What is the integral of displacement?

[Actaeus]

Formerly "Acceleration : Velocity :: V : Position :: Position : ???"

Seriously, what happens when you integrate a position function?

1. The position is that of a gate covering part of a hole in a large tank of water. The resulting function describes the amount that has flowed out by time t.

2. You are holding one end of rope low to the ground. Your friend holds the other, and you both run through a field of corn. You are drunk, so you weave side-to-side. The integral of your horizontal distance from him describes the amount of corn destroyed (as shown).
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[Charlie!]

So you're saying integrate x(t) of some particle?

Well, it'll be the average position, multiplied by the time you integrate over. If you say that the integral goes from 0 to t, you'll have the average position multiplied by t. Or, thinking about it another way, integrating gives you the total amount of position that you've "had." Neither of these are particularly useful concepts when talking about the world, so the integral doesn't show up much.

Even your examples don't directly deal with this. the running through the corn thing, for example, integrates a function x (how far you are apart) over another distance, let's call it the y axis, the total distance you've run, to give the very useful 2-dimensional area of the corn destroyed. Sure, you could go through all the trouble and express y in terms of t, but you're still integrating with x*dy.

[J Spade]

I think it gives you distance traveled, as opposed to displacement. I could be wrong.

[Xanthir]

I think it gives you distance traveled, as opposed to displacement. I could be wrong.

You would be, as a trivial look at the graph where I stand absolutely still at position y=1 would show you.

[Mathmagic]

I think it gives you distance traveled, as opposed to displacement. I could be wrong.

The units of the integral would be meters-seconds, not meters as distance traveled is.

[Soralin]

I think it gives you distance traveled, as opposed to displacement. I could be wrong.

The units of the integral would be meters-seconds, not meters as distance traveled is.

So how far away you are from some point for how long, multiplied together into some not very useful number.

Hey, there's a use, integrate 1/position² dt, and use it to determine total radiation exposure from some source.

[Ended]

One reason why this integral is not very useful is that the numerical value of position is essentially arbitrary, depending on where you choose your zero-point. For instance, if I stand still at x = 0, then the integral will be 0. If I stand still at x = 1, then the integral will be t. But these situations are physically identical - it's only our frame of reference which has changed. (Compare this with the integrals of velocity dt or acceleration dt, which are unchanged by a spatial translation.)

[idobox]

Sometimes, the origin can have a physic meaning, like spring equilibrium point, center of gravity of a celestial body, etc.
Also, speeds depend on the referential, and are essentially arbitrary too.

[Actaeus]

Sometimes, the origin can have a physic meaning, like spring equilibrium point, center of gravity of a celestial body, etc.
Also, speeds depend on the referential, and are essentially arbitrary too.

Exactly - it's only when you reach acceleration that the origin has an absolute universal meaning.

By the way, distance is the integral of speed, which is the absolute value of velocity.

a=∫j dt
v=∫a dt
s=∫v dt
?=∫s dt

d=∫|v| dt

[Charlie!]

Sometimes, the origin can have a physic meaning, like spring equilibrium point, center of gravity of a celestial body, etc.
Also, speeds depend on the referential, and are essentially arbitrary too.

Exactly - it's only when you reach acceleration that the origin has an absolute universal meaning.

By the way, distance is the integral of speed, which is the absolute value of velocity.

a=∫j dt
v=∫a dt
s=∫v dt
?=∫s dt

d=∫|v| dt

Well, kinda. Both are the right units, they just have different meanings. In the one-dimensional case, if you sometimes have positive velocity and sometimes have negative velocity, it's rather handy to have the integral of v be the distance you are from the starting point, rather than distance traveled.

To generalize, you'll see the integral of just v show up more when the equation depends only on the distance at the end (like gravitational potential), and the integral of the magnitude of v will show up when you do care about the path (like when calculating energy lost to friction).

[Swivelguy]

I would call that integral the "life story" of the particle.

[Actaeus]

Sometimes, the origin can have a physic meaning, like spring equilibrium point, center of gravity of a celestial body, etc.
Also, speeds depend on the referential, and are essentially arbitrary too.

Exactly - it's only when you reach acceleration that the origin has an absolute universal meaning.

By the way, distance is the integral of speed, which is the absolute value of velocity.

a=∫j dt
v=∫a dt
s=∫v dt
?=∫s dt

d=∫|v| dt

Well, kinda. Both are the right units, they just have different meanings. In the one-dimensional case, if you sometimes have positive velocity and sometimes have negative velocity, it's rather handy to have the integral of v be the distance you are from the starting point, rather than distance traveled.

