Why is torque defined as $\vec{r} \times F$?

Here I cannot convince myself myself that it is units because the torque is defined to be in units of Newton meter is a reiteration of the law stated above. Why was it not $$r^2 \times F$$ or $$r^3 \times F$$ or $$r^2 \times F^2$$ etc. The argument "in our experience how much something rotates depends on the lever length and the force applied" is really insufficient. Can someone outline a more rigorous proof or motivation?

• But all of physics comes from what we see in experiments. "Rigorous proof" deals with mathematical objects, which are products of human imagination and often have non-physical properties. Apr 24 at 12:19
• Does Why do we use cross products in physics? or Why is torque a cross product? answer your question? May 3 at 18:03

It's not like someone said "Ah ha, torque! What should the definition of torque be?" That doesn't make sense; you don't think up a term and then try to assign a definition.

Instead, it was found that this thing $$\mathbf r\times\mathbf F$$ turned out to be really useful in explaining physical phenomena. Particularly rotational dynamics of systems. So it got its own special definition.

If you want to define $$r^2F$$ or $$r^2F^2$$ as something, you can go right ahead. No one will stop you. The goal would then be to show that your new definition has physical significance.

P.S. I think the answers that have "Define torque as the time-derivative of angular momentum" or something like it is showing the physical significance of $$\mathbf r\times\mathbf F$$ by linking it to other things with physical significance (assuming you think angular momentum and its rate of change is significant). But I think they miss the point of showing that you can't derive or prove definitions and that questions like "why is insert physics term here not equal to insert modified definition here instead?" misunderstand why something would be defined in physics in the first place.

What you are doing here is not actually asking why is torque defined as $$\mathbf r\times\mathbf F$$, but rather (possibly unaware to yourself) you have some other notion of what torque is/should be, and you want to know why we can go from what you are thinking of to $$\mathbf r\times\mathbf F$$. Of course, we would need to know what your actual starting point is that you desire in order to get there. Other answers assume rate of change of angular momentum, or use virtual work and equilibrium, but that isn't necessarily what you might be thinking of. The fact that there are a variety of other answers here shows how flawed the question (not the post though) is.

• So you are saying that it is like that because it is useful? But why and how is it exactly useful? I don't understand your answer. Apr 24 at 6:53
• @bananenheld At the time I was writing my answer, two other answers had already covered why it is useful. I figured I wouldn't put the same information down a third time. Apr 24 at 13:59
• You wrote in your answer that the other answers you are referring to are missing the point so that was why I was confused. I still don't think the other answers really provide anything else then a derivation. Apr 24 at 14:05
• @bananenheld You left off the rest. I said miss the point of showing that you can't derive or prove definitions. They start from a different view of torque the OP did not specify, and get to $\mathbf r\times\mathbf F$. This is probably why it is unsatisfying; it's just a bunch of circular reasoning. My answer tries to break that by showing how the question is flawed in the first place. I added an addition to my answer to try and help Apr 24 at 22:20
• You can't prove definitions 'by definition'. However the question is why torque is defined the way that it is, why it is useful. You say that the question is flawed, however I don't think that the question is flawed but you are missing the point. I believe that the reason torque is useful must have to do with conservation of angular momentum which comes from noethers theorem/isotropy of space. Your answer seems to be 'because it is useful, and your question is flawed' which seems to me not appropriate and dismissing the question. May 7 at 9:26

Torque is change of angular momentum: $$\vec{\tau} = \frac{d\vec{L}}{dt}$$

Angular momentum is defined as $$\vec{L} = \vec{r} \times\vec{p}$$

Using the chain rule: $$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times\vec{p})$$

$$\frac{d}{dt}(\vec{r} \times\vec{p}) = \frac{d \vec{r}}{dt}\times \vec{p}+ \vec{r} \times\frac{d \vec{p}}{dt} = \vec{0} + \vec{r} \times\frac{d \vec{p}}{dt} \tag{1}$$

Remembering that $$|\vec{a} \times \vec{b} |= |\vec{a}||\vec{b}| \sin \theta$$ and that the velocity vector is parallel to the linear momentum vector we get:$$\frac{d \vec{r}}{dt}\times \vec{p} = \vec{v} \times \vec{p} = \vec{v} \times m\vec{v} = m|\vec{v}| |\vec{v}| \sin \theta \ \hat r=m|\vec{v}| |\vec{v}| \sin 0 \ \hat r = \vec{0}$$

$$\theta$$ is the angle between two vectors, which is $$0$$ for any vector with itself.

