7
$\begingroup$

As the title asks, what is the motivation for the definition of angular momentum and by extension torque? In all the books, be it undergrad or grad, the definition of the above-mentioned is just put out there without giving any context or background for the need for such quantity.

The other approach is to use lagrangians, but that seems quite artificial and unsatisfactory to me personally because angular momentum fits well within the Newtonian framework and thus should be motivated from it.

The most digestible motivation comes from Feynman when he derives work done by a body through an angle, and then defines the quantity multiplied by the change in angle as torque (Feynman lectures, Vol 1, chapter 18 pages 18-4). However the problem with this definition is again that if we use some other coordinate system then we would have got some other quantity as work done (that is components of force would be different), however, we don't go defining such quantities with special names like we did for torque.

So what I am asking through this question is the motivation for defining torque/angular momentum the way we do and if possible then please provide the timeline for the use of angular momentum and torque, that is their first use in the physics literature.

I know about archimedean experiments with levers and the conclusions drawn from them. However, if we treat these experiments as basic facts that is they can't be explained/motivated from $F=ma$, then it becomes another fundamental law along with Newton's three laws, which again becomes highly unsatisfactory because of the reasons mentioned above.

Just to be clear what I mean by motivation consider work done as an example. We know that the second law can be written as $F=mv\frac{dv}{dx}$. Thus it becomes natural to solve for $\int F\, ds=m\int vdv$ because of the structure of the equation and also because we know that fundamental forces (gravity and Columb force) in Newtonian mechanics depend on distance.

Also, this is for Moderators: This question is not a duplicate, neither it is asking for textbook resources, so please consider this before closing this question.

Edit: At the time when I had chat with ACat in the comments of this post, I was pretty convinced with their arguments, but looking back at those arguments and Feynman's derivation I realize that the essential point still remains, why does the concept of torque require the explicit usage of polar coordinates?

What Feynman essentially does is to calculate work done on a body in polar coordinates as given below $$\int\vec F\cdot d\vec r=\int\vec F\cdot (\vec r\times d\vec\phi)$$ Since in polar coordinates $d\vec r= \vec r\times d\vec\phi$, where $d\vec\phi$ is the angle of rotation. Thus $$\int\vec F\cdot (\vec r\times d\vec\phi) =\int(\vec F\times\vec r)\cdot d\vec\phi=\int\vec\tau\cdot d\vec\phi$$

$\endgroup$
7
  • $\begingroup$ How about the fact that when you express the condition for the equilibrium of a rigid body, Newton's laws take the form of torque-balance equations? $\endgroup$
    – user87745
    Apr 11, 2022 at 10:19
  • $\begingroup$ @DvijD.C. well, that is just another variation of archimedean experiments. The condition that you speak of is not motivated by the second law and thus has to be treated as a fundamental law of nature separate from Newtonian laws, thus raising the concerns mentioned in the post. $\endgroup$ Apr 11, 2022 at 12:30
  • 1
    $\begingroup$ Since you mention Feynman, I suppose you've already seen this but see Eq. 18.10 and 18.11 here. Forget about the fact that he says "Here a new concept, force, must be introduced. Let us inquire whether we can invent something which we shall call the torque (L. torque, to twist)" in the preceding paragraph. Simply notice that 18.10 and 18.11 directly follow from Newton's laws -- we haven't introduced any new machinery. $\endgroup$
    – user87745
    Apr 11, 2022 at 12:44
  • 1
    $\begingroup$ I don't exactly understand your objection that this is something coordinate-dependent. Yes, if you write things in actual numbers, a lot of things are coordinate dependent, force itself is coordinate dependent if we are to use language that way. The point is that there is a coordinate independent way to write $xF_y-yF_x$, namely, $\vec{x}\times\vec{F}$. $\endgroup$
    – user87745
    Apr 11, 2022 at 12:47
  • 1
    $\begingroup$ @DvijD.C. Now after rereading the passage in Feynman lectures and your comment, it is clear to me that I was making a mistake in my assumptions. You are correct about the coordinates and I was not. Now it all seems clear to me. Thanks a lot for your help. $\endgroup$ Apr 12, 2022 at 13:04

3 Answers 3

19
$\begingroup$

Angular momentum is a conserved quantity$^1$. This is an experimentally confirmed reality, and the equations you talk about naturally follow from this fact since one of them defines this conserved quantity and the other (for torque) describes how it might change when there is an external influence.

Conserved quantities are powerful in helping us solve physical problems and determining how certain systems evolve over time. That is the motivation. Analogous to when Newton found that the quantity $\bf mv$ is always conserved$^2$, the quantity defined by $$\bf L=r\times p=I\omega$$ is also conserved.

It is then quite natural to find that the timed rate in change of this quantity $$\bf\frac{dL}{dt}=I\frac{d\omega}{dt}=\tau$$ which is torque.

