# I Difference between vectors in physics and abstract vectors

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1. Aug 27, 2016

### Mr Davis 97

I am taking a linear algebra course and an introductory physics course simultaneously, so I am curious about the connections between the two when it comes to vectors.

In beginning linear algebra, you typically study vectors in $\Re^{2}$ and $\Re^{3}$. Are these the same vector spaces used in physics to represent physical quantities? Why does physics tend to emphasize the "magnitude and direction" aspect of the vector, while linear algebra emphasizes the "component" aspect of the vector?

2. Aug 27, 2016

### Staff: Mentor

Different physical quantities have different vector spaces. For example, velocity and acceleration have different vector spaces from each other. This is one reason why you cannot add an acceleration to a velocity.

As you get into more advanced physics courses you will get even more interesting vector spaces, particularly once you get to quantum mechanics where the state of a system is a vector of possibly infinite number of dimensions.

3. Aug 27, 2016

### Staff: Mentor

As long as you think of velocities or forces, yes. However, there are many, many more vector spaces used in physics: complex instead of real, infinite dimensional instead of the $2-$ or $3-$dimensional which you mentioned or simply those of higher finite dimensions. It depends on what quantities you deal with. Vector spaces are basically the objects that allow linearity: addition (subtraction) and stretching (compressing).
As linearity is easy to compute and often (not always, but often) can be taken as a good first approximation, it's an important concept. Also many of the symmetries in physics can be expressed with linear concepts.

The components are only the part of a vector with respect to a certain given direction, usually a Cartesian coordinate (5 in x-direction, 3 in y-direction or something like this). You need them to do calculations as you need rulers and their scales if you want to tell someone how long something is, and in which direction, e.g. length or height.
In the case of a force, the components might be the one pointing towards the ground and the one pointing in direction of movement. Even in physics a force will often be split into components, i.e. into the parts in certain directions.

4. Aug 27, 2016

### Mr Davis 97

Also, my physics book says that position vectors aren't "true" vectors because they only represent a point in space and cannot be moved around at will, while vectors like displacement and force are true vectors because they can move around and still be the same vector. What is meant by this?

5. Aug 28, 2016

### Staff: Mentor

If the position is only a point, then it is no vector.

However, if the position is a distance, e.g. from the origin of a coordinate system or any other certain point to another point, then it has direction, magnitude and applies at the origin, therefore it is a vector. But strictly speaking, in this case, it's not a position (point) anymore, it's a oriented distance (from to, a vector). From a physical point of view this might be regarded as being not a true vector, because it doesn't do anything: no force, no acceleration, no velocity. Mathematically, as an oriented distance, it is a vector.

What they call true vectors is simply an arrow, that can be moved around, i.e. applied at any point.
One may consider a vector more precisely as this pair {(point of application, arrow)} which is IMO not helpful at this stage. But if you read something about vector fields (e.g. here on PF), then these pairs are meant. But for now the physic textbook's description is good enough.

What is meant? Think again of a vector representing a force. Then you can apply this force to the top edge of a heavy box and it might cant. You could also apply the same force (equal direction, equal magnitude / strength) at the bottom of the box and it will move. The force itself - the vector - is the same. But even this simple example shows, that strictly speaking the pair {((starting) point, vector)} should be regarded. But it would complicate vector calculus unnecessarily.

6. Aug 28, 2016

### Mr Davis 97

Is the idea that a vector can be moved around (applied at any point) just a physical notion, or does it also make sense when strictly talking about linear algebra? For example, in linear algebra, at least in the beginning, we mainly only talk about vectors in $\Re ^2$ that emanate from the origin to a point. We don't really "move them around" or care that vectors that are not in the same position are still the same vectors. Why is this?

7. Aug 28, 2016

### Staff: Mentor

But, you move them! If you add $\vec{a}+\vec{b}$ then you actually calculate $(0,\vec{a}) + (0,\vec{b}) = (0,\vec{a}) + (a,\vec{b}) = (0,\vec{a}+\vec{b})$, i.e. you started $\vec{b}$ at the end $a$ of $\vec{a}$. See, this is far too complicated to carry the application point along all calculations. And nothing (no information) is gained here. However, the vector parallelogram means you have to kind of "move" one of the two vectors, such that the peek of one touches the tail of the other.

I only mentioned these pairs (point, vector) for two reasons:
1. It is important when dealing with vector fields. And they occur in physics. Also the point of application is in general very important in physics.
2. If I had not, you bet, someone here would have intervened and insisted on this description.

Another very important fact is the following: In physics you often deal with frames of reference. These are only different coordinate systems. And many calculations will be done by changing between two of them. Now, if you mathematically deal with vectors in one coordinate system and change to another one, then either you look at this situation by not moving the vectors and drawing new coordinate axis, or you keep the coordinates the same and, e.g. rotate the vectors instead. These are equivalent. In the latter case you have moving vectors.
Another simple example would be, if you stretch one coordinate. You could either use another scale, i.e. numbering on the axis, or compress your vectors instead.

8. Aug 28, 2016

### Staff: Mentor

Position is better represented by an affine space than a vector space. An affine space has a difference operation, and the difference between two elements of an affine space is an element of a vector space.

9. Aug 28, 2016

### sophiecentaur

A real vector would be Displacement, relative to an origin. I think.

10. Aug 30, 2016

### David Lewis

In mathematics, a vector is a directed line segment. In physics, vectors are more narrowly defined. A physical vector must not change sign when it's reflected. So, for example, a magnetic field and angular velocity, even though they have direction and magnitude, cannot be represented by vectors.

Likewise, some quantities that have direction and magnitude (such as electric current and angular displacement) are not vector quantities because they don't obey the rules of vector addition.

Last edited: Aug 30, 2016
11. Aug 31, 2016

### robphy

When one talks of "magnitude" of a vector, a dot-product [or inner-product] has been defined (probably implicitly) as additional structure to a vector space, which is where the math course begins. Vector spaces don't come with dot-products automatically... if one wishes to define an inner-product [since you aren't forced to], different choices of inner-products lead to different magnitudes.

As @Dale points out [no pun intended], to describe position it is better to use an affine space (a vector space that has forgotten its origin or doesn't distinguish its origin).

I think what your book is trying to say is that a position is really just a triple (or pair) of numbers used to locate a point in space.
While one can draw a vector with its tip at the location and its tail at a certain point (call it the origin), that vector changes when the choice of origin is changed.
Furthermore, the operation of "adding two position vectors" doesn't make sense because the result depends on the choice of origin.
What such a vector would represent is a "displacement vector from a distinguished (but likely arbitrary) point".
Given two such positions, one can form a "displacement vector from one point to another", obtained from a difference of points in the affine space... and this result doesn't change if the origin is changed.

As @fresh_42 raises, the idea of a vector being "moved around" is really the issue of "bound vectors" (which are tied to a point [e.g. applying a force at certain spot on an extended object]) vs. "free vectors" (which are not). In addition, your space must have enough symmetry in it to allow you to "move around" ("parallel transport") the vector.

Technically speaking (and this isn't often noted), the vector space of displacements is different from the vector space of forces. First they have different units and are never added or subtracted from each other [in the other space]. Often, we rescale one [like the force vectors or electric field vectors] to make the diagram nicer. When one describes the work done on a particle, one often uses a dot-product... but one has to do this carefully (with some additional structure) to get it right [with no handwaving] in a purely mathematical construction.

12. Aug 31, 2016

### Staff: Mentor

I am glad you mentioned this. I had forgotten it, but you are correct. The basic structure of a vector space are the scalar multiplication and vector addition operations.