What is a time-dependent symmetry in Hamiltonian mechanics?

Question

I've read something from John Baez which I don't understand:

If we consider a single nonrelativistic free particle - in one-dimensional space, to keep life simple - and describe its state by its position q and momentum p at t = 0, we see that the Galilei boost

t |→ t

x |→ x + vt

has the following effect on its state:

p |→ p + mv

q |→ q

In other words, a Galilei boost is just a translation in momentum space.

In nonrelativistic quantum mechanics this should be familiar, though somewhat disguised. Here it is a commonplace that the momentum observable p generates translations in position space; in the Schroedinger representation it's just -i hbar d/dx. But by the same token the position observable q generates translations in momentum space. As we've seen, a translation in momentum space is just a Galilei boost. So basically, the generator of Galilei boosts is the observable q.

Ugh, but there is a constant "m" up there in the formula for a Galilei boost. So I guess the observable that generates Galilei boosts is really mq. If we generalize this to a many-particle system we'd get the sum over all particles of their mass times their position, or in other words: the total mass times the center of mass.

Now this seems weird at first because it's not a conserved quantity! Wasn't Noether's theorem supposed to give us a conserved quantity? Well, it does, but since our symmetry (the boost) was explicitly time-dependent - it involved "t" - our conserved quantity will also be explicitly time-dependent. What I was just doing now was working out its value at t = 0.

If we work out its value for arbitrary t, I guess we get: the total mass times the center of mass minus t times the total momentum.

Using the fact that total mass is conserved we can turn this conserved quantity into something perhaps a bit simpler: the center of mass minus t times the velocity of the center of mass.

This sounds very interesting but I don't understand it. I try to explain why.

Noether's theorem is about two Hamiltonian flows on the same symplectic manifold $(M,\omega)$. One flow is generated by the Hamiltonian as a generator function, i.e. the symplectic gradient of the Hamiltonian is the velocity field of this flow. This flow is the time evolution of the mechanical system, so, let's call it the dynamical flow. The other flow is the symplectic action of a one-parameter group. Noether's theorem states that if this second flow is a symmetry, i.e. it preserves the Hamiltonian (i.e, if the hamiltonian is constant along its orbits), then the dynamical flow preserves the generator function of the symmetry flow, that is, the generator function of the symmetry flow is a first integral of the system.

But in Baez's example, we have 3 flows:

1. The dynamical flow
2. The symplectic group action of the one-parameter subgroup $v\mapsto g(v,t)$. where $g(v,t)$ is the boost belonging to $v$ and $t$
3. The symplectic group action of the one-parameter subgroup $t\mapsto g(v,t)$.¹

Explicitly, the boost group is $G=\{g(v,t):v,t\in\mathbb R\}\subseteq\mathrm{GL}_3(\mathbb R)$ where $$g(v,t) = \begin{pmatrix}1 & 0 & vt \\ 0 & 1 & v \\ 0 & 0 & 1 \end{pmatrix}\tag{1}\label{eq1}$$

The symplectic action of this group on the phase space is

$$A:G\times M\to M: \left(g(v,t), (x,p)\right)\mapsto (x+vt,\,p+mv)\tag{2}\label{eq2}$$

The velocity field of flow 2. in every point of $M$ is $(t,m)$, so the generator function of it is $f_{2,t}(x,p)=tp-mx$, because $\left(\frac{\partial}{\partial p}(tp-mx),-\frac{\partial}{\partial x}(tp-mx)\right)=(t,m)$. In the special case of $t = 0$, $f_{2,0}(x,p)=-mx$, in accordance with Baez (up to a minus sign)

The velocity field of flow 3. in every point of $M$ is $(v,0)$, so its generator function is $f_{3,v}(x,p)=vp$.

Since flow 2 doesn't preserve the Hamiltonian of the free particle $H=\frac{p^2}{2m}$, it isn't a symmetry, so its generator function $f_{2,t}$ isn't a first integral of the motion, however, it is a constant of motion as Baez says.

My problem is that I don't see the relationship between Noether's theorem and this fact. Noether's theorem is about the integrals of motions and Hamiltonian-preserving flows but we have now a flow that doesn't preserve Hamiltonian and a time-depending constant of motion. What have these to do with each other? Baez says that Galilean boost is a time-dependent symmetry. But what is the definition of "time-dependent symmetry" in this setting?

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¹As user1379857 pointed out, this isn't a subgroup when $v\neq 0$ because $g(v,0)$ isn't the identity matrix.

Answer 1

First allow me to type out some basic equations in a more pedestrian notation. The Gallilean symmetry acts as \begin{align} p &\mapsto p + m v \\ q &\mapsto q + vt. \end{align} This is generated by the time dependent boost charge $$ K(t) = -mq + pt. $$ (In fact, if one realizes that $p_0 = \gamma m$ in relativistic mechanics, in the $v \ll c$ limit we get $p_0 = m$. One might therefore recognize the boost charge $K(t)$ as exactly low velocity limit of the quantity which generates relativistic boosts. But I digress.)