To generalize, you'll see the integral of just v show up more when the equation depends only on the distance at the end (like gravitational potential), and the integral of the magnitude of v will show up when you do care about the path (like when calculating energy lost to friction).

I was using s as the displacement, and d as the total distance traveled.

[Yakk]

As others have noted, Position is the wrong term -- Accelleration is to Velocity as Velocity is to Displacement, not Position. We have to avoid the change in the arbitrary reference point from causing a change in the value we are talking about.

Going up the stack, we notice that Velocity suffers the same issue -- physics doesn't change when we do it on a speeding train (well, an inertial speeding train heh).

Is the same true of Acceleration? If you have a huge uniform acceleration field pulling you and everything else in your observable universe thataway at 1000 m/s^2, and pulling on everything else in the same way, does physics change?

[Xanthir]

Is the same true of Acceleration? If you have a huge uniform acceleration field pulling you and everything else in your observable universe thataway at 1000 m/s^2, and pulling on everything else in the same way, does physics change?

Of course it does. For one thing, over time it will take more energy to accelerate yourself further, as your velocity creeps closer to c.

[jmorgan3]

Is the same true of Acceleration? If you have a huge uniform acceleration field pulling you and everything else in your observable universe thataway at 1000 m/s^2, and pulling on everything else in the same way, does physics change?

Of course it does. For one thing, over time it will take more energy to accelerate yourself further, as your velocity creeps closer to c.

Would a local observer be able to tell, though?

[Actaeus]

Is the same true of Acceleration? If you have a huge uniform acceleration field pulling you and everything else in your observable universe thataway at 1000 m/s^2, and pulling on everything else in the same way, does physics change?

Of course it does. For one thing, over time it will take more energy to accelerate yourself further, as your velocity creeps closer to c.

Would a local observer be able to tell, though?

Try dropping something. Einstein's Principle of Equivalence states that uniform acceleration acts on a system just like uniform gravitational force.
If the universe was accelerating right at a high acceleration, we would all be "falling" left.

[Yakk]

Yes, you drop something. And it would fall.

But you'd be falling.

Both of you are in 'free fall', which looks a lot like an inertial frame of reference. I'm wondering if it looks exactly like an inertial frame of reference.

If so, then the "zero point of acceleration" isn't special either.

Going up to the next step, is the "zero point of jerk" special?

[Actaeus]

Yes, you drop something. And it would fall.

But you'd be falling.

Both of you are in 'free fall', which looks a lot like an inertial frame of reference. I'm wondering if it looks exactly like an inertial frame of reference.

If so, then the "zero point of acceleration" isn't special either.

Going up to the next step, is the "zero point of jerk" special?

Try an accelerating box, which you are in. You are definitely not in an inertial frame of reference. If you were falling, you would be accelerating w.r.t. to the frame, which makes it not much of a frame of reference at all.

The zero point of jerk is special, because it makes you not nauseous. Constant acceleration is fine, but when it changes... urrp.

[Yakk]

Try an accelerating box, which you are in. You are definitely not in an inertial frame of reference.

So, Alice is 1 billion ly away from a ridiculously huge gravity source in a box.

The difference in gravitational attraction over the volume of the box cannot be distinguished, as the curve is very flat at this distance.

Alice ise falling towards it, accelerating at a rate of 10 m/s^2.

Bob is floating in empty space, far away from anything, in a box.

There is no gravity gradient near Bob.

Describe the experiment Alice can perform in the box that will distinguish her situation from Bob's.

The zero point of jerk is special, because it makes you not nauseous. Constant acceleration is fine, but when it changes... urrp.

I think that is caused by non-uniform changes acceleration/force over the particles in your body.

...

Now, I believe that if you where orbiting in free-fall above a gravitational body, you could use frame dragging inside the box to determine if you where in empty space, or in orbit. But I'm not certain if that requires that the body you are orbiting be rotating.

I also believe you can distinguish between yourself spinning in space, and the universe spinning around you as you are stationary (I believe, however, the difference is smaller than one might think, but I'm less than confident in this particular claim).

[ThomasS]

Both of you are in 'free fall', which looks a lot like an inertial frame of reference. I'm wondering if it looks exactly like an inertial frame of reference.

If so, then the "zero point of acceleration" isn't special either.

Going up to the next step, is the "zero point of jerk" special?

