Is Newton's third law valid for rotation / Torque?

[jaumzaum]

Is third Newton law valid for rotation / Torque? I mean, can we say that for every torque there must be another torque with equal magnitude and opposite direction?

This can only be true for contact forces or radial forces, as these forces will create a reaction that will cancel the torque from the first one.
Reaction forces that are not radial or that has a different point of application will not cancel the torque and therefote third Newton Law would not be valid.

But we have 4 fundamental forces. And in my understanding all of them are radial. That said, is Third Newton law for rotation a fact always valid?

 

[etotheipi]

It depends on your definition of Newton's third law! The strong form of NIII between two particles states that $\vec{F}_{12} = -\vec{F}_{21} = \lambda(\vec{r}_1 - \vec{r_2})$ for some $\lambda \in \mathbb{R}$. Choose an origin $\mathcal{O}$, and then$$\vec{\tau}_{12} + \vec{\tau}_{21} = \vec{r}_1 \times \vec{F}_{12} + \vec{r}_2 \times \vec{F}_{21} = (\vec{r}_1 - \vec{r}_2) \times \vec{F}_{12} = \frac{1}{\lambda}\vec{F}_{12} \times \vec{F}_{12} = \vec{0}$$which means that $\vec{\tau}_{12} = -\vec{\tau}_{21}$. However, if we only insist upon the weak form of NIII (which does not constrain the forces to be parallel to the separation vector) then this is no longer necessarily true.

 

[andrewkirk]

Yes we can say that, because a torque is simply the cross product of a force with the location vector of the point of application relative to a designated centre of rotation. If object A applies force $\vec F$ to object B at a point that is displaced by vector $\vec r$ from point O, then A has applied a torque of $\vec r\times\vec F$ to B around point O. Newton's third law tells us that object B applies a force of $-\vec F$ to A, and the force is applied at the same point. So B applies a torque of $\vec r\times (-\vec F)$ to A around point O. That torque is equal in magnitude and opposite in direction to the first torque.

EDIT: Changed order of cross product to align with the directional convention cited by etotheipi. But of course that doesn't change the point of the post.

 

[etotheipi]

If object A applies force $\vec F$ to object B at a point that is displaced by vector $\vec r$ from point O, then A has applied a torque of $\vec F\times\vec r$ to B around point O.

Antisymmetry of the cross product means that order is important! Torque is $\vec{r} \times \vec{F}$, although this doesn't affect your main point.

 

[Dale]

But we have 4 fundamental forces. And in my understanding all of them are radial. That said, is Third Newton law for rotation a fact always valid?

When you start talking about the fundamental forces you are talking about fields and Newton’s 3rd law becomes substantially less useful. The generalization to fields is the conservation of momentum, and the rotational equivalent is the conservation of angular momentum.

From Noether’s theorem we know that conservation of angular momentum is associated with the isotropy of the laws of physics. So far, all is the evidence indicates that all of the fundamental forces share that feature. So it does seem that the conservation of angular momentum is always valid at a fundamental level.

 

[PhDeezNutz]

I don’t think Newton’s Third is necessarily true for torques. Problem 4.5 in the fourth edition of Griffiths Introduction to Electrodynamics deals with such a situation.

And here’s the solution

Torques are not the same.

 

[etotheipi]

I don’t think Newton’s Third is necessarily true for torques. Problem 4.5 in the fourth edition of Griffiths Introduction to Electrodynamics deals with such a situation.

I think this might be because the forces on each dipole due to the other are not parallel to the separation vector of the dipoles, i.e. in this case only the weak form of Newton III holds. So like what I said in #2, the torques are no longer negatives of one another

Edit: this is wrong

 

[PhDeezNutz]

This is because the forces on each dipole due to the other are not parallel to the separation vector of the dipoles, i.e. in this case only the weak form of Newton III holds. So like what I said in #2, the torques are no longer negatives of one another

Ah I see your post now. That makes perfect sense.

Also I was reading your signature earlier and laughing uncontrollably.

 

[A.T.]

I don’t think Newton’s Third is necessarily true for torques. Problem 4.5 in the fourth edition of Griffiths Introduction to Electrodynamics deals with such a situation.

View attachment 272504
And here’s the solution

View attachment 272505
Torques are not the same.

These two torques are not around the same point, but around two individual centers.

