Is Conservation of Linear Momentum subservient to conservation of Angular Momentum?

Question

When particles physically interact, they transfer linear momentum and angular momentum between one another via force .

When particle P1 exerts force on particle P2, P2 exerts an equal and vectorially opposite force on P1 . That way, any momentum leaving P1 is attached to P2, and total momentum is conserved.

"equal and opposite force" is a necessary and sufficient condition for conservation of linear momentum.

But linear momentum is uncaring about the direction of the force, so long as it is equal and opposite.

If P1 and P2 were on opposite sides of a chosen origin point, at x,y (1,0) and (-1, 0) , and P1 put a +Y ( clockwise ) force on P2, the vectorially opposite -Y force on P1 on the other side of the origin point would also be clockwise around the origin. That would perfectly satisfy "equal and opposite", and therefore satisfy conservation of linear momentum, but would violate conservation of angular momentum. Particles in our universe cannot do this.

Equal and opposite forces along a single straight line do not create torque or modify total angular momentum when viewed from any arbitrary origin point .

"Equal and Opposite force, along the path of separation" ( attraction or repulsion only ) is a necessary and sufficient condition for conservation of Angular Momentum to all origin points . ( It is possible to choose origin points to ignore various torques . But true conservation of Angular Momentum works from any and all origins )

So, from this point of view, it appears as if the more stringent "Equal and Opposite force, along the path of separation" of Angular Momentum automatically enforces the less stringent "Equal and Opposite force" of linear momentum .

So, is conservation of linear momentum a simplified consequence of conservation of Angular momentum ?

Here is a messy supporting argument :

If a finite number of particles were interacting in a finite sized cube ( perhaps 1 meter across ), and you chose an origin point at z = 1 million meters, and you measured the angular momentum around your chosen origin, you could get a pretty accurate estimate of the total linear (x,y) momentum of those particles by dividing the measured angular momentum by the vector to the center of the box. You could make your estimate arbitrarily accurate by increasing your distance.

Conservation of angular momentum from this far origin point would imply that linear momentum was arbitrarily close to being conserved.

I have never seen thes arguments stated anywhere in print or elsewhere.

As an undergraduate circa 1981, I briefly showed this argument to my favorite physics professor, who off-the-cuff said he thought that this was a known fact. I was confused as to why it was not covered in the Halliday and Resnick.

So, is this a correct or incorrect concept ?

Can anyone find a hole in my not-terribly-complicated math ?

Does it appear in current textbooks ?

Answer 1

Conservation of linear and angular momentum arise from fundamental symmetries of spacetime via Noether's theorem. Conservation of linear momentum comes from translational symmetry and conservation of angular momentum come from rotational symmetry. The two symmetries are not necessarily related so the two conservation laws are not necessarily related.

Your argument is based upon the notion that if we consider the entire system then both types of momentum are conserved, which is equivalent to saying that no external forces or torques are acting. But this is simply the statement that spacetime, i.e. the background on which the bodies move, obeys both symmetries. This is certainly true of flat spacetime but not necessarily true of curved spacetimes i.e. not necessarily true of the universe as a whole, though I concede that it does appear to be true for at least the bit of the universe we can see.

But in practice we are often considering systems in which we have a background field, for example motion in a $1/r$ potential ($1/r^2$ force) such as the gravitational field of a spherical mass. You would argue that this is not an inertial frame since we are taking the spherical mass to be fixed, and this is certainly true. However it is still the case that the symmetries of the system determine the conservation laws i.e. in this case the rotational symmetry means angular momentum is conserved while the lack of translational symmetry means linear momentum is not conserved. To argue that the two conservation laws are equivalent would be to ignore a property of the system that is very important for performing calculations.

Answer 2

I want to build on John Rennie's answer, which invokes Nother's theorem, because it actually clarifies your idea beautifully. I will assume we are considering Euclidean space throughout the entire post.

Now rotational symmetry does not necessarily say anything about translational symmetry. To quote John Rennie here,

The two symmetries are not necessarily related so the two conservation laws are not necessarily related.

