4
$\begingroup$

I've seen an interesting explanation for lots of what I previously thought were unmotivated definitions in Newtonian mechanics, namely that power is always defined as effort times flow. But when trying to define power in dynamics obviously you need to deal with vectors, hence my question : how is the dot product the good generalization of multiplication in $\mathbb{R}$ to space? Why not any other inner product on $\mathbb{R}^3$ that does reduce to multiplication on $\mathbb{R}$? I know it's sort of the canonical inner product on $\mathbb{R}^3$, has that something to do with it?

$\endgroup$
  • 1
    $\begingroup$ The fact that it works is not enough justification for you? $\endgroup$ – AccidentalFourierTransform Apr 22 '16 at 17:37
  • $\begingroup$ I recommend Gabriel Weinreich's Geometrical Vectors, Chicago Lectures in Physics (1998). He gives topological definitions, which provide more intuition. You can derive the Cartesian dot product by a geometric analysis of a projection; it is almost obvious in 2D, but you should work it out for yourself. Direction cosines are the key. $\endgroup$ – Peter Diehr Apr 22 '16 at 18:08
  • $\begingroup$ Your motivation of studying physics "for the sake of maths" does neither physics nor math justice. One has absolutely nothing to do with the other. Physics studies nature and its entire ontology is experimental (which you don't have an interest in), while mathematics tries to establish a classification of structure and it mostly asks about properties of the infinite, which doesn't exist in reality. Math is mental chess on the highest level, physics is down and dirty experimentation with matter. If math is what you are interested in, I suggest you forget physics, it is not going to be useful. $\endgroup$ – CuriousOne Apr 22 '16 at 19:02
  • $\begingroup$ @RaphaelPicovschi: Do me a favor and do the most simple experiment in math for me: enumerate the natural numbers. See you tomorrow. $\endgroup$ – CuriousOne Apr 23 '16 at 5:35
3
$\begingroup$

There is a little bit more thinking behind saying that $P=\vec F \cdot \vec v$ than it being a generalised multiplication in 3D. There are even cases where multiplication with scalar becomes a cross product when using 3D vectors. For example, torque $T=Fr$, becomes $\vec T = \vec r \times \vec F$. Whenever implementing vectors into existing scalar equations, you need to be careful into deciding how to implement "direction" into your equation.

For power, it is important to consider exactly what power is: power is equal to force multiplied by the velocity component in the same direction as the force. In purely 1D problems, the force is always in the same or opposite direction, so a plus or minus sign is all that's needed to consider direction. In higher dimensions, it's more complicated. Say we have vectors for force and velocity $\vec F$ and $\vec v$. Being a problem in dimensions higher than 1, the force and velocity vectors are not necessarily in the same direction. So, let's say that an angle $\theta$ separates the two vectors. If so, the scalar component of the velocity in the direction of the force, using trigonometry, is $\left| \vec v \right| \cos\theta$. So, this means power is equal to:

$$P = \left| \vec F \right| \left| \vec v \right| \cos\theta$$

Then, by looking at the right hand side, you can see how it matches a definition of the dot product: $\left| \vec a \right| \left| \vec b \right| \cos\theta = \vec a \cdot \vec b$, with $\theta$ being the angle between the two vectors. Hence, we have derived:

$$P=\vec F \cdot \vec v$$

So, it is by considering some aspect of the direction of the vectors of interest that we obtain the vector equation, rather than the dot product being a mathematical generalisation of scalar multiplication.

$\endgroup$
2
$\begingroup$

It's true that there are many inner products you can choose on $\mathbb{R}^3$. However, physics supplies the additional principle of rotational invariance: the result should not depend on our coordinate system. Now, any inner product of vectors $a$ and $b$ can be written as $$a \cdot b = a^T M b$$ for a matrix $M$. Rotational invariance tells us that $M$ should look the same in a rotated coordinate system, so $$M = R^T M R$$ for any rotation matrix $R$. It's straightforward to show only one matrix satisfies this condition: the identity, times a constant. To correctly reduce to the 1D case, the constant must be 1, uniquely specifying $$a \cdot b = a^T b.$$ This is the standard inner product.


Note that if we allow the product of vectors to output another vector, $M$ becomes a rank 3 tensor, and this same argument shows that $M$ is the Levi-Civita tensor, giving the cross product.

$\endgroup$