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theorems:noethers_theorems [2018/03/27 09:23]
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theorems:noethers_theorems [2021/03/31 18:43] (current)
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 //see also: [[theorems:​noethers_theorems:​fields]] // //see also: [[theorems:​noethers_theorems:​fields]] //
  
-<tabbox Why is it interesting?> ​ 
-Noether'​s most famous first theorem connects each symmetry of a system with a conserved quantity. ​ 
  
-<blockquote>Noether’s second theorem, ​is an old but somewhat underappreciated tool, which acquires new +<tabbox Intuitive 
-significance in light of recent developmentsBy combining ​the robustness ​of the path integral formalism +When an object ​is at rest it has no momentum. When it has a lot of momentum it changes its location quicklyHence there is a connection between momentum and the change ​of location. Similarly, if an object doesn'​t rotate it has no angular momentum. Hence there is a connection between rotations and angular momentum. In a similar spirit ​we can say that an object that does not change at all over time has no energy
-with the elegance of Noether’s second theorem ​we find that writing down Ward identities +
-for residual gauge symmetries becomes essentially automatic.<​cite>​https://​arxiv.org/​abs/​1510.07038</​cite></​blockquote>​+
  
-----+In a somewhat more formal way we say that momentum generates translations (= changes of the position), angular momentum generates rotations and energy generates temporal translations (= movements forward in time). Momentum is responsible that an object changes its location, angular momentum that it rotates and energy that it changes as time passes on.  
 + 
 +Noether'​s theorem tells us that there is also a connection the other way round. Namely starting from the transformations:​ change of location, rotations and movements forward in time; we can //derive// the quantities; momentum, angular momentum and energy.  
 + 
 +Whenever we have a physical system, we can transform this system and if the transformed system is indistinguishable from the original one, the transformation we performed is a [[basic_tools:​symmetry|symmetry]]. So, for example, if we rotate our system and it is indistinguishable to the unrotated system, we say it has rotational symmetry.  
 + 
 +  * Now, Noether'​s theorem tells us that whenever a system is rotational symmetric there is a conserved quantity that we can recognize as the usual angular momentum.  
 +  * Similarly, if a system does not change under translations we get a conserved quantity that looks exactly like the quantity we call momentum. 
 +  * And if a system does not change as time passes on its energy is conserved. ​
  
  
 +-----
  
-<tabbox Layman?> ​ 
  
 <​blockquote>​If we assume that the laws of physics are describable by a minimum principle, then we can show <​blockquote>​If we assume that the laws of physics are describable by a minimum principle, then we can show
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 ISBN 0-679-60127-9),​ Chapter 4, pages 97 to 100</​cite>​ ISBN 0-679-60127-9),​ Chapter 4, pages 97 to 100</​cite>​
 </​blockquote>​ </​blockquote>​
 +
 +<​blockquote>​The Lagrangian proof is also intuitive: suppose you have a trajectory x(t) which is the minimum of S. Then you can perform an infinitesimal symmetry transformation on the trajectory, and you get S(x+dx), which is S(x) plus an integral over time of dS/dx times dx, where d is the variational change in S, the variational derivative. The bulk path is an infinitesimal variation of the bulk path, so there is no change in action, by the principle that the variations of S are zero on the true trajectory. So the variational change can only depend on the endpoints, because the path is a minimum, so you get that there is a quantity A(x_i)dx - A(x_f) dx which is zero summed over the two endpoints, which means there is a constant quantity along the trajectory. Feynman explains the theorem this way in "The Character of Physical Law"​.<​cite>​http://​qr.ae/​TU1ODh</​cite></​blockquote>​
  
