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

• Physgeek64

## 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

## 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?

How would you show that ##|r_1|## is 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

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.

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 ?

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

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
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
To get back to @Physgeek64 : Is the reasoning by @Ray Vickson in post #4 clear to you ? And the completed proof (with the inclusion of what @Orodruin indicates in #9 ) ?

Just checking ... 