Invariant quantities of a lagrangian?

[KleZMeR]

Given a basic Lagrangian, how would I determine invariant quantities? My hunch says it would be quantities that do not depend on position or time? Saying that, perhaps using the Lagrange equation to solve for equations of motion and along the way whatever terms disappear would be my invariant quantities? This seems like a harsh oversimplification, but I can not tell? Any help understanding the invariant quantities would be appreciated. I don't want to solve the actual problem here so to speak, but rather understand the theory.

I did solve a problem a few weeks back showing that electromagnetic force was invariant under a gauge transformation, but this seems to be a different concept?

 

[voko]

You need to define "invariant", otherwise your question is meaningless. If that means "constant with respect to time", you might find it useful to consider that when ##x## is constant with respect to time, then $$ {dx \over dt} = 0 .$$ Would that remind you something in Lagrange equations?

 

[KleZMeR]

Yes, that is what I was thinking, conserved quantities. So my first term regarding some time derivative of a partial derivative of a generalized coordinate might give me zero. The wording I was missing was that the Lagrangian is invariant under symmetry.

 

[voko]

I am not sure what you actually mean now. Have you figured this out?

 

[KleZMeR]

No, I haven't. It asks: a Lagrange with kinetic energy in cartesian coordinates KE = [m/2]*(dx^2, dy^2, dz^2), and a potential in cartesian V = V(x^2 + y^2 , z) is invariant under some symmetry, find the corresponding invariant quantity. What I see is that the potential (V) is a function of some cylindrical shape. But I do not know what the "invariant quantities" are?

 

[voko]

Let's go back to the Euler-Lagrange equation: $$ {d \over dt} {\partial L \over \partial \dot q} - {\partial L \over \partial q} = 0. $$ Under what condition does it easily result in something that is conserved, i.e., whose time-derivative is zero?

 

[guitarphysics]

I think you should look into Landau's mechanics for inspiration. There, Landau shows that since the Lagrangian of a system of free particles is time-independent, energy is conserved. GIven translational invariance, we know that the total momentum is conserved. So all these symmetries lead to conservation laws.

In your problem, by "invariant quantities" they mean things like momentum and energy, which are conserved as a result of symmetries in the Lagrangian. They want you to find, given that the Lagrangian is invariant in some way, what quantity is conserved.

 

[KleZMeR]

Yes, this is where my confusion takes place! The word 'quantities' is a little vague to me. We did see that under a gauge transformation that the Lagrangian differs by a time derivative, and that the Electromagnetic force was invariant under these transformations. We saw that the equations of motion in the field was unchanged given the transformation.

But, here, we are not given any gauge transformation, but rather some seemingly general case. So my first instinct was to use the Lagrange equation. An important question is, my potential is a function of some cylinder, so what does my actual V term look like?

Regarding Landau's book, I hear nothing but good things about this, and I don't have it! :/ but I am using Goldstein, and coming across vague suggestions as well as horrible typos. Anybody who has Landau in PDF is welcome to send it my way :)

 

[guitarphysics]

Here's the Mechanics PDF, it's great :D
https://archive.org/details/Mechanics_541
And maybe you want to look into Emmy Noether's theorem (not in full detail, just to get an idea of it)- that's where all this wonderful stuff comes from, after all.

 

[voko]

Landau & Lifschitz is good stuff, yet I would suggest you should try to answer the question in #6.

 

[samalkhaiat]

No, I haven't. It asks: a Lagrange with kinetic energy in cartesian coordinates KE = [m/2]*(dx^2, dy^2, dz^2), and a potential in cartesian V = V(x^2 + y^2 , z) is invariant under some symmetry, find the corresponding invariant quantity. What I see is that the potential (V) is a function of some cylindrical shape. But I do not know what the "invariant quantities" are?

