, I was also misusing "exact". I said "exact" but what I meant was "well defined". The derivative is well defined, so we know exactly what it is. But the question is what things it is exactly or approximately equal to. Obviously anything is exactly equal to itself, so that is uninformative except as a definition.

 

[GeorgeDishman]

⇔

yes I am still learning the terminology. I mistakenly assumed the process of determining a derivative would be derivation. Oops.

No problem, learning is good.

Also I use the word ratio interchangebly with rate . I am beginning to suspect this may be semantically wrong in this context also?

Consider two hypothetical chemistry questions:

1) Tenth molar solutions of Sodium Chloride and Copper Sulphate are mixed in the ratio 7 parts NaCl to 2 parts CuSO₄. How much NaCl is left after the reaction has completed?

2) How would the rate of the above reaction change if the temperature were increased by 10°C?

When I made post thirty I only had a geometric understanding of instantaneous velocity ... You either did not read post 32 (above) or completely disregarded it ..

Yes, the comment was just on the history of how the disagreement had arisen in the forum, you revised your views later of course.

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

This statement was directed to the limits I was familiar with: asymptotic approach to c with constant acceleration, the "Slow clock transport" as velocity approaches 0, etc.

The problem is that none of those have anything to do with calculus and it was specifically calculus about which you asked the question. Calculus is not an approximation, it is exact.

I was clearly mistaken as a general statement about limits. But unless you have changed your mind and now state that constant acceleration to the limit at infinity, is exactly c I was not completey wrong.

I haven't changed my mind about anything, my response in post #46 is still valid, but again it has nothing to do with calculus.

Just kidding. I plan to carefully avoid any such provocative statements in the future.

I would suggest, rather than that, just plan to check your facts before telling people who are trying to help you that they are wrong.

 

[vin300]

A particle with an average velocity between two points may have never actually traveled any distance at all at that velocity.

Not true.Velocity, in all practicality, is continuous.
Edit:Sorry, your statement is true if the velocity vector changes direction

 

[Austin0]

No, I don't think that has been a source of confusion. As far as I recall the word "abstract" hasn't even come up in our conversation.
.

I think it is germane to the underlying question.The relationship of abstract values to the reality they describe.Calculus is a fantastic tool but is still simply a complex set of algorithms. It produces values but does not interpret them. .

Yes we can calculate an instantaneous velocity and this is a useful abstraction; eg. ICMIF's

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.
.

I interpret this, in the case of velocity, as f representing the worldline of the particle in question and the tangent approximating the velocity near (a) the point of exact value..

No. The tangent line is approximately equal to the position (to first order). You can take the derivative of the tangent line and get an approximation to the velocity, but it is constant (0th order), not linear (1st order). To get the best linear approximation to the velocity you would need to do a first order Taylor series on the velocity, or equivalently take the derivative of a second order Taylor series on the position..

I see what you are saying and that I was confused. But it appears to me that velocity as the derivative of position is merely a convention. That velocity as the derivative of time would produce the same slope and tangent. Correct?.

So in practicallity the tangent is a linnear approximation of both position and time to the same degree of accuracy and it would seem to follow that the slope of the tangent would also have the same degree of approximate accuracy regarding the same small intervals around the point of exact values. Yes?

So my conclusion from all of the above is that an instantaneous velocity regarding an accelerating particle is exact as an abstract value but not as a meaningful description of the real world motion of that particle.
But it is a meaningful and accurate approximate description of that motion over a very small interval.

In the first statement, the velocity is exactly equal to what? But in the second statement, it is approximately equal to what?

Btw, I was also misusing "exact". I said "exact" but what I meant was "well defined". The derivative is well defined, so we know exactly what it is. But the question is what things it is exactly or approximately equal to. Obviously anything is exactly equal to itself, so that is uninformative except as a definition.

I am not sure what you mean by well defined. Certainly as the term is applied to a function it is well defined. As a process a derivative is also well (precisely) defined.

Regarding the meaning of exact:

Not approximate.

Exact in the sense that it is accepted that the returns of most functions are exact values.

Exact in principle, Ignoring possible round off considerations etc

Exact as the return of v= u + at is taken to be an exact value.

