Absement: the Integral of Displacement.

[Agent Intellect]

I found this concept rather strange. Anyone familiar with simple calculus is aware that the derivative of displacement is velocity, the derivative of velocity is acceleration, and possibly familiar with further derivatives such as jerk, jounce (snap), crackle, pop, lock, and drop. Anyone a little more familiar will know that if given the formula for one of these, taking the integral will find the next higher (eg the integral of acceleration is velocity and the integral of velocity is displacement).

I always wondered if there was such thing as the integral of displacement, but they never mentioned it at school. Fortunately, I found a couple websites that talk about it, and it's called Absement (Absence and displacement). There are integrals above that, as well (Absity, Abseleration, Abserk, and Absounce, all portmanteau of the derivatives of displacement). In addition, there is the integral of reciprocal position called Presement (Presence and displacement).

What is Absement: one of the websites describes it like this:

Consider a 5-hour train ride that takes you 500 miles directly away from your home, in a straight line, to another destination where you stay for 5 hours and then return. Suppose you want to stay wirelessly glogged into your home computer at a "roaming" communications cost of $1/mile/hour. For simplicity, assume a linear long-distance rate, i.e. $1/hour when you're 1 mile away, $2/hour when you're 2 miles away, $3.14/hour when you're 3.14 miles away, etc.. The total cost of your online communications is $5000, since the absement (time-integral of displacement) is 5000 mile hours (1250 mile hours on the way to your destination, plus 500 miles * 5 hours stay = 2500 mile hours, plus 1250 mile hours of absement during the return trip).

The middle plot shows Displacement. The first 5 hours are spent in the train going at velocity 100mph (miles per hour) away from home. The area under this triangular part is 1/2 five times 500 mile hours, which is 1/2 times 2500 mile hours, i.e. 1250 mile hours. The next 5 hours are spent at your destination, 500miles from your home, where you pay $500/hour for 5 hours, for a cost of $2500. Staying online during your return trip costs you another $1250. Your total cumulative running cost is the area under the middle plot up to a particular point in time. This integral is called absement and is shown on the top plot.
Each of the three plots is the time-derivative of the plot above it:

• Displacement is the time-derivative of Absement
• Velocity is the time-derivative of Displacement.

I guess I don't have any real questions (except maybe what the hell Absity and Abseleration could be measuring), but I thought this was interesting.

EDIT: Forgot to link my sources:
http://wearcam.org/absement/Derivatives_of_displacement.htm
http://wearcam.org/absement/examples.htm

 

[Oblivious]

DAMN MATH YOU SCARY! Truly, Math is one of the few things I respect but do not understand.

I would start my line of thought by venturing that that absement would be the rate of change in absity and that absity is the rate of change in abseleration. Judging from the fact that these are higher level integrals rather then higher level derivatives.

At first I tried to think of a real life example of absement and remembered the kilowatt-hour from electrical engineering. Then I remembered that the watt is velocity, and not displacement. The kilowatt-hour would then be a form of displacement, and not absement.

Truly a strange and interesting concept. The real wall to break would to somehow to think of displacement as a rate, or am I totally off?

Theoretically, if you keep deriving an expression, there will come a time when the derivative would be zero and remain at zero. However, if you keep integrating the same expression, you would produce infinitely many expressions. What does this mean?

Well, geometically, the curve of the integral would reach infinity, and so as the number of integrations reach infinity, so too should the slope of the graph at any given point, assuming a continuous function.

I suppose that still does not answer anything though.

 

[Melllvar]

This was some good info. I personally only knew the names of the time derivatives up to jerk, and none of the integrals of x * dt. Even physicsforums.com only seemed to have one thread on the subject, and it was significantly less informative than this one. So thanks AI, although I really don't have many comments to add myself.

It seems that in physics integrations with respect to displacement (e.g. Work) are more common. Not sure if this is because they're actually more common or because of the applications physics education tends to focus on more. It seems there are at least useful applications of the derivatives/integrations you never hear names given for after all.



Edit: Can you believe they don't even have a wikipedia page on any of this stuff? I think this looks like a job for... INTPf Members!

Seriously, you want to write it or should I? I've never tried making a wikipedia page before. I rarely even edit/discuss. This could be interesting.

 

[pjoa09]

Don't know calculus. I guess it's how much of you went for how long. Some application in GPS? The only application I see right now is a proximity sensitive piano.

 

[Melllvar]

Don't know calculus. I guess it's how much of you went for how long.

