Find a transformation that leaves the given Lagrangian invariant

In summary, the conversation discusses the use of rotations as a symmetry in the Lagrangian, particularly in the 2D case. The components of the tangent vector field for rotations are derived and it is shown that rotations are indeed a symmetry of the action. However, the application of this idea to the given Lagrangian is causing difficulties.
  • #1
JD_PM
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Homework Statement
Find a transformation that leaves the given Lagrangian invariant.

Show explicitly that the transformation you have chosen is a symmetry of the action.
Relevant Equations
Please see the given Lagrangian below
The given Lagrangian is:

##L = \frac 1 2 m_1 ( \dot x_1^2 + \dot y_1^2 + \dot z_1^2) + \frac 1 2 m_2 ( \dot x_2^2 + \dot y_2^2 + \dot z_2^2) + G \frac{m_1 m_2}{\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}}##Please note: I have been inspired by the post: https://www.physicsforums.com/threa...rem-to-get-a-constant-of-motion.982721/page-2. Particularly in Orodruin's #39

OK the first is to propose a transformation 'candidate'. Let's pick rotation.

Defining rotation in 3D as follows:

$$x \to x \cos(\lambda) + y \sin(\lambda)$$

$$y \to -x\sin(\lambda) + y \cos(\lambda)$$

$$z \to z$$

By taking the partial derivative of ##x##, ##y## and ##z## with respect to ##\lambda## we get the components of the tangent vector field to such a transformation:

$$v^1 = \left.\frac{dx}{d\lambda}\right|_{\lambda = 0} = y, \qquad
v^2 = \left.\frac{dy}{d\lambda}\right|_{\lambda = 0} = -x, \qquad
v^3 = \left.\frac{dz}{d\lambda}\right|_{\lambda = 0} = 0$$

Thus the vector field is:

$$\vec v = \frac{d\vec x}{d\lambda} = y\vec e_1 - x \vec e_2$$

The point is to show if the following transformation is a symmetry of the action:

$$\vec x \to \vec x + \lambda \vec v + \mathcal O(\lambda^2)$$

The time derivative of it is

$$\dot{\vec x} \to \dot{\vec x} + \lambda \dot{\vec v} + \mathcal O(\lambda^2)$$

OK let me at this point drop masses 1 and 2 the constant G and introduce ##\gamma## (which has dimensions ##L^2/T##), so that we get a cleaner picture. We now have:

$$L = \dot x_1^2 + \dot y_1^2 + \dot z_1^2 + \dot x_2^2 + \dot y_2^2 + \dot z_2^2 + \frac{\gamma}{\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}}$$

It follows that

$$L = \dot{\vec x_1}^2 + \dot{\vec x_2}^2 + \frac{\gamma}{\sqrt{\dot{\vec x_1}^2 - \dot{\vec x_2}^2}}$$

But this does not look right; If we do the algebra we do not get ##L \to L + \mathcal O(\lambda^2)##

What am I missing?

Any help is appreciated.

Thank you.
 
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  • #2
Of course the transformations are
$$\vec{x}_1'=\vec{x}_1 + \lambda \vec{v}_1, \quad \vec{x}_2'=\vec{x}_2+\lambda \vec{v}_2$$
and analogously for ##\dot{\vec{x}}_1## and ##\dot{\vec{x}}_2##.

Of course everything is much simpler by just writing
$$\vec{x}_1'=\hat{O} \vec{x}_1, \quad \vec{x}_2'=\hat{O} \vec{x}_2, \quad \hat{O} \in \mathrm{SO}(3).$$
The infinitesimal version is
$$\hat{O}=\hat{1}+\hat{\epsilon}, \quad \hat{\epsilon}^{\text{T}}=-\hat{\epsilon}.$$
 
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  • #3
vanhees71 said:
Of course everything is much simpler by just writing
$$\vec{x}_1'=\hat{O} \vec{x}_1, \quad \vec{x}_2'=\hat{O} \vec{x}_2, \quad \hat{O} \in \mathrm{SO}(3).$$
The infinitesimal version is
$$\hat{O}=\hat{1}+\hat{\epsilon}, \quad \hat{\epsilon}^{\text{T}}=-\hat{\epsilon}.$$

Sorry Sr but I am not familiar with the terminology you used here. I am also not familiar with 3D rotational groups (I have checked its wiki article though https://en.wikipedia.org/wiki/3D_rotation_group).

