In introductory physics we are taught about position and velocity vectors. Though the representation of those vectors is identical (i.e. $x\hat{x}+y\hat{y}+z\hat{z}$) they must belong to separate vector spaces, as it would be nonsensical to add a position vector to a velocity vector. Now, when we multiply a velocity vector by time, what we get is a displacement, and this is addable to position, i.e. $\vec{x}=\vec{x}_0 + \vec{v}t$. This seems odd, as though somehow multiplying by time projects the velocity vector into the position vector space. But all we're doing is multiplying velocity by a scalar. But now we can add the result to a vector that was not in the original vector space. This is not how vector spaces work, per my understanding. What is going on when a velocity vector is multiplied by time, a scalar value, that allows the result to add with a position vector which was not part of the original vector space?


3 Answers 3


So my answer is that they are almost the same space $\mathbb R^3$ but not quite.

What is actually happening is that anything you regard as a physics scalar is actually an equivalence class of pairs $(m, s) \in \mathbb R\times\mathbb R_+$ where $(m, s)\equiv (M, S)$ just when $m/s = M/S$. The quantity $m$ is the measurement we actually see and the quantity $s$ is a scale factor which we attempt to determine through a process we call “calibrating our instruments.” (In fact there is also a third number here which we call our uncertainty about the measurement, that depends on both uncertainty in $m$ and $s$, but let's ignore that because we can kind of tack it on later just as easily as now and doing it rigorously requires Borel sigma-algebras or the like.)

So for example you measure something with a meter stick and it tells you 100, I measure the same thing with my tape measure and I get a very different number, 39 ⅜, we do not immediately think that the world is broken, rather we think that you are measuring in centimeters and I am measuring in inches. The two are equivalent once we calibrate the meter stick and the tape measure to each other, to arrive at some common base. Moreover, we expect the exact choice of base for this number to not matter much, so we kind of have in the back of our minds a scaling isomorphism separate from the equivalence relation where we could do all the same physics with $(m, \alpha s)$ for some $\alpha$.

Given this alone, $+$ is still well-defined as $P_1+P_2=\{(m_1+m_2,s): (m_1, s)\in P_1, (m_2,s)\in P_2\}$. So like if I had chosen $k s$ I would have ended up with $(k m_1 + k m_2, ks)\equiv (m_1+m_2, s),$ so my equivalence class property is not broken. But $\cdot$ is not because the same approach yields $(k^2 m_1 m_2, k s)$. There are other definitions which work like say $(tm_1m_2/(s_1s_2), t)$ but they violate the scaling isomorphism. To get back our precious multiplication operation, we have to define an even larger space.

Here we assume that there are $n$ fundamental “dimensions” for the fundamental scale factors and we consider an equivalence class of $\mathbb R \times \mathbb R_+^n$. The trick is that we don't have just one of these, we have countably infinitely many, parameterized by a $(q_1,\dots q_n)\in\mathbb Q^n$. So now $(m, (s_1, \dots s_n))\equiv (M, (S_1,\dots S_n))$ just when $${m_1\over s_1^{q_1}\cdot \dots s_n^{q_n}} \equiv{M_1\over S_1^{q_1}\cdot \dots S_n^{q_n}}.$$ Now each one space has an internal addition operation as before, but now also we have a multiplication operation which connects any two spaces, $$(m, \mathbf s)\cdot (M, \mathbf s) = (m \cdot M, \mathbf s)$$ now works precisely because this lives in $\mathbf q+\mathbf Q$ and, had you chosen $k s_i$, you would have $k^{q_i} m ~k^{Q_i} M/(s_1^{q_1+Q_1}\dots(k s_i)^{q_i+Q_i}\dots)$ and the scale factor divides out cleanly.

You may notice that this trick just requires $\mathbb N$ for the exponent rather than the full power of $\mathbb Q$ and that is true, but defining division requires $\mathbb Z$, and it is very nice to be able to invert from square meters back down to meters so closing this under $n^{\text{th}}$ roots is valuable. Closing under real roots has on the flip side no added value for physicists as far as I know. But especially in electrodynamics with Gaussian units you do get these strange square-roots-of-cm units and so forth.

So I will also say that in practice, most people use the SI units, which have $n=7$ and $q_1$ designates scaling with mass with $m_1=s_1=1$ for one kilogram, $q_2$ with length calibrated to meters, $q_3$ with time calibrated to seconds, $q_4$ with temperature calibrated to kelvins, $q_5$ with amount of substance calibrated to moles, $q_6$ with electric current calibrated to amperes, and $q_7$ with luminous intensity calibrated to candelas. So for example a speed has the vector $\mathbf q = (0, 1, -1, 0, 0, 0, 0)$ and if we want it in kilometers per hour then it has the scale factors $\mathbf s=(1, 1000, 3600, 1,1,1,1)$ and in practice we denote the both of these as $\text{km}/\text{hr}$ in one go rather than trying to write out these two abstruse 7-component vectors.

