I'm going through the derivations of Noether's theorems and I have several criticisms as to how they are presented in popular sources (note that I'm only referring to classical mechanics here and not interested in the theorems in the context of field theory). My comments are presented below:

The Hamiltonian is defined as $H=\sum \dot{q}_i \frac{\partial L}{\partial \dot{q}_i}-L$. I wont go through the details, but it can be shown that if the potential energy is only dependent on generalized coordinates (and not on velocities) and if the kinetic energy is a homogeneous quadratic function of $\dot{q}_i$ then the Hamiltonian is the total energy.

The fact $\frac{\partial L}{\partial t}=0$ only directly implies that $\frac{d}{dt}H=0$ and nothing else. So my criticism is: We cannot say in any absolute sense that time-translation symmetry implies conservation of energy; we can only say it implies conservation of the Hamiltonian, which may or may not be the total energy according to the conditions I posted above

On the other hand, it is often said that if there is space translation symmetry w.r.t. a certain variable, then the conjugate momentum is conserved. And this is shown rather simply by:

if $\frac{\partial L}{\partial \dot{q}_i}=0$ then $\frac{d}{dt}p_{{q}_i}=0$.

But this is only valid whenever the potential is independent of velocities, unless you accept $\frac{\partial L}{\partial \dot{q}_i} = P_{q_i}$ even when potentials are velocity dependent

So where am I screwing up here? Are my statements true but nevertheless useless since all potentials in the universe are velocity independent (which I think is false)? Is it always possible to find a coordinate system for which $H=E$?


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    $\begingroup$ I'm not really sure what you're talking about. Noether's theorem is the extremely general statement that every symmetry has a corresponding conservation law, and the theorem gives the general form of the conserved current. It doesn't state the Hamiltonian is always the energy. Also, the momentum $p_i$ corresponding to a coordinate $q^i$ is defined as $\partial L / \partial\dot{q}^i$, not sure why you claim this would only be valid for velocity-independent potentials. $\endgroup$
    – ACuriousMind
    Oct 24, 2015 at 18:53
  • $\begingroup$ Yes you could go by definition (w.r.t. the generalized momenta) but I was speaking about the more concrete sense of momentum where (if we had cartesian coordinates) $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$ iff $\frac{\partial U}{\partial \dot{x}}=0$. $\endgroup$
    – DLV
    Oct 24, 2015 at 18:56
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    $\begingroup$ For a mechanical system the total energy is that given by the Hamiltonian by definition. If there is another energy term that breaks the symmetry (and there are plenty of such terms in real situations), then you have simply made a mistake in formulating the system boundaries. $\endgroup$
    – CuriousOne
    Oct 24, 2015 at 18:57
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    $\begingroup$ Comment to the question (v1): The heart of the question is about the Lagrangian formalism (as opposed to the Hamiltonian formalism). Introducing the Hamiltonian formalism and possible singular Legendre transform is irrelevant and obfuscate the matter. Assuming this is purely a Lagrangian question, then what OP calls $H$ and $p$ are then Lagrangian definitions of energy and momentum, respectively. They are usually called the Lagrangian energy function and Lagrangian conjugate/canonical momentum, respectively. They do not necessarily correspond to other definitions of energy and momentum. $\endgroup$
    – Qmechanic
    Oct 24, 2015 at 19:07
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    $\begingroup$ For a case where canonical momentum is conserved but mechanical momentum is not, you might want to look at the case of a charged particle in a magnetic field under the Lagrangian formalism. In such a case, it's pretty obvious that $m \dot{v}$ shouldn't be a constant, but the conjugate momentum can still be conserved. There's a good discussion of this setup in "Thoughts on the magnetic vector potential" (Semon & Taylor, Am. J. Phys., Vol. 64, p. 1361.) $\endgroup$ Oct 24, 2015 at 20:47

2 Answers 2


As I see, maybe the problem is energy. So, What is energy?

The formal classical definition of energy is: Energy is a dynamical invariant of a system that came from time-translation symmetry. There is also a question here about it. If you want more references about it, let me know.