John Rennie rightfully invokes an example involving a system consisting of a spherically symmetric potential. For concreteness, we may consider the Coulomb potential $V(r) = A/r$ where $r$ is the distance from the origin and $A$ is some nonzero constant. This system has rotational symmetry about any axis going through the origin, and yet it very obviously has no translational symmetry.

However, there is a specialized case where a relationship between the two symmetries does exist. The key realization is that every translation in Euclidean space can be written as a composition of rotations. Because of this somewhat surprising fact, it follows that if we postulate rotational symmetry about every possible axis, then translational symmetry necessarily follows for free. In this specialized circumstance, there is a relationship between the two symmetries as you seem to be pointing out.

Again, I have to stress that this relationship only occurs in a special case. This explains why people seem to be talking past each other, and it also explains why no one is contradicting one another if you think about this carefully enough.

---

Returning back to my claim that every translation is a composition of rotations, I will demonstrate a case in two dimensional Euclidean space. The generalization is straightforward.

Lastly, it is crucial to understand that I will be talking about transformations in the active sense.

Suppose we have a translation of the form $T:(x, y)\mapsto (x+a, y)$. To write this in terms of rotations, define $R_{1}:(x, y)\mapsto (-x+a/2, -y)$ and $R_{2}:(x', y')\mapsto (-x'+3a/2, -y')$. Here $R_{1}$ is a counterclockwise $180^{\circ}$ rotation about the point $(a/4, 0)$, and $R_{2}$ is a counterclockwise $180^{\circ}$ rotation about the point $(3a/4, 0)$. The result, as one can check, is that $$ T = R_{2}\circ R_{1}. $$ We can even demonstrate this visually (again keep in mind these are active transformations acting on points).

To generalize, simply consider that any translation of the form $(x, y, z)\mapsto (x+a, y+b, z+c)$ is really just three translations $(x, y, z)\mapsto (x+a, y, z)\mapsto (x+a, y+b, z)\mapsto (x+a, y+b, z+c)$.

Answer 3

About linear momentum:
In order to have a definable linear momentum at all having a space with a single spatial degree of freedom to move along is already sufficient (obviously). Example: the physics demonstration that is known as an air rail.

About angular momentum:
In order to have a definable angular momentum the minimum is a space with two spatial degrees of freedom. The simplest case is two objects exerting a force upon each other, deflecting each other's motion. That motion is in a plane. (If the motion is in a space with three spatial degrees of freedom we can always transform to a coordinate system such that the plane of motion coincides with a plane of the coordinate system.) Demonstration is with an air table.

Here is a crucial property of any system with more than one spatial degree of freedom.

The motion of each of the participating objects can be decomposed in motion components, along the axes of the cartesian coordinate system that is used to describe the motion.

For example, the simple case of two objects of equal mass, attracting each other, circumnavigating the common center of mass in circular motion. Let the objects be named $0_1$ and $0_2$. For both $0_1$ and $0_2$ the motion can be decomposed in two perpendicular motion components. As we know, circular motion, when decomposed in two perpendicular motion components, results in two harmonic oscillations.

That is: if you project the two-dimensional circular motion onto a single degree of freedom the projected motion is harmonic oscillation. The projection of the force is according to Hooke's law. In all we have that each of the projected motions ($0_1$ and $0_2$ exerting a force upon each other, changing each other's momentum) satisfies conservation of linear momentum.

This generalizes to any form of interaction that acts from the center of mass of $0_1$ to the center of mass of $0_2$. Project the motions of the interacting objects onto two perpendicular axes. For each of the projections linear momentum is conserved

In that sense we can think of angular momentum as a combination of instances of linear momentum.

The connection between linear momentum and angular momentum can be represented in several different ways; this is one of them.

Answer 4

In Newtonian dynamics, "equal and opposite forces" has some technicalities to it that one should be aware of. The statement is that a force and its reaction are

1. equal in magnitude,
2. opposite in direction, and
3. lie on the same line.

For conservation of linear momentum, it's enough that (1) and (2) are true; (3) does not need to be true. But for conservation of angular momentum you need (3).