Could you please either briefly explain it or suggest a book I could read to learn?

Thanks.
 
  • #4
JD_PM said:
...

Please note: I have been inspired by the post: https://www.physicsforums.com/threa...rem-to-get-a-constant-of-motion.982721/page-2. Particularly in Orodruin's #39

...

I've been thinking that it would be nicer for you all if I explicitly showed Orodruin's post:

Orodruin said:
Instead of that Lagrangian, let us consider the two-dimensional case, and instead of translations, let us consider rotations (to get your mind away from fixating on translations, which I believe is stopping you from reaching insight).

The 2D Lagrangian of relevance is
$$
L = \dot x^2 + \dot y^2.
$$
We can add a rotationally invariant potential term to this, but let us keep from doing that at the moment.

A rotation is defined by ##x \to x \cos(\lambda) + y \sin(\lambda)##, ##y \to -x\sin(\lambda) + y \cos(\lambda)##. The components of the tangent vector field to this rotation is given by taking the derivative with respect to ##\lambda##:
$$
v^1 = \left.\frac{dx}{d\lambda}\right|_{\lambda = 0} = y, \qquad
v^2 = \left.\frac{dy}{d\lambda}\right|_{\lambda = 0} = -x.
$$
The vector field generating rotations is therefore given by ##\vec v = d\vec x/d\lambda = y\vec e_1 - x \vec e_2##. Note that, ##\lambda##, ##\vec x \to \vec x + \lambda \vec v + \mathcal O(\lambda^2)##.

Is this a symmetry of the action? The time derivative of ##\vec x## transforms as
$$
\dot{\vec x} \to \dot{\vec x} + \lambda \dot{\vec v} + \mathcal O(\lambda^2).
$$
It follows that
$$
L = \dot x^2 + \dot y^2 = \dot{\vec x}^2 \to [\dot{\vec x} + \lambda \dot{\vec v} + \mathcal O(\lambda^2)]^2
= \dot{\vec x}^2 + 2\lambda \dot{\vec x} \cdot \dot{\vec v} + \mathcal O(\lambda^2).
$$
We find
$$
\dot{\vec x} \cdot \dot{\vec v} = (\dot x \vec e_1 + \dot y \vec e_2) \cdot (\dot y \vec e_1 - \dot x \vec e_2)
= \dot x \dot y - \dot y \dot x = 0
$$
and thus
$$
L \to L + \mathcal O(\lambda^2).
$$
Thus, the derivative of the Lagrangian is zero at ##\lambda = 0## (and therefore everywhere) and it follows that the rotations indeed are a symmetry of the action.
...

My approach is using the idea above on the given Lagrangian, but I am stuck...

Thank you.
 
  • #5
Can you show us what you get when you inteoduce the transformation?
 
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  • #6
It would be cleaner to re-express the Lagrangian in terms of new independent variables, namely center-of-mass and relative coordinates, and reduced mass. Then the problem is obviously rotation-invariant with respect to the center-of-mass origin.

(Googling for the terms I mentioned above yields many learning resources.)
 
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  • #7
strangerep said:
It would be cleaner to re-express the Lagrangian in terms of new independent variables, namely center-of-mass and relative coordinates, and reduced mass. Then the problem is obviously rotation-invariant with respect to the center-of-mass origin.

(Googling for the terms I mentioned above yields many learning resources.)
That is a great advice. Let's do it.

Let's go step by step. The provided Lagrangian describes a binary system (rough sketch below):

Screenshot (1040).png


Where ##G## denotes the center of mass of the two body problem (in astronomy it is called barycenter if I am not mistaken) and ##O## the origin of cartesian coordinates.