I'll enjoy explaining this.

Calculus is mainly about derivatives and integrals (more precisely it's about limits and limitation, but I won't go into that right now - derivatives and integrals are both based off limits). The closest analogies are division for differentiation and multiplication for integration. Differentiating (finding the derivative of a function) gives you another function describing the instantaneous rate of change (i.e. slope) of the first function at every point on it, while integration (finding the anti-derivative) gives you another function describing the area under the curve of the function between any two points.

When the function is a straight line finding the slope or the area under the curve is very easy, and you don't really need calculus. For example, the slope between two points is the "rise over the run" or the change in y over the change in x:

The ∆y divided by the ∆x would give you the slope, which would be the exact same as the derivative of the function. The area under the curve along the ∆x would be the area of the triangle plus the area of the rectangle underneath it.

As an example:

The slope/derivative between x = 0 and x = 1 is (3 - 1) / (1- 0) = 2 / 1 = 2. The area between those same two points is (1 * 1) + (1/2) * (3 - 1) = 1 + 1 = 2. (It's a coincidence they're the same.)

For non-linear functions it's harder to find though, which is what calculus is for. Instead of looking at the delta x, delta y you look at the "instantaneous delta", i.e. the differential: dx, dy. The slope/derivative is then dy/dx instead of ∆y/∆x, while the area under the curve/integral is ∫y * dx.

So yeah, you're basically finding the area under the curve by using an abstraction on multiplication with integration, and finding the slope/rate of change of the curve by using an abstraction on division with differentiation. There are rules for doing each of these, and they're also inverse processes. Taking the derivative and then integrating again will leave you with the same thing you started with, much like dividing by a number and then multiplying by the same number will get you back to the same place.

As an example, if y = 2x^3 * cos(x), the derivative y' = 6x^2 * cos(x) + 2x^3 * -sin(x). The integral/anti-derivative Y = 2x^3 * sin(x) + 6x^2 * cos(x) - 12x * sin(x) - 12 * cos(x) + C, where C is an arbitrary constant. Barring any mistakes I made along the way. From there you'd find the numerical value of the slope at any given point by evaluating y' at that point (x = whatever) and the area under the curve between two points by evaluating Y at whatever two points and subtracting the first evaluated point from the second (see Fundamental Theorem of Calculus). You'd first find the value of the arbitrary constant by evaluating the Y function at some known point on the function - any known (x, y) point that lies on the curve, which allows you to solve for C.

Which is a long way of saying, yeah, it's basically how far you went for how long.

 

[Agent Intellect]

I think a better way to explain derivatives and integration is this:

A derivative is the slope of the line at a point. It's like (Y2-Y1)/(X2-X1) if the difference between X2-X1 approaches arbitrarily close to zero - you are finding the slope of the line when there is only an infinitesimal change in X. See difference quotient. The simplest derivative is a polynomial, because every term can be changed in the form of n*ax^(n-1). For instance, (5x^3) + 3 would become 3*5x^(3-1) = 15x^2.

An integral is finding the area under a curve f(x) by finding the area of n number of rectangles under f(x) from the x-axis to the line. This becomes more precise as the number of rectangles n approaches infinity. See Riemann sum. The simplest (indefinite) integral is a polynomial, because every term can be changed in the form (ax^(n+1))/(n+1) + C. For instance, 15x^2 would become (15x^(2+1))/(2+1)+C = (15/3)x^3+C = (5x^3) + C. This is the same term I took the derivative of above (except if we don't always know that the C = 3) which is why this is called the antiderivative.

As Mellvar said, a derivative is a division. For instance, the derivative of distance, we'll say meters, is meters/second. The derivative of this is meters/second^2 and so on. The integral of meters/second^2 would be like doing meters/second^2 * seconds, which gives you meters/second. The integral of this would be like meters/second * seconds, which gives you meters. So the integral of meters would give you meters*seconds, which is absement. Integrals can also measure things such as work, where if the integral of force is taken with respect to distance, you get F*d (eg Newtons*meters).

 

[pjoa09]

So it is a very complicated way of find the slope and the area underneath a curve?

I got lost after the algebra. But I think I have the very big picture. There is still a little time before I get into depth on my own.

 

[typus]

the C = 3

 

[Words]

Well, my background only consists of a few things in arithmetic, but I got enough unconventional logic anyways.