   * [[https://​gravityandlevity.wordpress.com/​2011/​01/​20/​problems-you-can-solve-just-by-looking-at-them-the-meaning-of-noethers-theorem/​|Problems you can solve just by looking at them: The meaning of Noether’s Theorem]] by Brian Skinner   * [[https://​gravityandlevity.wordpress.com/​2011/​01/​20/​problems-you-can-solve-just-by-looking-at-them-the-meaning-of-noethers-theorem/​|Problems you can solve just by looking at them: The meaning of Noether’s Theorem]] by Brian Skinner
   * http://​physics.stackexchange.com/​questions/​325916/​laymans-version-of-noethers-theorem-or-the-intuition-behind-it/​325920#​325920   * http://​physics.stackexchange.com/​questions/​325916/​laymans-version-of-noethers-theorem-or-the-intuition-behind-it/​325920#​325920
  
-<​tabbox ​Student>  +<​tabbox ​Concrete>  
-The Noether theorem tells us that when there is some transformation $q \rightarrow q+\epsilon \delta q$ that leaves the action unchanged, ​then there exists a conserved current ${d j \over dt} = 0$. +The Noether theorem tells us that when there is some transformation $q \rightarrow q+\epsilon \delta q$ that leaves the action unchanged, there exists a conserved current ${d j \over dt} = 0$. 
  
 This Noether current is defined as This Noether current is defined as
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 So formulated differently,​ Noether’s Theorem tells us that whenever there is a continuous symmetry of the action, there is a conserved quantity. So formulated differently,​ Noether’s Theorem tells us that whenever there is a continuous symmetry of the action, there is a conserved quantity.
 +
 +----
 +
 +**A concrete Example**
 +
 +Let's consider a projectile, which is described by the
 +Lagrangian
 +\begin{eqnarray}
 +L = {1\over 2} m (\dot{x}^2 + \dot{y}^2) - mgy \label{eq:​projectile}
 +\end{eqnarray}
 +This Lagrangian is unchanged under the transformation $x \rightarrow
 +x+\epsilon$,​ where $\epsilon$ is sufficiently small constant. (Here, we have $\delta q=1$ in
 +the notation from above). We can see this because $$x \rightarrow x+\epsilon \Rightarrow
 +\dot{x} \rightarrow \dot{x}.$$
 +
 +Therefore, Noether'​s theorem tells us  that $$j = {\partial L \over \partial
 +\dot{q}} \delta q = m \dot{x}$$ is conserved. ​ We can now recognize that $m\dot{x}$
 +is the usual momentum. ​
 +
 +Noether'​s theorem, therefore, tells us here that the invariance of the Lagrangian under translations $x \rightarrow x+\epsilon$ implies the conservation of momentum.
  
 ---- ----
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 **Derivation** **Derivation**
  
-We start with a [[frameworks:​lagrangian_formalism|Lagrangian]] $L=L(q,​\dot{q})$ and consider an infinitesimal (= tiny tiny tiny) transformation+We start with a [[formalisms:​lagrangian_formalism|Lagrangian]] $L=L(q,​\dot{q})$ and consider an infinitesimal (= tiny tiny tiny) transformation
 \begin{eqnarray} \begin{eqnarray}
 q \rightarrow q+\epsilon \delta q \, . q \rightarrow q+\epsilon \delta q \, .
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 In words, this means that the current $j$ is constant in time and therefore conserved.  ​ In words, this means that the current $j$ is constant in time and therefore conserved.  ​
  