This is all to say that when you create vector spaces on top of these scalars, you have a similar situation of networked vector spaces indexed by a $\mathbb Q^n$, and while there is a field of scalars with $\mathbf q=0$ which you can lean on now, this variable $t$ is not one of them, it has a non-zero $\mathbf q =(0,0,1,0,0,0,0)$ and therefore a scalar multiplication with it takes the space $(0,1,-1,0,0,0,0)$ to the space $(0,1,0,0,0,0,0)$ which is indeed a different vector space.

We do teach all of this but never at this level of rigor, you are much more likely to just see the named units and you are told that you cannot add two numbers unless those units are exactly the same, and you are told that they multiply like numbers but are not numbers, and you are hopefully told, although I am finding this less and less, that if you have an equation among units, say $1000\text{ m}=1\text{ km},$ or $3600\text{ s}=1\text{ hr},$ that you can use that to substitute: in this case if you had a car driving at $72 \text{ km}/\text{hr}$ you can substitute both of those units as $72\cdot(1000\text{m})/(3600 \text{s}) = 20 \text{m/s}.$

One advantage of this level of rigor is that I can tell you the fundamental idea of dimensional analysis, which is that because you can't add scalars with different $\mathbf q$, and because a Taylor series of an arbitrary smooth function would add terms with different powers, you can only take an arbitrary smooth function of a dimensionless quantity $\mathbf q=0$. This in turn creates a linear algebra problem: I know that these five quantities are the only ones that are physically relevant for this problem, I know each of their $\mathbf q_1\dots\mathbf q_5$, what are the independent linear combinations which are dimensionless? Then an arbitrary answer must be, always, some combination of the five parameters which has the right units, times an arbitrary function of the dimensionless combinations.

So for example if you want to know the radius of curvature of a particle moving with velocity $v$ and acceleration $a$, well you are asking for a length and I have a length per unit time and a length per unit squared time, it has to be $k v^2/a$ for some pure number $k$. Turns out $k=1/\sin\theta$ where $\theta$ is the angle between the vectors, dimensional analysis can't tell you that, but you got 90% of the solution “without thinking about it.”


What you have discovered is that "position" doesn't really live in a vector space, it lives in an affine space. Of course there are technical abstract definitions of the two, but what it boils down to is the origin.

Vector spaces don't really have one, though there is a unique null vector, $\vec 0$, that satisfies:

$$ \vec a + \vec 0 = \vec a $$

for instance. Meanwhile, the coordinate ${\bf O} = (0, 0, 0)$ is not special, and depends entirely on your choice of coordinates.

To define the position of a point, you need to define the origin and some displacement vector:

$$ {\bf A} = {\bf O} + \vec r_A $$

One can add a vector to a point to get a new point:

$$ {\bf A} +\vec r_{B-A} = {\bf B} $$

and you can subtract two points to get their vector difference:

$$ {\bf B} - {\bf A }=\vec r_{B-A} $$

but you cannot add two points:

$$ {\bf A} + {\bf B }=?$$

in a coordinate independent way.

  • $\begingroup$ Perhaps the question should be rephrased in terms of displacement vectors rather than position vectors, then? $\endgroup$ Commented Sep 19, 2020 at 5:58

Who says they are separate vector spaces? You are thinking of quantum mechanics where position and velocity (momentum) belong to separate vector spaces. In classical mechanics this is not the case and in the example you provide above, both vector spaces are the same. You do not add position to velocity because they are different quantities and have different physical dimensions.

What you are confusing are the abstract (Hilbert) vector spaces we deal with in QM. In classical mechanics we are dealing with real actual dimensions given by the unit vectors i j k which “span” (define) physical (not abstract) space. So that we can add velocity and position (by using t multiplication) within that very same space.

  • $\begingroup$ I must be fundamentally misunderstanding something, and if you could help me understand what I would appreciate it. If position/displacement and velocity are on the same vector space then then one must be addable to the other, by definition. What would be the interpretation of such an operation? Would higher derivatives also belong to the same vector space? $\endgroup$ Commented Sep 19, 2020 at 6:07
  • $\begingroup$ Yes. What you are confusing are the abstract (Hilbert) vector spaces we deal with in QM. In classical mechanics we are dealing with real actual dimensions given by the unit vectors I j k which span physical (not abstract) dimensional space. So that we can add velocity and position (by using t multiplication) within the very same vector space. $\endgroup$
    – joseph h
    Commented Sep 19, 2020 at 6:45

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