# Why aren't all 3 numbers a vector?

1. Feb 2, 2010

### LucasGB

Dear fellas, please have in mind that during this exposition I am thinking of simple 3D physical vectors:

I know not all 3 numbers form a vector. In addition to being 3 numbers, they must transform correctly under a change in coordinate systems. I can superficially understand why this is true. For example, my age and the age of my two brothers are numbers, which could be the components of a vector, but they are not, since they don't transform correctly under rotation, they don't get "mixed up". (I don't get older, and my brother doesn't get younger when the coordinate system changes.) But I'm trying to understand this on a deeper level. I don't want to merely say "vectors are 3 numbers that transform correctly under Lorentz transformations", I want to be able to find a deeper principle, such as, for example, "vectors are 3 numbers, which, by themselves, represent a quantity which has direction", or "vectors are three numbers that have some form of dependence to space" or something like that.

In summary: how can I state this idea (precisely) not in terms of "numbers that behave in a certain way under an operation" but in terms of "numbers that are such a way". I hope I expressed myself well enough to be understood. Thank you for your help!

2. Feb 2, 2010

The most precise definition for a vector, believe it or not, is to say that it's an element of a vector space. To know what your vector is and what it represents, you have to know all about the vector space. I suggest you look up a definition.

As for the three numbers, I don't quite understand what you're trying to say. For example, (3, 4, 5) is most certainly a vector in R^3.

3. Feb 2, 2010

### D H

Staff Emeritus
Transforming correctly is not a general characteristic of what constitutes a vector.

As radou already mentioned, a vector is simply an element of a vector space. In short, a space V is a vector space over a field F if
• Any two elements of V can be added. Vector addition is commutative and associative. Mathematically speaking,

\aligned \mathbf u + \mathbf v &\in V \\ \mathbf u + \mathbf v &= \mathbf v + \mathbf u \\ \mathbf u + (\mathbf v + \mathbf w) &= (\mathbf u + \mathbf v) + \mathbf w \endaligned

• There is a unique element of V, the zero vector, that when added to some other element of V yields that other element.

$$\boldsymbol 0 + \mathbf v = \mathbf v$$

• Every element of V has an additive inverse.

$$\mathbf v + (-\mathbf v) = \boldsymbol 0$$

• Any element of V can be scaled multiplicatively by a member of F. Scalar multiplication is fully distributive and is consistent with field multiplication.

\aligned a\mathbf v &\in V \\ (a+b)\mathbf v &= a\mathbf v + b\mathbf v \\ a(\mathbf u + \mathbf v) &= a\mathbf u + a\mathbf v \\ a(b \mathbf v) &= (ab)\mathbf v \endaligned

That is all there is to vectors in general. Nothing about length, nothing about angle, and most certainly nothing about transforming properly. Normed vector spaces add some concept of length to the overall concept. Inner product spaces add some concept of multiplying vectors to form a scalar. Combining these two concepts, norm and inner product, yields a way to identify the angle between two vectors.

Transforming properly is more a property of tensors than vectors.

4. Feb 2, 2010

### HallsofIvy

I believe LucasGB is asking specifically about "physics vectors" not the more general mathematical concept of "vector space". The best way to talk such vectors is not to talk of "numbers that behave in a certain way under an operation" or in terms of "numbers that are such a way", but of the vector itself, independent of its "numbers". If v is a vector then if the equation "v= 0" is true in one coordinate system, it is true in all coordinate systems. The "coordinate system" is where the numbers come from. If those numbers are all 0 in one coordinate system, then they are 0 in all coordinate systems. Of course, you can choose different coordinate system that correspond to a translation or rotation, etc.

And, if the equation u= v is true in one coordinate system, then u- v= 0 is true in that coordinate system and, as long as u and v are vectors, u- v= 0 so u= v in all coordinate systems. That immediately generalized to "tensors" of all orders.

5. Feb 2, 2010

### Fredrik

Staff Emeritus
I had teachers who used that definition too. It's insane that people are still using it. It's very awkward and weird compared to the modern definitions, and even worse, the people who use it never state it in a way that makes sense.

If you're only given three numbers, or three of anything really, there's no way to tell how they "transform". It doesn't even make sense to say that the transform, or that they don't. In order to be able to tell if you have a vector or not, you must have been given three real-valued functions for each coordinate system. If you have, then you can check how the three functions associated with one coordinate system are related to the three functions associated with another.

My teachers didn't tell me this, and it frustrated the hell out me. It still makes me angry when I think about it. (If I seem a bit irritated, it's because of that. It certainly isn't your fault).

The problem here is that there's no natural way to associate your ages with any of the coordinate systems, while you would need a way to associate three numbers with each coordinate system. You can however always define a vector as "the vector that has components (x,y,z) in this coordinate system". Now the components in all the other coordinate systems are determined by the transformation law.

What I said above is the key to understanding the absurd definition that your teachers are using. But there is no need to use their definition. The set $\mathbb R^3$ with the usual definitions of addition, and multiplication by a real number, satisfies the definition of a vector space. So it is appropriate to call the members of $\mathbb R^3$ "vectors" (as long as you stick to the standard definitions of addition and multiplication by a real number). How the vectors transform under rotations is fixed by the definition of a rotation as a norm-preserving linear operator on $\mathbb R^3$. (This means that a linear operator R is a rotation if and only if it's orthogonal. If we want to exclude reflections, and we usually do, we also require that the determinant of R is =1. Such a rotation is called a proper rotation).