If we define the Hamiltonian vector field $X_Q$ of a quantity $Q$ by its action on the coordinate functions $q$ and $p$ by \begin{align} X_Q(q) &= \frac{\partial Q}{\partial p} \\ X_Q(p) &= -\frac{\partial Q}{\partial q} \end{align} then we can implement finite symmetry transformations by solving the functions $(q(\lambda), p(\lambda))$ which satisfy the differential equations \begin{align} q'(\lambda) &= X_Q(q) \\ p'(\lambda) &= X_Q(p) \end{align} and have the initial conditions $(q(0), p(0)) = (q_0, p_0)$. If we then plug in $Q = K(t)$, for some fixed $t$, then \begin{align} q'(\lambda) &= X_{K(t)}(q) = t\\ p'(\lambda) &= X_{K(t)}(p) = m. \end{align} This is solved by \begin{align} q(\lambda) &= q_0 + \lambda t\\ p(\lambda) &= p_0 + \lambda m. \end{align} Identifying $\lambda = v$ we see that this does indeed implement a boost. More importantly, we can see that this boost transformation really is implemented at a fixed time.

By the way, note that, for any two functions $Q$ and $H$ $$ X_Q(H) = \{H, Q\} = -X_H(Q). $$

Okay. Now let's discuss Noether's theorem for a time dependent quantity. If we want to compute the time derivative of some quantity $K(t)$, we have \begin{align} \frac{d}{dt} K(q(t), p(t), t) &= \dot q \frac{\partial K}{\partial q} + \dot p \frac{\partial K}{\partial p} + \frac{\partial K}{\partial t} \\ &= - \{ H, K(t) \} + \frac{\partial K}{\partial t}. \end{align} If $\frac{\partial K}{\partial t} = 0$, then the above is just $-X_K(H)$, and so we see that $K$ is conserved if $H$ is constant along the flow of $K$, which one could call Noether's theorem. However that is not the case for us. Indeed, if $$ H = \frac{p^2}{2m} $$ then $$ \{ H, K(t) \} = p $$ while also $$ \frac{\partial K}{\partial t} = p $$ so $$ \frac{d}{dt} K(q(t),p(t),t) = - p + p = 0 $$ and is a constant of motion. (Intuitively, Noether's theorem is saying that a system with a boost symmetry must have its center of mass move at a constant velocity.)

So, this was just my shpiel on how the time dependent $K(t)$ fits into Noether's theorem. The time dependence modified the proof a bit, but the time dependent term $\frac{\partial K}{\partial t}$ ends up cancelling out the flow term $-X_K(H)$. I'm not sure I have phrased this result in symplectic geometry terms in a way that would satisfy you, but this is the basic calculus of what's going on. When you ask

Noether's theorem is about the integrals of motions and Hamiltonian-preserving flows but we have now a flow that doesn't preserve Hamiltonian and a time-depending constant of motion. What have these to do with each other?

the answer is that the time dependence and non Hamiltonian preservingness cancel out to give a constant of motion.

Now, let me address something else you brought up. I have been working with what you called the one parameter family $v \mapsto g(v,t)$. But you also brought up the fact that there is another one parameter family $t \mapsto g(v,t)$. You claimed that this is generated by the quantity $vp$, but that is not actually true. Note that a boost acts on momentum as $p \mapsto p + mv$. This is $t$ independent, i.e. $p$ does not flow parameterized by $t$. If we are fixing $v$, then $p$ just "jumps" to the value $p + mv$ and stays there as $q$ flows as $q + vt$. So it is not true that the 1 parameter group action $t \mapsto g(v,t)$ can be implemented by Hamiltonian flow, because that would imply that we would have the identity symplectomorphism for $t = 0$.

Answer 2

I've found that the flow of the one-parameter group of Galilean boosts for which we can apply Noether's theorem, should be defined on the extended phase space (the phase space extended by time and energy). This flow preserves the extended Hamiltonian, i.e, it is a symmetry of the extended phase space. By Noether's theorem, its generator function is a constant of motion.

