1
$\begingroup$

For example, for this problem:

Consider the object in Fig. 9-2. Invert the coordinate system so that $ x \rightarrow-x, y \rightarrow-y $ and $ z \rightarrow-z $. Clearly $ \overrightarrow{\mathbf{r}} \rightarrow-\overrightarrow{\mathbf{r}} $ under this transformation. What happens to $ \overrightarrow{\boldsymbol{\tau}} $ and $ \overrightarrow{\mathbf{F}} $ ?

(A) $ \overrightarrow{\boldsymbol{\tau}} \rightarrow \overrightarrow{\boldsymbol{\tau}} $ and $ \overrightarrow{\mathbf{F}} \rightarrow \overrightarrow{\mathbf{F}} $
(B) $ \overrightarrow{\boldsymbol{\tau}} \rightarrow \overrightarrow{\boldsymbol{\tau}} $ and $ \overrightarrow{\mathbf{F}} \rightarrow-\overrightarrow{\mathbf{F}} $
(C) $ \overrightarrow{\boldsymbol{\tau}} \rightarrow-\overrightarrow{\boldsymbol{\tau}} $ and $ \overrightarrow{\mathbf{F}} \rightarrow \overrightarrow{\mathbf{F}} $
(D) $ \overrightarrow{\boldsymbol{\tau}} \rightarrow-\overrightarrow{\boldsymbol{\tau}} $ and $ \overrightarrow{\mathbf{F}} \rightarrow-\overrightarrow{\mathbf{F}} $

This is the fig. 9-2: enter image description here

This is a solution that I found (not sure if it's correct though):

If $ \vec{r} $ became $ -\vec{r} $, that change dosen't affect $ \vec{F} $ because the force is independent so it remains the same. But it do affect torque because they are in direct relation: $$ \vec{\tau}=\vec{r} \times \vec{F} $$ So if we put minus sign before $ \vec{r} $ that changes direction of $ \tau . $ Magnitude doesn't change, it just changes direction to the opposite of the initial torque. $$ \boxed{\vec{F} \rightarrow \vec{F} \quad \vec{\tau} \rightarrow-\vec{\tau}} $$

In the above solution, I don't understand why $\vec{F}$ doesn't become $-\vec{F}$ when the coordinate systems are inverted. I'm thinking that if we move $\vec{F}$ to the origin, it's akin to $\vec{r}$, so why is it that $\vec{r}$ needs to be changed to $-\vec{r}$ while $\vec{F}$ doesn't? enter image description here

$\endgroup$
4
  • 2
    $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$
    – ACuriousMind
    May 16, 2021 at 15:17
  • 1
    $\begingroup$ For future questions, you might find tex.stackexchange.com/q/116863 helpful $\endgroup$
    – jng224
    May 16, 2021 at 16:55
  • $\begingroup$ You can also use that $\vec{F}=m \frac{d^2 \vec{r }}{dt^2}$ to see why the force changes under a coordinate inversion. $\endgroup$
    – Triatticus
    May 16, 2021 at 17:29
  • $\begingroup$ "Consider the object in Fig. 9-2. Invert the coordinate system so that π‘₯β†’βˆ’π‘₯, π‘¦β†’βˆ’π‘¦ and π‘§β†’βˆ’π‘§. Clearly π«β†’βˆ’π« under this transformation." It's not clear to me. Surely changing the co-ordinate system doesn't change the vector. $\endgroup$ May 16, 2021 at 17:50

3 Answers 3

3
$\begingroup$

You have 3 different "vectors" in this problem. $F_i$ is a vector. It doesn't have an origin, and it changes sign under coordinate inversion:

$$ P(F_i) \rightarrow -F_i $$

A point, $R_i$, doesn't live in a vector space, it lives in an affine space. The non-technical explanation is that it's a an origin, $O_i$, plus some definition of an origin, $O_i$. If you change the origin, the point doesn't change, but the vector does. To make it a vector, you look at the difference between it and the origin:

$$ r_i \equiv R_i-O_i $$

Being a vector, it changes sign under coordinate inversion:

$$ P(r_i) \rightarrow -r_i $$

The torque is not really a vector. It sometimes called a pseudo-vector and/or and axial-vector. The pseudo-vector name can mean that it depends on $O_i$...if you move that, its value changes.