 +----
 +
 +**Derivation of the conservation of energy**
 +
 +In the example above, we derived that momentum is conserved when the Lagrangian is invariant under spatial translations. Analogously,​ we can derive that energy is conserved when the Lagrangian is invariant under temporal translations. ​
 +
 +To see this we consider the quantity ​
 +\begin{eqnarray}
 +{d L \over dt} = {d \over dt} L (q,\dot{q}) = {\partial L \over
 +\partial q} {dq \over dt} + {\partial L \over \partial \dot{q}}
 +{d\dot{q} \over dt} + {\partial L \over \partial t} 
 +\end{eqnarray}
 +Now, because in most cases the Largangian $L$ does not depend explicitly on time, ${\partial L \over
 +\partial t} = 0$, we have
 +\begin{eqnarray}
 +{d L \over dt} = {\partial L \over \partial q}\dot{q} + {\partial L
 +\over \partial \dot{q}} \ddot{q} = \bigg({d\over dt} {\partial L \over
 +\partial \dot{q}} \bigg) \dot{q} + {\partial L \over \partial \dot{q}}
 +\ddot{q} = {d \over dt} \bigg({\partial L \over \partial \dot{q}}
 +\dot{q}\bigg) .
 +\end{eqnarray}
 +Here we used the [[equations:​euler_lagrange_equations|Euler-Lagrange equation]] to get to the second line. 
 +
 +We therefore have ${d L \over dt} = {d \over dt} \big({\partial L
 +\over \partial \dot{q}} \dot{q}\big)$,​ or
 +\begin{eqnarray}
 +{d \over dt} \bigg({\partial L \over \partial \dot{q}}\dot{q} - L
 +\bigg) = 0 .
 +\end{eqnarray}
 +
 +In general the [[formalisms:​lagrangian_formalism|Lagrangian]] for a non-relativistic system reads $L=T-V$ and therefore we have ${\partial L \over
 +\partial \dot{q}} = {\partial T \over \partial \dot{q}}$ because the potential $V$ is
 +usually a function of $q$ only. In most cases we have
 +\begin{eqnarray}
 +T \propto \dot{q}^2 \quad \Rightarrow\quad ​ {\partial L \over \partial \dot{q}} \dot{q} = 2T .
 +\end{eqnarray}
 +Thus, we can conclude
 +
 +$${\partial L \over \partial \dot{q}} \dot{q} - L = 2T
 +- (T-V) = T+V = E,$$
 +
 +which is the total energy of the system.  ​
 +
 +Here in this context the total energy $T+V \equiv H$ is also often called the [[formalisms:​hamiltonian_formalism|Hamiltonian]]. ​  
 +
 +We can understand the connection between the Lagrangian and the Hamiltonian better by defining that ${\partial L \over \partial \dot{q}} \equiv p$ is
 +the momentum of the system. ​ The relationship between the Lagrangian and the Hamiltonian then reads
 +\begin{eqnarray}
 +p \dot{q} - L = H .
 +\end{eqnarray}
 +
 +In mathematical terms, the relationship between the Lagrangian and the Hamiltonian is given by a [[advanced_tools:​legendre_transformation|Legendre Transformation]].
  
  
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 +----
  
 +**Noether'​s Theorem in the Hamiltonian formalism**
  
 +Noether'​s theorem is especially transparent in the [[formalisms:​hamiltonian_formalism|Hamiltonian formalism]]. ​
  
 +A symmetry is a transformation that leaves the Hamiltonian unchanged
  
 +$$ \delta H =0. $$
  
 +In general, the variation of $H$ is
  
-<​tabbox ​Researcher+$$ \delta H = \frac{\partial H}{\partial q_i} \delta q_i + \frac{\partial H}{\partial p_i} \delta p_i . $$ 
 + 
 +We can rewrite this using the [[equations:​hamiltons_equations|generalized Hamiltons equations]] as 
 + 
 +\begin{align} \delta H &= \frac{\partial H}{\partial q_i} \delta q_i + \frac{\partial H}{\partial p_i} \delta p_i \notag \\ 
 +& \alpha \frac{\partial H}{\partial q_i} \frac{\partial G}{\partial p_i} - \alpha \frac{\partial H}{\partial p_i} \frac{\partial G}{\partial q_i} + \mathcal{O}(\alpha^2)\notag \\ 
 +& \equiv \alpha \{ H,G\}, 
 +\end{align} 
 +where $ \{ H,G\}$ denotes the [[advanced_notions:​poisson_bracket|Poisson bracket]] and $G$ is a generating function. 
 + 
 +Now, we can see that $G$ generates a __symmetry__ if  $ \{ H,G\} =0$.  
 + 
 +In addition, we know that the equation of motion for $G$ is  
 + 
 +$$\dot G = \{ G,H\} $$ 
 +which simply means that the Hamiltonian generates time translations.  
 + 
 +Since the Poisson bracket is antisymmetric we can conclude that if $G$ is a symmetry: $ \{ H,G\} =0$ we automatically also have $ \{ G,H\} =0$ and thus 
 +$$\dot G = 0. $$ 
 + 
 +Therefore, if $G$ is a symmetry then $G$ is conserved. Equally, if $G$ is conserved we automatically know that it generates a canonical transformation.  
 + 
 +---- 
 + 
 +**Graphical Summary** 
 + 
 +The diagram below shows the relationship between symmetry and conservation in the Lagrangian formalism for the case of a linear point transformation,​ such as a rotation. For a more detailed explanation see [[https://​esackinger.wordpress.com/​|Fun with Symmetry]]. 
 + 
 +{{:​theorems:​sym_and_cons_lagrange.jpg?​nolink}} 
 + 
 +The diagram below shows the relationship between symmetry and conservation in the Hamiltonian formalism. For a more detailed explanation see [[https://​esackinger.wordpress.com/​|Fun with Symmetry]]. 
 + 
 +{{:​theorems:​sym_and_cons_hamilton.jpg?​nolink}} 
 + 
 + 
 + 
 +<​tabbox ​Abstract
  