(As the other answer mentioned, in one dimension, (3) is not an issue because all vectors lie on the same line.)

For example, think of a particle at position $x_1=(0,1)$, and another particle at position $x_2=(0,-1)$. If you have a force $(1,0)$ on particle $1$ and $(-1,0)$ on particle 2, this would satisfy conditions (1) and (2), so it preserves linear momentum, but the forces are not on the same line, so (3) is violated and angular momentum is not conserved.

If you have conservation of angular momentum across all points in space, then yes, you have conservation of momentum too. This is particularly easy to see when you think of conserved quantities as generated by symmetries, you can get a translation (which corresponds to momentum symmetry), by rotating at opposite angles at two different points.

Answer 5

As you very well explained, you need both when describing the laws of motion for rigid bodies.

1. "Equal and opposite force" on its own isn't sufficient because the location of the forces isn't specified. This just ensures the two bodies exchange an equal amount of momentum in terms of magnitude and direction. Let us specify the two forces $\vec{F}_1$ and $\vec{F}_2$ acting on the two bodies, and look at the change in total momentum of the system

   $$ \Delta \vec{p} = \underbrace{ \int \vec{F}_1 \,{\rm d} t}_\text{body 1} + \underbrace{ \int \vec{F}_2 \,{\rm d} t}_\text{body 2} $$

   So what is the condition for total momentum to be unchanged $\Delta \vec{p}=0$?

   Since both bodies experience the same time in classical mechanics, the condition here is $\vec{F}_1 = -\vec{F}_2$ (equal and opposite forces, Newton's 2nd law).

   But what if the forces do not act through the same point?

2. "Equal and opposite force along the same line of action" is sufficient for most cases, except for some special cases. A line in space is specified by a direction (given by the force vector) and any point along the line (given by the point of contact $\vec{r}$)

   $$ \Delta \vec{L} = \underbrace{ \int \vec{r}_1\times \vec{F}_1 \,{\rm d} t}_\text{body 1} + \underbrace{ \int \vec{r}_2\times\vec{F}_2 \,{\rm d} t}_\text{body 2} $$

   So what is the condition for total angular momentum to be unchanged $\Delta \vec{L} =0$?

   Here we need condition 1. from above, in addition to having a common point where the two forces act though. Actually, the condition is that each force acts through a point along a common line $(\vec{r}_1 - \vec{r}_2 )\times \vec{F} = 0$. This is called the line of action of the force and it results in defining the line of action of momentum, commonly referred to as axis of percussion.

Your question is, do we need both, or 2. above is sufficient? My answer is yes we need both, because there are scenarios where $\vec{r}_1 \times \vec{F}_1 + \vec{r}_2 \times \vec{F}_2 =0$, but $\vec{F}_1 + \vec{F}_2 \neq 0$, which means angular momentum is conserved, but linear momentum isn't.

3. Now for a special case. Imagine an American football that flies through the air with some velocity $\vec{v}$ and at the same time a parallel spin $\vec{\omega}$ which results in linear momentum $\vec{p}$ and angular momentum $\vec{L}$ to be parallel to each other. This helical motion is the most general motion called a screw twist.

   Can a single force/impulse stop all the motion of the football? Can a force along the axis of percussion produce the change in momentum and angular momentum to stop the ball?

   The answer is no. Conditions 1. and 2. above aren't totally sufficient here. A single force cannot provide the necessary momentum pair. You need a parallel torque in addition to the force to get

_{I removed the mathematical treatment of this condition for brevity.}

Answer 6

The two arguments you give in separate posts to prove the assertion in the OP, as well as the argument given in the OP, both rely on using the fact that the total angular momentum will be separately conserved around any reference point.

However, what you are doing here is assuming that the laws of physics don’t change with respect to different points in space.

It turns out, via Noether’s Theorem, that you don’t need to go this messy route of considering the total angular momentum at each point in space.

The assumption that the laws of physics are the same at every point in space (translational invariance) by itself is sufficient to show that total linear momentum is conserved.

