# What is a time-dependent symmetry in Hamiltonian mechanics?

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 $$f_{2,t}(x,p)=tp-mx$$. 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?

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

• – Qmechanic Apr 11 at 8:40
• I think I found the answer in "Invariant Mappings and Classically Conserved Quantities" by Arthur Komar (Phys. Rev. D 8, 1028 – Published 15 August 1973) – mma Apr 12 at 5:58
• Unfortunately, the above paper is written in a style that I don't understand. I prefer clear symplectic geometry instead of infinitesimal calculus. – mma Apr 16 at 5:01
• Consider to review the geometric interpretation of eq. (1) as a $3\times 3$ matrix. – Qmechanic Apr 16 at 5:15

## 1 Answer

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 $$t = 0$$.

• I like your phrase "non Hamiltonian preservingness" :) – Mateo Apr 16 at 11:31
• Nice and clear answer. But I thought originally that we can construct somehow from the actions of $v\mapsto g(v,t)$ and $t\mapsto g(v,t)$ a third flow that preserves the Hamiltonian so we can apply Noether's theorem for it. But as you pointed out, the $t\mapsto g(v,t)$ curve isn't a one-parameter subgroup, so its action isn't a flow. The other possibility for keeping the pure geometric view is perhaps the contact geometry on the extended phase space. – mma Apr 17 at 5:41
• I think that, to truly understand the symmetry properties of Gallilean boosts, it is actually easier to think of them as a limit of Lorentzian boosts. There, boosts and time evolution are two elements in one larger algebra. Oftentimes, we say that symmetries are transformations which commute with time evolution. However, that is too restrictive. Boosts don't commute with time translation because boosts physically change the direction that "time" is pointing. ...Perhaps you could make some argument that boosts commute with time evolution if change what you mean by "time" after the boost is done – user1379857 Apr 17 at 22:50
• I don't have any paper or textbook in mind. Frankly, I'm just referring to the Poincare Lie algebra itself. You could implement the Poincare transformations as canonical transformations on phase space in a direct way if you wanted to. But I was literally just referring to the fact that, in special relativity, boosts will change the $t$ axis to some other $t'$ axis pointing in a different direction. So, just as spacetime transformations, boosts do not commute with time evolution. – user1379857 Apr 19 at 2:29
• A while ago, I, like you, spent a lot of time trying to "get to the bottom" of boosts in Hamiltonian mechanics, and "how exactly" the Galilean algebra arises from the Poincare one. It's interesting to think about, but I found that it didn't really lead anywhere. There's a lot of literature relating to the subject, but none that I found really elucidated much. I don't want to discourage you from thinking about it, in fact I'm happy there are people who are very curious about it. But I'm pretty sure there are no amazing answers which other people know that you just haven't stumbled across yet. – user1379857 Apr 19 at 2:36