Invariance of a Lagrangian under Transformation

In summary, the conversation discusses the invariance of the Lagrangian \mathcal{L}=\frac{m}{2}\vec{\dot{r}}^2 \, \frac{1}{(1+g \vec{r}^2)^2} 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. The conversation includes an attempt at a solution, which involves calculating \tilde{r}^2 and applying
  • #1
Fragezeichen
5
0

Homework Statement


Show that the Lagrangian
[itex]\mathcal{L}=\frac{m}{2}\vec{\dot{r}}^2 \, \frac{1}{(1+g \vec{r}^2)^2}[/itex]
is invariant under the Transformation

[itex]\vec{r} \rightarrow \tilde{r}=\vec{r}+\vec{a}(1-g\vec{r}^2)+2g\vec{r}(\vec{r} \cdot \vec{a}) [/itex]

where b is a constant and [itex]\vec{a}[/itex] are infinitesimal parameters.



2. The attempt at a solution
[itex](1+g\vec{\tilde{r}}^2)^2=(1+g\vec{r}^2)^2 (1+4g(\vec{r} \cdot \vec{a}))[/itex]
[itex]\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}}}_{?}))[/itex]

Can you tell me, wheter this is OK so far?
 
Physics news on Phys.org
  • #2
Looks OK to me... What happens when you plug those things into the Lagrangian?
 
  • #3
Thanks for your answer.
[itex](\frac{d \vec{\tilde{r}}}{dt})^2[/itex] should be at least something like [itex]\vec{\dot{r}}^2(1+4g(\vec{a}\cdot \vec{r}))[/itex]

But i can´t get it into this form, especially the underbraced part of it makes me think there´s sth. wrong...
 
  • #4
Fragezeichen said:
Thanks for your answer.
[itex](\frac{d \vec{\tilde{r}}}{dt})^2[/itex] should be at least something like [itex]\vec{\dot{r}}^2(1+4g(\vec{a}\cdot \vec{r}))[/itex]

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
 
  • #5
Seems like i do the same mistake every time i try it...


[itex](\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[/itex]
[itex]=(\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[/itex]
How to deal with the underbraced term without expanding like this:



[itex]((\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}})) )[/itex]

But how to manage the remaining term?
 
Last edited:
  • #6
[tex]\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}}))[/tex]

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

[tex] \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)[/tex]
 
  • #7
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:
[itex]\vec{r} \longmapsto \vec{r}+\vec{a}\cdot \vec{\psi}(\vec{r})[/itex]

Therefore:
[itex]\vec{\psi}(\vec{r})=1-g\vec{r}^2+\underbrace{2g \vec{r}}[/itex]
But the underbraced term only produces a [itex]2g \vec{a}\cdot \vec{r}[/itex], but i need an additional [itex]\vec{r}[/itex] to get [itex]2g\vec{r}(\vec{a}\cdot \vec{r})[/itex].
 
  • #8
Fragezeichen said:
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:
[itex]\vec{r} \longmapsto \vec{r}+\vec{a}\cdot \vec{\psi}(\vec{r})[/itex]

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

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 [tex] \vec{r}\rightarrow \vec{r} + \psi(\vec{r}) \vec{a}. [/tex] However, this is not possible as you need also a component along [itex]\vec{r} [/itex]. Next possibility would be [tex] \vec{r}\rightarrow \vec{r}+ \vec{\psi}(\vec{r}) \times \vec{a} [/tex] but this doesn't work either, as the cross product is perpendicular to [itex]\vec{a} [/itex]
 
  • #9
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:

Related to Invariance of a Lagrangian under Transformation

1. What is the concept of invariance of a Lagrangian under transformation?

The concept of invariance of a Lagrangian under transformation refers to the property of a physical system's Lagrangian remaining unchanged even when the system undergoes a transformation. This means that the equations of motion and physical quantities derived from the Lagrangian will also remain unchanged.

2. What types of transformations can a Lagrangian be invariant under?

A Lagrangian can be invariant under various types of transformations, such as spatial translations, rotations, and time translations. It can also be invariant under more complex transformations, such as gauge transformations.

3. Why is invariance of a Lagrangian under transformation important?

Invariance of a Lagrangian under transformation is important because it allows for the use of powerful mathematical tools, such as Noether's theorem, to derive conservation laws and symmetries of a physical system. It also simplifies the analysis and understanding of a system's behavior.

4. How is the invariance of a Lagrangian under transformation mathematically represented?

The invariance of a Lagrangian under transformation is mathematically represented by the Lagrangian being equal to its transformed counterpart, up to a total derivative. This can be written as L = L' + dF, where L' is the transformed Lagrangian and F is a function of the system's coordinates and time.

5. What are some real-world applications of invariance of a Lagrangian under transformation?

The concept of invariance of a Lagrangian under transformation has many applications in physics, such as in classical mechanics, quantum mechanics, and field theory. It is used to derive conservation laws, symmetries, and equations of motion in various physical systems. It also plays a crucial role in the Standard Model of particle physics.

Similar threads

  • Advanced Physics Homework Help
Replies
3
Views
762
Replies
1
Views
742
  • Advanced Physics Homework Help
Replies
11
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
26
Views
4K
  • Advanced Physics Homework Help
Replies
4
Views
2K
  • Advanced Physics Homework Help
2
Replies
58
Views
5K
  • Advanced Physics Homework Help
Replies
11
Views
1K
  • Advanced Physics Homework Help
Replies
11
Views
1K
Back
Top