 There are actually two theorems by Noether about the consequences of an action integral under a continuous group of transformations. For a nice description,​ see section 2.2 and 2.3 in https://​arxiv.org/​pdf/​1510.07038.pdf and also the nice discussion in "​Symmetries in Fundamental Physics"​ by Sundermeyer. There are actually two theorems by Noether about the consequences of an action integral under a continuous group of transformations. For a nice description,​ see section 2.2 and 2.3 in https://​arxiv.org/​pdf/​1510.07038.pdf and also the nice discussion in "​Symmetries in Fundamental Physics"​ by Sundermeyer.
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 If the action is invariant under both $G_{\infty r}$ and $G_r$ (as a rigid subgroup), then the conservation laws that we get from Noether'​s first theorem are direct consequences of the generalized Bianchi identities that we derive using Noether'​s second theorem. ​ If the action is invariant under both $G_{\infty r}$ and $G_r$ (as a rigid subgroup), then the conservation laws that we get from Noether'​s first theorem are direct consequences of the generalized Bianchi identities that we derive using Noether'​s second theorem. ​
 +
 +----
 +
  
  
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 </​blockquote>​ </​blockquote>​
  
-<​blockquote>​Noether’s 1918 theorem [41] relating infinitesimal “global” symmetries to conservation laws, is a cherished cornerstone of modern theoretical physics; however, the second theorem (appearing in the same work) applicable to “local” symmetry remains somewhat obscure [43]. Our goal is to use Noether’s second theorem as a starting point for a general approach to [[advanced_notions:​ward_indentities|Ward identities]] for gauge symmetry. In particular, we are motivated by recent new Ward identities for large gauge symmetry in gravity and QED [4–6, 8–13], and recent discussions in [37]. (This assertion is based, in part, on informal discussions. An important exception is [[https://​arxiv.org/​abs/​hep-th/​0111246|44]],​ which introduced the authors+<​blockquote>​Noether’s 1918 theorem [41] relating infinitesimal “global” symmetries to conservation laws, is a cherished cornerstone of modern theoretical physics; however, the second theorem (appearing in the same work) applicable to “local” symmetry remains somewhat obscure [43]. Our goal is to use Noether’s second theorem as a starting point for a general approach to [[advanced_tools:​ward_indentities|Ward identities]] for gauge symmetry. In particular, we are motivated by recent new Ward identities for large gauge symmetry in gravity and QED [4–6, 8–13], and recent discussions in [37]. (This assertion is based, in part, on informal discussions. An important exception is [[https://​arxiv.org/​abs/​hep-th/​0111246|44]],​ which introduced the authors
 to Noether’s second theorem) to Noether’s second theorem)
  
