Intuition behind the definition of Continuous Symmetry of a Lagrangian (Proof of Noether's Theorem)

Suppose there is a one-parameter family of continuous transformations that maps co-ordinates $$q(t)\rightarrow Q(s,t)$$ where the $$s$$ is the continuous parameter. Also, for when $$s=0$$ the transformation is the identity, i.e. $$Q(0,t)=q(t)$$.

Then if we have a Lagrangian, $$L$$ which is invariant under the replacement of $$q\rightarrow Q$$, then why is it intuitively true that:

$$\frac{d}{ds}L|_{s=0}=0$$

In other words, why is the derivative taken at $$s=0$$?

• If you're asking the question, doesn't that mean it is not intuitively true? Are you asking why your book or why someone else described it as such? Is this from a particular textbook or lecture?
– hft
Commented Sep 7, 2022 at 20:26
• Yes, I am asking why it makes sense that a textbook wrote it so Commented Sep 7, 2022 at 21:23
• Which textbook? Which page? Commented Sep 7, 2022 at 21:57
• courses.physics.ucsd.edu/2010/Fall/physics200a/LECTURES/… Commented Sep 7, 2022 at 22:16
• End of page 1 and start of page 2 Commented Sep 7, 2022 at 22:17

Consider a one-parameter group of transformations $$Q=Q(q,s)$$ And define $$\tilde{L}(q,\dot{q},s):=L(q(Q,s), \dot{q}(Q,\dot{Q},s))\tag{1}$$ Where $$\dot{Q}(q,\dot{q},s)=\frac{\partial Q}{\partial q}(q,s)\dot{q}$$ Now, the condition in the statement of the theorem $$\frac{\partial}{\partial s}\tilde{L}(q,\dot{q},s)\lvert_{s=0}=0\quad\forall (q,\dot{q})\tag{2}$$ is not weaker than the condition $$\frac{\partial}{\partial s}\tilde{L}(q,\dot{q},s)=0\quad\forall(q,\dot{q}).\tag{3}$$ This stems from the fact that the initial state $$(1)$$ i.e. $$(Q(q,0),\dot{Q}(q,\dot{q},0))=(q,\dot{q})$$ is arbitrary and it can be chosen to be $$(Q(q,s_0),\dot{Q}(q,\dot{q},s_0))$$ for some $$s_0\in\mathbb{R}$$. In such case, define $$\hat{L}(q,\dot{q},s):=L(q(Q,s+s_0), \dot{q}(Q,\dot{Q},s+s_0))\tag{1'}$$ which is just the tranformed lagrangian with the new initial state (using group properties under composition). The usual condition now takes the form $$\frac{\partial}{\partial s}\hat{L}(q,\dot{q},s)\lvert_{s=0}=0\quad\forall (q,\dot{q})\tag{2'}$$ Of course differentiating $$\hat{L}$$ at $$s=0$$ in $$(2')$$ is the same as differentiating $$\tilde{L}$$ at $$s=s_0$$ as its a mere translation. The point $$s_0$$ was chosen arbitrarily,thus $$(3)$$ is true. We proved that $$(2)\implies(3)$$, the reverse is trivial.



$$^1$$Valter Moretti. Meccanica Analitica. Section 9.2.2

$$^2$$A. Fasano, S. Marmi. Analytical Mechanics. Section 4.9

Let there be a transformation $$q(t) \rightarrow Q(s,t)$$ such that $$L$$ remains invariant. There exists an identity point $$s_{0}$$ where:

$$$$Q(s_{0},t) = q(t)$$$$

$$L$$ is initially a functional of $$q$$, $$\dot{q}$$ and $$t$$, but after the transformation it becomes a functional of $$Q(s,t)$$ and $$\dot{Q}(s,t)$$. Furthermore, this is the only way that $$L$$ depends on $$s$$, so:

$$$$\frac{d}{ds} L[Q(s,t), \dot{Q}(s,t),t] = \int ds' \, \left \{ \frac{\partial Q(s',t)}{\partial s} \frac{\delta L[Q(s,t), \dot{Q}(s,t),t]}{\delta Q(s',t)} + \frac{\partial \dot{Q}(s',t)}{\partial s} \frac{\delta L[Q(s,t), \dot{Q}(s,t),t]}{\delta \dot{Q}(s',t)} \right \}$$$$

More specifically calculated at the identity point:

$$$$\frac{dL}{ds} \Bigg |_{s_{0}} = \int ds' \, \left \{ \frac{\partial Q(s',t)}{\partial s} \Bigg | _{s_{0}}\frac{\delta L[Q(s,t), \dot{Q}(s,t),t]}{\delta Q(s',t)} \Bigg|_{Q(s_{0},t)} + \frac{\partial \dot{Q}(s',t)}{\partial s} \Bigg|_{s_{0}}\frac{\delta L[Q(s,t), \dot{Q}(s,t),t]}{\delta \dot{Q}(s',t)} \Bigg|_{\dot{Q}(s_{0},t)} \right \}$$$$

$$\dot{Q}(s_{0},t)$$ here denotes the function $$\dot{Q}(s,t)$$ calculated at $$s_{0}$$, but it is easy to to show that it is also equivalent to the time derivative of $$Q(s_{0},t) = q(t)$$.

