Physical meaning of the angular momentum

Question

Still reading Classical Mechanics by Goldstein, I'm struggling on a very basic notion: angular momentum. I physically understand it as the momentum of an object rotating around something given a certain position. However, I can't give a physical explanation to the formula. Why do we multiply the linear momentum by the position? Why does the angular momentum is a function of the position?

Answer 1

Ultimately, what's special about angular momentum is this:

• Look up in the sky. A certain set of physical laws pertain in that direction.
• Look to the north. A certain set of physical laws pertain in that direction.
• Look to the west. A certain set of physical laws pertain in that direction.

Those physical laws: They're the same in all directions. There's an underlying conserved quantity whenever you find a symmetry like this. A similar concept applies if you take a trip to China, Proxima Centauri, the Andromeda galaxy, or even further. Here the laws of physics are the same regardless of translation. What about time? Blink your eyes and the laws of physics don't change. Age 80 years and the laws of physics don't change. The laws of physics are timeless.

The timelessness of the laws of physics means that energy is a conserved quantity. The translational independence of the laws of physics means that linear momentum is a conserved quantity. Finally, the rotational independence of the laws of physics means that angular momentum is a conserved quantity. These are all consequences of Noether's theorem. There are a number of other conserved quantities that result from Noether's theorem, and this turned out to be very important for quantum mechanics. This is obviously critical to classical mechanics as well. Noether's theorem explains exactly why those conserved quantities are conserved.

Angular momentum would be a rather useless concept if angular momentum was not a conserved quantity in the absence of external torques. It is a conserved quantity thanks to the rotational symmetry of space.

Answer 2

I physically understand it as the momentum of an object rotating around something given a certain position. However, I can't give a physical explanation to the formula. Why do we multiply the linear momentum by the position? Why does the angular momentum is a function of the position?

1) - Angular momentum($L = rmv $)

It is very simple: in the other question you have understood the concept of linear momentum, now you have only to join it to the concept of the lever.

Imagine that ball B is the same ball which was in the linear-momentum question $(m = 2 Kg)$ which was traveling at a velocity of three meters per second and had a momentum of six kilograms meters per second

We can think that it has a line in the direction in which it is moving and a hook hanging. This hook gets caught by a peg $F$. What will happen? $B$ will start to rotate around the fulcrum $F$ (sketch on the left). The direction of motion will be perpendicular to the radius (line), therefore the angle will be $90°$, and it's $\sin$ will be $1$.

Consider another scenario (sketch on the right; same as in a lever) the torque exerted depends also on the radius, the distance of the body from the fulcrum which is the arm of the lever. The magnitude of the torque depends on the value of $r$. A weight of $6 kg$ will exert a torque of $12$ $Nm$ at the distance of $2$ $m$, and you will have balance only if you put (on the other arm) a weight of $6 Kg$ at $2 m$ or weight of $12$ $Kg$ at $1 m$.

If you understand the concept of the lever, you can easily understand the physical explanation of the formula of the angular momentum. In the same way, if B ($m = 2$) is rotating anticlockwise at $v$ = $3 m/s$ (linear $momentum$ = $6$) at distance $2 m$ from the fulcrum it will have angular momentum (6 * 2 =) 12 Kg * m²/s). If the line hanging from B had been only $1 m$ long, the magnitude of $L$ would have been (6 * 1) = 6.

Likewise, if another body A ($m$ = $2$, $v$ = $3$, $p$ = $6$) is rotating clockwise on the other arm, there will not be equilibrium, even though mass, speed and linear momentum are the same; the same would happen if a force of $6N$ is applied at $r$ = $2m$ and another opposite force of $6N$ is applied at $r$ = $1m$. Note that B had angular momentum with reference to F even before it started to rotate around it all along its trajectory and it always was (p * r) = $12 Kg * m^2/s$.

2) - Definition of L

A body B with velocity (and linear momentum) has a potential rotational momentum L with reference to/around any point/body O which does not lie on its trajectory.

The magnitude of L can be found multiplying its linear momentum (p = m*v) by the distance of point O from the trajectory: $r$.

In the full formula: $L = m * [v * sinλ * d]$, L is obtained multiplying mass by tangential velocity $V_t = v * sinλ$ times distance $d$, but $d * sinλ$ is always equal to $r$

3) - Conservation of angular momentum

Angular momentum L is conserved if no external torque is exerted on the system and this property helps you understand the importance of radius. When body B is bound to O by a line/rod or by a non-contact force (like g) it starts rotating around it and acquires actual rotational momentum L.

If, while rotating around O, B impacts with a similar ball A ($m$ =2, $v$ = 0), B stops dead and A acquires same v/p/E, and potential L with reference to point F, if it collides with the bob of a pendulum A ($m$ = 2, $r$ = 2) it will acquire same v/p/L/E. If the line/rod of the pendulum $r_p = k$, p will be conserved, but $L_p$ will become $L \times \frac{k}{r}$.