Just to be clear what I mean by motivation consider work done as an example. We know that the second law can be written as $F=mv\frac{dv}{dx}$. Thus it becomes natural to solve for $\int F\, ds=m\int vdv$

And it's a natural analogy to extend this to the case of objects with angular momentum. Consider torque, which as discussed is $$\bf\tau=I\frac{d\omega}{dt} \ =I\frac{d\theta}{dt}\frac{d\omega}{d\theta}=I\omega \frac{d\omega}{d\theta}$$

$$\rightarrow \bf \tau \int d\theta=I\int\omega d\omega$$ or $$\bf W=\tau\Delta\theta=\frac 12 I(\omega^2_f-\omega^2_i)$$

This is the work done by a torque which is simply the analogue of work done by force. That is, the work done by a torque is equal to the change in rotational kinetic energy, just as the work done by a force is equal to the change in kinetic energy.

$^1$Conserved quantities like angular momentum, energy and linear momentum all arise from symmetries in nature. See Noether's theorem.

$^2$What Newton referred to as the quantity of motion, or (linear) momentum is conserved when there are no net forces acting on the object or system. The same applies to angular momentum conservation when there are no net torques.

$\endgroup$
11
  • $\begingroup$ I am afraid you misunderstood my question, I am not asking about the motivation for the definition of work done by a rotating body, instead, I am asking about the motivation for the formula of angular momentum $\endgroup$ Apr 10, 2022 at 17:55
  • 2
    $\begingroup$ Yes. And I explained that it is a conserved quantity. That’s its motivation. The quantity $I\omega $ is conserved. Similar to the motivation for momentum and kinetic energy. Your question is akin to why do we have quantities that are conserved. What is it about my answer that you object too? Thanks. $\endgroup$
    – joseph h
    Apr 10, 2022 at 20:07
  • 6
    $\begingroup$ But that's the point. It was not defined before we knew it was a conserved quantity, but the other way around. The "weird combination" of $\bf r\times\bf p=I\omega$ came as soon as we realized that that quantity is conserved and that's why it was "worth investigating". Make sense? And it is Newtonian mechanics but extended to rotating objects. I also explained that in my answer. Thanks. $\endgroup$
    – joseph h
    Apr 10, 2022 at 20:47
  • 1
    $\begingroup$ To ease OP's objections, I'd phrase "[I]t is Newtonian mechanics but extended to rotating objects." as "[I]t is Newtonian mechanics but applied to the specific case of rotating objects." cc: @GedankenExperimentalist $\endgroup$
    – user87745
    Apr 11, 2022 at 12:55
  • 1
    $\begingroup$ @josephh Eli is right when he states that $\vec{\omega} = \dot {\vec{\theta}}$ only works for small angles. Rotation is represented by rotation tensors $\mathbb{R} $, and omega comes from the time derivative of rotation tensor, that can be expressed with an antisymmetric tensor that can be written as $\vec{\omega}_{\times} = \dot{\mathbb{R}} \cdot \mathbb{R} ^T$. Only when you perform small-angle rotations, you can linearize this equation and get $\vec{\omega}_{\times} \sim \dot{\vec{\theta}}_{\times}$ $\endgroup$
    – basics
    Nov 18, 2022 at 13:06
7
$\begingroup$

Kepler's law of areas follows logically from the laws of motion, and as we know, Kepler's law of areas generalizes to the concept of angular momentum.

While linear momentum is a property of motion along a line, angular momentum is property of motion in a plane. It can be circular motion or spiralling motion, the point is that the minimum space you need in order to have an angular momentum at all is two spatial dimensions: a plane.

As we know, when we are dealing with two spatial dimensions we are dealing with area. There is a correlation between angular momentum and area.

Kepler had no way of judging whether the area law was an independent law, or whether it was an organic part of a larger logical system. Historically it was Isaac Newton who placed Kepler's law of areas into context.

Newton's derivation of the concept of angular momentum

The first derivation in the Newton's Principia is a derivation of Kepler's law of areas from first principles.

(I reproduce the text from wikipedia, because I am the author of that image, and I wrote that exposition)
During the first interval of time, an object is in motion from point A to point B. Undisturbed, it would continue to point c during the second interval. When the object arrives at B, it receives an impulse directed toward point S. The impulse gives it a small added velocity toward S, such that if this were its only velocity, it would move from B to V during the second interval. By the rules of velocity composition, these two velocities add, and point C is found by construction of parallelogram BcCV. Thus the object's path is deflected by the impulse so that it arrives at point C at the end of the second interval. Because the triangles SBc and SBC have the same base SB and the same height Bc or VC, they have the same area. By symmetry, triangle SBc also has the same area as triangle SAB, therefore the object has swept out equal areas SAB and SBC in equal times.

At point C, the object receives another impulse toward S, again deflecting its path during the third interval from d to D. Thus it continues to E and beyond, the triangles SAB, SBc, SBC, SCd, SCD, SDe, SDE all having the same area. Allowing the time intervals to become ever smaller, the path ABCDE approaches indefinitely close to a continuous curve.


The above geometric reasoning shows there is a conserved quantity. This conserved quantity is proportional to the area of the triangle that are swept out. This explains why the expression for angular momentum, $m r^2 \omega$, is a quadratic expression.

