# How the torque/moment-of-force can be mathematically defined?

Given the definition of torque/moment-of-force $\mathbf F$ applied in $P$ with respect to the pole $O$

$$\mathbf M_O=\vec{OP}\times\mathbf F$$

and given that the vectors $\vec{OP}$ and $\mathbf F$ belong to different vector spaces (we cannot add positions and forces, and they are not measured with the same units), how can we define the vector product between vectors of different vector spaces?

• Well... you are not adding position and force which wouldn't make sense. You're multiplying them, which produces a new unit Newton*meter. Mathematically, they both belong to $\mathbb R^3$, so they belong to the same space and that product is perfetly well defined. – MyUserIsThis Feb 4 '15 at 21:06
• @MyUserIsThis: this is not mathematically satisfactory. They do belong to the same vector space $\mathbb{R}^3$, but they cannot be added, it is contraddictory. – enzotib Feb 4 '15 at 21:09
• Yes it is, they are not being added, they are being multiplied. Do you have any trouble multiplying distance with the inverse of time to get a speed? This is the same thing. – MyUserIsThis Feb 4 '15 at 21:52
• Let us consider 1 dimensional vectors, i.e. scalars and let us take into consideration Ohm's law. How can we multiply a resistance with a current to get a voltage? They clearly belong to two different vector spaces, right? – Phoenix87 Feb 4 '15 at 23:52
• You all put too much attention on dimensions, I shouldn't have talked about that. Still from a mathematical point of view the problem exists, given that, as you admit, the object belong to different spaces (or sets). Well, right, the problem arises also for simpler formulas with scalar instead of vectors. – enzotib Feb 5 '15 at 11:25

Mathematicians usually are not worried about the conecpt of physical units. As such, a mathematician probably would argue that $\mathbf M_O$, $\vec{OP}$ and $\mathbf F$ belong to $\mathbb{R}^3$, as MyUserIsThis did in his comment.

If this is not satisfactory to you, you could consider three distinct fields of numbers, say $\mathbb{R}_F$ for forces, $\mathbb{R}_P$ for positions, and $\mathbb{R}_T$ for torques, and think of these fields as having the corresponding units associated with them.

Then, you can construct vector fields like $\mathbb{R}_P^D$ for $D$-dimensional positions from these fields, and endow them with the usual vector space operations, such as addition and scalar multiplication.

By definition, the vector spaces are not identical, but clearly they are isomorphic. To preserve the physical distinction between e.g. forces and torques, you would refrain from defining addition operations that take a pair $\left(\mathbb{R}_F^D,\mathbb{R}_T^D\right)$ to some other space. However, as a physicist, you would probably want to be able to form a cross product, i.e. a map $\left(\mathbb{R}_N^3,\mathbb{R}_P^3\right) \to \mathbb{R}_T^3$, just like you would desire to define a product $\left(\mathbb{R}_N,\mathbb{R}_P\right) \to \mathbb{R}_T$ that allows you to multiply scalar forces and distances to get something with the dimension of "space times force". (In fact, you probably would use the latter operation to define the former.)

This procedure has the advantage of being type-safe, to borrow a term from computer science, in the sense that you cannot add distances and forces "by accident", since there is no such addition operation defined. However, thinking along these lines may be tedious to some, and others may argue there is little mathematical reason to make the distinction between the different $\mathbb R$s, since they are isomorphic after all.

I hope you find these thoughts useful. I'd be happy to discuss them further if you so desire.

• This is interesting, also if I would have used general vector spaces and not $\mathbb{R}^3$, which already suggest an isomorphism. You say "... there is little mathematical reason...", probably it is better to say that there is little physical reason, but a big mathematical one. – enzotib Feb 5 '15 at 11:28
• @enzotib: It is not the fact that you're talking about $\mathbb{R}^3$ that induces an isomorphism, but the fact that you're talking about different copies $\mathcal{V}_i$ of some arbitrary vector space $\mathcal{V}$. The same problem would have arised with any choice other than $\mathbb{R}^3$. – RQM Feb 5 '15 at 12:12
• I was saying that $\mathbb{R}^3$ simplify (eccessively) things, because there is an obvious isomorphism between different copies of $\mathbb{R}^3$. Using arbitrary 3-dimensional vector spaces does not provide such an obvious isomorphism, unless you choose a basis for each of them. Also, I would not say that they are different copies of the same space, but different spaces. – enzotib Feb 5 '15 at 14:19