Free vs bound vectors and torque

Question

When considering basic Newtonian mechanics, we can treat vector as free and move their point of application at will. This is consistent with the affine nature of Euclidean space. However, when calculating torque on a body, we must treat forces as bound to their point of application. What is the mathematical reason for this? What does it imply for affine structure of Euclidean space*

Answer 1

This is because we do not distinguish between $T_p\mathcal{M}$ and $\mathcal{M} $ in the Euclidean space. Objects that you call the "free vectors" are vectors in the tangent space of a point $T_p\mathcal{M}$, so it can be freely transported by the Euclidean affine connection; however, the "bound vectors" originates from $\mathcal{M} = \mathbb{E}^3 $ (i.e. the position vectors $\vec{r}$, etc.) and have a specific reference point (the origin $O$) which must be specified in all calculations.

So, in the perspective of modern differential geometry or the notion of relativity (we are considering Newtonian mechanics so the relativity principle would be the Galilean one), the position vectors are not "vectors": they can't be true vectors because it is not invariant under arbitrary coordinate transformations, i.e. they change when we change our origin of the coordinate system. The "true" vectors start from velocity, not position. The homogeneity and isotropy of space forces the physical entities to be invariant under translation and $SO(3)$ group action; so, we only have potential energies depending on relative distances among particles, kinetic energy $\frac{1}{2} m v^2$, etc. as physical entities, not the one like potential energies depending on individual coordinates of the particles.

However, we sometimes meet equations that involve the "bound vectors." This is not problematic because the occurrence of the "bound vectors" does not affect the heart (the physical meaning) of the equation, but only reflects our particular choice of a coordinate system. Such "loss of generality (referring to particular coordinates)" disappears when the form invariance of the equation during transformations from one coordinate system to another is established. (It is analogous to the structure of gauge transformations.) For example, $\vec{\tau}$ and $\vec{L}$ are both "bound vectors" and their value differs by the choice of reference point ($O$), but $\vec{\tau} = \frac{d}{dt} \vec{L}$ holds in any frames. [To be more accurate, $\vec{\tau} = \frac{d}{dt} \vec{L}$ holds only in inertial frames; in frames those are not inertial, the "inertial terms" will be involved (think about inertial forces). If you have experiences with gauge theories, this is analogous to terms that arise from "covariant correction" (connections); for example, in Dirac Lagrangian, $\partial_\mu \rightarrow D_\mu = \partial_\mu -ie A_\mu$ will be the "covariant correction" and $A_\mu$ will be the "inertial term." Specific values would differ among the choice of gauge but the essence is maintained.]

Answer 2

You are right. Torques are not free vectors, the same way that linear velocity is not a free vector and it is associated with a specific point. When the point moves, the components of torque and velocity change according to the same transformation law

$$ \begin{align} \vec{v}_B & = \vec{v}_A + (\vec{r}_{A}-\vec{r}_B) \times \vec{\omega} \\ \vec{\tau}_B & = \vec{\tau}_A + (\vec{r}_{A}-\vec{r}_B) \times \vec{F} \end{align} $$

This is because the geometry of motions and loadings represent lines in space in plücker coordinates (6 components). The torque and linear velocity vectors represent the moment about the line and force and rotational velocity vectors the direction of the line. Together they can be used to describe full properties of the line of action for forces, and line of motion for velocities.

See this answer for more details.

To be exact, Chasles' Theorem states that motions are described by not only a line in space, but by a magnitude and a pitch representing a screw motion. Similarly loadings represent a screw (called a wrench) in space.

Answer 3

A pure vector (in $\mathbb{R}^3$) is just a magnitude and a direction. There is no geometry (point, plane or line in space) associated with it.

Some examples are

Quantity	Symbol	Description
Rotational Velocity	$\vec{\omega}$	Rotation of a body
Linear Momentum	$\vec{p}$	Momentum of a body
Force	$\vec{F}$	Net force on a body

All of the above quantities can be considered properties of the body (or frame) shared by all points on the body. You can move these vectors anywhere in space it it won't change the nature of the problem.

Now there is a corresponding list of vector quantities that are not pure vectors, but are defined as the moment-of something. These quantities usualy do not stand on their own, but require a qualifier for where in space they are measured.

Quantity	Symbol & Definition	Description
Linear Velocity	$\vec{v}_A = \vec{r}_A \times \vec{\omega}$	Velocity of a point (moment of rotation)
Angular Momentum	$\vec{L}_A = \vec{r}_A \times \vec{p}$	Ang. momentum about a point (moment of momentum)
Force Moment	$\vec{\tau}_A = \vec{r}_A \times \vec{F}$	Torque at a point (moment of force)

You can only slide the above vectors along the direction of the corresponding pure vector.

In general you need the pure vector to define the corresponding moment vector. But there are special cases of the above becoming pure vectors. These cases below also do not have any geometry associated with them

Quantity	Symbol	Condition	Description
Pure Linear Velocity	$\vec{v}$	$\vec{\omega}=0$	Pure translation
Pure Angular Momentum	$\vec{L}$	$\vec{p}=0$	Rotation about CM
Pure Force Moment	$\vec{\tau}$	$\vec{F}=0$	Pure Torque

The usefulness of the moment vectors is that they retain the geometry of the problem at hand. They describe where in space the line of action, or the rotation axis is which is additional information from the direction and magnitude the pure vectors contain.

3D Mechanics is intimately connected to important lines in space, and the pair of (pure vector, moment vector) has a name in geometry as the Plücker coordinates of a line

Line	Magnitude	Direction	Position	Pitch
Rotation axis	$$\omega = \| \vec{\omega}\| $$	$$ \hat{e} = \frac{\vec{\omega}}{\omega} $$	$$ \vec{r}_{\rm axis} = \vec{r}_A + \frac{ \vec{\omega} \times \vec{v}_A}{\omega^2} $$	$$h= \frac{ \vec{\omega} \cdot \vec{v}_A}{\omega^2} $$
Percussion axis	$$p = \| \vec{p}\| $$	$$ \hat{e} = \frac{\vec{p}}{p} $$	$$ \vec{r}_{\rm axis} = \vec{r}_A + \frac{ \vec{p} \times \vec{L}_A}{p^2} $$	$$h= \frac{ \vec{p} \cdot \vec{L}_A}{p^2} $$
Line of action	$$F = \| \vec{F}\| $$	$$ \hat{e} = \frac{\vec{F}}{F} $$	$$ \vec{r}_{\rm axis} = \vec{r}_A + \frac{ \vec{F} \times \vec{\tau}_A}{F^2} $$	$$h= \frac{ \vec{F} \cdot \vec{\tau}_A}{F^2} $$

Notice how when adding up all the independent quantities the total is 6

• Magnitude, 1 (scalar)
• Direction, 2 (unit vector)
• Position, 2 (direction along axis does not matter, vectors can slide)
• Pitch, 1 (scalar)

And the quantities described by the pair (pure vector, moment vector) is also 6. There is a 1:1 correspondence between the representations above. This is completed with the general transformation law from the special line to any point in space.

$$ \text{(moment vector at A)} = (\vec{r}_A - \vec{r}_{\rm axis}) \times \text{(pure vector)} + h \, \text{(pure vector)} $$

Answer 4

Torque is defined as the cross product of two vectors: $r$ and $F$. $r$ points from the pivot point to the point where the force $F$ is applied. Both vectors are not bound and stay the same vectors no matter where you put them.

However, if you combine them via mathematical operations, you have to stick to their rules (here: $T_1=r_2\times F_3-r_3\times F_2$ etc...). There are no implications on the affine structure of Euclidian space.