Can Absolute Velocity be Measured?

Tags: absolute, measured, velocity
PF Gold
P: 4,087
Hi Austin, you're right about nxn-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 ...
P: 255
 Quote by Austin0 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??
 Quote by Mentz114 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.
Mentor
P: 17,271
Sorry I have been delayed. I was travelling for business very heavily. Some of your questions may have already been resolved, so I apologize for any redundancy.

 Quote by Austin0 Here you have questioned my logic. Fair enough , it may be flawed. So lets 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.

 Quote by Austin0 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.

 Quote by Austin0 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.

 Quote by Austin0 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.

 Quote by Austin0 Are you changing your mind about the value being c???
No. The limit is exactly c.

 Quote by Austin0 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:

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

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.

 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. That requires intelligence and understanding.
I agree. I think you have the required intelligence, and hopefully you now have the required understanding also.
P: 1,162
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?
[QUOTE=GeorgeDishman;3899076]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?

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

[QUOTE=GeorgeDishman;3894814]
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
P: 1,162
 Quote by DaleSpam 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 occured 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.
 Quote by DaleSpam 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
 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...
 Quote by DaleSpam 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.

 Quote by Austin0; 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???

 Quote by DaleSpam 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; 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?
 Quote by DaleSpam 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]
P: 255
 Quote by 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.
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???

 Quote by GeorgeDishman 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.
Mentor
P: 17,271
 Quote by Austin0 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.
P: 643
 Quote by Austin0 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 ...
Mentor
P: 17,271
 Quote by Austin0 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.
P: 1,162
 Quote by DaleSpam 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
 Mentor P: 22,294 Where did I say that?!! Perhaps the issue here is you are confusing "a line" with "the slope of a line"?
Mentor
P: 17,271
 Quote by Austin0 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:
$$f(x_0)+f'(x_0)(x-x_0)\ne f'(x)$$

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).
P: 643
 Quote by DaleSpam The tangent line is: $$f(x_0)+f'(x_0)(x-x_0)\ne f'(x)$$ 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.
P: 255
 Quote by Austin0 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 we to take it that you now agree with Yuiop's statement in post#28?
Mentor
P: 17,271
 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.
 P: 512 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.
P: 1,162
 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
P: 1,162
 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

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