How do you show |r_1-r_2| is rotationally invariant

Homework Statement

How do you show |r_1-r_2| is rotationally invariant

The Attempt at a Solution

So i get that we need to show that it is invariant under the transformations

$r_1 \rightarrow r_1 + \epsilon (n \times r_1)$
$r_2 \rightarrow r_2 + \epsilon (n \times r_2)$

but that gives

$|r_1 + \epsilon (n \times r_1) -r_2 - \epsilon (n \times r_2)|$
and i can't see what this equals $|r_1 -r_2|$

Many thanks

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Mark44
Mentor

Homework Statement

How do you show |r_1-r_2| is rotationally invariant

The Attempt at a Solution

So i get that we need to show that it is invariant under the transformations

$r_1 \rightarrow r_1 + \epsilon (n \times r_1)$
$r_2 \rightarrow r_2 + \epsilon (n \times r_2)$

but that gives

$|r_1 + \epsilon (n \times r_1) -r_2 - \epsilon (n \times r_2)|$
and i can't see what this equals $|r_1 -r_2|$

Many thanks
What is n?

BvU
Homework Helper
2019 Award
How would you show that $|r_1|$ is invariant ?

Ray Vickson
Homework Helper
Dearly Missed

Homework Statement

How do you show |r_1-r_2| is rotationally invariant

The Attempt at a Solution

So i get that we need to show that it is invariant under the transformations

$r_1 \rightarrow r_1 + \epsilon (n \times r_1)$
$r_2 \rightarrow r_2 + \epsilon (n \times r_2)$

but that gives

$|r_1 + \epsilon (n \times r_1) -r_2 - \epsilon (n \times r_2)|$
and i can't see what this equals $|r_1 -r_2|$

Many thanks
So, you want to show that $|r_1' - r_2'| = |r_1 - r_2|,$ where $r_1' = r_1 + \epsilon (n \times r_1)$ and $r_2' = r_2 + \epsilon (n \times r_2).$ If $s = r_1 - r_2$ and $s' = r_1' - r_2'$, we have that
$$s' = s + \epsilon (n \times s),$$
so
$$s'^2 = s' \cdot s' = s^2 + \epsilon (n \times s) \cdot s + \epsilon s \cdot (n \times s) + \epsilon^2 (n \times s) \cdot (n \times s).$$
Since $n \times s \: \perp s$, the two middle terms (linear in $\epsilon$) vanish. So, you need only worry about the $\epsilon^2$ term.

It is clear that the result you want is not true, but I think that is because you misrepresented how a rotation acts on a vector. In other words, $r \to r + \epsilon (n \times r)$ is incorrect.

BvU
Homework Helper
2019 Award
The reasoning by Ray shows that an infinitesmal rotation conserves $\ \left | \vec r_1 - \vec r_2\right |$. I suppose that is sufficient to show it for a finite rotation too ?

Again: @Physgeek64 : how would you prove that $\ \left | \vec r_1 \right |\$ is invariant ?

Ray Vickson
Homework Helper
Dearly Missed
The reasoning by Ray shows that an infinitesmal rotation conserves $\ \left | \vec r_1 - \vec r_2\right |$. I suppose that is sufficient to show it for a finite rotation too ?

Again: @Physgeek64 : how would you prove that $\ \left | \vec r_1 \right |\$ is invariant ?
OK: the transformation shown sort-of-works for small $|\epsilon|$, but it fails for large values of $|\epsilon|$.

The point is that the OP's transformation equation $r \to r + \epsilon (n \times r)$ is just not correct! It describes taking a vector r and then adding something perpendicular to it on to its end. That will give a right triangle with sides $|r|$ and $|\epsilon (n \times r)|$, whose hypotenuse is not just $|r|$, but something bigger---only very slightly bigger if $|\epsilon|$ is small.

The OP needs to find a linear transformation $T = T(\epsilon)$ with the property that for small $|\epsilon|$ we have $Tr \approx r + \epsilon (n \times r)$.

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So, you want to show that $|r_1' - r_2'| = |r_1 - r_2|,$ where $r_1' = r_1 + \epsilon (n \times r_1)$ and $r_2' = r_2 + \epsilon (n \times r_2).$ If $s = r_1 - r_2$ and $s' = r_1' - r_2'$, we have that
$$s' = s + \epsilon (n \times s),$$
so
$$s'^2 = s' \cdot s' = s^2 + \epsilon (n \times s) \cdot s + \epsilon s \cdot (n \times s) + \epsilon^2 (n \times s) \cdot (n \times s).$$
Since $n \times s \: \perp s$, the two middle terms (linear in $\epsilon$) vanish. So, you need only worry about the $\epsilon^2$ term.

It is clear that the result you want is not true, but I think that is because you misrepresented how a rotation acts on a vector. In other words, $r \to r + \epsilon (n \times r)$ is incorrect.
But is this not the transformation one makes in lagrangian mechanics to then say $\frac{\partial L}{\partial \dot{q}} (n \times r) = constant$ by Noether's theorem?

Many thanks

Ray Vickson
Homework Helper
Dearly Missed
But is this not the transformation one makes in lagrangian mechanics to then say $\frac{\partial L}{\partial \dot{q}} (n \times r) = constant$ by Noether's theorem?

Many thanks
You need to distinguish between the differential relationship and the "integrated" relationship. In other words, if $\vec{r}(t) = T(t) \vec{r}_0$ we have
$$\frac{d\vec{r}}{dt} = \vec{\omega} \times \vec{r},$$
but the relationship between $\vec{r}(t)$ and $\vec{r}_0$ is not quite that simple. It is certainly true that
$$\vec{r}(t+\Delta t) \approx \vec{r}(t) + \Delta t (\vec{\omega} \times \vec{r}(t))$$
for small $|\Delta t|$, but you cannot just let $\Delta t$ become "large" in this equation and still have a correct relationship.

Note added in edit: If all you want is to show that $|\vec{r}(t)|$ remains constant (at least for constant $\vec{\omega}$) you do not actually need to find $\vec{r}(t)$ at all. Showing that $|\vec{r}|$ is constant is a lot easier than finding $\vec{r}$ itself.

Last edited:
BvU
Orodruin
Staff Emeritus
Homework Helper
Gold Member
OK: the transformation shown sort-of-works for small $|\epsilon|$, but it fails for large values of $|\epsilon|$.

The point is that the OP's transformation equation $r \to r + \epsilon (n \times r)$ is just not correct! It describes taking a vector r and then adding something perpendicular to it on to its end. That will give a right triangle with sides $|r|$ and $|\epsilon (n \times r)|$, whose hypotenuse is not just $|r|$, but something bigger---only very slightly bigger if $|\epsilon|$ is small.

The OP needs to find a linear transformation $T = T(\epsilon)$ with the property that for small $|\epsilon|$ we have $Tr \approx r + \epsilon (n \times r)$.
$\epsilon$ is never intended to be large. The relationship given is the one of an infinitesimal rotation. It is sufficient to show it to first order in $\epsilon$ as it is a direct result that the derivative is zero and therefore it extends directly to the integrated version.

BvU
BvU