Details

Let's regard a single particle moving in one dimension with Hamiltonian $H$. The phase space of this system is the two-dimensional symplectic vector space $(M,\omega)$ where $$M = \mathbb R^2 = \{(q,p):q,p\in \mathbb R\}\tag{1} $$ and $$\omega:\mathbb M\times M\to\mathbb R: ((q_1,p_1),(q_2,p_2)\to q_1p_2-q_2p_1\tag{2}$$ The Hamiltonian flow on $M$ is $$\varphi:\mathbb R\times M\to M:(t,(q,p))\mapsto c_{(q,p)}(t)\tag{3}$$

where $$c_{(q,p)}:\mathbb R\to M\tag{4}$$ is the orbit of the Hamiltonian flow across $(q,p)\in M$, that is, $c_{(q,p)}$ is the unique integral curve of the Hamiltonian vector field $$\mathrm{sgrad} H = \left(\frac{\partial H}{\partial p},-\frac{\partial H}{\partial q}\right),\tag{5}$$ for which $c_{(q,p)}(0)=(q,p)$, where $H:M\to \mathbb R$ the Hamiltonian of the system.

$$-.-$$

In order to describe the flow of Galilean boost, we should consider the extended phase space instead of the phase space above. The extended phase space for this system is a four-dimensional symplectic vector space

$$\hat M=M\times \mathbb R^2\simeq\mathbb R^4=\{(q,p,\lambda,\epsilon):q,p,\lambda,\epsilon\in \mathbb R\}\tag{6}$$

with symplectic form $$\hat \omega:\hat M\times \hat M\to\mathbb R:((q_1,p_1,\lambda_1,\epsilon_1),(q_2,p_2,\lambda_2,\epsilon_2))\to q_1p_2-q_2p_1+\lambda_1\epsilon_2-\lambda_2\epsilon_1.\tag{7}$$ The extended Hamiltonian on $\hat M$ is$\!\phantom{|}^1$ $$\hat H:\hat M\to \mathbb R:(q,p,\lambda,\epsilon)\mapsto H(q,p)+\epsilon.\tag{9}$$ where $H:M\to \mathbb R$ is the Hamiltonian of the system on $M$.

The Hamiltonian vector field on $\hat M$ is

$$\mathrm{sgrad} \hat H=\left(\frac{\partial \hat H}{\partial p},-\frac{\partial\hat H}{\partial q},\frac{\partial \hat H}{\partial \epsilon},-\frac{\partial \hat H}{\partial\lambda}\right)=\left(\frac{\partial \hat H}{\partial p},-\frac{\partial\hat H}{\partial q},1,0\right),\tag{10}$$

and the flow determined by this vector field is

$$\hat \varphi:\mathrm R\times \hat M\to \hat M\simeq M\times\mathbb R^2:(t,(q,p,\lambda,\epsilon))\mapsto (c_{(q,p)}(t),\,\lambda+t,\,\epsilon)\tag{11}$$ From this we see that $\lambda$ is the time and on the level set $\hat H=0$, $-\epsilon$ is the total energy of the system.

$$-.-$$

The group action of the one-parameter group $G$ of Galilean boosts on $\hat M$ is

$$\hat A:G\times \hat M\to \hat M: \left(g(v), (q,p,\lambda, \epsilon)\right)\mapsto \left(q+v\lambda,\,p+mv,\,\lambda,\,\epsilon-pv-\frac{mv^2}{2}\right)\tag{12}$$ i.e. the action of Galilean boosts is the flow

$$\hat \psi:\mathrm R\times \hat M\to \hat M:(v,(q,p,\lambda,\epsilon))\mapsto \left(q+v\lambda,\,p+mv,\,\lambda,\,\epsilon-pv-\frac{mv^2}{2}\right)\tag{13}$$ The velocity field of flow $(13)$ by its parameter $v$ is the section of $T\hat M \simeq \hat M\times\hat M$

$$(q,p,\lambda,\epsilon)\mapsto ((q,p,\lambda,\epsilon),(\lambda,m,0,-p)),\tag{14}$$ since $$\left.\frac{d(q+v\lambda)}{dv}\right|_{v=0}=\lambda,\quad \left.\frac{d(p+mv)}{dv}\right|_{v=0}=m,\quad \left.\frac{d\lambda}{dv}\right|_{v=0}=0,\,\text{and }\\ \left.\frac{d}{dv}\right|_{v=0}\left(\epsilon-pv-\frac{mv^2}{2}\right)=\left.(-p-mv)\right|_{v=0}=-p \tag{15}$$ so the generator function of $\hat\psi$ is $$f:\hat M\to\mathbb R:(q,p,\lambda,\epsilon)\mapsto \lambda p - mq,\tag{16}$$

since

$$\mathrm{sgrad}f = \left(\frac{\partial f}{\partial p},\,-\frac{\partial f}{\partial q},\,\frac{\partial f}{\partial \epsilon},\,-\frac{\partial f}{\partial \lambda}\right)=(\lambda,\,m,\,0,\,-p)\tag{17}.$$ We can easily check that for a free particle, i.e, for $H=\frac{p^2}{2m}$, $\hat \psi$ preserves $\hat H$, so by Noether's theorem, the value of $f$ is constant along the orbits of $\hat\varphi$, i.e. $f$ is a constant of motion. As conclusion, we can say that a time-depending symmetry is a symmetry on the extended phase space.

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$\phantom{|}^1$ Cf. Libermann-Marle, Symplectic Geometry and Analytical Mechanics p. 492