How do you take the inner product of a vector whose components have different units?
For example, what is the inner product of $\langle1m, 1s\rangle$ and $\langle2m, 3s\rangle$?
How do you take the inner product of a vector whose components have different units?
For example, what is the inner product of $\langle1m, 1s\rangle$ and $\langle2m, 3s\rangle$?
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The set of lengths forms a one-dimensional real vector space $L$, and the set of durations forms a one dimensional real vector space $T$. It follows that the cartesian product $L\times T$ forms a two-dimensional real vector space $V$ in a natural way. A basis of $V$ is given, e.g., by the two vectors $e_1:=(1\,{\rm m},0)$ and $e_2:=(0,1\,{\rm sec})$.
A scalar product $\langle\cdot,\cdot\rangle$ on $V$ is a bilinear function $$B:\quad V\times V\to{\mathbb R},\qquad (x,y)\mapsto\langle x,y\rangle$$ satisfying certain axioms. In order to install such a product you have to prescribe the numbers $$g_{ik}:=\langle e_i,e_k\rangle\in{\mathbb R}\qquad(i, k=1,2)\ .$$ This means that you have to prescribe the values $$g_{11}=\langle(0,1\,{\rm m}),(0,1\,{\rm m})\rangle,\quad g_{12}=g_{21}=\langle(1\,{\rm m},0),(0,1\,{\rm sec})\rangle,\quad g_{22}=\langle(0,1\,{\rm sec}),(0,1\,{\rm sec})\rangle.$$ Whether this will lead to meaningful things depends on the application you have in mind.
While one might (wrongly) think this question is meaningless, you can have a sensible meaning involving mixed-unit vectors as follows:
The inner product is always a bi-linear form where you have a vector and a conjugate vector (in linear algebra, a row vector and a column vector). As such, you need a (linear) way to go from a vector to a vector in conjugate space. This "way" is what we think of as a metric tensor.
For your problem to make sense, the metric tensor has to have units. This is not so strange; think of special relativity, where the metric tensor looks like $$ g_{11} = g_{22} = g_{33} = 1; g_{44} = -c^2 $$ where $c$ is the speed of light (say in meters per second).
Then $$ <1 \mbox{ m} , 1 \mbox{ s}> \cdot <2 \mbox{ m} , 3 \mbox{ s}> = ( 2 - 3 c^2\mbox{ s}^2) \mbox{ m}^2 $$