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A physics textbook that I use mentions that "If a physical quantity has magnitude as well as direction but doesn't add up according to the triangle rule, it will not be called a vector quantity". The paragraph further explains that electric current is not a vector quantity because there is no meaning of triangle law there. While I understand why electric current is a scalar quantity, I am unable to comprehend how definite the rule is. For all I think, in the case of a right angled triangle, the rule is just calculating the hypotenuse (resultant vector). In dimensions two and above, it would make complete sense. However, in one dimension there isn't any angle involved and each point depends on only one real number. The two directions: left and right can be dealt with two sets of real numbers, i.e., negative and positive.

I want to know why vectors in one dimension can't be written as scalars. Why they must be written in the form of $x\hat{\imath}+0\hat {\jmath}+0\hat k$ when they can be mentioned simply as the scalar number on the $x$ axis.

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  • $\begingroup$ If $V$ is a $K$-vecor space of dimension $1$, then $V\cong K$, where $K$ is a field (of scalars). $\endgroup$ Aug 9, 2023 at 14:51
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    $\begingroup$ Physicists and mathematicians often disagree in definitions, scalars (either real or complex, or even in any field) are obviously vectors in the scalar field when that field is seen as a vector space over itself of dimension one. Is this useful? Maybe not. $\endgroup$
    – William M.
    Aug 9, 2023 at 14:58
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    $\begingroup$ Your text points out a specific case of when something is not a vector quantity. Does it also have a precise definition of what it does consider a VQ? $\endgroup$
    – user170231
    Aug 9, 2023 at 15:19
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    $\begingroup$ What do they mean by current not adding according to the triangle rule? $\endgroup$ Aug 9, 2023 at 15:45
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    $\begingroup$ In physics one likes to consider how quantities transform under rigid transformations (rotations, reflections) of the coordinate system. Scalars (like mass) are invariant. Vectors (like velocity) transform in a particular way; for instance, when ${\bf x} \rightarrow -{\bf x}$, then ${\bf v} \rightarrow -{\bf v}$. From that perspective, even in one dimension, vectors and scalars should be distinguished. $\endgroup$
    – mjqxxxx
    Aug 9, 2023 at 15:49

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I think the crux of the issue lies in the different purposes of mathematics and physics as fields. Let me explain.

In mathematics everything is crystal clear. A vector $v$ is an element of a vector space $V$ which to qualify as such must obey certain axioms. Associated with a vector space there is a scalar field $K$ whose elements are known as scalars. Again, scalars are required to satisfy certain properties by both the field and vector space axioms. From the mathematical perspective a vector is an element of $V$ and a scalar is an element of $K$. Full stop. Nevertheless, it is true that one can build a vector space as $\mathbb{R}$ whose field is $\mathbb{R}$ itself. Therefore, at least in this sense, in mathematics it possible to think of real numbers as vectors.

The purpose of physics is to build mathematical models which behave as objects in reality. Now the choice of model is far from unique. It is a quote of Feynman that a good theoretical physicists should always have multiple distinct models of the very same physical phenomenon at disposal. Therefore, one can build a mathematical model of reality in which electrical current is represented by a 1D vector. Similarly, one can build an analogous model in which electrical current is represented by a scalar. The difference is often simply usefulness. A model might be better at explaining something or more intuitive to understand than others but no model is intrinsically better than other as long as they both agree with empirical data.

Now to your question. Your book has a very practical definition of vector which current does not satisfy. When two electrical currents meet at an angle they do not behave as vectors. Instead they add as scalars. To calculate the resulting current out of a junction we simply perform an algebraic sum of the current distinguishing only between currents going into the junctions and out of the junction (irrespective of the angles). Therefore, according to the definition of your book, current is not a vector. Nevertheless, nothing stops you from building a different model of reality in which current is represented by a 1D vector. Such a model would distinguish current going into the junction with positive real numbers and current going out of the junction with negative ones. You might find that this model is better for some things and worse for others, but it is certainly possible.

