Mentor

Can Absolute Velocity be Measured?

 Quote by Whovian 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.
 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.

 Quote by russ_watters 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 : $$f'(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+h)-f(x)}{\Delta h}$$ 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

 Quote by GeorgeDishman 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?

 Quote by GeorgeDishman 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.

 Quote by Austin0; 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
 Quote by Austin0; 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

 Quote by GeorgeDishman 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.

 Quote by GeorgeDishman 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

 Quote by DaleSpam 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 calculting 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.

 Quote by DaleSpam Mathematically f'(x)$\neq$f'(x+$\Delta$x) in general, for $\Delta$x$\neq$0.
 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

Mentor
 Quote by Austin0 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.

 Quote by Austin0 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.

 Quote by Austin0 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.

 Quote by Austin0 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.

 Quote by Austin0 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

 Quote by DaleSpam 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. .
 Quote by Austin0 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. .
 Quote by Austin0 Yes we can calculate an instantaneous velocity and this is a useful abstraction; eg. ICMIF's
 Quote by DaleSpam 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. .
 Quote by Austin0 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..

 Quote by DaleSpam 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?

 Quote by Austin0 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.
 Quote by DaleSpam 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.
 Quote by DaleSpam 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

 Quote by Austin0 .. 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).

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

 Quote by Austin0 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.

 Quote by Austin0 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.

 Quote by Austin0 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.

 Quote by 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.

 Quote by Austin0 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.

 Quote by Austin0 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. ...

 Quote by 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. .
 Quote by DaleSpam 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?

 Quote by austin0 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..

 Quote by DaleSpam 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

 Quote by austin0 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. .
 Quote by DaleSpam 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 lets 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??

 Quote by DaleSpam 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???

 Quote by DaleSpam 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

Mentor
 Quote by Austin0 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.

 Quote by Austin0 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).

 Quote by Austin0 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).

 Quote by Austin0 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.

 Quote by Austin0 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.

Recognitions:
Gold Member
 Quote by DaleSpam 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$$\frac{a(t)}{\left(1 - \frac {v(t)^2} {c^2} \right)^{3/2}}$$
 Mentor 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.

 Quote by Austin0 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.

Mentor
I am not certain what you are asking, but I will do my best guess.
 Quote by Austin0 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.

 Quote by Austin0 Case-----sub-atomic particle and rocket.
This is no different from the previous case. Simply make the substitution cannonball -> particle.

 Quote by Austin0 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.

 Quote by Austin0 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.

 Quote by Austin0 Are you now talking here about real word measurement of a particle with unknown motion?
Yes.

 Quote by Austin0 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.

 Quote by Austin0 DO you disagree with the above???