Let's analyse when both masses are aligned with ##G## (rough sketch below):

Screenshot (1041).png


We can use the definition of center of mass to get:

$$\mathbf R(m_1 + m_2) = m_1 \mathbf r_1 + m_2 \mathbf r_2 \ \ \ \ (1)$$

Now let's analyse when both masses are not aligned with ##G## (rough sketch below):

Screenshot (1042).png


Based on the diagram we can get two more equations for ##\mathbf r_1## and ##\mathbf r_2##

$$\mathbf r_1 = \mathbf R + \mathbf r_1' \ \ \ \ (2)$$

$$\mathbf r_2 = \mathbf R + \mathbf r_2' \ \ \ \ (3)$$

We now can get equations for ##\mathbf r_1'## and ##\mathbf r_2'## in function of ##\mathbf r## only by applying some algebra: solve equation ##(1)## for ##\mathbf R## and plug it into ##\mathbf r_1' = \mathbf r_1 - \mathbf R## to get

$$\mathbf r_1' = \frac{m_2 \mathbf r}{m_1 + m_2} \ \ \ \ (4)$$

Now plugging the solved-for-##\mathbf R## equation ##(1)## into ##\mathbf r_2' = \mathbf r_2 - \mathbf R## we get

$$\mathbf r_2' = -\frac{m_1 \mathbf r}{m_1 + m_2} \ \ \ \ (5)$$

Where I have used the definition

$$\mathbf r := \mathbf r_1 - \mathbf r_2 = \mathbf r_1' - \mathbf r_2' \ \ \ \ (6)$$

OK so far! :)

Now we note that the given Lagrangian is a function of ##\mathbf r_1##, ##\mathbf r_2##, ##\mathbf {\dot r_1}## and ##\mathbf {\dot r_2}## (AKA ##L(\mathbf r_1, \mathbf r_2, \mathbf {\dot r_1}, \mathbf {\dot r_2})##). After googling a bit I found that I should go for a Lagrangian ##L(\mathbf R, \mathbf r)## (see for instance the attached file, beginning of page 8).To do so we need to apply the following equations into the given Lagrangian (which is ##L = \frac 1 2 m_1 \mathbf {|\dot r_1|^2} + \frac 1 2 m_1 \mathbf {|\dot r_2|^2} + G \frac{m_1 m_2}{|\mathbf r_1 - \mathbf r_2|}##):

$$\mathbf r = \mathbf r_1 - \mathbf r_2$$

$$\mathbf r_1 = \mathbf R + \mathbf r_1$$

$$\mathbf r_2 = \mathbf R + \mathbf r_2$$

Noticing that

$$\mathbf {|\dot r_1|^2} = \mathbf {|\dot R|^2} + \frac{m_2^2}{(m_1 + m_2)^2} \mathbf {|\dot r|^2} \ \ \ \ (7)$$

$$\mathbf {|\dot r_2|^2} = \mathbf {|\dot R|^2} + \frac{m_1^2}{(m_1 + m_2)^2} \mathbf {|\dot r|^2} \ \ \ \ (8)$$

Thus we get

$$L = \frac 1 2 \Big( m_1 \mathbf {|\dot R|^2} + \frac{m_1 m_2^2}{(m_1 + m_2)^2} \mathbf {|\dot r|^2} + m_2 \mathbf {|\dot R|^2} + \frac{m_2 m_1^2}{(m_1 + m_2)^2} \mathbf {|\dot r|^2} \Big) + G \frac{m_1 m_2} {|\mathbf r|}$$

Simplifying we get

$$L = \frac 1 2 ( m_1 + m_2 ) \mathbf {|\dot R|^2} + \frac{m_1 m_2}{2(m_1 + m_2)} \mathbf {|\dot r|^2} + G \frac{m_1 m_2} {|\mathbf r|}\ \ \ \ (9)$$
 

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  • #8
Orodruin said:
Can you show us what you get when you inteoduce the transformation?

Yes. It is true I should have included it in #1.