 <​cite>​https://​arxiv.org/​pdf/​1510.07038.pdf</​cite></​blockquote>​ <​cite>​https://​arxiv.org/​pdf/​1510.07038.pdf</​cite></​blockquote>​
 +
 +----
 +
 +**Noether'​s theorem in the Hamiltonian formalism**
 +
 +The analogue to Noether'​s theorem in the Lagrangian formalism is the moment map in the Hamiltonian formalism.
 +
 +These moment maps provide a way of encoding of how the Lie group $G$ acts on the [[basic_tools:​phase_space|phase space]]. In addition, the moment maps give us a way to find the observables corresponding to the conserved quantities/​generators of the symmetry $G$.
 +
 +  * see also [[https://​www.math.columbia.edu/​~woit/​wordpress/?​p=7146|Use the Moment Map, not Noether’s Theorem]] by P. Woit
 +
 +The crucial object in the Hamiltonian formalism is the [[advanced_notions:​poisson_bracket|Poisson bracket]] of two observables,​ which can be thought of as the rate of change of the first along the flow given by the second.
 +
 +So when we use the Hamiltonian $H$ as the generator, we end up with time evolution. Therefore for an observable $A$ without explicit time dependence we have
 +
 +$$\frac{\mathrm{d}A}{\mathrm{d}t} = \{A, H\}.$$
 +
 +Therefore, $A$ is a constant of motion (= does not change as time passes on) only if $\{A, H\} =0$.
 +
 +However, we can equally consider the reverse Poisson bracket $\{H, A\}$. Here $A$ appears in the second slot and therefore this Poisson bracket represents the rate of change of $H$ along the flow generated by $A$. So if $A$ generates a symmetry of a system we expect that $\{H, A\}$  is true.  This comes about because if the Hamiltonian remains unchanged by the flow generated by $A$ then also the [[equations:​hamiltons_equations|equations of motion]]: $\dot q = \frac{\partial H}{\partial p}, \quad \dot p = - \frac{\partial H}{\partial q}$ remain unchanged under the same flow. 
 +
 +
 +  * If $\{H, A\} =0$ which means we have a symmetry of the Hamiltonian generated by $A$, then since the Poisson bracket is antisymmetric we automatically have
 +  * $\{A, H\} =0$, which states that $A$ is a conserved quantity. ​
 +
 +For example, if the Hamiltonian is invariant under translations,​ we have $\{H, p\}=0$ since $p$ generates translations. In turn, we automatically also have  $\{p, H\}=0$ which means that momentum is conserved. ​
 +
 +
 +In general, the infinitesimal evolution under the transformation generated by any element $g$ of the Lie algebra of observables on our [[basic_tools:​phase_space|phase space]], (parametrized by an abstract "​angle"​ $\varphi$) is given by
 +
 +$$ \partial_\phi F = \{G,F\}.$$
 +
 +This has immediately important consequences because it means that if the Poisson bracket of $F$ and $G$ vanishes, they describe an infinitesimal transformation that is a symmetry for the other. So if the Poisson bracket of an observable with the Hamiltonian vanishes it means that this observable is invariant under time translations,​ i.e. a constant in time. Similarly, if the Poisson bracket of an observable and the generator of rotations (= angular momentum $L^i = \epsilon^{ijk}x^jp^k$) vanishes it is invariant under rotations.
 +
 +More precisely, consider a "​generator"​ $\delta G$ and some quantity $A$.  The infinitesimal transformation,​ generated by $\delta G$ is:
 +
 +$$A \to A+\delta A,​\quad\quad \delta A = -\{\delta G, A\}. $$
 +
 +
 +
 +-->​Momentum#​
 +$\delta G = \epsilon_i p_i \quad \delta A = -\epsilon_i\{p_i,​ A\} = \epsilon_i\frac{\partial A}{\partial q_i},​$  ​
 +$\quad \Rightarrow \quad A(p_i,​q_i)\to A(p_i,​q_i)+\epsilon_i\frac{\partial A}{\partial q_i} = A(p_i,​q_i+\epsilon_i)$  ​
 +Therefore momentum is the generator of translations.
 +<--
 +
 +-->​Angular Momentum#
 +$$\delta G = \epsilon_i e_{ijk} p_jq_k \quad \delta A = -\epsilon_i e_{ijk} \{p_jq_k, A\}$$
 +
 +Here $e_{ijk}$ is the Levi-Civita symbol. Expanding yeilds ​  
 +$\delta A = -\epsilon_i e_{ijk} \sum_\alpha \left(\frac{\partial p_jq_k}{\partial q_\alpha}\frac{\partial A}{\partial p_\alpha}-\frac{\partial p_jq_k}{\partial p_\alpha}\frac{\partial A}{\partial q_\alpha}\right) = -\epsilon_i e_{ijk}\left(p_j\frac{\partial A}{\partial p_k}+q_j\frac{\partial A}{\partial q_k}\right)$
 +
 +$\quad \Rightarrow \quad A(p_i,​q_i)\to A(p_i,​q_i)-\epsilon_i e_{ijk}\left(p_j\frac{\partial A}{\partial p_k} +  q_j\frac{\partial A}{\partial q_k}\right) = A(R_{ij}p_j,​R_{ij}q_j)$  ​
 +For infinitesimal rotations $R_{ij}$
 +
 +So we conclude that angular momentum is the generator of rotations.
 +
 +<--
 +
 +-->​Energy#​
 +$\delta G = \epsilon H \quad \delta A = -\epsilon\{H,​ A\} = \epsilon\frac{d A}{d t},$   
 +
 +$\quad \Rightarrow \quad A(p_i(t),​q_i(t))\to A(p_i(t),​q_i(t))+\epsilon\frac{d A}{d t} = A(p_i(t+\epsilon),​q_i(t+\epsilon))$  ​
 +
 +Therefore energy is the generator of time evolution.
 +<--
 +
 +----
  