If you are in sealed container in free fall, the only difference between an inertial frame of reference is a higher order (read, typically small) tendency for objects in the ensemble to move towards each other. e.g. if you drop 2 balls, they move very slowly towards each other because they are both heading for the center of the earth. This is, in some sense, the curvature of gravitation.

On a related note, have you ever mixed a container of orange juice or similar by shaking in a rotational manner? e.g. hold it at the center of gravity and make your wrist rotate back and forth in place. It mixes even if the liquid has consistent density and no air bubbles. Is this reference frame independent? To first order, linear acceleration, jerk, and so on do not mix a homogeneous fluid. The idea behind Mach's principle is that even this portion of a reference frame should be dependent on some aspect of the reference frame created by the stars, which leads to the concept of frame dragging.

The question is also interesting in electrodynamics, if a charged particle is accelerating it tends to emit photons. Subject to a certain amount to quantum mechanics and relativity incompatibility, I think it is likely that "accelerating" means accelerating relative to those particles which will receive the photons. However, there be great dragons here involving retarded potentials and similar scary stuff. Because well, how does the electron know it is being accelerated so that it can throw off a photon, without a zero point to acceleration.

[Yakk]

If you are in sealed container in free fall, the only difference between an inertial frame of reference is a higher order (read, typically small) tendency for objects in the ensemble to move towards each other. e.g. if you drop 2 balls, they move very slowly towards each other because they are both heading for the center of the earth. This is, in some sense, the curvature of gravitation.

Sure, but that depends on the gravity varying over the volume of the box by a detectable amount.

Ie, the direction gravity is pulling an object in free fall towards the earth varies between one side of the box to the other. Make the earth far enough away and massive enough, and the direction gravity is pulling gets more and more parallel, and the gradient in gravitational attraction magnitude shrinks from the far to the near side. And given a massive enough gravitational source you can make this as smooth and flat as you like, yet still result in you being pulled at 10 m/s^2.

Of course, as you make the source larger, the gravitational force at the event horizon goes down ... which means that in order for a 1 g pull to be sufficiently smooth, eventually you'll have to be inside the black hole. Which makes the experiment hard to do and report back from.

That generates a possibly interesting curve -- how closely can an accelerating reference frame at a certain rate in a room of a given size approximate flat space internally before it must be within the event horizon of a black hole?

On a related note, have you ever mixed a container of orange juice or similar by shaking in a rotational manner? e.g. hold it at the center of gravity and make your wrist rotate back and forth in place. It mixes even if the liquid has consistent density and no air bubbles. Is this reference frame independent? To first order, linear acceleration, jerk, and so on do not mix a homogeneous fluid. The idea behind Mach's principle is that even this portion of a reference frame should be dependent on some aspect of the reference frame created by the stars, which leads to the concept of frame dragging.

*nod*, except as it happens if you rotate the entire universe in a similar way around the OJ, the OJ also mixes. The differences apparently exist, but I think (with low certainty) that they are not trivial to detect. (ie, frame dragging is a very weak effect). Then again, maybe with the entire universe rotating, frame dragging would be a much stronger effect...


The question is also interesting in electrodynamics, if a charged particle is accelerating it tends to emit photons. Subject to a certain amount to quantum mechanics and relativity incompatibility, I think it is likely that "accelerating" means accelerating relative to those particles which will receive the photons. However, there be great dragons here involving retarded potentials and similar scary stuff. Because well, how does the electron know it is being accelerated so that it can throw off a photon, without a zero point to acceleration.[/quote]

[Actaeus]

I see your point. If everything in the frame of reference is accelerating, I suppose you wouldn't be able to notice.
Hmm...

There is the stronger point that if just the "box" was moved at constant velocity, you wouldn't notice; this isn't true of acceleration. That's really what I meant.

[Yakk]

If the box is moving at a constant velocity, and you are not, you notice.

It just gets rectified pretty quickly. With an ouch.

[Actaeus]

If the box is moving at a constant velocity, and you are not, you notice.

It just gets rectified pretty quickly. With an ouch.

If the box is your reference frame, you would be starting with an initial velocity. After the collision, you would be at rest WRT the box and vis-versa.
This isn't true of an accelerating box; you would continue to feel the effects of the acceleration.

[Yakk]

If the box is moving at a constant velocity, and you are not, you notice.

It just gets rectified pretty quickly. With an ouch.

If the box is your reference frame, you would be starting with an initial velocity. After the collision, you would be at rest WRT the box and vis-versa.

*nod*, if something causes your velocity to change, then you'd no be stationary relative to the box.

Assuming the box is 'solid', that happens after the collision.