After rewriting the Lagrangian as strangerep suggested, Is it best if I introduce the transformation to the simplified Lagrangian

$$L = \frac 1 2 ( m_1 + m_2 ) \mathbf {|\dot R|^2} + \frac{m_1 m_2}{2(m_1 + m_2)} \mathbf {|\dot r|^2} + G \frac{m_1 m_2} {|\mathbf r|}$$

Or to the given Lagrangian?

$$L = \frac 1 2 m_1 \mathbf {|\dot r_1|^2} + \frac 1 2 m_1 \mathbf {|\dot r_2|^2} + G \frac{m_1 m_2}{|\mathbf r_1 - \mathbf r_2|}$$

Your advise is appreciated.

Thanks.
 
  • #9
JD_PM said:
[...] Simplifying we get $$L = \frac 1 2 ( m_1 + m_2 ) \mathbf {|\dot R|^2} + \frac{m_1 m_2}{2(m_1 + m_2)} \mathbf {|\dot r|^2} + G \frac{m_1 m_2} {|\mathbf r|}\ \ \ \ (9)$$
The next thing to realize is that (by translation invariance) you can change the global coordinate system so that the center of mass (at R) becomes the origin 0 in the new coordinates. In those coordinates, ##\dot R=0##, so L simplifies even more.

(##r = r_1-r_2## remains unchanged because both ##r_1## and ##r_2## are being translated by the same amount.)
 
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  • #10
strangerep said:
The next thing to realize is that (by translation invariance)
The original rotation, which is defined in the first frame, is not equivalent to rotations in the new frame, which is the frame in which OP wanted to show rotational invariance. For the purpose of this problem, it is much easier to ignore any sort of change of variables and just use ##\vec r_i \to \hat R \vec r_i##, where ##\hat R## is the rotation operator. Or one could just pick overall translations as the symmetry and do the same.

A rotation in the CoM frame is equivalent to rotation+translation in the original one. Of course, we know that translations leave the Lagrangian invariant - but the OP could then just have chosen translations.
 
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  • #11
You are making it much harder than it needs to be. The important thing to recall is that, for rotations, the transpose is the inverse.

So if you have a vector ##\vec{x}## and you want to take the square, in matrix notation it's like so.

$$ [x, y, z]
\left[ \begin{matrix}
x \\
y \\
z
\end{matrix} \right]
$$

So if ##M## is the matrix giving the rotation, then the square of a vector is just ##M \vec{x}##. So then the square of the rotated vector is just this.

$$ [x, y, z] M^T
M \left[ \begin{matrix}
x \\
y \\
z
\end{matrix} \right]
$$

From that it's pretty easy to see your Lagrangian is invariant under rotation.
 
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  • #12
Orodruin said:
[...] not equivalent to rotations in the new frame, which is the frame in which OP wanted to show rotational invariance. [...]
Ah yes, I see now -- I was "answering" the wrong question. Thanks for the correction.
 
  • #13
Orodruin said:
The original rotation, which is defined in the first frame, is not equivalent to rotations in the new frame

Why not?

Orodruin said:
For the purpose of this problem, it is much easier to ignore any sort of change of variables and just use ##\vec r_i \to \hat R \vec r_i##, where ##\hat R## is the rotation operator.

OK So you propose I stick to the given Lagrangian

$$L = \frac 1 2 m_1 \mathbf {|\dot r_1|^2} + \frac 1 2 m_2 \mathbf {|\dot r_2|^2} + G \frac{m_1 m_2}{|\mathbf r_1 - \mathbf r_2|}$$

And then apply the transformation (rotation for instance) to it.

Orodruin said:
Can you show us what you get when you introduce the transformation?