So.. when Bob write, $E = T + V$ in dissipative systems (damped OHS for instance), Bob is thus formally wrong, because the quantity $E$ clearly varies with time, thus is not a dynamical invariant, thus can't be the energy of this system.

Also, you said that: $$ \frac{\partial L}{\partial t} = \frac{dH}{dt} $$

That does not proves Noether Theorem. That only indicates when $H$ is conserved. The complete proof of Noether Theorem has to do with the least action principle. When you take the action $A$, and make it vary infinitesimally $\delta A$ by time translation $\delta t$ and spatial translation $\delta q_i$, and after a few calculations, you will arrive in: $$ \delta A = \delta\int L(q_i, \dot q_i, t) dt = \int\left(\frac{\partial L}{\partial q_i} - \frac{d}{dt}\frac{\partial L}{\partial\dot q_i}\right) \left(\delta q_i - \dot q_i\delta t\right)dt + \left[\frac{\partial L}{\partial\dot q_i}\delta q_i - H\delta t\right] $$

The spatial translation $\delta q_i$ is associated with the momentum $p_i$, and the time translation $\delta t$ is associated with the hamiltonian $H$.

Now we apply the least action principle: $\delta A = 0$, and we get: $$ \left(\frac{\partial L}{\partial q_i} - \frac{d}{dt}\frac{\partial L}{\partial\dot q_i}\right) \left(\delta q_i - \dot q_i\delta t\right)dt = -\frac{d}{dt} \left[\frac{\partial L}{\partial\dot q_i}\delta q_i - H\delta t\right] $$

Here we identify Euler-Lagrange equations, which must be satisfied, and thus are zero (be careful with the dot products). Then we have a conserved quantity for each symmetry of the action: $$ p_i\delta q_i - H\delta t = cte $$

For a time symmetry, $H$ is conserved, and thus $H$ is the energy of the system, by definition. There is no such thing proven that $T+V$ is the conserved, thus, they may not be dynamical invariants, thus they may not be the energy.

In summary, the current formal definition of classical energy was motivated thanks to Noether Theorem.

  • $\begingroup$ It seems counterintuitive though. Can one obtain work from $H$ like in thermodynamics? One can from $T+V$. Maybe references would be nice here, like you say. I'd like to know more about this. Also, one can still talk about energy when it isn't conserved, would $H$ still be the definition of energy in such a case? Thanks! $\endgroup$
    – DLV
    Oct 29, 2015 at 3:42

There are at least two generalizations of Noether's theorem.

1) Assume that the Hamiltonian system with Hamiltonian $H(z),\quad z=(p,q)$ has a one-parameter symmetry group $\{g^s_F(z)\}$ which is generated by a Hamiltonian system with Hamiltonian $F$. Then $F$ is a first integral for $H:\quad \{F,H\}=0$, moreover if $dF\ne 0$ then there are local canonical coordinates $P,Q$ such that in these coordinates (these coordinates can be built by quadratures provided $g^s_F$ is given) Hamiltonian $H$ does not depend on $Q_1$ and $(P,Q)\mapsto g^s_F(P,Q)=(P_1,\ldots,P_n,Q_1+s,Q_2,\ldots,Q_n)$.

2) Consider a nonholonomic system with Lagrangian $L=L(q,\dot q)$ and the constraints equation $a_i^j(q)\dot q^i=0,\quad j=1,\ldots,k<n$. Assume that there exists a one-parameteer group $\{g^s(q)\},\quad L(q,\dot q)=L(g^s(q),d g^s(q)\dot q)$ and $a_i^j(q) v^i(q)=0$. Here $v$ is the vector field which generates $g^s$. Then the system has the first integral $f=\frac{\partial L}{\partial \dot q^l}v^l.$ One can also apply the rectification theorem to $v$

  • $\begingroup$ +1. Concerning a Hamiltonian version of Noether's theorem, see also this Phys.SE post. $\endgroup$
    – Qmechanic
    Nov 12, 2015 at 1:00

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