# Lack of physical understanding of Noether's theorem

1. Apr 7, 2015

### Coffee_

Let me first give a quick sketch of how Noether's theorem was stated in class and then explain what is not very clear to me.

Consider for simplicity the Lagrangian of a single coordinate $L(q,\dot{q},t)$. Now, if there exists a variation of the coordinate $\delta q$ for which at any time the corresponding variation in the Lagrangian $\delta L=\frac{dF(q,t)}{dt}$ for some function $F(q,t)$, then :

$\frac{\partial L}{\partial \dot{q}} \delta{q} - F(q,t) = constant$

I understand the math between the steps, I also can apply this to simple cases like conservation of total momentum and angular momentum for central forces and stuff like that. When I start to think about the meaning I fail. In 1) I state my physical understanding of the law and in 2) I state what confuses me about statements I find on things like wikipedia.

1) How I understand it, the statemlent $\delta L=\frac{dF(q,t)}{dt}$ is fully equivalent with saying that $L'=L+\delta L$ gives the same mechanical equations as $L$. So my best understanding of the law at this point is exactly this:

I take a function $\delta{L}$ that corresponds to certain variations of the coordinates, and add it to the Lagrangian $L$, resulting in the new lagrangian $L'$.If the mechanical differential equations are the same for $L'$ and $L$, then the expressions I mentioned earlier is conserved.

2) However I have often encountered statements about this law that ''If the laws of motion don't change after a certain change in the system, then there is a conserved quantity''. What are these ''laws of motion''? IF they are the differential equations describing the system, then obviously the expression $\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q}$ won't change, no matter what you do to the coordinates. So to me, the statement above doesn't make sense. No matter what you do to the coordinates, the mechanical equation of your original Lagragian stay the same.

To conclude: I don't understand the physical meaning of the new lagrangian $L'$ and how it relates to the ''physical laws not changing under a rotation, or translation..''. The E-L lagrange equations clearly do hold for my system at any configuration, so that law remains unchanged no matter what. Because I understood that this is a very important law I'm trying to think really deep about this but it confuses me to the point that I have troubles expressing what exactly is not clear.

Last edited: Apr 7, 2015
2. Apr 7, 2015

### ShayanJ

Your statement of the theorem in 1 seems right to me.
And about 2, You're misunderstanding the theorem. The equation $\frac{d}{dt}\frac{\partial L}{\partial \dot q}=\frac{\partial L}{\partial q}$ doesn't depend on the coordinates but it only means that whatever coordinates you use to formulate the problem and write the Lagrangian in terms of, the equations of motion are obtained by the same form of equation, but you should notice that the equations of motions may have different forms in terms of different coordinate systems. For example, the equation of motion of a point mass m under the gravitational force of a mass M is given in Cartesian coordinates by the equations $m\ddot x=-\frac{GMmx}{(x^2+y^2)^2}$ and $m\ddot y=-\frac{GMmy}{(x^2+y^2)^2}$ and in polar coordinates, by the equations $m \ddot \rho=m\rho\dot\varphi^2-\frac{GMm}{\rho^2}$ and $m \rho^2 \dot\varphi=const$. As you can see, the equations are quite different but they all can be found from $\frac{d}{dt}\frac{\partial L}{\partial \dot q}=\frac{\partial L}{\partial q}$ .

3. Apr 7, 2015

### Coffee_

Then what do statements like ''The laws are invariant under for ex. translations, and thus this conserved quantity can be associated with this symmetry'' mean? What is meant by ''the laws are invariant''? Also what is the physical meaning of $L'$? (Of this new Lagrangian where you add the variation of the Lagrangian)

4. Apr 7, 2015

### ShayanJ

If you rotate the coordinate system, the equations are still $m\ddot x=-\frac{GMmx}{(x^2+y^2)^2}$ and $m\ddot y=-\frac{GMmy}{(x^2+y^2)^2}$, $m \ddot \rho=m\rho\dot\varphi^2-\frac{GMm}{\rho^2}$ and $m \rho^2 \dot\varphi=const$, but only in terms of the primed coordinates. This is what we call invariance under the mentioned transformations. But translation will change these equations because they're derived under the assumption that the origin of the coordinates is the centre of the force.
$L$ and $L'$ are the Lagrangians in the two different coordinate systems. For the above example, they're the same but it may be that they're different by a total time derivative term which is still called symmetry because it doesn't change the equations of motion.

5. Apr 7, 2015

### Coffee_

Thanks I'm feeling that I'm getting closer to having the 'click' where it all makes sense. I will write out my understanding of things until now based on your explanation, and would love feedback.

A Lagrangian of a system $L(q,\dot{q},t)$ can have a different forms of the function itself (not value) based on your frame. As long as Lagrangians from two different frames have the same function-shapes(/forms?) , up to a total time derivative, then the frames can be considered as equivalent. So when I'm looking at a system of N particles, and translate them all by the same vector $\vec{a}$ what I'm doing is switching form my frame to the frame that is translated by $-\vec{a}$. The Lagrangian $L'$ is then the Lagrangian that one would construct starting from this new translated frame. Saying that ''laws are invariant'' means that the differential equations in the coordinates themselves, are the same from both frames. Since translating the frame by $-\vec{a}$ or translating the full system by $\vec{a}$ is equivalent I conclude that translating the system by $\vec{a}$ keeps the laws invariant.

6. Apr 7, 2015

### ShayanJ

Yeah, right. The only thing is that your conclusion about translations in not true in general, as I mentioned in my last post.

7. Apr 7, 2015

### Coffee_

Oh yeah I meant this for central force interactions between the particles with no outside field only. One last comment about this and then I think I'm ready to go on to the next chapter ^.^ - In this reasoning above I (you earlier actually) defined ''law invariance'' as something in regards to changing reference frames. Now for this very simple example I have managed to equate the translation of the system to a change of reference frames. However, what if I consider variations in the coordinates that can't be equated to changing frames? Like doing a variation of the coordinates of one specific particle while keeping all the rest constant. Then the reasoning above with ''law invariance'' as changing frames will not be usable anymore.

8. Apr 7, 2015

### ShayanJ

As you mentioned before, e.g. a rotation can be considered as an active or a passive transformation. Now about translating only one particle of the system, its true that it doesn't correspond to a change of coordinates so it can't be considered as a passive transformation. But I think we can consider it as an active transformation. So we actually take only one particle of the system and move it and see what happens.

9. Apr 7, 2015

### Coffee_

When doing an active transformation on a mass in a central potential field, do we move the potential source as well? So for example consider a mass on a spring, and we do an active coordinate transformation, does the equilibrium position of the spring translate as well or only the mass?

10. Apr 7, 2015

### ShayanJ

If you translate them, they'll be translated and if you don't translate them, they won't be translated! It depends on what transformation you're doing.

11. Apr 8, 2015

### vanhees71

Put a bit differently, the Euler-Lagrange equations are invariant under general coordinate transformations (diffeomorphisms). The Lagrangian behaves as a scalar field under these transformations, i.e.,
$$L'(q',\dot{q}')=L(q,\dot{q})=L[q(q'),\dot{q}(q',\dot{q}')].$$
Now a (special class of) symmetry transformations is if you don't only have this general covariance, but if the Lagrangian itself is invariant, i.e., if
$$L'(q',\dot{q}')=L(q',\dot{q}').$$

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