OK so the transformation we are dealing with is

$$\vec r_i \to \hat R \vec r_i$$

Where ##\hat R## is the rotation operator. More explicitely

$$x \to x\cos\theta - y\sin\theta$$

$$y \to x\sin\theta + y\cos\theta$$

Where ##x## corresponds to ##r_1## and ##y## to ##r_2##

And their derivatives wrt ##\theta## are

$$\dot x \to -x\sin\theta - y\cos\theta$$

$$\dot y \to x\cos\theta - y\sin\theta$$

Thus the vector field is ##\vec v = y\vec e_1 - x \vec e_2##

After the transformation I get the Lagrangian

$$L = \frac 1 2 m_1 |\dot y|^2 + \frac 1 2 m_2 |\dot x|^2 + G\frac{m_1 m_2}{|x-y|}$$

I think this is equivalent to the Lagrangian previous to the transformation. Thus rotation leaves the Lagrangian invariant.

But I feel like I am being sloppy (I prefer to be formal like you did in your example in the other thread we discussed). How can I improve my answer?

Thank you.
 
  • #14
strangerep said:
The next thing to realize is that (by translation invariance) you can change the global coordinate system so that the center of mass (at R) becomes the origin 0 in the new coordinates. In those coordinates, ##\dot R=0##, so L simplifies even more.

(##r = r_1-r_2## remains unchanged because both ##r_1## and ##r_2## are being translated by the same amount.)

OK so if we move to CoM frame we get that (by definition of CoM)

$$m_1 \mathbf r_1 + m_2 \mathbf r_2 = 0 \ \ \ \ (1')$$

So based on the reasoning #7 we end up with the Lagrangian

$$L = \frac{m_1 m_2}{2(m_1 + m_2)} \mathbf {|\dot r|^2} + G \frac{m_1 m_2} {|\mathbf r|}$$

Is this what you meant?
 
  • #15
JD_PM said:
Is this what you meant?
Since I've already made one boo-boo in this thread, it's probably best if I leave further explanation to Orodruin. :oldshy:
 
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  • #16
JD_PM said:
Why not?
Because boosts and rotations do not commute.
JD_PM said:
OK So you propose I stick to the given Lagrangian

$$L = \frac 1 2 m_1 \mathbf {|\dot r_1|^2} + \frac 1 2 m_2 \mathbf {|\dot r_2|^2} + G \frac{m_1 m_2}{|\mathbf r_1 - \mathbf r_2|}$$

And then apply the transformation (rotation for instance) to it.
OK so the transformation we are dealing with is

$$\vec r_i \to \hat R \vec r_i$$

Where ##\hat R## is the rotation operator. More explicitely

$$x \to x\cos\theta - y\sin\theta$$

$$y \to x\sin\theta + y\cos\theta$$

Where ##x## corresponds to ##r_1## and ##y## to ##r_2##

And their derivatives wrt ##\theta## are

$$\dot x \to -x\sin\theta - y\cos\theta$$

$$\dot y \to x\cos\theta - y\sin\theta$$

Thus the vector field is ##\vec v = y\vec e_1 - x \vec e_2##

After the transformation I get the Lagrangian

$$L = \frac 1 2 m_1 |\dot y|^2 + \frac 1 2 m_2 |\dot x|^2 + G\frac{m_1 m_2}{|x-y|}$$

I think this is equivalent to the Lagrangian previous to the transformation. Thus rotation leaves the Lagrangian invariant.

But I feel like I am being sloppy (I prefer to be formal like you did in your example in the other thread we discussed). How can I improve my answer?

Thank you.
How can it be the transformed Lagrangian if you do not even have the same number of parameters?

You do not need to use the infinitesimal form of the transformation when you show invariance. All you need to show is that the Lagrangian remains the same when you replace ##\vec r_i## by ##\hat R \vec r_i##. You do not need to specify the rotation, just use the properties of rotations, such as preserving inner products.
 
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  • #17
strangerep said:
Since I've already made one boo-boo in this thread, it's probably best if I leave further explanation to Orodruin. :oldshy:
It was not wrong, just a very non-direct way to show that rotations are a symmetry of the Lagrangian when all you are after is a symmetry. Especially as it requires to show that boosts are a symmetry as one of the steps and then you would already have a symmetry. :wink:
 
  • #18
Orodruin said:
You do not need to use the infinitesimal form of the transformation when you show invariance. All you need to show is that the Lagrangian remains the same when you replace ##\vec r_i## by ##\hat R \vec r_i##. You do not need to specify the rotation, just use the properties of rotations, such as preserving inner products.