In other words, your argument makes it seem like total linear momentum is conserved because the total angular momentum is conserved, but the actual crux of the issue is that the total linear momentum is conserved because you are assuming physics is the same about every “reference point.”

Of course, Noether’s Theorem is only about sufficient conditions. You could try to make the point that you are using weaker assumptions.

Answer 7

Building on the debate on this subject, using classical mechanics, given FULL Conservation of Angular Momentum around all origin points, I can now offer a direct derivation of Conservation of Linear Momentum for N particles .

For N particles in a finite space for which Angular Momentum is fully conserved,
label the particles P1 .. PN
choose an arbitrary origin point P0
choose an instant in time
label the total force on each P1..PN as $\overrightarrow{f1}..\overrightarrow{fN}$
label vector from-P0-to-P1 as $\overrightarrow{r1}$
label each vector from P(n-1) to P(n) as $\overrightarrow{rn}$
AM is conserved around P0
AM is conserved around P1
(torque around P0) $= 0$ :
$0 = ((\overrightarrow{r1}) \times \overrightarrow{f1}) + ((\overrightarrow{r1} + \overrightarrow{r2}) \times \overrightarrow{f2}) + ... + ((\overrightarrow{r1} + ... + \overrightarrow{rN}) \times \overrightarrow{fN})$
(torque around P1) $= 0$ :
$0 = ((\overrightarrow{r2}) \times \overrightarrow{f2}) + ((\overrightarrow{r2} + \overrightarrow{r3}) \times \overrightarrow{f3}) + ... + ((\overrightarrow{r2} + ... + \overrightarrow{rN}) \times \overrightarrow{fN})$
(torque around P0) equals (torque around P1) :
$((\overrightarrow{r1}) \times \overrightarrow{f1}) + ... + ((\overrightarrow{r1} + ... + \overrightarrow{rN}) \times \overrightarrow{fN})$
$= ((\overrightarrow{r2}) \times \overrightarrow{f2}) + ... + ((\overrightarrow{r2} + ... + \overrightarrow{rN}) \times \overrightarrow{fN}$)
( cross product is distributive, pull out all $\overrightarrow{r1}$ terms )
$(\overrightarrow{r1} \times \overrightarrow{f1}) + ... + (\overrightarrow{r1} \times \overrightarrow{fN})$
$+ ((\overrightarrow{r2}) \times \overrightarrow{f2}) + ... + ((\overrightarrow{r2} + ... + \overrightarrow{rN}) \times \overrightarrow{fN})$
$= ((\overrightarrow{r2}) \times \overrightarrow{f2}) + ... + ((\overrightarrow{r2} + ... + \overrightarrow{rN}) \times \overrightarrow{fN})$
( cancel $\overrightarrow{r2}...\overrightarrow{rN}$ terms )
$(\overrightarrow{r1} \times \overrightarrow{f1}) + ... + (\overrightarrow{r1} \times \overrightarrow{fN}) = 0$
( group terms )
$\overrightarrow{r1} \times ( \overrightarrow{f1} + ... + \overrightarrow{fN} ) = 0$
We can choose any arbitrary P0, and therefore choose any $\overrightarrow{r1}$ .
The only way the previous equation is true for all values of $\overrightarrow{r1}$ is :
$\overrightarrow{f1} + ... + \overrightarrow{fN} = 0$
therefore Linear Momentum is conserved