As far as I can see none of the usual definitions of exact are predicated on what the term is equal to.

in the abstract you can deal with the exact calculated values, . However, the exact values of abstract theory...
.

So a derivative is both well defined and exact as the abstract return of a function.

The specific value of that velocity is also exact as a description of the motion of an inertial particle but is not exact as a desription of the motion of an accelerating particle.

In the first case it exactly describes the motion and exactly predicts real world measurements of position and time along the path.
In the second case it does not describe the motion outside of a dimensionless point nor does it predict real world measurements of position and time on the path outside of approximate predictions within an extremely limited domain around that point.

Hopefully we might find agreement on these definitions.
Thanks

 

[GeorgeDishman]

.. it appears to me that velocity as the derivative of position is merely a convention. That velocity as the derivative of time would produce the same slope and tangent. Correct?.

dx/dt=v but dt/dt=1 so "no".

Exact as the return of v= u + at is taken to be an exact value.

Correct.

So a derivative is both well defined and exact as the abstract return of a function.

Correct, other than that it is no more abstract than the notion of "position".

The specific value of that velocity is also exact as a description of the motion of an inertial particle but is not exact as a desription of the motion of an accelerating particle.

No. Velocity is a function of time v(t) so at any time t it has only one value v(t). The derivative gives that value exactly, it is not an approximation to that value.

This is no different to saying that x(t)=vt+x₀ is the exact location of a particle moving at constant speed v with initial location x₀.

Hopefully we might find agreement on these definitions.

Sorry, not yet. Classically, velocity has a specific value at any specific time which is given exactly by the derivative of the location vector.

Average velocity is an approximate value when averaged over a non-zero duration (unless the velocity is constant over the period of course).

 

[Dale]

Oops, clearly the word abstract had come out in the conversation, but I still don't think it is a point of any confusion.

I see what you are saying and that I was confused. But it appears to me that velocity as the derivative of position is merely a convention.

Of course it is a convention, all definitions are conventions.

Nontheless, the definition of velocity is unambiguous and ubiquituous, there is no point in trying to change it. You will only confuse yourself and others trying to communicate with you. It is a very bad idea to even attempt it.

That velocity as the derivative of time would produce the same slope and tangent. Correct?.

Not even close. The derivative of time would be 1, as GeorgeDishman has pointed out. Do not pursue this line of thought any further. The definition of velocity is clear, use it and don't try to change it. It serves no useful purpose to do so.

So in practicallity the tangent is a linnear approximation of both position and time to the same degree of accuracy and it would seem to follow that the slope of the tangent would also have the same degree of approximate accuracy regarding the same small intervals around the point of exact values. Yes?

No, the approximation is one order worse for the velocity than for the position. If you want a first-order approximation of the velocity then you need a second-order approximation of the position.

Regarding the meaning of exact:
...
As far as I can see none of the usual definitions of exact are predicated on what the term is equal to.

OK, if you don't wish to help communicate clearly then there will be limits to how much I can help.

The specific value of that velocity is also exact as a description of the motion of an inertial particle but is not exact as a desription of the motion of an accelerating particle.

I disagree, and so does modern physics. Unless you can explain what you think the velocity of an accelerating particle is not exactly equal to then I cannot do anything more than simply disagree.

In the second case it does not describe the motion outside of a dimensionless point nor does it predict real world measurements of position and time on the path outside of approximate predictions within an extremely limited domain around that point.

On the contrary, it does accurately predict the values of many real world measurements. That is the whole point. Momentum, kinetic energy, Doppler shift, etc. all depend on the velocity, as usually defined.

The position and the velocity are mathematically orthogonal to each other. A point measurement of the velocity does not give any information about position, nor vice versa. Expecting that velocity tell you about position is like expecting stock tips from your thermometer.

However, you are using the word "motion" which includes position, velocity, acceleration, and all higher order derivatives. Knowledge of the exact position at a point in time allows you to predict the motion at other times to 0th order accuracy. Knowledge of the exact position and velocity at a point in time allows you to predict the motion at other times to 1st order accuracy. Knowledge of the exact position, velocity, and acceleration at a point in time allows you to predict the motion at other times to 2nd order accuracy. ...

