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I was reviewing notes of physics, and i realized that something about the mathematics of vectors was wrong in my head.

Suppose a vector is $\vec{A}=5\textbf{i} + 3\textbf{j}$, and other $\vec{B}=7\textbf{i}+3\textbf{j}$. Then $\vec{A}-\vec{B}=\vec{C}=-2\textbf{i}$.

Now, take for example a point charge vector field:

$${\mathbf {E}}({\mathbf {r}})= \ q_{i}{\frac {{\mathbf {r}}-{\mathbf {r}}_{i}}{|{\mathbf {r}}-{\mathbf {r}}_{i}|^{3}}}$$ (Dropped the constants)

This kind of vectors have a new implementation, they are attached to a point in space.

I do not know if it is the same with velocity vector.

I just want to know if the idea is wrong and what should I review.

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  • $\begingroup$ The $\vec{r}$ is position vector of a point in space. Although vectors themselves are not affected by translation, but fixed vectors like the position vector have various applications like specifying the position of a point wrt a fixed point (origin). Consider the case of force being applied on a rigid body, in that case the point of application of force does matter, as it may produce some torque. So when it comes to application of vectors, we may fix their tails in some situations. Same is the case for velocity of a rolling rigid body. $\endgroup$
    – jonsno
    May 19, 2017 at 14:02
  • $\begingroup$ If i put r (position vector) or i put x,y,z (if i am dealing with cartesian coordinates) is the same? I mean, it is the point what is important.. And something else.. Vectors are not attached to a point in space, but physicists do that for practical purposes? $\endgroup$
    – user153036
    May 19, 2017 at 14:31
  • $\begingroup$ It seems like your question depends on coordinate systems - vectors are independent of your choice of coordinate system, you can think of them as representing a change from a reference point. So if your vector adds 2 to the x position, in one system you can go from x to x+2 and (x-1) to (x+1) and the vector is the same because the displacement is too $\endgroup$
    – PhysicistAbroad
    May 19, 2017 at 14:35
  • $\begingroup$ Actually $P(x,y,z)$ is the coordinate, but $\vec{r}$ is position vector of that coordinate. Thus $\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}$. Vectors are sometimes treated as being fixed, in physics as I earlier said, and in mathematics too, for example consider the equation of line in 3 dimensions ($\vec{r} = \vec{r_o} + \lambda \vec{b}$). Here, $\vec{r}$ and $\vec{r_o}$ are taken to be fixed as they tell about location of a point in a coordinate system. $\vec{b}$ is not fixed, as it is used to scale along the line. $\endgroup$
    – jonsno
    May 19, 2017 at 14:40
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    $\begingroup$ Mathematically, you're looking for the concept of affine space. This may help you imagine things better. $\endgroup$
    – Martino
    May 19, 2017 at 16:06

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When you say "they are attached to a point in space" you are implicitly translating the concept of "function" in to "attached". The field is a function that relate a vector with another, in terms of mathematics this two vector are still free vectors like the one you talked before. It is the interpretation, the meaning you give to this vectors that leads to that sort of difference, the first, the domain vectors, are interpreted as position, while the image vector represent forces, so you are binding forces to a position, but the two vectors themselves, taken separately from the relation have nothing different from velocity vectors, it's the relation, the function that creates that "binding", a function is indeed binding couple of elements.

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  • $\begingroup$ I do not agree, position is not a vector..it can be a point. I think the vector E(r) and E(x,y,z) are the same. Why not? $\endgroup$
    – user153036
    May 20, 2017 at 0:00
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    $\begingroup$ What you call a point is generally described through vectors in physics. Velocity is a vector right? Is is defined as the derivative of another vector....position. You do the same when you write the decomposition of a point in his coordinate and versors rapresentation, or change frame. And even thinking about it as a "point" (that is not like a different entity from vectors in the same environment, it is the same concept outside of algebra) it is still the definition of a function that create the binding and not a difference in the nature of the vectors- $\endgroup$
    – Claudio P
    May 20, 2017 at 0:11
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In purely mathematical sense, you can put the origin of the electric field vector $\textbf{E}$ everywhere in the space. This is ok, actually.

However, the object which your are considering is not only mathematical but physical: Electric field vector. Remember that electric field is defined as a vector field that associates to each point in space the Coulomb force that would be experienced per unit of electric charge at a point,by an infinitesimal test charge at that point, so each E vector has its defined origin to identify the point which the source act on. That is to say, the origin of the E vector has its own physical meaning.

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