Honestly I am afraid I am not following your explanation.

I think the best I could do is look for a good book on symmetries and Lagrange invariance, as it is clear I do not have enough knowledge yet (all I know about it is the discussion we had in the thread I linked).

What book should I go for? I’ve seen third edition of Griffiths has a symmetry chapter but of course there may be other options.

If you prefer, you could give me an example like you did in the other thread.

Thank you.
 
  • #19
JD_PM said:
If you prefer, you could give me an example like you did in the other thread.
Example: Translations in one dimension for the free particle Lagrangian ##L(x,\dot x) = m\dot x^2/2##.

Translations are defined by ##\hat T_\ell x = x + \ell##. Since ##d(\hat T_\ell x)/dt = \dot x##, this implies that ##L(\hat T_\ell x, d(\hat T_\ell x)/dt) = L(x+\ell,\dot x) = m\dot x^2/2 = L(x,\dot x)## and the Lagrangian is invariant under translations.
 
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  • #20
JD_PM said:
Honestly I am afraid I am not following your explanation.

I think the best I could do is look for a good book on symmetries and Lagrange invariance, as it is clear I do not have enough knowledge yet (all I know about it is the discussion we had in the thread I linked).

What book should I go for? I’ve seen third edition of Griffiths has a symmetry chapter but of course there may be other options.

If you prefer, you could give me an example like you did in the other thread.

Thank you.
Well, in this case you have a Lagrangian that is explicitly written with scalar products and lengths of vectors only. That implies at the first glance that it is invariant under rotations (around the center of the chosen reference frame).

Formally you see this by the fact that an arbitrary rotation matrix fulfills ##\hat{R} \hat{R}^{\text{T}}=\hat{R}^{\text{T}} \hat{R}=\hat{1}##. Vectors transform as ##\vec{V}'=\hat{R} \hat{V}##. That implies
$$\vec{V}_1' \cdot \vec{V}_2' =\vec{V}_1^{\prime \text{T}} \vec{V}_2'=\vec{V}_1^{\text{T}} \hat{R}^{\text{T}} \hat{R}^{\text{T}} \vec{V}_2 = \vec{V}_1 \cdot \vec{V}_2.$$
By the same argument from ##|\vec{V}|=\sqrt{\vec{V} \cdot \vec{V}}## it follows that also the length (modulus) of a vector is invariant under rotations.
 
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  • #21
Orodruin said:
Example: Translations in one dimension for the free particle Lagrangian ##L(x,\dot x) = m\dot x^2/2##.

Translations are defined by ##\hat T_\ell x = x + \ell##. Since ##d(\hat T_\ell x)/dt = \dot x##, this implies that ##L(\hat T_\ell x, d(\hat T_\ell x)/dt) = L(x+\ell,\dot x) = m\dot x^2/2 = L(x,\dot x)## and the Lagrangian is invariant under translations.

Thank you! :) I am almost there.

OK so let's make the analogy.

Orodruin said:
Example: Translations in one dimension for the free particle Lagrangian ##L(x,\dot x) = m\dot x^2/2##.

In this case we have the Lagrangian

$$L(r_1, r_2, \dot r_1, \dot r_2) = \frac 1 2 m_1 \mathbf {|\dot r_1|^2} + \frac 1 2 m_1 \mathbf {|\dot r_2|^2} + G \frac{m_1 m_2}{|\mathbf r_1 - \mathbf r_2|}$$

Which is equivalent to

$$L = \frac 1 2 m_1 ( \dot x_1^2 + \dot y_1^2 + \dot z_1^2) + \frac 1 2 m_2 ( \dot x_2^2 + \dot y_2^2 + \dot z_2^2) + G \frac{m_1 m_2}{\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}}$$

Orodruin said:
Translations are defined by ##\hat T_\ell x = x + \ell##

Rotations are defined by (checking wiki article https://en.wikipedia.org/wiki/Rotational_invariance):

$$\hat R_{\theta} x = x \cos\theta - y \sin\theta$$

$$\hat R_{\theta} y = x \sin\theta - y \cos\theta$$

$$\hat R_{\theta} z = z$$

Orodruin said:
...##d(\hat T_\ell x)/dt = \dot x##...