The axial-vector moniker indicates that it not really vector; rather, it's the non-zero parts of an antisymmetric tensor. The tensors is formed from the available dyads:

$$ T_{ij} = r_iF_j - r_jF_i $$

That tensor only has 3 independent components, so for connivence, we write it as:

$$ \tau_i = \frac 1 2\epsilon_{ijk}T_{jk} \equiv (\vec r \times \vec F)_i $$

Being a rank-2 tensor underneath, it is even under parity:

$$ P(\tau_i) \rightarrow +\tau_i $$

A deeper dive into vectors and axial vectors leads into something called geometric algebra, and the generalization to $\mathbb{R}^N$ is handled with Clifford algebras.

$\endgroup$
0
$\begingroup$

In general, under a transformation $\mathbf{T}$ that maintains the handedness of the coordinate system you have the identity

$$ \mathbf{T} ( \boldsymbol{a} \times \boldsymbol{b} ) \equiv (\mathbf{T} \boldsymbol{a}) \times ( \mathbf{T} \boldsymbol{b} ) \tag{1}$$

The transformation of the cross product of two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ equals to the cross product of two transformed vectors. This is true only when the transformation is orthonormal and does not change the lengths of vectors $\mathbf{T}^{-1} = \mathbf{T}^\top$.

Additionally, all vector quantities to be used in mechanics need to be expressed on the same basis vectors. So all quantities need to transform to maintain this consistency.

Additionally, the cross-product implies a certain handedness to the coordinate system. Equations involving $\times$ must also all be in the same right-hand (or left-hand) rule. But in this case, the flip transformation $\mathbf{T}$ changes the handedness of the cross product. As a result

$$ \mathbf{T} ( \boldsymbol{a} \times \boldsymbol{b} ) \equiv -(\mathbf{T} \boldsymbol{a}) \times ( \mathbf{T} \boldsymbol{b} ) \tag{2}$$

or more simply $\mathbf{T}=-1$ and $\boldsymbol{r}' = -\boldsymbol{r}$, $\boldsymbol{F}' = - \boldsymbol{F}$ and check the cross product

  • Preserve Handeness (1) $$ -(\boldsymbol{r}\times\boldsymbol{F}) = (-\boldsymbol{r}) \times (-\boldsymbol{F}) = \boldsymbol{r} \times \boldsymbol{F} \;\;\; \boxtimes$$
  • Reverse Handeness (2) $$ -(\boldsymbol{r}\times\boldsymbol{F}) = - (-\boldsymbol{r}) \times (-\boldsymbol{F}) = -(\boldsymbol{r} \times \boldsymbol{F}) \;\;\; \checkmark$$

So under a flip transformation (2) is true, and the signs of the cross-product components are flipped.

I think maybe the op confused $-\boldsymbol{r} \times \boldsymbol{F}$ with $(-\boldsymbol{r}) \times \boldsymbol{F}$ implying that force does not flip. But there is a double negative here, with the torque being $-(-\boldsymbol{r})\times(-\boldsymbol{F})$.

$\endgroup$
0
$\begingroup$

The cross product of two ordinary vectors $\vec A$ and $\vec B$, $\vec A \times \vec B$, is not itself an ordinary vector since its direction changes when we change the handedness of the coordinate system. An ordinary vector that has a direction independent of the coordinate system is called a polar vector. A vector whose direction depends on the handedness of the coordinate system is called an axial vector, or a pseudovector. Torque is an axial vector, as is angular velocity. Position and force are polar vectors.

See a good physics mechanics textbook for further details, such as Goldstein Classical Mechanics or Symon Mechanics.

$\endgroup$

Your Answer

By clicking β€œPost 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.