--

The N=2 version is very understandable :
torque around P0 $= 0 = ((\overrightarrow{r1}) \times \overrightarrow{f1}) + ((\overrightarrow{r1} + \overrightarrow{r2}) \times \overrightarrow{f2})$
torque around P1 $= 0 = \overrightarrow{r2} \times \overrightarrow{f2}$
(torque around P0) equals (torque around P1)
$((\overrightarrow{r1}) \times \overrightarrow{f1}) + ((\overrightarrow{r1} + \overrightarrow{r2}) \times \overrightarrow{f2}) = (\overrightarrow{r2}) \times \overrightarrow{f2}$
( cross product is distributive, pull out all $\overrightarrow{r1}$ terms )
$(\overrightarrow{r1} \times \overrightarrow{f1}) + (\overrightarrow{r1} \times \overrightarrow{f2}) + (\overrightarrow{r2} \times \overrightarrow{f2}) = \overrightarrow{r2} \times \overrightarrow{f2}$
( cancel $\overrightarrow{r2}$ terms )
$(\overrightarrow{r1} \times \overrightarrow{f1}) + (\overrightarrow{r1} \times \overrightarrow{f2}) = 0$
( group terms )
$\overrightarrow{r1} \times (\overrightarrow{f1} + \overrightarrow{f2}) = 0$
We can choose any arbitrary P0, and therefore choose any $\overrightarrow{r1}$ .
The only way the previous equation is true for all values of $\overrightarrow{r1}$ is :
$\overrightarrow{f1} + \overrightarrow{f2} = 0$
therefore Linear Momentum is conserved

--

So, unless someone can point out an error in my logic, I think this is the answer .

FULL Conservation of Angular Momentum on N Particles proves and enforces Conservation of Linear Momentum .

( Conservation of Linear Momentum does not enforce Conservation of Angular Momentum )

( These proofs do not work for systems that conserve AM only around a single point )

Answer 8

Not only does conservation of Angular Momentum $\vec{L}$ imply and enforce conservation of Linear Momentum $\vec{p}$, but you can calculate the value of $\vec{p}$ from $\vec{L}$ :

$$ \vec{p} = ( ( \frac {\partial \vec{L}}{\partial z} \times \vec{i} ) \times -\vec{j} ) + ( ( \frac {\partial \vec{L}}{\partial x} \times \vec{j} ) \times -\vec{k} ) + ( ( \frac {\partial \vec{L}}{\partial y} \times \vec{k} ) \times -\vec{i} ) $$

Proof:

Given N bodies with positions $\vec{r1} .. \vec{rN} ,$

at any instant of time, Angular Momentum around ( variable origin point ) $\vec{G} : $

$ \vec{L}(\vec{G}) $$ = \sum (\vec{r}_i - \vec{G}) \times \vec{p}_i $

( consider the partial derivative varying $\vec{G}$ by $\partial z ) $

$ \frac {\partial \vec{L}(\vec{G})}{\partial z} = \sum (\frac {\partial \vec{r_i}}{\partial z} - \frac {\partial \vec{G}}{\partial z } ) \times \vec{p}_i$

( $\vec{r_i}, \vec{p_i} $ are each constant with respect to variable $\vec{G}$ )

$ \frac {\partial \vec{L}(\vec{G})}{\partial z} = - \sum ( \frac {\partial \vec{G}}{\partial z } ) \times \vec{p}_i $

( $\frac {\partial \vec{G}}{\partial z } = [0,0,1]$ )

$ \frac {\partial \vec{L}(\vec{G})}{\partial z} = - \sum [0,0,1] \times \vec{p}_i $

$ \frac {\partial \vec{L}(\vec{G})}{\partial z} = - \sum [- {p}_{iy}, p_{ix}, 0] $

$ \frac {\partial \vec{L}(\vec{G})}{\partial z} = \sum [{p}_{iy}, - p_{ix}, 0] $

$ \frac {\partial \vec{L}(\vec{G})}{\partial z} = [{p}_{y}, - p_{x}, 0]$

( concentrate on $p_x$ )

$ \frac {\partial \vec{L}(\vec{G})}{\partial z} \times \vec{i} = [ {p}_{y} , -p_{x} , 0 ] \times \vec{i} = [0, 0, p_x]$

$ ( \frac {\partial \vec{L}(\vec{G})}{\partial z} \times \vec{i} ) \times (-\vec{j}) = [0, 0, p_x] \times (-\vec{j}) = [p_x, 0, 0 ] $

( I spare you the separate derivation of $p_y, p_z ) :$

$ ( \frac {\partial \vec{L}(\vec{G})}{\partial x} \times \vec{j} ) \times (-\vec{k}) = [0, p_{y}, 0]$

$ ( \frac {\partial \vec{L}(\vec{G})}{\partial y} \times \vec{k} ) \times (-\vec{i}) = [0, 0, p_{z}]$

$ \vec{p_{total}} = ( ( \frac {\partial \vec{L}(\vec{G})}{\partial z} \times \vec{i} ) \times -\vec{j} ) + ( ( \frac {\partial \vec{L}(\vec{G})}{\partial x} \times \vec{j} ) \times -\vec{k} ) + ( ( \frac {\partial \vec{L}(\vec{G})}{\partial y} \times \vec{k} ) \times -\vec{i} ) $

Proven .

