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Why is the high-school definition of a vector as "a quantity with a magnitude and a direction" incomplete? For example, Griffiths Introduction to Electrodynamics book says:

The definition of a vector as "a quantity with a magnitude and direction" is not altogether satisfactory. (section 1.1.5 "How vectors transform")

However, I am not very satisfied with his chain of arguments.

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    $\begingroup$ Who says it is? Personally, I think it's a fine definition. If you want to do rigorous math with vectors, you need to do a little more work to define them, but the math definition of a vector is just a axiomatization of the intuitive definition. $\endgroup$
    – d_b
    Oct 8 at 18:23
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    $\begingroup$ @mithusengupta123 Vectors don't change under coordinate transformations; their components do. Applying the high school definition of a vector, if you have an arrow with a magnitude and direction, and if you transform the coordinate axes (say, scaling or rotating them), then the components of the vector must transform correspondingly. $\endgroup$
    – d_b
    Oct 8 at 18:34
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    $\begingroup$ A "better" definition would not benefit very many high schoolers. The definition is a suitable approximation of a better definition. It's perfectly adequate for the job at hand. Don't we do that every waking minute as a physicist? $\endgroup$
    – garyp
    Oct 8 at 18:43
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    $\begingroup$ "However, I am not very satisfied with his chain of arguments." We could help you with that if you quoted or summarized them. $\endgroup$
    – J.G.
    Oct 8 at 19:27
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    $\begingroup$ I'd say it's a definition that is incomplete, because you're not told what a direction is as a mathematical object. The high school student knows about $\mathbb{R}^2$ and will make a link between "direction" and the notion of an angle wrt some axis, but this is very restrictive. $\endgroup$
    – Joce
    Oct 8 at 19:56

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It's almost perfectly satisfactory. First, there are three ways to introduce vectors:

  1. Geometric
  2. Algebraic
  3. Transformational

All three definitions are important and it's important to understand how they relate to each other. The first two ways are common in high school mathematics. The geometric way is magnitide with direction and the second is axiomatically as an element of a vector space.

Griffiths says that the geometric definition is unsatisfactory so as to introduce the transformational definition. This is important because this is how tensors are traditionally introduced. This is not often simple to see as at the same time tensors are introduced as tangent tensors, so there is additional structure that complicates the picture. But at this level the structure is not neccessary.

However, in Griffiths electrodynamics tensors are not used. So why does Griffith introduce the transformational definition? This is because he wants to introduce axial vectors otherwise also known as polar vectors or pseudo-vectors. The magnetic vector is not an ordinary vector but an axial vector. In fact, angular momentum and torque are also axial vectors.

To define an axial vector actually requires expanding the notion of a vector to include a representation of the vector space. The representation required for an axial vector is called the sign representation, the simplest representation.

So Griffiths is right: the definition of a vector - and not just the 'high-school definition' but all three definitions - is not sufficient, an expansion is neccesary to include a representation to properly model physical phenomena like angular velocity, angular momentum and the magnetic vector. This is often not made explicit and results in quite a bit of confusion.

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    $\begingroup$ I certainly wish someone had explained this to me in Electricity and Magnetism. $\endgroup$
    – phoog
    Oct 9 at 10:09
  • $\begingroup$ (+1) minor correction: polar vectors are the usual vectors with direction and magnitude, axial vectors reverse direction under inversions. By "representation of the vector space" here I think you are referring to a representation of the vector space automorphism group, correct? Then polar vectors transform in a trivial representation and axial vectors transform in a sign representation (I think that makes sense) $\endgroup$
    – Kai
    Oct 9 at 14:30
  • $\begingroup$ @kai: If you check the Wikipedia entry for polar vectors you will see that they describe them as I have done. What's your source for describing as the same as usual vectors? I agree with your second point, that I ought to have mentioned the group represented. Thus your final statement should say ordinary vectors transform in the trivial representation and axial and polar vectors transform in the sign representation. $\endgroup$ Oct 10 at 0:40
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Why is the high-school definition of a vector as "a quantity with a magnitude and a direction" incomplete?

The high school definition of vectors applies to Euclidean vectors, or, more generally, to vectors within an inner product space. However, the mathematician’s definition of vectors and vector spaces allows for vectors that possess neither magnitude nor direction.

