Vector Field Derivatives: Why & How?

[Lajka]

Hi

I thought about putting this topic in physics subforum, but I think it's overall more fitting here.

So, I'm having problems understanding some basic stuff, and I'm kinda embarassed. I'm an engineer and I'm trying to sort things I already know but on a more rigorous mathematic foundation, and then I get confused from time to time, such as now.

Anyway, let's say we have a parametric vector-valued function $$r(t)=(x(t),y(t))$$

Nothing too fancy, just your everyday normal trajectory. Now, if we define a derivative of this function as

$$\dot{r}(t)=\frac{dr(t)}{dt}=(\dot{x}(t),\dot{y}(t))$$

it is said that this represents the velocity of a function in all the points of the curvature $$r(t)$$.

Therefore, it must be a vector field, by its nature. And now that I think about it, I can't seem to understand this. For something to be a vector field, it must have an input vector and an output vector. For me, $$\dot{r}(t)$$ is just another parametric vector-valued function, with a scalar as an input, that's all I see.
Moreover, let's recall the definition of a derivative
$$\dot{r}(t)=\frac{dr(t)}{dt}=lim_{\Delta t \rightarrow 0} \frac{r(t + \Delta t) - r(t)}{\Delta t}$$

But what exactly is $${r(t + \Delta t) - r(t)$$? Is it this?

or this?

I'm inclined to say the latter, because substraction of two vectors must be a vector in that same vector space. So, in my mind, $$\dot{r}(t)$$ looks like an ordinary parametric vector-valued function, and not a vector field. Just a function, where all the tails of vectors are at the origin, just like $$r(t)$$. If it were a vector field, it would have to use coordinates $$x$$ and $$y$$ as a part of its definition, right? You would have to see it clearly from a definition, it has to be $$\dot{r}(x,y)$$ in order to be a vector field. But it doesn't, so it's not.
So, what am I missing here?

--

Also, this could be an indication of a bigger problem I'm having, namely, an understanding of a substraction of vectors. You see, when I think of a vector space in $$R^{3}$$, I see this

So, a substraction of two vectors is

but one of the most common pictures of substraction is

I'm having problems with this picture, and it's the most normal thing we use in engineering or physics all the time. What exactly is this $$v_{1}-v_{2}$$ in the second picture?
It clearly doesn't belong to the vector space where $$v_{1}$$ and $$v_{2}$$ belong, otherwise its tail would be at the same starting point. It looks like it's detached from all the other vectors, like it's an element of a new vector space, whose starting point is a tip of a vector $$v_{1}$$. How do you explain this mathematically?
I hope I made at least a little sense, and thanks in forward for all the help I can get.

 

[olivermsun]

A vector just specifies a length and a direction, so it doesn't matter whether you draw it sticking out of the origin (as by convention) or as the third side of a triangle (in vector addition/subtraction).

If you go the other way and think of v2 + (v1 - v2) = v1, you can see why the triangle construction works (you can draw v2, then start (v1 - v2) from the end of v2 and you should get v1). By the way, your v1 - v2 looks backwards (v2 - v1) to me.

The first question about the derivative -- well if the derivative is a vector that depends only on the parameter t, then it's still parametric, right? (The function r'(t) only makes sense along the single curve r(t) ).

 

[pergradus]

The first question about the derivative -- well if the derivative is a vector that depends only on the parameter t, then it's still parametric, right? (The function r'(t) only makes sense along the single curve r(t) ).

Another way of saying this is that the domain of r'(t) is only the points that lie on the curve C = r(t).

 

[I like Serena]

Hi Lajka!

Nothing too fancy, just your everyday normal trajectory. Now, if we define a derivative of this function as

$$\dot{r}(t)=\frac{dr(t)}{dt}=(\dot{x}(t),\dot{y}(t))$$

it is said that this represents the velocity of a function in all the points of the curvature $$r(t)$$.

Therefore, it must be a vector field, by its nature. And now that I think about it, I can't seem to understand this. For something to be a vector field, it must have an input vector and an output vector. For me, $$\dot{r}(t)$$ is just another parametric vector-valued function, with a scalar as an input, that's all I see.

It's not a vector field for the very reason that you mention.

Moreover, let's recall the definition of a derivative
$$\dot{r}(t)=\frac{dr(t)}{dt}=lim_{\Delta t \rightarrow 0} \frac{r(t + \Delta t) - r(t)}{\Delta t}$$

But what exactly is $${r(t + \Delta t) - r(t)$$? Is it this?

or this?

It is both.
A vector is not about "where" the vector is.
It's about a direction and a magnitude.

Also, this could be an indication of a bigger problem I'm having, namely, an understanding of a substraction of vectors. You see, when I think of a vector space in $$R^{3}$$, I see this

So, a substraction of two vectors is

but one of the most common pictures of substraction is

Same thing.

 

[Lajka]

Thank you all for your responses.

It's not a vector field for the very reason that you mention.

Okay, I give up, what is it then?

A vector just specifies a length and a direction, so it doesn't matter whether you draw it sticking out of the origin (as by convention) or as the third side of a triangle (in vector addition/subtraction).