$--$

Also, this leads to the equation ( which I numerically verified) : $$\vec{L}(\vec{r}) = \vec{L}(0,0,0) - \vec{r} \times \vec{p}$$

$--$

For a demonstration, I generated random numbers in Excel for 3 bodies :
B1 : 87 kg, $\vec{r1} = [ 56, 72, 41] m ,\; \vec{v1} = [-13,-69, 2] m/s $
B2 : 24 kg, $\vec{r2} = [-85,-99, 76] m ,\; \vec{v2} = [ 46, 80,-93] m/s $
B3 : 75 kg, $\vec{r3} = [ 43,-23, 78] m ,\; \vec{v3} = [-22, 37,-26] m/s $

$ \vec{p1} = [-1131,-3864, 144] kg m/s $
$ \vec{p2} = [ 1104,-6800, 9207] kg m/s $
$ \vec{p3} = [-1650, 1591, 598] kg m/s $

$\vec{p}_{total} = [-1677,-9073,9949] kg m/s $

$L(0,0,0) = [-363753, 657650, 582807]$
$L(1,0,0) = [-363753, 667599, 591880]$
$L(0,1,0) = [-373702, 657650, 581130]$
$L(0,0,1) = [-372826, 659327, 582807]$
kg $m^2$ / s

( partial derivative is constant )

$\frac {\partial \vec{L}}{\partial x} = L(1,0,0) - L(0,0,0) = [ 0, 9949, 9073]$
$\frac {\partial \vec{L}}{\partial y} = L(0,1,0) - L(0,0,0) = [-9949, 0,-1677]$
$\frac {\partial \vec{L}}{\partial z} = L(0,0,1) - L(0,0,0) = [-9073, 1677, 0]$
kg m / s

$\frac {\partial \vec{L}}{\partial x} \times \vec{j} = [-9073, 0, 0]$
$\frac {\partial \vec{L}}{\partial y} \times \vec{k} = [ 0, 9949, 0]$
$\frac {\partial \vec{L}}{\partial z} \times \vec{i} = [ 0, 0, -1677]$

$( \frac {\partial \vec{L}}{\partial x} \times \vec{j} ) \times -\vec{k} = [ 0, -9073, 0]$
$( \frac {\partial \vec{L}}{\partial y} \times \vec{k} ) \times -\vec{i} = [ 0, 0, 9949]$
$( \frac {\partial \vec{L}}{\partial z} \times \vec{i} ) \times -\vec{j} = [-1677, 0, 0]$

$ (( \frac {\partial \vec{L}}{\partial x} \times \vec{j} ) \times -\vec{k}) + (( \frac {\partial \vec{L}}{\partial y} \times \vec{k} ) \times -\vec{i}) + (( \frac {\partial \vec{L}}{\partial z} \times \vec{i} ) \times -\vec{j})$
$ = [-1677, -9073, 9949]$
$ = \vec{p} $
( Same numeric answer as above - so it works )

So Linear Momentum is made of the ( re-ordered ) ( constant ) partial derivatives of Conserved Angular Momentum !

Please check my work, find any errors, and do the stackExchange "is this useful or interesting ? " thing .

Answer 9

You can not estimate linear momentum from the angular momentum, no matter how far you put origin.

If you put your origin at z=1 million meters, you can estimate only x and y component of linear momentum from the angular momentum calculated around these origin.

Similarly, for the particles colliding in z-direction, the conservation of angular momentum is trivially satisfied because the origin is located in the same line of the collision. and just because the angular momentum is conserved here, it can not be said that linear momentum is also conserved.