One(or at least "I") would think, based on the immediate pattern available, that the integral of something is the simpler(contemporary) idea; but this absement idea clearly demonstrates an increase in complexity. If I assume that "displacement" is simply a type of measurement of space, a tangible facet of reality, then, via this directly observable property which displacement measures, displacement separates itself from every integral or derivative of itself and the rest of the continuation of every other derivates/integrals that originates from it. Absement, Velocity, Acceleration and stuffs measure something that is not tangible or physical. Time is also not observable.

How about drawing a graph that shows the relationship between displacement and absement?

 

[Melllvar]

So it is a very complicated way of find the slope and the area underneath a curve?

Yeah, at least for real-valued functions of a single variable. You can also apply this stuff to multi-variable functions or complex-valued functions, in which case the slope/area analogy becomes slightly less applicable at more complicated levels.

One(or at least "I") would think, based on the immediate pattern available, that the integral of something is the simpler(contemporary) idea; but this absement idea clearly demonstrates an increase in complexity.

Since they're mainly inverse processes defined using limits, I'm not really sure which idea could/can be considered "simpler." Derivatives are generally easier to find solutions to than integrals; finding integrals can be quite a bit more difficult, even impossible sometimes.

If I assume that "displacement" is simply a type of measurement of space, a tangible facet of reality, then, via this directly observable property which displacement measures, displacement separates itself from every integral or derivative of itself and the rest of the continuation of every other derivates/integrals that originates from it. Absement, Velocity, Acceleration and stuffs measure something that is not tangible or physical. Time is also not observable.

I don't think you can consider displacement as being more 'real' than velocity, acceleration, absement, etc. Time and distance/displacement are relative to the observer measuring them, velocity can be relative (for most things) or it can be non-relative (for light/electromagnetic waves). Honestly acceleration seems like the most physically 'real' concept out of any of them, as that basically depends only on the curvature of your worldline as you move through spacetime, and the curvature of spacetime is absolute for all observers. An accelerating observer is fundamentally different from a non-accelerating one - there can really be no disagreement over who has accelerated and who hasn't, unlike with displacement or velocity.

That's just from a physical point of view though. From a mathematical point of view you're just describing the relationships between the variables of a function (I think - I'm sure someone more familiar with advanced calculus/real analysis could give a better/more precise answer). The order of the derivative or anti-derivative is pretty much just a relationship between different functions, e.g. 2x will always be the first derivative of x^2, while 2 will always be the first derivative of 2x and the second derivative of x^2. The physical point of view only arises when you use those functions as a mathematical model for something not-really-mathematical.

How about drawing a graph that shows the relationship between displacement and absement?

It would just be visualized as the area under the curve between two points on a distance vs. time graph. If you chose a specific point as one of those two, you could make another graph describing that area as a function of the other arbitrary point (you would still have to specify a specific function - e.g. the area under y = x will be different from the area under y = x^2). You could also do it using a three-dimensional graph (i.e. a function of two variables, so that both limits of integration could be treated as variable), although you'd still need to pick a specific function.

If I had a computer algebra system available (I don't) it might be fun to make a few f(y,z) graphs in three dimensions describing the areas under the curve of various functions f(x). That's basically the only way I could really visualize this, I think. (and again, it would be specific to the function f(x) that was chosen.)

 

[Latro]

An interesting thing about the "difficulty" of finding derivatives vs. finding integrals is that vastly more functions are integrable than are differentiable, and yet many of them are not integrable in closed form or the closed form is difficult to find. Integrability in the Riemann sense is considerably weaker than continuity, with continuity "almost everywhere" (you can look online for a precise definition of this term) being enough to ensure integrability. Integrability in the Lebesgue sense is even weaker than that; there are functions which are nowhere continuous which are Lebesgue integrable.

 

[Melllvar]

@Latro:
I have to wonder about this. Unless every anti-derivative leads to a continuous function (seems unlikely), then as inverse processes, shouldn't the derivative exist as the original function, despite the function it is derived from being non-continuous? That shouldn't be true though because it would violate the requirement for continuity of differentiable functions. Do you know what the explanation is?

 

[Words]

Y
I don't think you can consider displacement as being more 'real' than velocity, acceleration, absement, etc. Time and distance/displacement are relative to the observer measuring them, velocity can be relative (for most things) or it can be non-relative (for light/electromagnetic waves). Honestly acceleration seems like the most physically 'real' concept out of any of them, as that basically depends only on the curvature of your worldline as you move through spacetime, and the curvature of spacetime is absolute for all observers. An accelerating observer is fundamentally different from a non-accelerating one - there can really be no disagreement over who has accelerated and who hasn't, unlike with displacement or velocity.