 

[etotheipi]

These two torques are not around the same point, but around two individual centers.

Oh, okay I was a little bit off then, in this case. Nice catch!

 

[PhDeezNutz]

These two torques are not around the same point, but around two individual centers.

I interpreted OP's question as "do two bodies exert the same torque on each other" and in that problem they don't. Maybe I misunderstood the OP.

 

[jbriggs444]

I interpreted OP's question as "do two bodies exert the same torque on each other" and in that problem they don't. Maybe I misunderstood the OP.

But as @A.T. points out, they do exert the same torque on each other as long as one is not cheating and judging the two torques by two different standards.

 

[etotheipi]

But as @A.T. points out, they do exert the same torque on each other as long as one is not cheating and judging the two torques by two different standards.

I think you're right (it makes sense from a physical perspective, in terms of conservation of angular momentum), but how do we show this? I tried, let $\vec{R}_1$ be the centre of the dipole $\vec{p}_1$, and $\vec{r}_{1a}$ and $\vec{r}_{1b}$ be the positions of $+q$ and $-q$ w.r.t. the centre. Then, the torque on the dipole about some other arbitrary point is just$$\begin{align*}
\vec{\tau}_1 &= (\vec{R}_1 + \vec{r}_{1a}) \times \vec{F}_{1a} + (\vec{R}_1 + \vec{r}_{1b}) \times \vec{F}_{1b} \\

&= \vec{R}_1 \times (\vec{F}_{1a} + \vec{F}_{1b}) + (\vec{r}_{1a}\times \vec{F}_{1a} + \vec{r}_{1b} \times \vec{F}_{1b}) \\

&= \vec{R}_1 \times \vec{F}_{1} + \vec{\tau}'
\end{align*}$$with $\vec{\tau}'$ the torque w.r.t. the centre, and $\vec{F}_{1}$ the net force on $\vec{p}_1$. This is equivalently$$\vec{\tau}_1 = \vec{R}_1 \times (\vec{p}_1 \cdot \nabla) \vec{E}_2 + \vec{p}_1 \times \vec{E}_2$$Now we need the field from the second dipole,$$\vec{E}_2 = \frac{1}{4\pi \varepsilon_0} \left( \frac{3[\vec{p}_2 \cdot (\vec{r} - \vec{R}_2)]}{|\vec{r} - \vec{R}_2|^5} - \frac{\vec{p}_2}{|\vec{r} - \vec{R}_2|^3} \right)$$All of this goes for the other dipole, as well, just with indices juggled. This expression scares me. I wonder if anyone has worked through the algebra before in general case?

 

[Dale]

I interpreted OP's question as "do two bodies exert the same torque on each other" and in that problem they don't. Maybe I misunderstood the OP.

To determine if two torques are the same or not they have to be judged about the same point.

What you are doing is analogous to what my 4 year old likes to do when he claims that he is taller than me while he is standing on the stairs and I am on the floor.

 

[jbriggs444]

how do we show this?

We have four third-law pairs. In each case the points of application of the two partner forces are separated by a displacement parallel to the direction of the force pair.

Likely we have already worked through the one-force case and convinced ourselves that with such a force pair, angular momentum is conserved -- the torques on each body about any arbitrary reference point are equal and opposite.

Since each force pair individually obeys the "third law for torque" about any chosen reference point, all the forces acting together must also obey the "third law for torque" about that point.

 

[etotheipi]

We have four third-law pairs. In each case the points of application of the two partner forces are separated by a displacement parallel to the direction of the force pair.

Likely we have already worked through the one-force case and convinced ourselves that with such a force pair, angular momentum is conserved -- the torques on each body about any arbitrary reference point are equal and opposite.

Since each force pair individually obeys the "third law for torque" about any chosen reference point, all the forces acting together must also obey the "third law for torque" about that point.

Ah, perfect! Yes I forget, that at the end of the day we can sort of ignore the dipole formalism, and just consider this a system of interacting point charges. Thanks!

 

[jbriggs444]

Ah, perfect! Yes I forget, that at the end of the day we can sort of ignore the dipole formalism, and just consider this a system of interacting point charges. Thanks!