Here we have

$$\frac{d(\hat R_\theta x)}{dt} = \dot x \cos\theta - x \dot \theta \sin \theta - \dot y \sin\theta - y \dot \theta \cos\theta$$

$$\frac{d(\hat R_\theta y)}{dt} = \dot x \sin\theta + x \dot \theta \cos \theta - \dot y \cos\theta + y \dot \theta \sin\theta$$

$$\frac{d(\hat R_\theta z)}{dt} = \dot z$$

Here I guess we have to set ##\theta = 0##. Thus we get

$$\frac{d(\hat R_\theta x)}{dt}|_{\theta=0} = \dot x$$

$$\frac{d(\hat R_\theta y)}{dt}|_{\theta=0} = -\dot y$$

$$\frac{d(\hat R_\theta z)}{dt}|_{\theta=0} = \dot z$$

We know that

$$|r|^2 = \Big(\frac{d(\hat R_\theta x)}{dt}|_{\theta=0} \Big)^2 + \Big(\frac{d(\hat R_\theta y)}{dt}|_{\theta=0} \Big)^2 + \Big(\frac{d(\hat R_\theta z)}{dt}|_{\theta=0} \Big)^2 = \dot x^2 + \dot y^2 + \dot z^2$$

This explains the invariance of the kinetic energy part of the Lagrangian.

Let's now go for the potential part of the Lagrangian.

We also notice that

$$\hat R_{\theta} x|_{\theta=0} = x$$

$$\hat R_{\theta} y|_{\theta=0} = - y$$

$$\hat R_{\theta} z|_{\theta=0} = z$$

Thus we get

$$|r_1 - r_2| = \sqrt{(x_1-x_2)^2 + (y_2-y_1)^2 + (z_1-z_2)^2}$$

Oops the second term should be ##(y_1-y_2)^2## in order to get roational invariance...

Mmm What am I missing?
 
  • #22
You are making things way more complicated than they need to be.

1. You do not need to write out the Lagrangian in components. This just complicates things. Vector notation was invented for a reason.

2. You cannot set ##\theta = 0## arbitrarily. As this corresponds to the identity transformation it tells you absolutely nothing about the invariance of the Lagrangian. It is obvious that you should get the same Lagrangian with the identity transformation.

3. Things such as realising that ##(y_2-y_1)^2 = (y_1-y_2)^2## neef to be second nature at this point.
 
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  • #23
Orodruin said:
You are making things way more complicated than they need to be.

1. You do not need to write out the Lagrangian in components. This just complicates things. Vector notation was invented for a reason.

I agree. It is true that my approach is not the neatest. This is of course not on purpose; I worked out the Lagrangian in components because I understand it better and also because I am afraid I do not see how to generalize it to vector notation at the moment.

Orodruin said:
3. Things such as realising that ##(y_2-y_1)^2 = (y_1-y_2)^2## neef to be second nature at this point.

Big time oops XD. Of course we have

$$y_2^2 -2y_2y_1 + y_1^2 = y_1^2 -2y_1y_2 + y_2^2$$

OK so at this point I would say I have shown that the Lagrangian is invariant under rotations.

But:

Orodruin said:
2. You cannot set ##\theta = 0## arbitrarily. As this corresponds to the identity transformation it tells you absolutely nothing about the invariance of the Lagrangian. It is obvious that you should get the same Lagrangian with the identity transformation.

Honestly I set it to zero because we also did so in the other thread (in that case was ##\lambda = 0##).

Why in there was OK to set it to zero and here not?
 
