Can Absolute Velocity be Measured?

Austin0

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

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



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.

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?

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?

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

A derivative is a specific limit. The derivative of x² can be found by working out
[tex]
\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
[/tex]
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

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 [itex]\Delta[/itex]t ---->[itex]\infty[/itex] is not a derivative? because of the infinity???

In the velocity equation [tex]v=\lim_{\Delta t\to 0} \, \frac{ x(t+\Delta t )- x(t)}{\Delta t}=dx/dt[/tex]
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'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.

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

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

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

[tex]v=\lim_{\Delta t\to 0} \, \ ...[/tex]

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

DaleSpam

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.

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.

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.

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.

In general the velocity does not equal the average velocity.

No. The limit is exactly c.

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

[tex]f'(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+x)-f(x)}{\Delta x}[/tex]

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

[tex]\lim_{t \to \infty}\,v(t)[/tex]

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

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

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 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 [tex]v=\lim_{\Delta t\to 0} \, \frac{ x(t+\Delta t )- x(t)}{\Delta t}=dx/dt[/tex]
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?

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 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

Austin0

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.

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

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.

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???

. 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

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

Yes, that is incorrect.

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.

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.

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.

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

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).

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

No, it is explicity the derivative.

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

I already answered that:

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.

DaleSpam

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

And derivatives are exact ...

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.

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. 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"?

DaleSpam

The tangent line is:
[tex]f(x_0)+f'(x_0)(x-x_0)\ne f'(x)[/tex]

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 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

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.

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.

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

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