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  • $\begingroup$ I guess most of us were told that there is a unique vector associated to each pair of points. So what we really consider in high school are euclidean spaces, i.e affine spaces with an inner product on the translation space. $\endgroup$
    – Filippo
    Oct 8 at 20:32
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    $\begingroup$ This answer is true, although I'm not sure it captures the point that Griffiths is making. $\endgroup$ Oct 8 at 21:25
  • $\begingroup$ @Filippo If vectors are defined in terms of pairs of points, then they form a group that is acting on Euclidean space, with translation being the group action. $\endgroup$ Oct 9 at 3:01
  • $\begingroup$ Adding onto what @kleingordon has said, I think only Mozibur Ullah from this answer section has answered from Griffith's POV $\endgroup$ Oct 10 at 6:33
  • $\begingroup$ @Acccumulation As far as I understand, we agree on the following: An affine space consists of three things, a set $A$, a group $G$ and a right action $A\times G\to A$. $\endgroup$
    – Filippo
    Oct 10 at 9:21
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What Griffiths is getting at is that a vector can be thought of as more than just a single list of coordinates that specifies a magnitude and direction. Rather, it is an object that exists independent of a chosen coordinate system. As such, it can be represented by an infinite number of different lists of coordinate values, depending on the chosen basis (I'm assuming we're working with vector spaces over infinite fields like the real numbers). Viewed this way, it is the knowledge of how these different coordinate lists transform into each other to describe the same object that defines the vector. This notion of coordinate transformations is not typically emphasized in the so-called high school description.

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This was originally supposed to be a comment but I think in the context of the question, specifically Griffiths Introduction to Electrodynamics , it may well be the answer also.

Intuitively, yes a vector is "a quantity with a magnitude and a direction" but there's an issue rising due to the nature of the vector that i believe no one has pointed out so far— Vectors do not add up following algebraic rules. They have their own vector addition rules that needs to be followed.

I will present a counter-example to the "high-school definition", but before that lets take a classic "high school" example of vectors. Consider two forces acting on an object. Let one of them be 4N and the other 3N. If the forces are parallel, perpendicular and antiparallel then the Net force on the object would be 7N, 5N,1N respectively.

This example is important, because it illustrates the angle depends of the vectors. In other words the effect of a vector always changes of the angles is changed.

Now, let's take an example in electrodynamics that Griffith would approve. Let's consider a wire carrying current 5A from South to North and another wire carrying current 4A from East to West.

Now let the wires meet at a junction and then join together in the North-West direction (ill add a pic later). What is the current carried by the resulting wire? Without a doubt it is always 9A. This is because current adds linearly as a result of Kirchoffs current rule rather than "vector"ly. Actually, regardless of where the resulting wire is pointing to, the resultant current will always be 9A. If it is not, then that would imply that the conservation of Charge is violated! The conservation of charge is after all the basis of Kirchoffs first rule.

Does the current(s) taken here have a magnitude? Yes. Do they have a direction? Yes. Are they vectors? No, because this particular physical quantity does not depend on the angles as charge is a scalar and it should always be conserved. I suspect this is what Griffiths was coming to when he said the high school defintion is incorrect.

In conclusion, a vector following vector addition is important. It implies that the nature of the vector depends on the direction also. If the quantity does not follow vector addition, then it is not a vector even if we can assign a direction to it.

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    $\begingroup$ Very nice counterexample! $\endgroup$
    – Prallax
    Oct 9 at 6:59
  • $\begingroup$ I think it's a bad counterexample, if not outright wrong. Current is a vector (well, a vector field). It's the $\mathbf J$ in Maxwell's equations. Kirchhoff's rule comes from the integral form of the continuity equation, in the special case that it can be split into discrete parts corresponding to wires. It's not an accident that current has magnitude and direction. It is a vector. Also, why would thinking of vectors as magnitude and direction interfere with understanding vector addition via the triangle rule? $\endgroup$
    – benrg
    Oct 10 at 0:50
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    $\begingroup$ Current is a scalar. Current density is a vector. Current has only a magnitude (and a sign). Current density has a magnitude and direction. $\endgroup$
    – Andrew
    Oct 10 at 1:33
  • $\begingroup$ @Andrew You're right of course, but the answer seems to want to have it both ways. The current through an oriented loop is a scalar, but it doesn't have a well defined direction beyond "through the loop". The flow of charge has a well defined direction at each point, but it is described by a vector at each point. There isn't any notion of current/current density that both has a vector-like direction and isn't a vector. $\endgroup$
    – benrg
    Oct 10 at 5:54
  • $\begingroup$ @benrg well I gotta first do the compulsory "I am a HS student, so apologize if I got something wrong". Secondly, isn't J referring to current density (which i agree is a vector) Thirdly, I learnt that Kirchhoff's current/junction rule comes from conservation of charge. I will look up the continuity equation too since you mentioned it. As for your last statement, My teacher taught me current isn't a vector as it doesn't obey vector addition. $\endgroup$ Oct 10 at 6:41
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The main problem with "magnitude and direction" is the follow-up question: "What's a direction?" It's something that can uniquely be defined by a unit vector. Not very satisfactory, but no worse than, "A vector is an element of a vector space".