It is both.
A vector is not about "where" the vector is.
It's about a direction and a magnitude.

Same thing.

This all would mean, if I'm not mistaken, that $$\dot{r}(t)$$ is a free vector? Position vector $$r(t)$$ is a bound vector, how come then that the substraction of two bound vectors gives a free one?

Another way of saying this is that the domain of r'(t) is only the points that lie on the curve C = r(t).

Yeah, that makes perfect sense, but is there a way I could actually see this from some formula? All I see now is that the domain of $$\dot{r}(t)$$ is a set of parameter $$t$$. Could i possibly rewrite $$\dot{r}(t)$$ somehow so that I can explicitly see something like $$\dot{r}(x,y), (x,y) \in C=r(t)$$?

 

[I like Serena]

Okay, I give up, what is it then?

It is just another parametric vector-valued function, with a scalar as an input.

The vector-value in this case is the velocity in direction and magnitude.

This all would mean, if I'm not mistaken, that $$\dot{r}(t)$$ is a free vector? Position vector $$r(t)$$ is a bound vector, how come then that the substraction of two bound vectors gives a free one?

There's no real difference between a bound vector and a free vector.

It's only that the position vector $$r(t)$$ has the special property that if you let it start at the origin, its end point will describe the position of a particle.

Equivalently the derivative vector $$\dot{r}(t)$$ has the special property that if you let it start at the position coordinates, its end point will predict where the particle will be 1 second later. In that sense it's a bound vector as well!

Another way of saying this is that the domain of r'(t) is only the points that lie on the curve C = r(t).

This does not seem to be correct to me.
The domain of r'(t) is the time interval for which it is defined, which is the same as the domain of r(t).

Yeah, that makes perfect sense, but is there a way I could actually see this from some formula? All I see now is that the domain of $$\dot{r}(t)$$ is a set of parameter $$t$$. Could i possibly rewrite $$\dot{r}(t)$$ somehow so that I can explicitly see something like $$\dot{r}(x,y), (x,y) \in C=r(t)$$?

Sure.

Since $$(x,y) = r(t)$$ we can define

$$\dot{r}(x, y) = \dot{r}( r(t) )$$ which is only defined for (x, y) in C.

 

[Lajka]

Yeah, I guess that's it, I just have to accept it. I always thought that there is a big difference, formally speaking, between bound and free vectors, but I guess they're matematically the same, it's just that we are giving them physical meaning by prioritizing some of their positions more than the others, thus making them 'bound'.
Thank you very much for your help :)

 

[I like Serena]

Any time! :)

 

[olivermsun]

Yeah, I guess that's it, I just have to accept it. I always thought that there is a big difference, formally speaking, between bound and free vectors, but I guess they're matematically the same, it's just that we are giving them physical meaning by prioritizing some of their positions more than the others, thus making them 'bound'.
Thank you very much for your help :)

They are not quite the same thing, but maybe I can give you a "formal" explanation of the difference.

Consider at some "free" vector u in the plane. As you know, it can be defined by a direction and a magnitude, or equivalently, by its x- and y- components (x, y). That's two components. Of course, the coordinate system (and hence the origin) also need to be specified.

If you have a bunch of vectors, say a collection forces acting at a certain point, the net force is the vector sum of all the forces. They apply simultaneously at one point. Writing the vector sum by lining up the beginning of each vector with the end of the last is just a geometrical convenience. You don't literally apply one force, go to the end of its vector, add the next force, etc.

Now the "bound" vector you referred to has everything the "free" vector has, plus an origin or initial position (x0, y0). In other words, it specifies a beginning and an i. To express this information, you need four components: (x0, y0) + (x, y). Equivalently, the bound vector has clear beginning and end points, (x0, y0), (x1, y1).

While the position of an object in space does not have a "beginning" and an "end," the movement of some object during a finite time interval [t0, t1] does have a beginning and an end. You can connect this to your parametric curve problem and to calculus in general in the following way: r(t1) = r(x(t1), y(t1)) is the end position and r(t0) = r((x(t0), y(t0)) might be the initial position. Then, the change in position would be a free vector $$\Delta r = r(t1) - r(t0) = \int_{t0}^{t1} \mathbf{r}'(t) dt .$$
However, you wouldn't know exactly where the object was after t1 unless you know the initial position r(t0), since r(t1) = r(t0) + ∆r. Hence you see that the absolute position is a "bound" vector, while the relative position is "free."

However, there's maybe a potential for confusion here because we started with the convention that the position r(t) was only specified by two components x(t), y(t). As a previous poster said, that's because position vectors are actually bound to the origin (for convenience).

I don't know, maybe this confuses more than it helps, but keep asking if you have more questions.

 

[Lajka]

Okay, firstly, may I just ask you about the notation r(x(t), y(t))? It kinda confuses me, because I always thought that the order pair (x(t), y(t)) is an output of the $$r(\cdot)$$ function, it belongs to its co-domain. In other words,
$$r:R \rightarrow R^{2}$$
$$r(t) = (x(t),y(t))$$

So if you write r(x(t), y(t)), it looks like the ordered pair (x(t), y(t)) is actually an input of this function. Maybe I'm missing something here?

