# I Noether's theorem

1. Feb 26, 2017

### Silviu

Hello! I looked over a proof of Noether theorem and I am a bit confused about the last step. So they got that $\delta q(t) p(t)$ is constant (I just took the one dimensional case here) where $\delta q$ is a variation of the q coordinate and p is the momentum conjugate of q. I am not sure I understand how does this imply that p is conserved (I guess I don't understand very well the meaning of $\delta q(t)$). Can someone please explain? Thank you!

2. Feb 28, 2017

### Orodruin

Staff Emeritus
Please provide the reference you are using or a more extended summary of its argumentation.

3. Feb 28, 2017

### Silviu

I am sorry. The book is called Geometry, Topology and Physics by M Nakahara, Second Edition, page 4. I attached the proof from the book. I understand it, except for the last part when they conclude that that relation implies that p is conserved

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4. Feb 28, 2017

### Orodruin

Staff Emeritus
He is saying that you have an expression of the form $f(t_1) = f(t_2)$ for arbitrary $t_1$ and $t_2$. Naturally, this means that the quantity is conserved. Note that it is not $p$ that is conserved, it is $p \,\delta q$.

5. Feb 28, 2017

### Silviu

I understand that the product is conserved. However, I attached you the part before what I sent previously. They first prove that if the Lagrangian doesn't depend on a coordinate, the generalized momentum associated with that coordinate is conserved. And after they show that by just using the Euler-Lagrange equation, they say "This argument can be mathematically elaborated as follows". So, by this I understand that they give another more mathematical proof of the above statement, but the final result should be the same -p is conserved. So I assume that if they prove the same thing using 2 different methods the results should be equivalent i.e. $p \delta q$ implies p is constant. Where am I wrong about this?

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6. Feb 28, 2017

### Orodruin

Staff Emeritus
The case when the Lagrangian does not depend on the coordinate itself is a special case. It corresponds to the symmetry with $\delta q$ being constant. In order to find the actual conserved quantity, you need to insert what the symmetry of the Lagrangian is, i.e., you have to specify what $\delta q$ is for the symmetry. In the case with a Lagrangian that is not dependent on $q$, you can make the transformation $q \to q + s$, where $s$ is an infinitesimal symmetry parameter, leading to $\delta q = 1$. Of course, this leads to $p \, \delta q = p$ as your conserved quantity, but the argument given is far more general.

On a side note, I do not think Nakahara is the best book to start learning this stuff from - it is not even in the first edition, which is the one I have, and it really is not the focus of the book. If the second edition is organised anything like the first, the first chapter is going through stuff that you should already know before taking on the rest of the text.

7. Feb 28, 2017

### Silviu

Thank you so much for the reply. It makes sense now. I want to read the book more for the topology and differential geometry part, but I wanted to read the first chapter too. Does it also require previous knowledge for topology and differential geometry, or for these parts they start with the basics?