  • #24
JD_PM said:
I agree. It is true that my approach is not the neatest. This is of course not on purpose; I worked out the Lagrangian in components because I understand it better and also because I am afraid I do not see how to generalize it to vector notation at the moment.

It was already in vector notation. You do not need a specific rotation, just the properties of rotations. The Lagrangian is invariant under any rotation. The only property you need is that rotations preserve inner products.

OK so at this point I would say I have shown that the Lagrangian is invariant under rotations.

No, you have not. You have shown that the Lagrangian is invariant under the identity transformation.

Honestly I set it to zero because we also did so in the other thread (in that case was ##\lambda = 0##).

Why in there was OK to set it to zero and here not?

We looked at infinitesimal transformations where only the linear contribution mattered. This is not necessary here. We certainly did not set the transformation parameter to zero.
 
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  • #25
OK so the quickest way I see to argue that the given Lagrangian is invariant under rotation is as follows:

Noticing that ##|\mathbf {\dot r}| = \sqrt{ \mathbf {\dot r} \cdot \mathbf {\dot r}}## we can rewrite the given Lagrangian

$$L = \frac 1 2 m_1 \mathbf {|\dot r_1|^2} + \frac 1 2 m_1 \mathbf {|\dot r_2|^2} + G \frac{m_1 m_2}{|\mathbf r_1 - \mathbf r_2|}$$

As follows

$$L = \frac 1 2 m_1 (\mathbf {\dot r_1} \cdot \mathbf {\dot r_1}) + \frac 1 2 m_2 (\mathbf {\dot r_2} \cdot \mathbf {\dot r_2}) + G \frac{m_1 m_2}{\sqrt{ (\mathbf {\dot r_1} - \mathbf {\dot r_2}) \cdot (\mathbf {\dot r_1} - \mathbf {\dot r_2})}}$$

As vanhees71 and Orodruin pointed out, rotations preserve dot products.

Thus the given Lagrangian is invariant under rotations.
 
  • #26
Yes, it does not have to be harder than that.
 
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  • #27
I have to say I find these kind of exercises really interesting, and I would like to learn more about it.

There are still several assertions I do not understand, like

Orodruin said:
A rotation in the CoM frame is equivalent to rotation+translation in the original one. Of course, we know that translations leave the Lagrangian invariant

Or

Orodruin said:
Because boosts and rotations do not commute.

I found that Griffiths (yeah I really like the way he explains xD) has a section about vector analysis: chapter one, section 1.1.5 'How vectors transform'. I think that is a good start (there is an exercise proving that rotations preserve dot products! :smile:) but what book should I go for next, once I finish that section?

Should I ask this question in Science and Math textbooks Forum?
 

Related to Find a transformation that leaves the given Lagrangian invariant

What is a Lagrangian?

A Lagrangian is a mathematical function that describes the dynamics of a physical system in terms of its position and velocity.

Why is it important to find a transformation that leaves the Lagrangian invariant?

It is important to find a transformation that leaves the Lagrangian invariant because it ensures that the physical laws governing the system remain the same even after the transformation is applied. This allows for a deeper understanding of the system's behavior and can lead to new insights and predictions.

What does it mean for a transformation to leave the Lagrangian invariant?

A transformation that leaves the Lagrangian invariant means that the Lagrangian remains unchanged when the transformation is applied. In other words, the equations of motion derived from the Lagrangian before and after the transformation are equivalent.

How do you find a transformation that leaves the Lagrangian invariant?

There is no one set method for finding a transformation that leaves the Lagrangian invariant. It often involves trial and error, and may require a deep understanding of the system and its symmetries. Some common techniques include using Noether's theorem or the principle of least action.

What are some real-world applications of finding a transformation that leaves the Lagrangian invariant?

Finding a transformation that leaves the Lagrangian invariant has many applications in physics, such as in classical mechanics, quantum mechanics, and field theory. It is also used in engineering and other fields to model and analyze complex systems. Some specific examples include studying the motion of particles in a magnetic field and analyzing the behavior of fluid flow in a pipe.

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