 

[Austin0]

Regarding the meaning of exact:
...
As far as I can see none of the usual definitions of exact are predicated on what the term is equal to. .

OK, if you don't wish to help communicate clearly then there will be limits to how much I can help. .

Don't you think this is a bit unfair to imply that I don't wish to communicate when I gave you a whole list of definitions
You seem to have some idea in mind that is different to standard meanings perhaps you could give an example?

The specific value of that velocity is also exact as a description of the motion of an inertial particle but is not exact as a desription of the motion of an accelerating particle..

I disagree, and so does modern physics. Unless you can explain what you think the velocity of an accelerating particle is not exactly equal to then I cannot do anything more than simply disagree. .

It appears to me that physics does define an explicit description of motion; A coordinate charting of change of position over some time interval is the exact description of the motion over that interval.
Would you disagree??

So perhaps the question could be ; Does a velocity regarding an accelerating particle , intrinsically provide the information
needed to plot such a chart??
Regarding an inertial particle it clearly does.
I hope you will take this question as it is explicitely stated

In the second case it does not describe the motion outside of a dimensionless point nor does it predict real world measurements of position and time on the path outside of approximate predictions within an extremely limited domain around that point. .

On the contrary, it does accurately predict the values of many real world measurements. That is the whole point. Momentum, kinetic energy, Doppler shift, etc. all depend on the velocity, as usually defined. .

1) If you will look at my statements above I think you will see that your response had no correlation to my statements.
You were countering arguments I never made.

2) And your counter argument had been repeatedly stated by me from the beginning. Velocity was exact as input for other calculations.
But all those evaluations depend on using the exact abstract value within the mathematical structure not as a description outside of it and not as motion over time..

3) Since you brought it up let's look at the question of the result in these cases

Case ----elastic collision of a macro system.

Supposing an accelerating system (rocket) collides with a cannonball.
Would the resulting vectors be exactly the same as those consequent to the same interaction with the momentarily co-moving inertial system??
It seems to me they would not . That the propagation of momentum during collision would necessarily require some finite time interval during which the momentum propagating through the system from the thrust would have an effect. The magnitude of the effect would be dependent on the masses,velocity and magnitude of acceleration involved.

Would you agree??.

Case-----sub-atomic particle and rocket.
Looking at the specific atom at impact; The momentum propagating away from thrust continues along that vector as long as it encounters additional mass. So at any instant the atom in question has arriving momentum to transmit to the particle, in addition to the momentum from velocity..As the particle has substantially less mass, the effect of that added momentum would have proportionately greater effect .

Do you think this would not affect the resulting vector??

Case------inelastic collision
I would think that the preceding would apply equally in this case, both to the resulting vectors and to the kinetic energy. DO you see any compelling reason to think there woud not be any effect?

Case------- Doppler shift
Certainly under most circumstances it would seem that emission can be considered instantaneous.
But what about high velocities and magnitudes of acceleration with long wavelength emissions??
Would you maintain that under all circumstances the measured Doppler shift would be exactly equal to the same signal from the MCIRF??

The position and the velocity are mathematically orthogonal to each other. A point measurement of the velocity does not give any information about position, nor vice versa...... Expecting that velocity tell you about position is like expecting stock tips from your thermometer. .

Are you now talking here about real word measurement of a particle with unknown motion?

Any description of motion which says nothing about positions or times would seem to be, self evidently, an abstraction, not a description of the real world.

Now if you are saying that a velocity regarding an accelerating particle should not be expected to directly describe it's motion in reality , I would not argue..

In the case in point , (constant acceleration) an expression of acceleration would be needed to explicitly and exactly describe that motion.

If it was non-uniform acceleration under consideration, then it would be back to the same situation and an instantaneous acceleration evaluation would be an approximation and it would require some expression of jerk to exactly describe the motion and I would imagine with a limited range of exactness..
DO you disagree with the above?

However, you are using the word "motion" which includes position, velocity, acceleration, and all higher order derivatives. Knowledge of the exact position at a point in time allows you to predict the motion at other times to 0th order accuracy. Knowledge of the exact position and velocity at a point in time allows you to predict the motion at other times to 1st order accuracy. Knowledge of the exact position, velocity, and acceleration at a point in time allows you to predict the motion at other times to 2nd order accuracy. .

