Invariance of a Lagrangian under Transformation

[Fragezeichen]

Homework Statement

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?

 

[clamtrox]

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

 

[Fragezeichen]

Thanks for your answer.
(\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...

 

[clamtrox]

Thanks for your answer.
(\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...

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

 

[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 remaining term?

 

[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)

 

[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}).

 

[clamtrox]

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}).

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}

 

[Fragezeichen]

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

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