The physics we're talking about is formulated in Euclidean space, aka ${\mathbb R}^3$, and physical laws are presented as relationships between geometric objects that represent the symmetries of that space. There are several objects that do that, the scalar for instance. Since it is unchanged by rotations in space, its representation is trivial.

The simplest non-trivial geometric object is the vector, in fact in the form of spherical vectors, which are eigenstates of rotations, it is the fundamental representation of the rotational symmetries of $SO(3)$, the group of rotations in ${\mathbb R}^3$. That's a bit much for high school, so we're introduced to them in their Cartesian form, $(\hat x, \hat y, \hat z)$, where they have the interpretation as orthogonal unit arrows from which any and all vectors can be constructed. Foregoing that as too mathematical, we're left to describe them as things with magnitude and direction, and that does capture their essence.

Magnitude is not a difficult concept: if you have a velocity, $\vec v$, then the idea of $2\vec v$ is intuitive. Every student should have an intuitive idea of "direction", and the fact that if you rotate 180 degrees, you're now in "the opposite" direction, or if you rotate a full 360 degrees, your direction is unchanged is indeed a defining property of vectors. Moreover, it's easy to see why there are 3 independent basis vectors.

Compared with other geometric objects that represent the symmetries of space, that's no so bad. How would you describe a natural 2nd rank tensor? It has an alignment, but not a direction, it has magnitude....it can also have bulge. Moreover, there are 5 basis tensors. Not very intuitive.

Meanwhile, spinors have a magnitude and a direction...and a sign, so it's more than just a vector....but there are only 2 basis spinors? How does that work out?

So "magnitude and direction" isn't so bad.

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  • $\begingroup$ Why do you say that a 2nd rank sensor has 5 basis elements? I'd say 9 (3x3) if we are in 3D space $\endgroup$
    – Prallax
    Oct 9 at 7:28
  • $\begingroup$ @Prallax Natural form means trace-free and symmetric so a rank-$l$ natural form tensor as $2l+1$ DoF. Which is 5 for $l=2$. The Trace (1 DoF) transforms as a scalar (isotropic $\delta_{ij}$), and the antisymmetric part rotates like a vector (cross product). This is why you'll see: ${\bf 3}\otimes {\bf 3}={\bf 5}\oplus{\bf 3}\oplus{\bf 1}$ sometimes. $\endgroup$
    – JEB
    Oct 9 at 13:53
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    $\begingroup$ For me, I think the main deficiency is that there is no reference to the parallelogram rule for adding vectors, as mentioned in other answers and comments. A vector space has this parallelogram rule before additional structure (like a dot-product) provides a magnitude to vectors. Certainly one can scalar multiply and compare two vectors along the same line without the assignment of a real-valued magnitude to a vector. But without additional structure, one cannot compare magnitudes of vectors in different directions. I don't mind using "magnitude"... but I must have the parallelogram rule. $\endgroup$
    – robphy
    Oct 9 at 16:21
  • $\begingroup$ @robphy I think the magnitude comes before the dot product, as one can declare: $a\cdot b \equiv \frac 1 4[||a+b||^2 - ||a-b||^2]$ as long as there is some $||a||$. $\endgroup$
    – JEB
    Oct 9 at 18:37
  • $\begingroup$ Sure. I'm not trying to specify the minimal structure needed. My main point a vector space is not equipped to define a "magnitude of a vector", extra structure must be imposed... somehow, it doesn't matter. And, in general, different structures lead to different assignments of magnitudes... as we see in special relativity vs Euclidean geometry. $\endgroup$
    – robphy
    Oct 9 at 18:54
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I don't have the book to read Griffiths arguments, but what causes confusion on that high school definition is when we start to use vectors in curved surfaces. For example, if a sailor keeps the same speed and the same route according to the compass (say $221^\circ$), its velocity vector apparently is always the same, because it has the same magnitude and direction. But it is not true, what is more evident close to one pole.

The notion of basis vectors, and how they can change point to point is also important for their concept. However, when first presented in cartesian coordinates, the notion of basis vectors seem superfluous.