.
What exactly is the meaning of 1st order accuracy?..
How does a position of itself form a basis for evaluating motion or predicting motion at other points? I am clearly missing something here.
A last question;
What is your interpretation of the unqualified expression (37meters/second)? What is it's intrinsic meaning??
Thanks

 

[Dale]

Don't you think this is a bit unfair to imply that I don't wish to communicate when I gave you a whole list of definitions

No, I don't think it is unfair. I don't understand your meaning (sometimes you say velocity is an approximation and sometimes you say it is exact) and I repeatedly asked a specific question in order to understand your meaning (velocity is approximately equal to what) which you repeatedly avoided and explicitly refused to answer. To me that seems both deliberate and designed to hinder communication. The list of definitions was not an attempt to help me understand your meaning but rather an attempt to justify not answering the question yet again.

Why do you not want to explain what you think velocity is an approximation to and what it is exactly equal to? To me, the hesitation to answer the question seems evasive.

You seem to have some idea in mind that is different to standard meanings perhaps you could give an example?

Yes. I just want a clear pair of statements of the form "I think that the velocity is approximately equal to ...". And "I think that the velocity is exactly equal to ...". It would be best if you cast them mathematically so that I could be sure if you are talking about velocity as a function v(t) or velocity at a point v(t₀).

It appears to me that physics does define an explicit description of motion; A coordinate charting of change of position over some time interval is the exact description of the motion over that interval.
Would you disagree??

The position, x(t), is a complete description of the motion of some point particle in a given coordinate system. From x(t) you can get a complete set of all higher order derivatives in that same coordinate system, including x'(t)=v(t) and x''(t)=a(t). All inertial coordinate systems will agree on a(t) although they will generally disagree on x(t) and v(t).

So perhaps the question could be ; Does a velocity regarding an accelerating particle , intrinsically provide the information needed to plot such a chart??

I assume that by "a velocity" you mean the velocity at a single point. I.e. v(t₀), not v(t). If so, then the answer to the question is "no".

However, the answer is also "no" to the related question "Does a position regarding an accelerating particle , intrinsically provide the information needed to plot such a chart". If you are taking instantaneous positions and velocities then you need all higher order derivatives, as I said earlier.

IMO, the distinction you are trying to draw between position and velocity is artificial, and is actually a distinction between point values and functions. If you have x(t) then you also know v(t). If you have v(t) then you also know x(t) minus some constant offset which has no physical significance. If you have only v(t₀) then you don't have much information at all, and if you have only x(t₀) then you have just as little information. Obviously, x(t) has more information than v(t₀). Similarly, v(t) has more information than x(t₀). That doesn't imply that either x(t₀) or v(t₀) are approximations.

1) If you will look at my statements above I think you will see that your response had no correlation to my statements.

You were countering arguments I never made.

That is what happens when you don't clarify. People will interpret things differently than you had intended.

I clearly don't understand what you are saying and you don't intend to clarify, so I don't see the use of continuing. For completeness I will try to respond to your cases later in the day, although I think it is an exercise in futility and doubt that it will lead anywhere. I still don't understand what you think about velocity is approximate.

 

[DrGreg]

All inertial coordinate systems will agree on a(t)...

Not particularly relevant to the discussion, but actually this isn't true. All inertial coordinate systems will agree on the proper acceleration which, for one-dimensional motion, is<br /> \frac{a(t)}{\left(1 - \frac {v(t)^2} {c^2} \right)^{3/2}}<br />

 

[Dale]

You are, of course, correct. Thanks for the clarification. Also, now that you mention it, all coordinate systems whether inertial or non-inertial agree on the proper acceleration.

 

[GeorgeDishman]

It appears to me that physics does define an explicit description of motion; A coordinate charting of change of position over some time interval is the exact description of the motion over that interval.
Would you disagree??

I would say the motion of the particle was described by a plot of its position over time but "motion" is an ill-defined term. When you say "change of position", do you mean the rate of change or the displacement from the initial position?