If you know the symmetry transformations, (the transformations which leave the Lagrangian invariant), you can easily deduce the corresponding constants of motion. The term “invariant quantities” is not accurate.
Consider your Lagrangian
[tex]L = \frac{ 1 }{ 2 } \left( ( \dot{ x }_{ 1 } )^{ 2 } + ( \dot{ x }_{ 2 } )^{ 2 } + ( \dot{ x }_{ 3 } )^{ 2 } \right) - V ( x_{ 1 }^{ 2 } + x_{ 2 }^{ 2 } , x_{ 3 } ) .[/tex]
By inspection, we see that both KE and PE are invariant under rotations in the [itex]x_{ 1 } x_{ 2 }[/itex]-plane. Therefore, the corresponding constant of motion is given by 3rd component of the angular momentum [itex]L_{ 3 } = x_{ 1 } p_{ 2 } - x_{ 2 } p_{ 1 }[/itex].
To see this, we need the explicit form of the transformations, i.e.,
[tex]\bar{ x }_{ 1 } = x_{ 1 } \cos ( \theta ) - x_{ 2 } \sin ( \theta ) ,[/tex]
[tex]\bar{ x }_{ 2 } = x_{ 2 } \cos ( \theta ) + x_{ 1 } \sin ( \theta ) ,[/tex]
and
[tex]\bar{ x }_{ 3 } = x_{ 3 } .[/tex]
To find the constant of motion, it is sufficient to consider the infinitesimal version of the above transformations. We obtain this by setting [itex]\cos ( \theta ) = 1[/itex] and [itex]\sin ( \theta ) = \theta[/itex],
[tex]\bar{ x }_{ 1 } = x_{ 1 } - x_{ 2 } \theta , \ \ \Rightarrow \ \ \delta x_{ 1 } = - \theta x_{ 2 } , \ \ (1a)[/tex]
[tex]\bar{ x }_{ 2 } = x_{ 2 } + x_{ 1 } \theta , \ \ \Rightarrow \ \ \delta x_{ 2 } = \theta x_{ 1 } , \ \ (1b)[/tex]
and
[tex]\delta x_{ 3 } = 0 . \ \ \ \ \ (1c)[/tex]
From these infinitesimal transformations, we find the following important relation
[tex]\frac{ d }{ d t } ( \delta x_{ i } ) = \delta ( \frac{ d }{ d t } x_{ i } ) \equiv \delta \dot{ x }_{ i } . \ \ \ \ (2)[/tex]
The fact that the transformations leave the Lagrangian invariant (i.e., symmetry transformations) mean that
[tex]0 = \delta L = \sum_{ i = 1 }^{ 3 } \frac{ \partial L }{ \partial x_{ i } } \delta x_{ i } + \sum_{ i = 1 }^{ 3 } \frac{ \partial L }{ \partial \dot{ x }_{ i } } \delta \dot{ x }_{ i } . \ \ \ (3)[/tex]
Now, if we use the Euler-Lagrange equation of motion in the first tern, and equation (2) in the second term, we find
[tex]\frac{ d }{ d t } \left( \sum \frac{ \partial L }{ \partial \dot{ x }_{ i } } \delta x_{ i } \right) = \frac{ d }{ d t } ( \sum p_{ i } \delta x_{ i } ) = 0 .[/tex]
Or, using the transformations (1a)-(1c),
[tex]\frac{ d }{ d t } ( x_{ 1 } p_{ 2 } - x_{ 2 } p_{ 1 } ) \equiv \frac{ d L_{ 3 } }{ d t } = 0 .[/tex]
Thus rotational invariance means that [itex]L_{ 3 } = x_{ 1 } p_{ 2 } - x_{ 2 } p_{ 1 }[/itex] is time-independent on the physical trajectories (constant of motion corresponding to the invariance under rotations in the xy-plane).

Sam

 

[voko]

By inspection, we see that both KE and PE are invariant under rotations in the ## x_{ 1 } x_{ 2 }##-plane. Therefore, the corresponding constant of motion is given by..

It seemed to me that KleZMeR wanted to understand the theoretical foundation for this "therefore" step.