This is a simple example in which the body is considered a point mass rotating on the circumference, if the mass is distributed along the radius, then we must apply a different formula $ L = I * \omega$, where $ω = v/r$ and $I = m *r^2$. P is not conserved, but KE and L are, in this way we can work out the outcome of the collision. You can find a simple example of conservation of L here

Answer 3

Why does the angular momentum depends on the position?

Angular momentum is always defined relative to a reference point, say $\mathbf r_0$, (which is often, but not necessarily the origin).

If the system is invariant under rotation around this reference point the quantity that we call "angular momentum with respect to $\mathbf r_0$" is conserved. (Note, that if only a rotation around a certain axis leaves the system unchanged, only this component of the angular momentum is conserved).

So since the angular momentum depends on a point of reference it is not a surprise that the angular momentum explicitly depends on the position.

How does this conservation of angular momentum work?

The abstract answer deals with the Noether theorem and the Lagrangian of the system you are looking at. For simplicity, let's just look at a single point particle moving on a straight line.

Note that, even a free particle moving on a straight line has a non-zero angular momentum with respect to certain points of reference. In fact the angular momentum is only zero, if the momentum and the connection between the point of reference are parallel (i.e. the point of reference is on the path of the particle).

Noether Theorem for a free particle under rotation

Let's use this free particle to see where this conservation of $\mathbf x\times \mathbf p$ comes from. The Lagrangian is here just the kinetic energy. If we rotate the coordinates around the origin and along the fixed axis $\mathbf n$ by the angle $\varphi$. The kinetic energy (the Lagrangian) should not depend on the angle of rotation.

Let the rotated positions be given by $\mathbf x'$. The kinetic energy is $\frac 12 m \dot {\mathbf x'}^2(\varphi)$, so our condition that the kinetic energy is independent of $\varphi$ can be written as:

$$\frac {\mathrm d(m \dot {\mathbf x'}^2(\varphi))}{\mathrm d\varphi}= \mathbf p \frac{\mathrm d \dot {\mathbf x'}(\varphi)}{\mathrm d \varphi} =0 \,,$$

since there are no forces acting on the free particle ($\dot{\mathbf p}=0$), we can write this as: $$\frac{\mathrm d}{\mathrm dt}\left(\mathbf p \frac{\mathrm d {\mathbf x'}(\varphi)}{\mathrm d \varphi}\right) = 0\,.$$

How does $\mathbf x'$ change with the angle? Take a look at the chapter on infinitesimal rotations and you should find something like $$\mathbf x' = \mathbf x + \varphi (\mathbf n \times \mathbf x)$$

Computing the derivative and plugging it in the equation above leads to:

$$\frac {d}{dt}\left(\mathbf p\frac {d({\mathbf x + \varphi (\mathbf n \times \mathbf x)})}{d \mathbf \varphi}\right) = \frac {d}{dt}\mathbf p \cdot (\mathbf n \times \mathbf x) = \frac {d}{dt}\mathbf n \cdot (\mathbf x \times \mathbf p)=0$$

Which tells us, that the $\mathbf n$-component of $\mathbf x \times \mathbf p$ does no change over time (i.e. it is conserved). In this case this is true for arbitrary axes,which means that the angular momentum $\mathbf L = \mathbf x \times \mathbf p$ is conserved.

Answer 4

Consider something like a door. A piece of wood with a hinge on one edge. Maybe it is one meter tall and three meters long.

Now say that you're trying to hold the door in place, at the position half a meter from the hinge, while someone else throws a baseball at the other side of the door.

If the baseball hits the hinge, you don't have to push at all.

If the baseball hits right in front of where you're pushing the door, you have to push a good amount.

If the baseball hits right on the three meter mark of the door away from the hinge, you'll have to push much harder. (think getting your finger pinched in a door)

Even though the momentum of the baseball was the same in all three cases, in the first case (if $r=0$ corresponds to the hinge) you didn't have to apply any torque$\cdot$time. In the second you had to apply a small torque$\cdot$time. In the third you had to apply a large torque$\cdot$time.

This can be phrased, mathematically, as cancelling out the angular momentum of the system. Of course things get less intuitive if you don't choose the hinge to be your origin, so you have to work and do some math to prove that the physical results are the same. Really, though, this is just as intuitive as linear momentum being dependent on reference frame (by viewing the system in a frame with a different velocity) and, in my opinion, miles more intuitive than energy being dependent on your reference frame!

Answer 5

There are several ways to describe a particle's motion. For example, in 2 dimensions, you could use cartesian $x,y$ coordinates or polar $r,\varphi$ coordinates.