So perhaps the question could be ; Does a velocity regarding an accelerating particle , intrinsically provide the information needed to plot such a chart??
Regarding an inertial particle it clearly does.
I hope you will take this question as it is explicitely stated

Yes. If you mean "rate of change", that is simply the velocity by definition. If you mean displacement from the initial position, there's an extra step involved. Since velocity is defined as the derivative of position, you have to integrate the velocity to get the position. You then normally need a way to define the constant of integration but since "change of position" is relative to some starting point, the delta is simply the integral.

Case ----elastic collision of a macro system.

Case-----sub-atomic particle and rocket.

Case------inelastic collision

The same argument as above applies to acceleration and velocity, given an initial velocity and position at some instant, thereafter the velocity at any time can be obtained by integrating the acceleration, and the position by integrating the velocity. All three are then exact. Your problem if a spacecraft is hit by a cannonball is finding the exact acceleration applied to each piece of debris and of course the acceleration is not going to be constant either in magnitude or direction.

Case------- Doppler shift

Treat the emitted signal as being amplitude times the sine of a linearly time-dependent phase, then the location at which any particular phase occurs is given exactly by the process described above, i.e. the second integral of the acceleration (with initial values). That value of signal is received at exactly the time it was emitted plus the propagation time to the receiver from the emission location. There are no approximations involved.

If it was non-uniform acceleration under consideration, then it would be back to the same situation and an instantaneous acceleration evaluation would be an approximation and it would require some expression of jerk to exactly describe the motion and I would imagine with a limited range of exactness..
DO you disagree with the above?

If the acceleration is not constant, then as you say you could find it as the integral of the jerk. However, you have said yourself several times that derivatives are exact and since integration is the reciprocal process, it too is exact so why do you suggest there is some approximation involved? There is no approximation in anything you have listed, you seem to be contradicting yourself.

 

[Dale]

I am not certain what you are asking, but I will do my best guess.

Case ----elastic collision of a macro system.

Supposing an accelerating system (rocket) collides with a cannonball.
Would the resulting vectors be exactly the same as those consequent to the same interaction with the momentarily co-moving inertial system??
It seems to me they would not

I agree, the velocity and momentum vectors of an accelerating rocket colliding with a cannonball are obviously not the same as those of a coasting rocket colliding with a cannonball. In the first case there are three objects involved in the collision (cannonball, rocket, exhaust) and in the second case there are only two (cannonball, rocket). I certainly never claimed that a set of two vectors would ever be equal to a set of three, nor is such a claim logically implied from any of my claims.

However, in both cases at each moment, the instantaneous momentum of each object is a function of the instantaneous velocity, and the total system momentum is conserved from moment to moment (in an inertial frame). The velocity is never an approximation to the value needed for the momentum, even for an accelerating rocket.

Case-----sub-atomic particle and rocket.

This is no different from the previous case. Simply make the substitution cannonball -> particle.

Case------inelastic collision
I would think that the preceding would apply equally in this case, both to the resulting vectors and to the kinetic energy. DO you see any compelling reason to think there woud not be any effect?

I agree, the preceeding does apply equally.

Case------- Doppler shift
Certainly under most circumstances it would seem that emission can be considered instantaneous.
But what about high velocities and magnitudes of acceleration with long wavelength emissions??
Would you maintain that under all circumstances the measured Doppler shift would be exactly equal to the same signal from the MCIRF??

Yes, although I am not certain of the relevance. If you analyze the same measurement of the same signal from any reference frame then you will get the same output. That is required by the diffeomorphism invariance of the laws of physics.

Are you now talking here about real word measurement of a particle with unknown motion?

Yes.

Any description of motion which says nothing about positions or times would seem to be, self evidently, an abstraction, not a description of the real world.

Any description, even with positions and times, is an abstraction. The only thing which is not an abstraction is the thing itself. Any description thereof, even a completely accurate description, is an abstraction.

DO you disagree with the above?

No, in fact, I already stated the same several posts ago in my comments about requiring information about all derivatives.

What exactly is the meaning of 1st order accuracy?

It means that the errors in the approximation (Taylor series expansion) are less than k |t-t₀| for all |t-t₀|<j. Second order accuracy means that the errors in the approximation are less than k |t-t₀|² for all |t-t₀|<j. Where j and k are some arbitrary positive (finite) numbers.