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Does the unit vectors carry any unit information ?

For example, when we define a vector in position space $\vec r =(2 m)\hat x + (3 m)\hat y$, the length of the vector has some unit, in this case meters, but when we talk about the unit vectors, say $\hat x$, does it have any unit ? If so, how does not affect unit of $\vec r$ ? If it doesn't have any unit, how can it has any meaning ? I mean, saying "1 in the direction of $\vec x$" does not have any meaning.

Our
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  • What is the reason for the close vote ? – Our Oct 09 '16 at 10:29
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    (1) no context; (2) how can vectors be associated with information in the sense of entropy or data? – Parcly Taxel Oct 09 '16 at 10:46
  • @ParclyTaxel (1)I think the question has the necessary context, it is a clear question.(2)For example, in a position space, w can define a position $\vec r= (2 m )\hat x + (3 m)\hat y$, so I think in this example a vector is associated with information if I understand what you mean by saying associated. – Our Oct 09 '16 at 10:51
  • (1)Even if I'm mistaken, the purpose of asking question is to learn something that you don't know, so you can say what is wrong or missing in order to answer this question, so that I can understand the insight of my question better. – Our Oct 09 '16 at 10:58
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    What kind of "unit information" are you talking about? What sort of "unit vectors" are you concerned with? I guess they must be part of some normed vector space, but how abstract/general do you want to take this? – user642796 Oct 09 '16 at 11:31
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    @arjafi I hope the last edit answers your question. – Our Oct 09 '16 at 11:40
  • If a vector has length $5$ meters, then you need to divide it by five meters to get a unit vector. In that sense a unit vector is void of any units - it's length is a pure, dimensionless number $1$. – Jyrki Lahtonen Oct 09 '16 at 11:48
  • @JyrkiLahtonen Could you justify your answer ? – Our Oct 09 '16 at 11:49
  • Hard to justify in ways other than reiterating that the length of a unit vector is a pure number. A physics teacher may be able to make it clear. The point here is in the sense that a unit vector should be indifferent to a choice of units. In other words, users should be able to agree on unit vectors even if one uses meters and the other feet to measure physical distances. I'm optimistic about somebody else being able to make this clearer. – Jyrki Lahtonen Oct 09 '16 at 11:55
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    @Leth: I'm not aware of any mathematical theory of physical units (although it probably exists), but I'd say the unit vector, being unit-less, lives in a different vector space than the space vector. I think it is possible to consider unit-carrying scalar quantities as one-dimensional vector spaces over $\mathbb R$, and then consider vectors with physical units as tensor product of $\mathbb R^3$ and the physical unit vector space. That probably can be made rigorous, using equivalence classes to deal with things like inverse units, but not in the space of a comment. – celtschk Oct 09 '16 at 12:02
  • I think this is a better suitable question for physics stackexchange.
    That said, I disagree with Jyrki. Putting units on the scalars would feel weird for some reasons in my mind. For example, we could multiply a "length" vector $x$ by $2$ and get $2x$ having $m^2$ as unit. Furthermore, multiplication by scalar feels "non-unit-dependent". Scaling something by $2$ does not depend on units. That said, a unit vector is a vector divided by a scalar. Therefore, in my head, it has unit, the same as the vectors. However, I am speaking on my gut only.     – Aloizio Macedo Oct 09 '16 at 22:07
  • http://math.stackexchange.com/questions/571050 formalizes something similar to @celtschk 's comments. – zyx Oct 10 '16 at 17:57

1 Answers1

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In what, IMO, is the best way to think about this sort of thing....

Scalars are dimensionless

The important point being that the product of two scalars is a scalar, and if we are attaching units to things, being dimensionless is the only way this can work.

Types of units are one-dimensional vector spaces

For example, you might have a one-dimensional vector space of lengths. Some elements of this vector space would be "meter", "kilometer", "inch", and so forth. Of course, you have scalar multiplication, so you also have elements like "3 meters".

The tensor product lets you multiply vectors in general, and that is how you combine units. e.g. the tensor product of a space of times with a space of speeds would be the space of lengths. e.g. on individual vectors,

$$ \mathrm{second} \otimes (\text{meters per second}) = \mathrm{meter} $$

This combines well with scalar multiplication:

$$ (3 \text{ second}) \otimes (10 \text{ meters per second}) = (3 \cdot 10) \text{ meter} $$

You really don't want to combine units and unit vectors

For example, you might decide that $\left( \frac{3}{5} \text{meter}, \frac{4}{5} \text{meter} \right)$ should be a unit vector, because $\sqrt{(\frac{3}{5})^2 + (\frac{4}{5})^2} = 1$.

However, $\left( \frac{3}{5000} \text{kilometer}, \frac{4}{5000} \text{kilometer} \right)$ is the same vector, and you probably don't want to call that a unit vector.

Vectors with dimensions still make sense

If you have the usual (dimensionless) copy of $\mathbb{R}^3$, you could take the tensor product with, say, lengths, and then you'd something you could think of as a three dimensional space of vector distances.

Because the space of lengths is one-dimensional, the tensor product acts a lot like scalar multiplication. For example, if $v \in \mathbb{R}^3$ is a vector, then $v \otimes \text{meters}$ would be a vector distance; of course, we usually just write the product with juxtaposition rather than $\otimes$: i.e. as $v \text{ meters}$.

Unlike vectors with dimension, dimensionless unit vectors are a reasonable notion to talk about (if we equip $\mathbb{R}^3$ with an inner product).