This isn't true of an accelerating box; you would continue to feel the effects of the acceleration.

B``1 and B``2, and A``1 and A``2.

You are A``, the box is B``. The experiment begins with everything being stationary relative to each other.

B``1 is a stationary box. A``1 is being acted upon by some unknown uniform force that is accelerating it at amount x.

B``2 is a box being accelerated by amount -x. A``2 is floating there.

(Very similar to the B`1/A`1 and B`2 /A`2 situation, where you differ with respect to who has a velocity, and B1/A1 B2/A2 situations where in one the box is offset 1 foot to the right, and in the other you are offset by 1 foot to the left).

Produce the experiment to distinguish between the B1/A1 and B2/A2 situations while in the box.

As an aside, I don't know the answer with any certainty.

You can repeat that for further derivatives of position with respect to time.

Note that the `` case is strange, in that it involves a uniform acceleration -- some force that varies in proportion to the mass of the thing being effected. It also requires that the force selectively apply to one or the other body in the system.

The Jerk case (```) might cause the problem of something akin to gravity waves, if we presume the change in force must propagate (is that required?)

I'm poking at this because it is an attempt to understand how relative things are. If we want to go up the rabbit hole to the other side of displacement, and the rabbit hole continues being relative as far down the rabbit hole we go, we would want something that stays relative in the other direction...

[Hydraulist]

Absement is the integral of displacement.
Absity is the integral of absement, etc..

Thus we have:

... abseleration, absity, absement, displacement, velocity, acceleration, ...

See http://wearcam.org/absement/

Proceedings of the 14th annual ACM international conference on Multimedia
Santa Barbara, CA, USA
Pages: 519 - 528
Year of Publication: 2006
ISBN:1-59593-447-2
ACM: Association for Computing Machinery
SIGMULTIMEDIA: ACM Special Interest Group on Multimedia

Publisher
ACM Press New York, NY, USA

[Hydraulist]

Here are some other examples that we designed for teaching the principle of absement:
http://wearcam.org/absement/examples.htm

Here's a paper on teaching absement to children,
published March 13, 2008, by Harbourfront Centre in Toronto:
http://wearcam.org/absement/Harbourfron ... Theory.pdf

[You, sir, name?]

"Absement" seems very artificial. Not every integral corresponds to a meaningful quantity. To make my point clear, let's define an utterly non-sequitur integral.

Let's first define the clown-nose field. At the center of every clown noise is a point "charge", so the clown-nose field

\vec{N} = \sum_{n \in noses} R_n\frac{1}{r^2_n}\hat{r}

where R_n is the radius of the clown nose, and \vec{r}_n is the coordinates to the clown nose.

Let us next define burrito-giraffe space. Burrito-giraffe space is a pseudo-cylindrical coordinate system oriented after a given burrito, and which end of them points closest to a giraffe. The positive z-axis is defined as the direction in which the symmetry line comes closest to a giraffe. The radial axis is defined normal to the burrito surface, and the angular axis is defined as though burrito-giraffe space was a regular right handed coordinate system.

Finally, let's find a dog, say a Labrador, and integrate \vec{N} \cdot d\vec{A} over it's surface, and get the clown-nose flux of a labrador in burrito-giraffe space. (Actually, you could use it to determine if there are clown-noses in your Labrador)

Where was I? Oh, yeah, just because you can integrate it doesn't mean it corresponds to some meaningful quantity that deserves a wikipedia article.

[Hydraulist]

Yes, I agree that not all integrals are useful. But the integral of displacement is incredibly useful! In many situations involving water, pumps, liquids collecting in buckets, etc., absement comes up often. Hydraulophones (musical instruments that make sound from liquids) are usually absement-responsive, and a 2-stage hydraulophone responds to absity (the integral of absement, i.e. integrates displacement twice).

Recently a number of researchers have made other connections to absement. For example, there is a connection between absement and the memristor. See
http://arxiv.org/pdf/1201.1032.pdf

Memory Elements: A Paradigm Shift in Lagrangian Modeling of Electrical Circuits


The abstract of this paper mentions absement:

Although time-integrated charge is a somewhat unusual quantity in
circuit theory, it may be considered as the electrical analogue of a mechanical quantity called
absement. Based on this analogy, simple mechanical devices are presented that can serve as
didactic examples to explain memristive, meminductive, and memcapacitive behavior.

Thus absement is useful in a wide range of mechanics, fluid mechanics, and electrical engineering problems, or at least more useful than the clown-nose and Labrador dog integral.