# Homework Help: Invariance of a Lagrangian under Transformation

1. Jan 8, 2013

### Fragezeichen

1. The problem statement, all variables and given/known data
Show that the Lagrangian
$\mathcal{L}=\frac{m}{2}\vec{\dot{r}}^2 \, \frac{1}{(1+g \vec{r}^2)^2}$
is invariant under the Transformation

$\vec{r} \rightarrow \tilde{r}=\vec{r}+\vec{a}(1-g\vec{r}^2)+2g\vec{r}(\vec{r} \cdot \vec{a})$

where b is a constant and $\vec{a}$ are infinitesimal parameters.

2. The attempt at a solution
$(1+g\vec{\tilde{r}}^2)^2=(1+g\vec{r}^2)^2 (1+4g(\vec{r} \cdot \vec{a}))$
$\frac{d \vec{\tilde{r}}}{dt}=\vec{\dot{r}}-2g\vec{a}(\vec{r}\cdot \vec{\dot{r}})+2g(\underbrace{\vec{\dot{r}} (\vec{a}\cdot \vec{r})+\vec{r}(\vec{a}\cdot\vec{\dot{r}}}_{?}))$

Can you tell me, wheter this is OK so far?

2. Jan 9, 2013

### clamtrox

Looks OK to me... What happens when you plug those things into the Lagrangian?

3. Jan 9, 2013

### Fragezeichen

$(\frac{d \vec{\tilde{r}}}{dt})^2$ should be at least something like $\vec{\dot{r}}^2(1+4g(\vec{a}\cdot \vec{r}))$

But i can´t get it into this form, especially the underbraced part of it makes me think there´s sth. wrong...

4. Jan 9, 2013

### clamtrox

Yeah that's what you get.. It's not terribly difficult to see either... Two of the cross terms vanish because they are proportional to a2. Then two of the terms proportional to a are equal with opposite signs so they cancel, and you're left with exactly what you were looking for

5. Jan 9, 2013

### Fragezeichen

Seems like i do the same mistake every time i try it...

$(\frac{d \vec{\tilde{r}}}{dt})^2=(\vec{\dot{r}}^2-2 \vec{a} g(\vec{r}\cdot \vec{\dot{r}})+2g \vec{\dot{r}}(-\vec{a}\cdot \vec{r})+2g\vec{r}(\vec{a}\cdot \vec{\dot{r}}))^2$
$=(\vec{\dot{r}}-2g \underbrace{(\vec{a}(\vec{r}\cdot \vec{\dot{r}})-\vec{\dot{r}}(\vec{a}\cdot \vec{r})-\vec{r}(\vec{a}\cdot \vec{\dot{r}})}_{?} )^2$
How to deal with the underbraced term without expanding like this:

$((\vec{a}(\vec{r}\cdot \vec{\dot{r}})-\vec{\dot{r}}(\vec{a}\cdot \vec{r})-\vec{r}(\vec{a}\cdot \vec{\dot{r}}))^2=\underbrace{(\vec{a}(\vec{r} \cdot \vec{\dot{r}}))^2+(\vec{\dot{r}}(\vec{a}\cdot \vec{r}))^2+(\vec{r}(\vec{a}\cdot \vec{\dot{r}}))^2}_{vanishes}-2(((\vec{a}(\vec{r}\cdot \vec{\dot{r}}))\cdot (\vec{\dot{r}}(\vec{a}\cdot \vec{r})) )-2((\vec{a}(\vec{r}\cdot \vec{\dot{r}}))\cdot(\vec{r}(\vec{a}\cdot \vec{\dot{r}})) )+2( (\vec{\dot{r}}(\vec{a}\cdot \vec{r}))\cdot (\vec{r}(\vec{a}\cdot \vec{\dot{r}})) )$

But how to manage the remaing term?

Last edited: Jan 9, 2013
6. Jan 11, 2013

### clamtrox

$$\frac{d \vec{\tilde{r}}}{dt}=\vec{\dot{r}}-2g\vec{a}(\vec{r}\cdot \vec{\dot{r}})+2g(\vec{\dot{r}} (\vec{a}\cdot \vec{r})+\vec{r}(\vec{a}\cdot\vec{\dot{r}}))$$

When you calculate $\tilde{r}^2$, all the cross terms between things containing $\vec{a}$ vanish. Therefore

$$\left(\frac{d \vec{\tilde{r}}}{dt} \right)^2 = \vec{\dot{r}}^2 + 2\vec{\dot{r}}\cdot(-2g\vec{a}(\vec{r}\cdot \vec{\dot{r}})+2g(\vec{\dot{r}} (\vec{a}\cdot \vec{r})+\vec{r}(\vec{a}\cdot\vec{\dot{r}})) + \mathcal{O}(a^2)$$

7. Jan 15, 2013

### Fragezeichen

Thank´s for your answer, now i get it. :)
Now i´m trying to apply Noether-Theorem.

At first i´d try to get the transformation in such a form:
$\vec{r} \longmapsto \vec{r}+\vec{a}\cdot \vec{\psi}(\vec{r})$

Therefore:
$\vec{\psi}(\vec{r})=1-g\vec{r}^2+\underbrace{2g \vec{r}}$
But the underbraced term only produces a $2g \vec{a}\cdot \vec{r}$, but i need an additional $\vec{r}$ to get $2g\vec{r}(\vec{a}\cdot \vec{r})$.

8. Jan 15, 2013

### clamtrox

You, or at least your notation, are going badly wrong here. Note that your transformation has to be a vector, and what you are suggesting is a scalar product between two vectors, so a scalar. Likewise your suggestion for ψ contains vectors and scalars added together, and this is clearly incorrect.

You could think that maybe your transformation is of the form $$\vec{r}\rightarrow \vec{r} + \psi(\vec{r}) \vec{a}.$$ However, this is not possible as you need also a component along $\vec{r}$. Next possibility would be $$\vec{r}\rightarrow \vec{r}+ \vec{\psi}(\vec{r}) \times \vec{a}$$ but this doesn't work either, as the cross product is perpendicular to $\vec{a}$

9. Jan 15, 2013

### Fragezeichen

I´m sorry i meant your first option.
Do you have any clue how to get ψ ?

Or at least a little hint... :-)

Last edited: Jan 15, 2013