If you really want to understand the "transforms as" stuff, you're going to have to read a chapter or two in a book on differential geometry, but you won't really need it until you study general relativity. The basic ideas are these:

* A manifold is a generalization of the concept of a "surface" to n dimensions.
* A manifold M has a tangent space TpM at each point p, which can be given a vector space structure in a natural way.
* A coordinate system x is a function from an open subset of M into $\mathbb R^n$ (its job is to assign n numbers (coordinates) to each point).
* A coordinate system can be used to construct a basis for the tangent space.
* When a tangent vector v is expressed in terms of a basis constructed from a coordinate system, the coefficients of the basis vectors are called the components of that tangent vector in that coordinate system.
* A change of basis implies a transformation of the components.

Note that the important quantity here, the tangent vector v, is coordinate independent. The laws of physics are always relationships between coordinate independent quantities, so if we choose to express them in terms of components associated with a specific coordinate system, and we don't always use the same coordinate system, then we have to talk about how those components transform.

Last edited: Feb 2, 2010
6. Feb 2, 2010

### Tac-Tics

Who says they don't? A list of three people's ages is a perfectly good vector. It occupies a weird (possibly useless) vectorspace, but you can define pretty much whatever the hell you want in math. The catch is once you choose a definition, you have to stick with it.

You can even define "rotations" on this age vector. Rotating the vector doesn't make you younger, no. But rotating a camera doesn't make a person smaller either. You age or the height of the person have a preferred frame to measure from.

The fact whatever operation you define doesn't "feel" like it should be a rotation doesn't matter (as long as it obeys the definition). There are a few really good examples of this in math. Here's a good example.

In audio analysis, we can think of a sound wave as a function from time to volume (decibels). Knowing the waveform is useful for a few things, but what's REALLY nice to know is the breakdown of the frequencies. There are all sorts of practical uses, including noise-reduction methods used on cell phones and lossy compression used on MP3's. The mathematics of transforming from "time-domain" (the waveform) to "frequency domain" (breakdown of the frequencies) is called Fourier Analysis. It's pretty interesting stuff.

Now, that makes sense to anyone who's ever owned a stereo with mixer controls. But what's interesting is you can also apply Fourier Analysis to digital images. You look at a picture and ask yourself, "what the hell does 'frequency' even mean for an image."

Well, from a mathematical standpoint, an image looks like a higher-order sound wave. Instead of "time" you have two parameters, "width" and "height". And instead of volume, you have three components, "red", "green", and "blue". But if you've done any linear algebra, you know adding extra dimensions usually isn't so hard to do.

It turns out that when you break down an image into its base frequencies, it looks like garbage. Our brains didn't evolve the software to extract useful meaning out of two-dimensional image frequencies. However, we can still run the usual filters on it, such as noise reduction and low- and high-pass filters on it. This is why JPEGs are so small compared to BMP images -- they cut out all the low and high frequency image data that our minds can fill in. The "meaning" of the image is contained in those nice mid-level frequencies.

The moral of the story: you can define unintuitive operations on mathematical things, and sometimes, they aren't so crazy after all.

7. Feb 3, 2010

### LucasGB

I'd like to thank you all for your kind (and very thorough) replies. The reason I'm stating this question is because, in my study of electromagnetism, I've come across the del operator, and I would like to know how one can determine wether the gradient and the curl are indeed vectors.

Let's take the example of the gradient. It's three functions are the partial derivatives of the scalar field with respect to x, y and z. How would I proceed from here?

Should I follow Richard Feynman's steps? In his Lectures, he makes use of geometry to show that these three functions transform under a coordinate system change in the same way as coordinates x,y and z transform. Therefore, he concludes, the gradient is indeed a vector.

Is this the procedure one must adopt when trying to decide wether a quantity is a vector? If he finds the three functions transform in the same way as coordinates, then they are vectors? What is the underlying principle behind this procedure?

PS. Thank you very much for the insight provided in this paragraph I quoted. It totally makes sense to speak of funtions instead of numbers.

8. Feb 5, 2010

### Fredrik

Staff Emeritus
Actually you can speak of numbers too, but you would still need a specification of three of them for each coordinate system. (Plus, we're usually interested in functions, not numbers, and numbers can be thought of as constant functions anyway). It's the "for each coordinate system" part that's important. The three "things" can be members of any vector space. (If they're not members of a vector space, the transformation law doesn't make sense).

About the gradient...I don't remember how this is usually done in the context of the obsolete definition of "vectors". We clearly need to associate partial derivative operators with coordinate systems somehow (and define what a coordinate system is) before we can proceed. I suggest we do it like this:

Define a coordinate system to be a smooth bijection from $\mathbb R^3$ into $\mathbb R^3$. I will write the ith partial derivative of a function f at a point p as $f_{,i}(p)$. Now, for each coordinate system x, we define the associated partial derivative operators at p by

$$\frac{\partial}{\partial x^i}\bigg|_p f=(f\circ x^{-1})_{,i}(x(p))$$

See #5 in this thread to see how to compare the derivative operators associated with two different coordinate systems. Also read #3 to see the definitions I'm using there.

Last edited: Feb 5, 2010