# A Noether's Theorem: mistake in my teacher's script?

1. Sep 8, 2016

### JulienB

Hi everybody! I'm currently studying Noether's theorem, but I'm a bit stuck around a stupid line of calculation for the variation of the symmetry. The script of my teacher says (roughly translated from German, equations left as he wrote them):

"V.2. Noether Theorem

How does the action change under arbitrary variations of the path and of the time?

$t' = t + \tau (t)$; $q_a ' (t') = q_a (t) + \Delta q_a (t)$

where $\tau$ and $\Delta q_a$ are arbitrary functions, which must not be equal zero at the boundaries. For the variation of $q_a$ by a fixed time $q_a ' (t') = q_a (t) + \delta q_a (t)$, we find a relation between the variations $\delta q_a$ and $\Delta q_a$:

$q_a ' (t') = q_a ' (t + \tau (t) = q_a ' (t) + \tau (t) \dot{q}_a (t)$
$= q_a (t) + \delta q_a (t) + \tau (t) \dot{q}_a (t)$
$\overset{!}{=} q_a (t) + \Delta q_a (t)$
$\implies \delta q_a (t) = \Delta q_a (t) - \tau (t) \dot{q}_a (t)$"

However, here is what I got when trying to repeat his steps:

$\delta q (t) = q ' (t') - q (t)$
$= q' (t + \tau (t)) - q(t) = q'(t) + \tau (t) \dot{q} ' (t) - q(t)$
$= q(t) + \Delta q(t) + \tau (t) \dot{q} ' (t) - q(t)$
$= \Delta q(t) + \tau (t) \dot{q} ' (t)$

That's close but not quite the same. He (very!) often makes little mistakes in his script, so I wondered if that was one of them. If not, I would be very grateful if someone could explain me when I went wrong :)

Julien.

2. Sep 8, 2016

### Orodruin

Staff Emeritus
These two definitions are incompatible (or equivalent - meaning $\Delta q_a = \delta q_a$ by definition). I assume that you intend for one to have $q_a'(t)$ on the LHS (or $t'$ instead of $t$ on the RHS). Without knowing which is which, it will be difficult to help you.

3. Sep 8, 2016

### JulienB

Hi @Orodruin and thank you for your answer. Those definitions unfortunately come straight out of my teacher's script... I would assume that my teacher made a mistake and -without much certainty- that $q_a ' (t) = q_a (t) + \Delta q_a (t)$ (then it would be the definition of a translation) and that $\delta q_a (t) = q_a ' (t') - q_a(t)$ (the symmetry variation). Does that seem correct to you? Unfortunately, I already used the known definitions of translation and time translation in my calculations, and I got a $+$ instead of a $-$.

Julien.

4. Sep 8, 2016

### Orodruin

Staff Emeritus
Well, then the result depends on which of the $\Delta q$ and $\delta q$ you change the definition for.

5. Sep 8, 2016

### Orodruin

Staff Emeritus
I mean, judging from your teacher's script, he has $q'(t) = q(t) + \delta q$ and $q'(t') = q(t) + \Delta q$. This also rhymes with the statement that $\delta q$ is the fixed time variation. You are using the opposite definition.

6. Sep 8, 2016

### JulienB

@Orodruin In my teacher's script, both definitions are written as $q' (t')$... Okay, I rewrite my whole development following your advice:

We consider the following transformation:
$t' = t + \tau (t)$, $q' (t) = q(t) + \delta q(t)$

We want to know what is the symmetry variation, defined as:
$\Delta q = q' (t') - q(t)$
(transformed coordinate - original coordinate)

So:

$\Delta q = q' (t + \tau) - q(t)$
$= q'(t) + \tau \cdot \dot{q} ' (t) - q(t)$
$= q(t) + \delta q(t) + \tau (t) \dot{q} (t) - q(t)$
$= \delta q(t) + \tau (t) \dot{q} (t)$

$\implies \delta q(t) = \Delta q(t) - \tau (t) \dot{q} (t)$

Indeed, that works now.. Raah sorry for that, I get it now.

Julien.

7. Sep 8, 2016

### Orodruin

Staff Emeritus
Yes, but the text itself "For the variation of qa by a fixed time" and the way he is using it in the equations suggests the definition in my previous post is what is intended.