I had a moment of panic when I started to think about "perfect dipoles". Since a perfect dipole has zero extent, how can any of these arguments work? After calming myself down, I realized that a perfect dipole is the limit of increasingly huge equal and opposite charges with an increasingly tiny separation. If angular momentum is conserved as the limit is approached, this should be upheld in the limit as well [that is a handwave, not rigorous logic, but I feel good about it in this case].

We end up with a dipole with negligible moment of inertia subject to a non-negligible torque about its own center. Sure hope somebody brought something to anchor it in place.

 

[PhDeezNutz]

But as @A.T. points out, they do exert the same torque on each other as long as one is not cheating and judging the two torques by two different standards.

To determine if two torques are the same or not they have to be judged about the same point.

What you are doing is analogous to what my 4 year old likes to do when he claims that he is taller than me while he is standing on the stairs and I am on the floor.

That actually makes sense. I see now. We can't compare torques against each other unless we have the same "pivot point".

Now that I've re-read the question I see the emphasis on about its own center

 

[Dale]

That actually makes sense. I see now. We can't compare torques against each other unless we have the same "pivot point".

Now that I've re-read the question I see the emphasis on about its own center

There are configurations where the EM fields produced by some isolated configuration of charges produces a non-zero net torque on those charges. However, those cases involve a change in the angular momentum of the EM field itself. So angular momentum is being transferred between the charges and the field.

 

[PhDeezNutz]

There are configurations where the EM fields produced by some isolated configuration of charges produces a non-zero net torque on those charges. However, those cases involve a change in the angular momentum of the EM field itself. So angular momentum is being transferred between the charges and the field.

I must admit that is a little bit beyond my level but interesting nevertheless. Something to keep in mind as I progress. I'm trying to think of such a situation but my intuition is not quite there yet.

 

[jaumzaum]

It depends on your definition of Newton's third law! The strong form of NIII between two particles states that $\vec{F}_{12} = -\vec{F}_{21} = \lambda(\vec{r}_1 - \vec{r_2})$ for some $\lambda \in \mathbb{R}$.

Thanks. I didn't know that there was this "strong" version that stays that the action and reactions must be radial. What I was trying to say is:

Consider there is a pair of action/reaction forces like the ones below:

They have equal modulus and opposite directions, but they do not conserve angular momentum nor produces opposite torques around any point. So the real question is, why these forces can't be pairs of action/reaction?

My understanding is that from all 4 fundamental forces that we discovered, all of them are radial. So we cannot have this configuration. But is there any other simpler explanation? An explanation where we don't talk about fundamental particles?

Another thing we could say is that angular momentum of the universe is conserved because angular momentum is a "fundamental property". But then we are talking about fundamental particles with fixed angular momentum being interchanged to generate a torque. Is there another explanation? Or better, is there an instance where angular momentum is not conserved in a process, and instead, is transformed to anything else? We know rest mass could be considered a fundamental property, but we also know that rest mass is not conserved in the universe, because an electron can annihilate another positron.

I could have said many wrong things above, sorry if that happened. My current understanding in Particle Physics is veeeery limited. But I really want to know why Third Newton Law for rotation must be true, with no expeptions?

 

[jbriggs444]

But I really want to know why Third Newton Law for rotation must be true, with no expeptions?

@Dale alluded to Noether's theorem in #5. It is pretty deep, but pretty compelling.

Ultimately, the universe behaves as it behaves. We attempt to find rules that the behavior conforms with.

Noether's theorem describes the consequences of some general classes of rules. In particular, if the laws of physics are "isotropic" (the same in every direction) and if some other conditions hold then it follows that angular momentum is conserved.

As far as we can tell, the laws of physics are indeed isotropic.

Edit: Let us try to make this more specific. Why should the fact that the laws of physics are isotropic imply conservation of angular momentum? That seems to be quite a stretch...

Consider the diagram that you had presented above in #21. We have two particles interacting. The force between them does not point along the axis between them. How was the direction of that force determined by the laws of physics?

If the laws of physics determine the direction of that force then they can either point in a particular arbitrary direction (in which case they are not isotropic) or they can point along the axis between the two particles (in which case they are isotropic).

 

[Dale]

So the real question is, why these forces can't be pairs of action/reaction?

As @jbriggs444 said, in principle they could, but such a force law would need to have a preferred direction in space. We don't know of any laws of physics that behave like that.

 

[Ganesh Mahadevan]

I stand corrected and shall delete my reply. Thanks