 

[samalkhaiat]

It seemed to me that KleZMeR wanted to understand the theoretical foundation for this "therefore" step.

Did you read ONLY this paragraph? The detailed proof of "therefore step" was given just after that paragraph.

 

[KleZMeR]

Thank you for the book guitarphysics! Ok, so I'm beginning to understand this I hope. I am curious about the explicit transformations shown by Sam, are those a general form of some rotational symmetry I would use after seeing by inspection that my potential is dependent on a rotation in the x, y plane? I am just expressing rotational symmetry?

If this is the case, I am guessing that although I was not given an explicit function of V, I should still be able to detect the motion of my system?
Is this called a cyclic coordinate? I understand I can explicitly put my Lagrangian into the Lagrange equations and find what is not conserved, but I was not given an explicit function.

From equation 3 to the next step, why was the partial derivative in respect to position dropped?

 

[KleZMeR]

Oh wait, I see that [tex]P_j = dL/d \dot q_j[/tex] , being the canonical momentum.

Ok, so why isn't my tex working? Firs timer here, trying to get with the program...

 

[KleZMeR]

Oh wait, I see that [tex]P_j = dL/d \dot q_j[/tex] , being the canonical momentum.

Ok, so why isn't my tex working? Firs timer here, trying to get with the program...

oh, it's working, happy!

 

[KleZMeR]

Also, what is the difference of finding these quantities in cartesian and cylindrical coordinates? Perhaps my [tex]R x L [/tex] could be written in cylindrical? I'm not sure.

 

[voko]

Did you read ONLY this paragraph? The detailed proof of "therefore step" was given just after that paragraph.

A problem with that detailed proof is that it focuses on the details of that particular Lagrangian, and leaves open the general question. Trees vs forest, so to speak.

KleZMeR, a cyclic coordinate is one that is not present in the Lagrangian (but its time-derivative may), so $$ {\partial L\over \partial q} = 0, $$ and the E. L. equation then gives you a conserved quantity.

Practically, change coordinates in your problem so that one them is cyclic, and you have a conserved quantity.

(You can go without changing coordinates like samalkhaiat demonstrated, but that is far less obvious.)

 

[KleZMeR]

I see, so there are symmetry shifts in a translation, rotation, or time, as well as many kinetic and potential functions which can be conserved under symmetry.

I am wondering about the differences in cartesian and cylindrical coordinates? Would I just then convert my [tex](x, y, z)[/tex] accordingly to [tex](r, theta, z)[/tex].

 

[voko]

Try that and see what happens with the Lagrangian. Pay attention to what I said in #18.

 

[KleZMeR]

If I am setting up my Lagrangian correctly, here in cylindrical, I think I see immediately what will happen to [tex] dL \over dz [/tex]

[tex] L = 1/2 * m * (-theta^.*r*sin(theta)^2 + (theta^.*r*cos(theta)^2 + (z^.)^2 - mg*sqrt((r*cos(theta))^2 + (r*sin(theta))^2) [/tex]

Also, my browser isn't showing the tex sometimes, is this on my end or is the site bogging? It makes me think I'm doing my tex wrong but other's tex is also bogging.

 

[KleZMeR]

[tex]
L = 1/2 *m* (- \dot \theta*r*sin(\theta)^2 + \dot \theta*r*cos(\theta)^2) + (z^.)^2 - mg*\sqrt(((r*cos(\theta))^2 + (r*sin(\theta))^2)
[/tex]

and putting into my Euler Lagrange, taking my generalized coordinate to be [tex] \dot r, r [/tex] I get:

[tex]d \over dt [/tex][tex] (m*r*\dot\theta^2 * cos(2\theta)) - mg[/tex]

 

[voko]

I am sure you can simplify ## \sin^2 \theta + \cos^2 \theta ##.

 

[KleZMeR]

I am sure you can simplify ## \sin^2 \theta + \cos^2 \theta ##.

Yes, I did that, so my potential is just mg, and my time derivative IS dependent on [tex]r[/tex] or [tex]\dot r[/tex]

So my potential is NOT dependent on r?