I guess it makes sense that perception muddles reality. I searched a bit and I've arrived at these so-called differing frames of reference, which I don't really get. Regardless, it seems generally easier for people to look at a reality that's personally attuned to their own er...spacetime. It's relative specific-wise, but the nature of the differing perceptions is still identical. I guess my point is that, a "collective human frame of reference" is available by pointing at its 3D property, which, if i'm making any sense, is not really a new idea.

That's just from a physical point of view though. From a mathematical point of view you're just describing the relationships between the variables of a function (I think - I'm sure someone more familiar with advanced calculus/real analysis could give a better/more precise answer). The order of the derivative or anti-derivative is pretty much just a relationship between different functions, e.g. 2x will always be the first derivative of x^2, while 2 will always be the first derivative of 2x and the second derivative of x^2. The physical point of view only arises when you use those functions as a mathematical model for something not-really-mathematical.

yeah. I think mathematics generalizes reality in a way that, aside from the differing levels of perceived complexity, it doesn't treat physics-related relationships as having any special qualities as compared to any other relationship like number of cows and number of dung over time. But If there is no mathematically inherent difference between one relationship of variables(say, a derivative of x-relationship) over another relationship of variables(an anti-derivative of x-relationship), then I think it's safe to say that the derivatives and integrals 'originating' from any relationship are infinite. (which, of course, contradicts with the meaning behind your earlier link.)

functions are integrable than are differentiable,

This is out of the topic but is there a relationship between integration/differentiation(math) and cellular integration/differentiation?

 

[Agent Intellect]

My turn to ask calculus questions:

From watching Khan Academy videos, I understand in an abstract way (eg I can do the calculations), but what practical purpose (physics application?) does Green's theorem serve? Or Stoke's theorem, for that matter?

What would it mean, in a practical application sense, for a path integral to be independent? Or dependent? Or for a vector field to be conservative?

 

[Melllvar]

what practical purpose (physics application?) does Green's theorem serve? Or Stoke's theorem, for that matter?

I'm not sure if this is the answer you want, but the main application I'm aware of is Maxwell's equations, to relate the differential and integral forms of the laws. The four equations describe the divergence and curl of the electric and magnetic fields (divergence and curl being properties of vector fields). Stokes' theorem equates the volume integral of the divergence to a surface integral of the field in the direction normal to the surface enclosing that volume, and the surface integral of the curl along a surface enclosing a volume to a path integral of the field tangent to the surface.

Green's theorem is a special case of Stokes' theorem in two dimensions, that's generally used in Gauss's law (Maxwell's first equation) to relate the divergence of a field over a given volume to the flux through the surface enclosing that volume.
Apparently you can also use it to find centroids (i.e. centers of mass for a plane of uniform density [I'd guess Stokes' theorem could be applied to the 3-d version] - sort of the geometric equivalent of actual 'center of mass').

I suppose the practical purpose would be to make calculations easier by allowing one to do line and surface integrals instead of surface and volume integrals. I don't know though, it's been a long time since I solved any of those problems. Maxwell's equations themselves are usually expressed in the differential form using the gradient, dot product and cross product (the combination of which gives the divergence and curl operators).

Additionally, the reason divergence depends on volume while curl depends only on the surface is that only the components of the vector field along the surface add to the curl. You can just as well consider it ('it' being any equations involving curl of a vector field in three dimensions) to be the curl of the entire volume enclosed, but the interior components of the curl will cancel each other out so that only the ones along the surface are left as contributions. I'm not entirely sure how it degenerates to a path integral along the surface from there though, instead of depending on the components of the field across the entire surface.

What would it mean, in a practical application sense, for a path integral to be independent? Or dependent? Or for a vector field to be conservative?

A vector field would be conservative if path integrals over it were independent of the path chosen (path independent), while being nonconservative if path integrals did depend on the path chosen (path dependent). You usually hear these referred to as conservative and non-conservative forces in physics. E.g. gravity, electricity, magnetism are all conservative, i.e. their vector fields are conservative (and ignoring the fact that relativistically gravity isn't a force or a field). Hence the reason that work done by gravity does not depend on the path chosen, only on the initial and final positions of the mass. Friction on the other hand would be a non-conservative force: if I slide something along a table and then move it back to the start along the same path, I do more work on the object than if I slide it along the table and then pick it up and set it back down at the beginning.