 **Reading Recommendations:​** **Reading Recommendations:​**
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   * https://​physics.stackexchange.com/​questions/​19847/​whats-the-interpretation-of-feynmans-picture-proof-of-noethers-theorem   * https://​physics.stackexchange.com/​questions/​19847/​whats-the-interpretation-of-feynmans-picture-proof-of-noethers-theorem
   ​   ​
-<​tabbox ​Examples+<​tabbox ​Why is it interesting?>​  
 +<​blockquote>​In 1915, Emmy Noether demonstrated that differentiable symmetries give rise to conservation laws. Her work is a foundational document in quantum theory because it verifies the ancient insight that what is most important in any physical system is what remains the same in the system as the system is changing.<​cite>​http://​inference-review.com/​article/​woits-way</​cite></​blockquote>
  
---> Example1#+Noether'​s most famous first theorem connects each symmetry of a system with a conserved quantity. ​
  
-  +<blockquote>Noether’s second theorem, is an old but somewhat underappreciated tool, which acquires new 
-<-- +significance in light of recent developments. By combining the robustness of the path integral formalism 
- +with the elegance of Noether’s second theorem we find that writing down Ward identities 
---Example2:# +for residual gauge symmetries becomes essentially automatic.<cite>​https://​arxiv.org/​abs/​1510.07038</​cite></​blockquote>​
- +
-  +
-<--+
  
 <tabbox FAQ> ​ <tabbox FAQ> ​
Line 274: Line 463:
 </​blockquote>​ </​blockquote>​
 <-- <--
-  ​+ 
 +-->Is there a symmetry associated to the conservation of information?#​ 
 +see https://​physics.stackexchange.com/​questions/​41765/​is-there-a-symmetry-associated-to-the-conservation-of-information 
 +<--
 <tabbox History> ​ <tabbox History> ​
  
  
-See Chapter 1 in The Noether Theorems: Invariance and Conservation Laws in the Twentieth Century+  * [[https://​arxiv.org/​abs/​1902.01989|Colloquium:​ A Century of Noether'​s Theorem]] by Chris Quigg 
 +  * See also Chapter 1 in The Noether Theorems: Invariance and Conservation Laws in the Twentieth Century
 by Yvette Kosmann-Schwarzbach by Yvette Kosmann-Schwarzbach
 </​tabbox>​ </​tabbox>​
theorems/noethers_theorems.1522135397.txt.gz · Last modified: 2018/03/27 07:23 (external edit)