$\endgroup$
1
  • 2
    $\begingroup$ Physics is not mathematics. The law came first, driven by observation. It was not derived from axioms. The rationalization (conservation of angular momentum) followed. Then, the rationalization proved useful in other contexts. So, it became a definition. $\endgroup$
    – John Doty
    Apr 9, 2022 at 22:40
0
$\begingroup$

Angular momentum is the Plücker coordinates that defines the location in space where the line of momentum acts (also known as the axis of percussion).

Conservation of angular momentum means that momentum is not only conserved in magnitude and direction, but also in the line it acts in space also.

Geometry

There are various ways of describing 3D lines algebraically, and one of them is using the so-called Plücker coordinates of a line. A ray with direction $\boldsymbol{v}$ and origin $\boldsymbol{g}$ defines the points along the line as $$\boldsymbol{p}(t) = \boldsymbol{g} + t\,\boldsymbol{v}$$ where $t$ is an arbitrary value.

Now take the moment of the points if the line to get

$$\require{cancel} \boldsymbol{m} = \boldsymbol{p}(t) \times \boldsymbol{v} =( \boldsymbol{g} + \cancel{t\, \boldsymbol{v}}) \times \boldsymbol{v} = \boldsymbol{g} \times \boldsymbol{v} $$

which is a constant value, not dependent on the choice of the arbitrary value of $t$. We use this vector together with the direction vector to define a line in space.

  • The 6 Plücker coordinates of a line with direction $\boldsymbol{v}$ that goes through a point $\boldsymbol{g}$ are $${\rm L} = ( \boldsymbol{v},\, \boldsymbol{g}\times \boldsymbol{v} )$$

From Plücker coordinates of a line (pdf) but using different notation for the vectors. fig1

Given a line ${\rm L}=( \boldsymbol{v},\,\boldsymbol{m})$, you can recover the direction of the line from $\boldsymbol{v}$, and you can recover the point on the line closest to the origin by $$\boldsymbol{g}_0 = \frac{ \boldsymbol{v} \times \boldsymbol{m} }{ \boldsymbol{v} \cdot \boldsymbol{v} }$$

The proof for this is not important right now. Just the idea that the 6 coordinates of $(\boldsymbol{v},\,\boldsymbol{m})$ are sufficient to describe the geometry of a line in space if $\boldsymbol{m} = \boldsymbol{g} \times \boldsymbol{v}$ is the moment of the direction vector.

Forces

Now consider a force $\boldsymbol{F}$ acting in space along its line of action, and we know some point $\boldsymbol{r}$ that this force is going through. The torque of this force about the origin is $\boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F}$ regardless of where along the line the point is.

The 6 coordinates $(\boldsymbol{F},\,\boldsymbol{\tau})$ are sufficient to describe the geometry of the force system since $\boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F}$ is the moment of the force.

The line of action passes through a point closest to the origin defined by

$$\boldsymbol{r}_0 = \frac{ \boldsymbol{F} \times \boldsymbol{\tau} }{ \| \boldsymbol{F}\|^2 }$$

Note that $\boldsymbol{F} \cdot \boldsymbol{F} = \| \boldsymbol{F} \|^2$, which is the square of the magnitude of the vector.

Momentum

Now imagine a single particle of mass $m$ traveling along a straight line with velocity $\boldsymbol{v}$. This line may or may not go through the origin.

The momentum vector of the particle is $\boldsymbol{p} = m \boldsymbol{v}$.

The momentum vector, together with the moment of momentum (angular momentum) defines the 6 Plücker coordinates of the line the particle is traveling along $( \boldsymbol{p},\, \boldsymbol{L}) $ given that $\boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}$.

  • Momentum magnitude $\| \boldsymbol{p} \|$ defines the quantity of momentum.
  • Momentum direction $\boldsymbol{p} / \| \boldsymbol{p} \|$ defines the direction of the momentum axis (percussion axis).
  • Angular momentum $\boldsymbol{L}$ defines the a point this axis goes through from $$\boldsymbol{r}_0 = \frac{ \boldsymbol{p} \times \boldsymbol{L} }{ \| \boldsymbol{p} \|^2}$$

So the motivation for defining angular momentum $\boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}$ and torque $\boldsymbol{\tau} =\boldsymbol{r} \times \boldsymbol{F}$ is to convey the geometry of a physical situation.


Appendix I

There is another quantity needed besides the moment-of vector to completely map between the 6 components of $(\boldsymbol{F},\,\boldsymbol{\tau})$ for example.

If you had a torque that was applied in a sense parallel to the force $\boldsymbol{F}$ then $ \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F}$ isn't sufficient you need some other scalar value $h$ we call the pitch to write $$ \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} + h\,\boldsymbol{F}$$

Note that the pitch value is in units of length always, and it describes the amount of parallel torque per unit of axial force applied.

To recover the pitch from the 6 components you do

$$ h = \frac{ \boldsymbol{F} \cdot \boldsymbol{\tau}}{ \| \boldsymbol{F} \|^2} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.