To each coordinate, we can associate a 'quantity of motion' or 'generalized momentum'. If a given coordinate corresponds to a symmetry of the system, the corresponding quantity is conserved by Noether's theorem.

Because physics works the same way 'over here' as it does 'over there' (ie changing $x$ or $y$), linear momentum is conserved. Because physics also works the same way no matter the orientation of the system (ie changing $\varphi$), angular momentum also is a conserved quantity.

However, the radial component doesn't correspond to a symmetry (chaning the $r$ coordinates results in distortion), so radial momentum is generally not conserved.

Now, back to your actual question:

Why does is the angular momentum is a function of the position?

Intuitively, position contributes to angular momentum because changing the angular coordinate will result in quite different 'amounts of motion' depending on the radial distance from the origin.

Answer 6

This is an abstract answer, but I find it extremely helpful to the kind of "basic nature" question you seem to be groping for. Think of two things: Noether's theorem and a thought experiment "what if we had evolved as unsighted but clever beings?".

As in David Hammen's Answer, it is Noether's theorem that would tell us that if our physical laws are invariant with respect to rotations of our co-ordinate axes, then we would still infer the existence of three conserved quantities. Technically, if you write down your laws as defining the path of least action by minimising a Lagrangian, Noether's theorem tells you that there is always one conserved quantity for every "continuous symmetry" of the Largrangian - i.e. every transformation that can be made of smaller transformations of the same kind by addition like real numbers (think of adding angles: two rotations about an axis sum to a rotation and you sum the angles) and still leave the Lagrangian unchanged. Our laws are indeed independent of rotations of the co-ordinate axes, for the latter are merely part of our description of physics, not the physics itself. These are Copernican notions,and Noether's theorem indeed gives the formula you state for the AM.

As David points out, it is the conservation that makes AM useful, not the idea of something spinning. This is well to keep in mind when you move on to studying the spin of quantum particles such as electrons. We tend to get very overwrought by trying to imagine these things spinning, indeed Wolfgang Pauli initially rejected outright the idea of the electron spin because a little ball would need to be spinning with its boundary far exceeding the speed of light to explain the observed spin of the electrons. The notion of "something spinning" as we intuitively conceive it would not hinder our sightless beings, who would nonetheless deduce the existence of AM through Noether's theorem: "spin" is a particular visual experience that happens to be helpful in spotting motion of something against its surroundings. So our evolution makes that a strong visual experience - it is crucial to the survival in our world of both predator and prey sighted animals, and we are both. Useful for survival of our evolutionary forebears, but not too useful for fundamental modern physics: it tripped even the great Wolfgang Pauli up.

Answer 7

Have a look at the video here

I hope that when you see the man spinning around and moving the weights (changing $r$) you can see that $r$ is important.

Remember $r$ is the distance from each part of a rotating object to the axis of rotation, (which is not exactly the same as the position)

Answer 8

The angular momentum is a concept analogous with the linear momentum p = mv, in which m is the mass of the body and v its velocity.

Now, see where the angular momentum comes from. Consider for simplicity a body moving on a circle around some axis, and let ω be the angular velocity, i.e. the angle by which the object rotates, in a unit time. The relationship between linear and angular velocity is

$$ v = \omega r. \tag{1}$$

The energy of movement of the object is $E = (1/2)mv^2$, and using (1),

$$E = \frac{1}{2}mr^2 \omega^2.\tag{2}$$

Then, it is convenient to define a quantity named momentum of inertia, $I$

$$I = mr^2, \tag{3}$$

and obtain a formula in which we use angular, not linear velocity

$$E = \frac{1}{2}I\omega^2.\tag{4}$$

Going on with the concept of momentum of inertia we get angular momentum, $L$

$$L = I\omega,\tag{5}$$

similar with the linear momentum $p = mv$, i.e. as $\omega$ replaces $v$, $I$ replaces $m$ . In this formula substituting $I$ from (3),

$$L = mr^2 \omega = mvr.\tag{6}$$

This is the answer to your question, why we multiply by $r$. But, what I said above is valid when the movement is circular. For a more general movement,

$$\vec{L} = m\vec{v} \times \vec{r},\tag{7}$$

where $\vec{L}$, $\vec{v}$, and $\vec{r}$ are vectors, and $\times$ indicates vector-product.

Good luck

Answer 9

Maybe you can see it this way:

The modulus of a vector multiplication is like this:

$$|\mathbf{L}|=|\mathbf{r}\times\mathbf{p}|=rp\sin{\hat{rp}}$$

where you can see the main feature of angular momentum: position and linear momentum of the matter considered need to be both proportional to $L$ and inversely related to each other.

That is how you guarantee describing phenomena like the dancing skater rotating: when arms are drawn to the body, the rotation increases conserving $\mathbf{L}$.