# Units in vector spaces

I am confusing myself about where physical quantities become mathematical objects. Where does one end and the other begin?

E.g. displacement $$\vec{s}$$. A displacement is a physical quantity that can be measured. In contrast to a mathematical vector which is an element of a vector space, which is a set V over a field F (with accompanying properties(+, $$\cdot$$ ) and axioms). In this case, are the elements of the set V the physical displacements? In that case, the (chosen) basis vectors $$\vec{b}_i$$ would also displacements and a general displacement can be written as a linear combination of basis vectors as

$$\vec{s} = s_i\vec{b_i}$$

The scalars are then elements of the field F which must be the set of real numbers. If the basis vectors are orthonormal (cartesian) then the norm/length of the vector is

$$|\vec{s}|^2 = s_i s_i$$

but this is just a number with no units. So the units are hidden in the basis vector $$\vec{b_i}$$. So maybe we should write this as

$$\vec{b_i} = (b \cdot [1\text{m}])\hat{e_i}$$.

Here $$[1\text{m}]$$ represents our unit of measure. But now $$\hat{e_i}$$ is a vector in a different vector space (the set of orientations, i.e. not a displacement). And $$(b \cdot [1\text{m}])$$ can't be a scalar of the vector space of displacements because we already said that was the set of real numbers So the actual "measurement" $$(b \cdot [1\text{m}])$$ should maybe be thought of as a map between two vector spaces (orientation and displacement)?

And you can't bake in the units in the field F either because if you scale a vector with a scalar twice (which should be allowed in a vector space) you get the unit squared. For the same reason you can't bake it into the scalar field used for the vector space of orientations.

So when we write

$$\vec{s} = (0.1 m)\hat{x} + (0.2 m) \hat{y}$$

maybe we should think about the thing in the parentheses as a map between two vector spaces (displacement and orientation) and the number there is a product of the scalar field F (no units) and the length of the basis vector (with units). (This makes the covariance of the units and the contravariance of the components somewhat intuitive also....maybe).

In summary: confused!

In summary #2: Maybe this boils down to: If you want to think about physical quantities in a the language of mathematical vector spaces, where should you build in the units?

• The physical quantity displacement contains two pieces of information: magnitude (i.e. distance between two points using the Pythagorean theorem) and direction. Hence the physical quantity displacement is a vector. All notions in physics that are represented as vectors are actually vectors "in real life" and notions that are represented as scalars are actually scalars in "real life." For instance, forces are vector quantities because they contain magnitude (strength) and direction while energy is a scalar because it contains only magnitude.
– user261609
Commented Feb 8, 2022 at 17:43
• Btw the notion of force vectors follow the notions of mathematical vectors. Vector spaces contain rules and these rules are vector addition, subtraction, etc.
– user261609
Commented Feb 8, 2022 at 17:45

There is an excellent blog post by Terry Tao on the mathematical formulation of dimensionsal analysis.

The easiest answer to your particular question is probably to take his second approach: A dimensionful quantity $$Q$$ is modeled by an associated 1-dimensional real vector space $$V_Q$$. A choice of isomorphism of vector spaces $$u_Q : \mathbb{R}\to V_Q$$ is a choice of units for our quantity, with $$u_Q(1)\in V_Q$$ being "one" of the chosen unit. Multiplication of two quantities $$Q_1,Q_2$$ corresponds to the universal map $$V_{Q_1}\times V_{Q_2} \to V_{Q_1Q_2} = V_{Q_1}\otimes V_{Q_2}$$ into the tensor product.

Now, an $$n$$-dimensional vector that represents a vector-like quantity with dimension $$Q$$ is just an element of $$V_Q^n$$, and the choice of unit $$u_Q$$ induces a choice of basis in $$V_Q^n$$ (namely $$(u_Q(1),0,\dots,0)$$, $$(0,u_Q(1),\dots,0)$$, $$\dots$$, $$(0,\dots,0,u_Q(1))$$). If we call these induced basis vectors $$u_i$$, then we can expand every $$v\in V_Q^n$$ as $$v = \sum_i v_i u_i$$ and the $$v_i\in\mathbb{R}$$ are the numerical values in the chosen unit for the components of the vector.

• How is it ensured in this formalism that the norm of the vector-like quantity with dimension $Q$ is a scalar-like quantity with dimension $Q$? $V^n_Q$ is a real vector-space so its norm would belong to $\mathbb{R}$ and not $V_Q$, right? I might be missing something trivial (or might have just misunderstood the formalism completely).
– user87745
Commented Feb 8, 2022 at 23:29
• @DvijD.C. Just concatenate the usual $\mathbb{R}$-valued norms with $u_Q^{-1}$ to get $V_Q$-valued norms. Commented Feb 8, 2022 at 23:38
• Ah, yes. Thanks. One related question: the isomorphism $u_Q$ naturally induces a multiplication operation on $V_Q$ (i.e., $\forall v,w\in V_Q, v\cdot w\equiv u_Q(u_Q^{-1}(v)\cdot u_Q^{-1}(w))$). Equipped with this multiplication, $V_Q$ is a field. So one can just define the vector-space of $n-$dimensional vector-like quantities with dimension $Q$ to be a vector-space over the field $V_Q$ instead of $\mathbb{R}$? PS: Of course, this "multiplication" would not correspond to the usual multiplication of quantities which is a map from the cartesian product to the tensor product like you said.
– user87745
Commented Feb 9, 2022 at 0:31
• @DvijD.C. Note that we do not fix a single $u_Q$ - $u_Q$ is a choice of units that we can change at any time, and so this multiplication structure (in particular its identity) depends on the choice of $u_Q$. The whole idea is based on not identifying $V_Q$ with $\mathbb{R}$ in general by fixing $u_Q$, and representing multiplication of quantities by the tensor product instead. Commented Feb 9, 2022 at 8:46
• Thank you for a great answer. I guess I was on to something when I concluded that a measurement could be thought of as a mapping. So going to my original example we could say that I choose the 'meter stick' (when that was the SI-unit) and call that $m$ such that $u_s(1) = m = u_i$ and I can then write a displacement $s \in V_s$ of twice that length as $s= 2 u_i = 2 ~\text{m}$. Commented Feb 9, 2022 at 10:54

Maybe this boils down to: If you want to think about physical quantities in a the language of mathematical vector spaces, where should you build in the units?

I will copy from my answer here:

Physics theories are mathematical models that fit current observations/data and are predictive of new ones. Prediction is the lynch pin that separates a Physics theory from a mathematical map of the data.

The mathematical part is rigorous and the variable phase space it can explore is much larger than the phase space that physical observations define. Laws ( or postulates) are extra axioms to the mathematical axioms, to pick up the subset of the functions which are compatible with the physical data to be modeled.

It is the postulates and principles and laws ( and tables, as the tables of the standard model of particle physics) at the head of each physics theory that pick up the solutions that can model the measurements, and important, be predictive of new ones. These physics "axioms" carry the units and definitions needed to identify physical quantities with a specific mathematical formula or structure.