 

[KleZMeR]

wait, but [tex]\dot r[/tex] is zero, so everything is conserved here?

 

[samalkhaiat]

Thank you for the book guitarphysics! Ok, so I'm beginning to understand this I hope. I am curious about the explicit transformations shown by Sam, are those a general form of some rotational symmetry I would use after seeing by inspection that my potential is dependent on a rotation in the x, y plane? I am just expressing rotational symmetry?

They are just a change from the [itex](x,y)[/itex] corrdinates to a rotated [itex]( x^{ ' } , y^{ ' } )[/itex] coordinates, i.e., rotation by an angle [itex]\theta[/itex] in the xy-plane about z. You use them because they do not change your potential, i.e., when your potential DOES NOT depend on such rotation.

If this is the case, I am guessing that although I was not given an explicit function of V, I should still be able to detect the motion of my system?
Is this called a cyclic coordinate? I understand I can explicitly put my Lagrangian into the Lagrange equations and find what is not conserved, but I was not given an explicit function.

See bellow.

From equation 3 to the next step, why was the partial derivative in respect to position dropped?

Because I used the Euler-Lagrange equation for x.
Ok, start with the general Lagrangian
[tex]L = \frac{ m }{ 2 } \dot{ r }^{ 2 } - V( r )[/tex]
In cylindrical coordinates, the Lagrangian becomes
[tex]L = \frac{ m }{ 2 } \left( \dot{ \rho }^{ 2 } + \rho^{ 2 } \dot{ \theta }^{ 2 } + \dot{ z }^{ 2 } \right) - V( \rho , \theta , z ) .[/tex]
If the symmetry in your problem is such that [itex]V[/itex] does not depend on (for example) [itex]\theta[/itex], then
[tex]\frac{ \partial V }{ \partial \theta } = 0.[/tex]
This implies (because KE also does not depend on [itex]\theta[/itex]) that
[tex]\frac{ \partial L }{ \partial \theta } = 0.[/tex]
But, the Lagrangian gives you
[tex]\frac{ \partial L }{ \partial \dot{ \theta } } = m \rho^{ 2 } \dot{ \theta } .[/tex]
So, the Euler-Lagrange equation for [itex]\theta[/itex] becomes
[tex]\frac{ d }{ d t } ( m \rho^{ 2 } \dot{ \theta } ) = 0 .[/tex]
This means that the z-component of the angular momentum is conserved (constant of motion):
[tex]L_{ z } = m \rho^{ 2 } \dot{ \theta } = m ( x \dot{ y } - y \dot{ x } ) .[/tex]

 

[samalkhaiat]

A problem with that detailed proof is that it focuses on the details of that particular Lagrangian, and leaves open the general question. Trees vs forest, so to speak.

No sir. What I have done is called Noether Theorem. It applies to ANY Lagrangian and ANY continuous symmetry transformation of that Lagrangian.

(You can go without changing coordinates like samalkhaiat demonstrated, but that is far less obvious.)

Equation (1a) and (1b) are called coordinates transformations. So, I did change the coordinates.

 

[voko]

No sir. What I have done is called Noether Theorem.

What you did is a sketch of the proof of a special case of said theorem, just like I said earlier.

It applies to ANY Lagrangian and ANY continuous symmetry transformation of that Lagrangian.

The theorem, yes. What you did, no.

Equation (1a) and (1b) are called coordinates transformations. So, I did change the coordinates.

Did you notice the word cyclic in the sentence immediately prior to the one you quoted? You did not convert to such coordinates, so you had all of the (unnecessary) complexity in your treatment of a simple special case.

 

[voko]

Yes, I did that, so my potential is just mg

Is it? Where did ##r## go?

 

[KleZMeR]

Is it? Where did ##r## go?

Ahh was I to use the chain rule on my [tex]d\over dt[/tex] term?

 

[voko]

The Lagrangian you got in polar coordinates can be simplified before you compute derivatives. Do that properly.