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  • $\begingroup$ "But it is not true, what is more evident close to one pole" I don't understand, can you please explain? $\endgroup$
    – Filippo
    Oct 11 at 17:28
  • $\begingroup$ Suppose you are very very close, say 100m from the North pole. And start moving west all the time at constant speed. While you are clearly in a circular motion, and the velocity vector is changing continuously, the route (west) points always to $270^\circ$ $\endgroup$ Oct 11 at 17:45
  • $\begingroup$ Thank you! If we consider some mass in a circular motion and decrease the radius while keeping the velocity constant, then the acceleration increases, i.e. the direction of the velocity changes faster. Is this what you were trying to explain? $\endgroup$
    – Filippo
    Oct 11 at 18:15
  • $\begingroup$ It is more than that. The velocity vector has always the same components, but the basis vectors (North and East) are changing all the time during the route. So, the velocity vector is changing indeed, because it is a function of the components and of the basis vectors. $\endgroup$ Oct 11 at 19:05
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$\def\A{{\bf A}} \def\N{{\bf 0}}$Here is a simple-minded reason that this definition is not satisfactory (having nothing to do with transformations): the zero vector has no direction. Note that if $\A\ne\N$, then $\A = |\A|\hat \A$, where $|\A|$ is the magnitude of $\A$ and $\hat \A = \A/|\A|$ is the unit vector in the $\A$ direction, i.e., where $\hat \A$ is the direction of $\A$. (Note that $|\hat\A| = 1$ by this definition.) Thus, every nonzero vector is a quantity with magnitude and direction. In fact, it is the product of these quantities. Now note that since $|\N| = 0$, the direction of the zero vector is undefined.

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  • $\begingroup$ While not strictly a bad answer, zero vector doesn't have any physical purpose/significance and I heard it is a purely mathematical term $\endgroup$ Oct 9 at 14:46
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    $\begingroup$ @AdilMohammed: A velocity or net force of zero is not physically significant? $\endgroup$
    – user26872
    Oct 9 at 15:14
  • $\begingroup$ Is there even acceleration, if it is zero? I can't find the original post, but there was a very good one revolving around that idea $\endgroup$ Oct 10 at 13:21
  • $\begingroup$ @AdilMohammed: Did it happen to be written by @SirIsaacNewton? ;) $\endgroup$
    – user26872
    Oct 12 at 0:26
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Good question; many basic discussions of vectors do not clearly explain this, in my opinion. The definition you provide is for a "free vector"; specifically, one that is dependent only on direction and magnitude, and not its absolute location relative to the origin of a coordinate system. Engineering developments distinguish among free, bounded, and sliding vectors. But, physics developments generally treat vectors as free; for example, in developing the equations of motion in a non-inertial frame that undergoes both translation and rotation relative to an inertial frame, the vectors are free vectors. Of course the effect of a vector depends on its location; for example, for a force to act on a particle the force vector must be located at the position of the particle. The text Introduction to Vector Analysis by Harry Davis has a good discussion of vectors in physics treated as free vectors. Also the text Mechanics by Symon discusses free vectors used on physics.

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    $\begingroup$ Ahah, of course there are specialized uses of the notion of vector. We could also say that a vector may be a virus...! No offense meant, but I believe the OP is rather clear that his question is about the word vector as in vector space, that is, what you call a free vector in engineering. $\endgroup$
    – Joce
    Oct 8 at 20:11
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In addition to this answer , here is another simple argument why definition of vector is not satisfactory. Electric current in simple circuits has both magnitude (Amps in SI units) and direction (positive to negative terminal or from high to low potential). But electric current does not satisfy either the triangle law of vector addition nor the parallelogram law of vector addition. Thus it cannot be regarded as a vector.

Similarly there are many quantities (Higher dimensional) which have magnitude and direction but don't follow the usual triangle law of vector addition.

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  • $\begingroup$ Isn't electric current just a one-dimensional vector? (There are no angles in one dimension, so only trivial triangles.) $\endgroup$ Oct 9 at 23:24
  • $\begingroup$ Current is a scalar (a signed, real-valued quantity). Current density is a vector. $\endgroup$
    – Andrew
    Oct 10 at 1:34
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The definition is not satisfactory because vector spaces do not come equipped with a notion of direction or magnitude. The matrix $$ \begin{pmatrix} 1 & 1 \\ 0&1 \end{pmatrix} $$ defines an automorphism of $\mathbb{R}^2$ as a vector space, but preserves neither magnitude nor direction (understood as the angle between vectors).

To have a notion of magnitude and direction, we need a metric (inner product), i.e. a non-degnerate symmetric bilinear form.

The confusion comes from always thinking of vector spaces as having a natural inner product. But often there is no natural or even meaningful inner product. For example, the set of pairs $(\text{\$ spent on apples}, \text{\$ spent on oranges})$ is a perfectly good and even useful vector space. What should the magnitude of such a vector be? There's no meaningful answer.

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  • $\begingroup$ "To have a notion of magnitude and direction, we need an inner product" The vector space $L^2$ comes with an inner product, but I have never seen someone claim that a wave function has a direction. $\endgroup$
    – Filippo
    Oct 11 at 17:25
  • $\begingroup$ We can speak of the angle between two wavefunctions (though one seldom does), but yes, there are spaces which have an inner product and nevertheless lack an especially meaningful notion of magnitude or direction. $\endgroup$ Oct 11 at 17:29

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