Therefore, to answer your question, yes, you can think of current as a 1D vector but there is no obvious advantage in doing so and it should only be a matter of convention and definitions.

PS: I realize this is somewhat of a rewording of a previous answer. I will still leave it as an alternative to it.

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    $\begingroup$ A tiny correction: I think that you want "rewording" rather than "rewarding" in your final paragraph. I guess that a spellchecker thought that "rewarding" was more likely. (I upvoted it anyway.) $\endgroup$
    – badjohn
    Aug 10, 2023 at 8:30
  • $\begingroup$ You are right, I definitely want "rewording". I will edit it. Thanks! $\endgroup$ Aug 10, 2023 at 16:00
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This is one of those things that is always confusing, and you will find lots of examples where the way people refer to certain objects is ambiguous, particularly so in Algebra and its sub fields.

A vector is a member of a vector space over a field (numbers, things you can multiply vectors by which also have an inverse, apart from the zero element). This is essentially an abstract construction meant to formalize a notion of adding, subtracting, scaling them up or down, in a way that mimics displacement or position vectors in analytical geometry, the little arrows that you are introduced to as vectors. The "triangle law" which your textbook refers to is the addition rule for these kinds of vectors, and it says that the vector sum of two displacement vectors is the hypothenuse composed of "following one vector, then the other". But this of course only works when you can visualize your vectors in some cartesian space and its a bit ambiguous seeing as triangles only exist from dimension two onward.

This is not the right answer; a scalar is not something which doesn't follow the "triangle law". A scalar is a member of the field over which you have constructed your vector space. So the distinction of scalars and vectors is somewhat arbitrary; it depend on what you decided to call your vector space, and over which field you've decided to construct it on. For example, in highschool electromagnetism, you construct a three-dimensional cartesian vector space over the reals; this means that you want your position, velocity, acceleration vectors to be able to be multiplied by real numbers. So in this context we see that electric current is not a vector, seeing as it is a real number and we have decided to construct a vector space over the reals; so $I$ is a scalar in this theory, in this context; or rather, you've decided to model it as a scalar, as the rate of passage of charge by a certain area over a certain time.

But of course nothing stops you from constructing a vector space of real numbers over themselves; in this context, you could call $I$ a vector, seeing as for all intents and purposes it belongs to a vector space. This would just be another way to model the problem, to formulate it.

So which is it, scalar or vector? This isn't the right question. What you want to ask is, what better suits the system I'm studying? To build a displacement vector space over the reals and call $I$ a scalar or to build a vector space with the reals over themselves and call $I$ a vector? And you will most likely find that the first one is better, in the context you're studying. But I stress this, it is a matter of context. In higher level electromagnetism, the electric current is modeled as a three dimensional kind displacement vector, by following the paths of charges and seeing how fast they cross a certain area (and these charges follow three dimensional paths, so).

In conclusion, how do you know what is a scalar or a vector? Does it have a meaningful way of multiplying the things you decided were vectors, for example, velocity, position? Then those are scalars. But as you can see, it is mostly a matter of subtext.

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Just using matrix multiplication if we take two column vectors $v,u$ then $v^Tu$ should be a $1 \times 1$ matrix. However this is also exactly the same calculation as the Euclidean inner product, or dot product. So while technically a field $\mathbb{F}$ and the $1 \times 1$ matrices $M_1(\mathbb{F})$ are different there is an obvious isomorphism $\varphi: \mathbb{F} \rightarrow M_1(\mathbb{F}) $ given by $\varphi (x) = [x]$. The additional notation for $[x]$ makes $x$ the prefered way to write them, so as scalars.

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Yes!

No of componets of a tensor of rank r are $3^r$. A scalar is zero rank tensor so it has one component, a vector has 3, matrix has 9 and tensor has 27 components to be put in a Rubik's cube.

Similarly, a vector in 1 dimension has $1^1=1$ components and a scalar in any dimension has $(dim)^0=1$ component. So they are the same type.

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