Also, thank you for your in-depth explanation on this topic. If I understood correctly, everything you said is in agreement with what Serena (and myself in the last post) said?

Of course, you explained it more thoroughly and I enjoyed reading it, but we agree on the notion that vectors are vectors, and the notion of 'freedom' and 'boundedness' is something that is added to them in a physical sense, not mathematical one?

What I mean to say is, if I understood correctly, you can't spot the difference between a bound vector with coordinates, say, [3 4]^T, and a free one with the same coordinates, at least not in a mathematical way. Am I correct?
Of course, in a physical way, it makes all the difference, if I move the velocity vector from its point, he is now just a geometrical object because I removed its physical meaning, namely, the velocity of a particle at that point. That's what makes him 'bound'.

Angular velocity, on the other hand... I guess I can't shift that vector wherever I want, it still has the same meaning as before, pointing in the direction of an axis of rotation of some object along with magnitude of that velocity. That's what makes him 'free'.

While the position of an object in space does not have a "beginning" and an "end,"

Well yeah, that position (let's call it P) is just a point, right? But if I choose some origin O, I can identify every point with its radius vector. Of course, I can shift that vector wherever I want, because that vector OP is just one particular representation of that vector in general (that vector is, mathematically speaking, 'free' to be anywhere else), but if I do that, it loses its physical meaning, it doesn't show anymore the position of that particle (because its head is not the point P anymore), so that vector is, physically speaking, 'bounded' to be right there and nowhere else.

Please tell if I'm not making any sense :D

 

[I like Serena]

Everything you say is *right on*!

Angular velocity, on the other hand... I guess I can't shift that vector wherever I want, it still has the same meaning as before, pointing in the direction of an axis of rotation of some object along with magnitude of that velocity. That's what makes him 'free'.

Actually, angular velocity is represented by a 'bound' vector.

Do note that angular velocity is of its own not a vector.
It's the combination of scalar angular velocity and an axis around which a particle turns.
The axis is a combination of a point and a directional vector of the axis.

By convention angular velocity is represented by a vector "bound" to the center of the circular motion, in the direction of the axis, and with a magnitude that corresponds to the actual angular velocity.
Since there are 2 choices for the direction, we use the "right hand rule" to define the direction.

A nice side effect of this convention is that combinations of angular velocities can be handled more or less as vector additions.

 

[Lajka]

Ah, I see, thanks! I remember I saw the 'free' explanation of angular velocity vector http://samizdat.mines.edu/tensors/ShR6b.pdf" (page 5), and that's why I memorized it like that. I guess that means that the example of 'free' vector from this pdf is wrong?

 

[I like Serena]

Ah, I see, thanks! I remember I saw the 'free' explanation of angular velocity vector http://samizdat.mines.edu/tensors/ShR6b.pdf" (page 5), and that's why I memorized it like that. I guess that means that the example of 'free' vector from this pdf is wrong?

Ah well, we're getting into semantics here I think.
You can get into lengthy discussions, talking about what is intended exactly with a word or concept.

I'll try to limit myself to what "makes sense".

A positional vector of a particle is meaningless without a point of reference.
Take another point of reference and the vector changes in direction and length.
So this is a nice example of a bound vector.

For an angular velocity vector to have meaning, it does not require a specific point of reference. We could choose the center of the earth, or we can choose the north pole, and we'll be defining the same rotation.
So in that sense it is more "free".
However, the location of the axis (a line) is relevant, because with another axis, we'll have a different rotation.

The article uses this to illustrate the difference between a bound and a free vector in a physical context.
As such I think it is a good example.

 

[olivermsun]

Okay, firstly, may I just ask you about the notation r(x(t), y(t))?

You are right, it was a mistake. The correct thing is r(t) = (x(t), y(t)).

...we agree on the notion that vectors are vectors, and the notion of 'freedom' and 'boundedness' is something that is added to them in a physical sense, not mathematical one?

I'm saying that "bound" vectors contain additional information over "free" vectors. A "free" vector contains the same information as a vector which is "bound" to the origin of your coordinate system.

Since vectors of either type are just mathematical objects, you can associate them with any physical meaning you choose.

What I mean to say is, if I understood correctly, you can't spot the difference between a bound vector with coordinates, say, [3 4]^T, and a free one with the same coordinates, at least not in a mathematical way. Am I correct?

A bound vector has a clearly defined beginning and end, requiring two pairs of coordinates. Your example has one one pair of coordinates -- is that a beginning or an end?

In the earlier example of the moving point along r(t), a free vector is sufficient to sufficient to describe the velocity r'(t). The position is specified by r(t), which is a separate, but related object.

If one were to define a beginning and an end for a "bound r'(t)" then what would you specify as the beginning and end points? (x0,y0) in units of length and (x1,y1) = (x0,y0) + (x', y') in units of velocity? (Something conceptually similar is done for graphing vector fields, but I think these aren't the objects you're looking for).