Next, we Taylor expand the Lagrangian around $$Q(s_{0},t)$$ as such:

$$$$\begin{split} L[Q(s,t), \dot{Q}(s,t),t] &= L[Q(s_{0},t), \dot{Q}(s,t),t] + \int ds' \, [Q(s,t)-Q(s_{0},t)] \frac{\delta L[Q(s,t), \dot{Q}(s,t),t]}{\delta Q(s',t)} \Bigg |_{Q(s_{0},t)} \\ &+ \frac{1}{2} \int ds' \int ds'' \, [Q(s,t)-Q(s_{0},t)]^{2} \frac{\delta ^{2} L[Q(s,t), \dot{Q}(s,t),t]}{\delta Q(s',t) \, \delta Q(s'',t)} \Bigg |_{Q(s_{0},t)} + \ldots \end{split}$$$$

Similarly we may expand around $$\dot{Q}(s_{0},t)$$ for the first term:

$$$$\begin{split} &L[Q(s,t), \dot{Q}(s,t),t] = L[Q(s_{0},t), \dot{Q}(s_{0},t),t] \\ +\int ds' &\, \left \{ [Q(s,t)-Q(s_{0},t)] \frac{\delta L[Q(s,t), \dot{Q}(s,t),t]}{\delta Q(s',t)} \Bigg |_{Q(s_{0},t)} + [\dot{Q}(s,t)-\dot{Q}(s_{0},t)] \frac{\delta L[Q(s,t), \dot{Q}(s,t),t]}{\delta \dot{Q}(s',t)} \Bigg |_{\dot{Q}(s_{0},t)} \right \} \\ + &\frac{1}{2} \int ds' \int ds'' \, \Bigg \{ [Q(s,t)-Q(s_{0},t)]^{2} \frac{\delta ^{2} L[Q(s,t), \dot{Q}(s,t),t]}{\delta Q(s',t) \, \delta Q(s'',t)} \Bigg |_{Q(s_{0},t)} \\ & \quad \quad \quad \quad \quad \quad + [\dot{Q}(s,t)-\dot{Q}(s_{0},t)]^{2} \frac{\delta ^{2} L[Q(s,t), \dot{Q}(s,t),t]}{\delta \dot{Q}(s',t) \, \delta \dot{Q}(s'',t)} \Bigg |_{\dot{Q}(s_{0},t)} \Bigg \} + \ldots \end{split}$$$$

Since $$Q(s_{0},t) = q(t), \, \dot{Q}(s_{0},t) = \dot{q}(t)$$ and the Lagrangian is invariant under the transformation:

$$$$L[Q(s,t), \dot{Q}(s,t),t] = L[q(t), \dot{q}(t),t] \, , \quad \forall \, s,t$$$$

then the first term of the Taylor expansion is equal to the Lagrangian on the lhs. By extension, the infinite amount of terms from the Taylor expansion(s) must vanish. The two cases which ensure this are $$Q(s,t) = Q(s_{0},t), \, \dot{Q}(s,t) = \dot{Q}(s_{0},t)$$ i.e. the trivial case in which we transform $$q(t)$$ into itself, and the case of the 1st order functional derivatives vanishing:

$$$$\frac{\delta L[Q(s,t),\dot{Q}(s,t),t]}{\delta Q(s',t)} \Bigg |_{Q(s_{0},t)} = \frac{\delta L[Q(s,t),\dot{Q}(s,t),t]}{\delta \dot{Q}(s',t)} \Bigg |_{\dot{Q}(s_{0},t)} = 0$$$$

Discarding the trivial case, by virtue of the third equation we find that indeed:

$$$$\frac{dL}{ds} \Bigg|_{s_{0}} = 0$$$$

In your case you have $$s_{0} = 0$$, but it could be anything based on the particular transformation at hand. Hence the choice of $$s=0$$ for the derivative of the Lagrangian is "intuitive" in the sense that you implicitly need to Taylor expand $$L$$ around the functions at that specific value. That leads to the $$s$$-independent term of the expansion to be the same as the Lagrangian before the transformation, leading to the conclusions discussed above.