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

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

• I think its convention since a world with choosing $F\times r$ would be the same. Since $F\times r$=$-r\times F$. But I might be wrong Commented Dec 5, 2018 at 15:42
• @ArthurMorgan, I agree. At some point, physicists realized that the torque vector should be perpendicular to the plane of motion of the spinning object that the torque is being applied to. The question became: should the torque vector point "up" or "down" (assuming a horizontal plane of spin)? Physicists had to choose one orientation or the other, and the convention became "fixed". Commented Dec 5, 2018 at 16:34
• Are you using right-handed or left-handed coordinate systems? Commented Dec 5, 2018 at 19:07
• Right handed systems. Commented Dec 5, 2018 at 21:10

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.

• +1 for pinning down the arbitrary sign to the case of 2D polars and trying to make the 3D case look as similar as possible. Commented Dec 5, 2018 at 18:03
• Just wondering whether the 2D polar direction might be related to the right hand (as in the "right hand rule"). Presumeably the right hand was used more by people, which might be of relevance. Another thought I had, perhaps the way people wrote could have a role as well. Commented Dec 5, 2018 at 18:24
• @user1583209 It is certainly possible. It would be very reasonable in early mathematics to have practical matters in mind. Instinctively, when I explore angular space with my right hand, I acquire the "standard" directions rather naturally. I can't say which one lead the other, but I could definitely see your way happening. Commented Dec 5, 2018 at 19:40
• Isn't it that angular momenta and subsequent stuff were defined after torque? I think the convention of torque would've set the definitions for the others. I am, though, confused, because I don't have a definitive knowledge of 'what came first'. Commented Dec 5, 2018 at 21:12
• @AabeshGhosh Archimedes did play with torque and moments, though torque as we know it did not exist until the late 1800s. I can't say much about the order things are taught in school, but when we look at building up a conceptual view of reality, the concepts of angles and rates of change of angles is considered more "fundamental" than the interaction of forces with these rates of changes of angles. The angles and rates of change of angles find more applications (such as phase angles in electric power) Commented Dec 5, 2018 at 22:01

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).

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.

• The arbitrary convention is the cross product ($\times$ operator) since it can be defined with either right-hand or left-hand rule, making $r \times_{\rm rh} F = F \times_{\rm lh} r$. Commented Dec 5, 2018 at 17:19
• @ja72 whilst the direction of a cross product is arbitrary, so is the definition of torque / angular momentum. Arguably, not just up to a sign but to an entire multiplicative constant. Of course, $\pm1$ seems the most natural. Commented Dec 5, 2018 at 17:21
• @jacob1729 - the multiplicative constant is not arbitrary because the cross product conveys geometrical information (the moment arm). You can recover $\boldsymbol{r}$ using $$\boldsymbol{r} = \frac{ \boldsymbol{M} \times \boldsymbol{F} }{ \| \boldsymbol{F} \|^2}$$ Commented Dec 5, 2018 at 17:44
• @ja72 I meant rather that we need to pick what we mean by angular momentum. And up to an overall constant, that is "the Noether charge conserved due to rotational invariance". We then choose the constant such that the moment of inertia of a point mass $m$ comes out nicely but there is really not much reason to do so. Commented Dec 5, 2018 at 18:02

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!

• This is not correct. If you use the same $F$ and $r$, the $M$ in the second definition must have a different direction than the first. Cross product is not a commutative operation. Commented Dec 5, 2018 at 17:25
• No, this is still part of the definition. It was to show that the direction of $\boldsymbol M$ or the order of the cross product per se doesn't change the physics behind it, as long as you are consistent! The second and the third paragraph use two different definitions, but yield the same result (the absolute value in this case). Commented Dec 5, 2018 at 17:28
• I agree they yield the same physics, but you have to redefine $M$, $F$, $r$, or $\times$ in order for it to point upwards. You should mention which one of these you did, or change $\mathbf{M}$ to point downwards. Commented Dec 5, 2018 at 17:44

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 $$F \times r$$.

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).

• The order of the cross product doesn't have anything to do with causation.
– user4552
Commented Dec 5, 2018 at 16:54
• I just wanted to let you know that I voted your post down because of: grammar, equation formating (please use LaTeX), and because of Ben Crowell's comment. Please don't shy away from sharing other ideas and answers! Commented Dec 5, 2018 at 17:13