So we get for torque, using the result in $$\text{(1)}$$:

$$\vec{\tau} = \frac{d\vec{L}}{dt} = \vec{r} \times\frac{d \vec{p}}{dt} = \boxed{\vec{r} \times\vec{F}}, \text{ since } \vec{F}=\frac{d \vec{p}}{dt}$$

But the question then reduces to why angular momentum is defined as $$\vec{L} = \vec{r} \times\vec{p}$$. I think this has to do with Noether's theorem, that this quantity is conserved when a system stays the same under a change of angle.

I personally prefer a derivation using the principle of virtual work where the formula of torque directly comes out. While angular momentum is a natural property to consider for a spherically symmetrical problem, this alternative approach shows its relevance for statics of rigid bodies even when this symmetry is not present.

Take a set of points indexed by $$i$$ at position $$\vec r_i$$, on which are applied respectively the forces $$\vec F_i$$. This gives first formula of vitual work for a general displacement: $$\delta W = \sum \vec F_i \cdot \delta\vec r_i$$ Furthermore, lets assume the points are rigidly constrained and can only rotate around the origin. Any allowed differential displacement can thus be written as $$\delta\vec r_i =\delta\vec \phi \times \vec r_i$$ where $$\vec \phi$$ is the differential angular displacement. Injecting in the work you get: $$\delta W = \sum \vec F_i \cdot (\delta\vec \phi \times \vec r_i)$$

$$\delta W = (\sum \vec r_i \times\vec F_i ) \cdot \delta\vec \phi$$

So static equilibrium is equivalent to a vanishing virtual work for any relevant virtual displacement, hence $$\sum \vec r_i \times\vec F_i$$, the torque naturally pops out. It also explains also the useful power formula for rotation (with angular velocity $$\vec \omega$$):

$$P = (\sum \vec r_i \times\vec F_i ) \cdot \vec \omega$$

Hope this helps and tell me if you find some mistakes.

• Best explanation, I wish I had written it! Apr 24 at 16:00
• why is $\partial \vec{r_i} = \partial\vec{\phi} \times \vec{r_i}$? Apr 24 at 17:21
• It’s because these transformations generate all the rotations. Abstractly, the antisymmetric matrices are the Lie algebra of $SO(3)$ which can be identified to $\mathbb R^3$ via the cross product. Check out this as a starter en.m.wikipedia.org/wiki/Euler%27s_rotation_theorem
– lpz
Apr 24 at 18:05

Consider a point particle of mass $$m$$ with velocity $$\vec v$$. The particle is located at some position $$\vec r$$ with respect to the origin $$O$$.

I will start with the angular momentum calculated about $$O$$. The angular momentum is $$\vec L = I\vec \omega$$, where $$I=mr^2$$.

Since we know $$v=\omega r$$, you can work out that $$\vec \omega = \dfrac{\vec r\times \vec v}{r^2}$$.

Next, we can substitute that into our angular momentum to get $$\vec L = m \vec r\times \vec v.$$

Define torque as the time-derivative of angular momentum, and we have that $$\vec \tau \equiv \dfrac {d\vec L}{dt}=m\vec r\times \dfrac{d\vec v}{dt}.$$

Since $$\vec F=m\vec a$$, we then have $$\boxed{\vec \tau =\vec r \times \vec F}$$

If we consider a small arc of length dl (almost 0) in a circle of radius R which subtends a small angle du As arc is almost 0 thus angle subtended will also be approximately 0

As dl is very small thus dl length can be assumed as straight line (as we can assume circle is made of very small straight line kept at some angle) So it forms a triangle with 2 sides as radius and one as dl And considering one angle as 0 the other 2 angle will be 90°

Using sin rule: [sin(du)/dl] = [sin(90)/R]

As we know sin(y)=y if y is very small So sin(du) =du

Thus dl = R(du)

Now differentiating with respect to time we get (angular-velocity)=(linear-velocity)(Radius)

Differentiating with respect to time again (angular-acceleration)=(linear-acceleration)(Radius)

Multiplying by mass We get (angular acc)(mass)=torque And (linear acc)(mass)=force