Why is torque defined as $r × F$ and not $F × r$?

Question

Is it merely due to popular convention or does it supply any special clarification regarding other physical quantities?

Answer 1

Torque is associated to angular momentum ($L$) via $\tau=\frac{dL}{dt}$. Angular momentum is, of course, associated with angular velocity ($\omega$) via $L=I\omega$. This means that if we used $F\times r$, we would either need to put a minus sign in one of those definitions, or change the direction of angular velocity. Nobody wants to put a minus sign where one isn't needed, so we can ignore that case and focus on the idea of angular velocity being in a different direction.

Angular velocity is, of course $\omega = \frac{d\theta}{dt}$. We don't want a minus sign here either, so the way we define $\theta$ would have to change directions.

And here it is, indeed, arbitrary. Mathematicians define polar coordinates such that the positive x axis maps to $\theta = 0$ and the positive y axis maps to $\theta=\frac{\pi}{2}$. They could have chosen a different convention, but this is the one that was indeed chosen. To the best of my understanding, that convention was chosen by Indian astronomers about 2,300 years ago!

The one option I didn't list here was to change the meaning of cross product. Cross product was invented in 1773 by Joseph Louis Lagrange to study tetrahedra. I can't find any further links, so that might be where an arbitrary decision was made.

Answer 2

A couple of things to consider first.

1. It depends on the definition of $\boldsymbol{r}$. If $\boldsymbol{r}$ points to the force then moment is $\boldsymbol{r} \times \boldsymbol{F}$, but if it points to the measuring point (like the origin) it is $\boldsymbol{F} \times \boldsymbol{r}$.

2. There is a convention built in the $\times$ operator. It has to do with the right-hand rule and the fact that it does not have the commutative property. This means that $\boldsymbol{r} \times \boldsymbol{F} = -(\boldsymbol{F} \times \boldsymbol{r}) = (-\boldsymbol{F}) \times \boldsymbol{r} = \boldsymbol{F} \times (-\boldsymbol{r})$. This means that torque can be interpreted as the moment of force acting on a line far from the orgin, and the order at which you take the vector cross product matters.

3. There is a definition symmetry between torque, angular momentum and linear velocity

   $$ \begin{aligned} \boldsymbol{v}_A & = \boldsymbol{v}_B + \boldsymbol{r}_{AB} \times \boldsymbol{\omega} & & \text{velocity transformation} \\ \boldsymbol{L}_A & = \boldsymbol{L}_B + \boldsymbol{r}_{AB} \times \boldsymbol{p}& & \text{momentum transformation} \\ \boldsymbol{M}_A & = \boldsymbol{M}_B + \boldsymbol{r}_{AB} \times \boldsymbol{F}& & \text{torque transformation} \end{aligned} $$

   Combined with the other two points above, it speaks of the geometry of the problem at hand, as all of the quantities (rotation $\boldsymbol{\omega}$), (momentum $\boldsymbol{p}$) and (force $\boldsymbol{F}$) exist on a line in space. It is called the axis of rotation, the percussion axis and the line of action (respectively).

   Also, see this answer to a very similar question.

In summary, for the math to work out, and physics to be consistent the cross product needs to be defined with either right-hand-rule or left-hand-rule and the equations for torque, angular momentum, and linear velocity be consistently described with the handedness chosen. Finally, make sure your location vectors $\boldsymbol{r}$ are consistently defined (like originating from a common point, the origin).

Answer 3

It's an arbitrary convention. If we changed the convention, there are some other ones we would have to change as well, such as $\textbf{L}=\textbf{r}\times\textbf{p}$ for the angular momentum of a particle.

Answer 4

Maybe we can make a quick comparison of both.

Let us define the torque $\boldsymbol M = \boldsymbol r \times \boldsymbol F$. Let us further consider $\boldsymbol r$ and $\boldsymbol F$ lying in a plane and $\boldsymbol M$ facing upwards. The absolute value of $\boldsymbol M$ becomes $M=r\cdot F\cdot \sin\theta$ with $r=|\boldsymbol r|$, $F=|\boldsymbol F|$, $M=|\boldsymbol M|$, and $\theta$ the angle between $\boldsymbol r$ and $\boldsymbol F$. Now we use $\boldsymbol M = \boldsymbol r \times \boldsymbol F = -\boldsymbol r \times \boldsymbol F$ and realise that $M$ remains unchanged.

Let us define now the torque as $\boldsymbol M = \boldsymbol F \times \boldsymbol r$. Again, $\boldsymbol M$ is facing upwards from the plane. And again, $M$ has the same value as before.

What do we learn from this? The order does not matter. It is just a convention. If you'd like to use another convention, feel free to do so, but make sure to be consistent!

Answer 5

Let's understand this $r \times F$ with an example.

Let a disc hinged from the centre, we applied a force $F$ at a distance $r$ from the centre in such a way that it rotates in anti-clockwise direction.

Now due to force, angular velocity is increasing and so kinetic energy is increasing. So work done should be positive.

We can write work done as

$$ \int F \,ds = \int F r \,dθ = \int \tau \left( T \right) \, dθ \tag{1} $$

So work needs to be positive. So the angle between $T$ and $\theta$ needs to be $0^{o}$. So $T$ and $\theta$ are in the same direction.

We know that if the disc is moving in the anti-clockwise direction then $\theta$ is out of the plane (by the right-hand rule).

So $\tau \left( T \right)$ needs to be out of the plane here. For that, we do $r \times F$, rather than $F \times r$.

Answer 6

I don't know the correct explanation but I think that $\boldsymbol{r}\times \boldsymbol{F}$ signifies that the $\boldsymbol{r}$ vector rotates because of the $\boldsymbol{F}$ vector. To produce the same result of a screw coming out when we apply a force- This is the direction of torque, whereas $\boldsymbol{F}\times \boldsymbol{r}$ will mean that the $\boldsymbol{F}$ vector rotates because of the $\boldsymbol{r}$ vector, which would be the opposite direction to which the screw comes out (downward).