Canonical vs. Noether momentum for longitudinal waves on a 1D chain

Consider longitudinal vibrations of particles on a line connected by springs. Setting all constants to one, the Lagrangian is $$L = \frac12 \sum_i \dot{\phi}_i^2 - (\phi_i - \phi_{i-1})^2.$$ Here, $\phi_i$ is the displacement of particle $i$ from its equilibrium position, and the canonical momenta are $\pi_i = \dot{\phi_i}$. The system has translational symmetry, $$\phi_i \to \phi_i + a$$ and the resulting conserved quantity is the total canonical momentum $\sum \pi_i$, which makes sense.

Now suppose we take the continuum limit, giving Lagrangian density $$\mathcal{L} = \frac12 \dot{\phi}^2 - \frac12 (\partial_x \phi)^2.$$ This system is supposed to be basically the same, but there are now two symmetries, $$\phi(x) \to \phi(x+a) \quad \text{and} \quad \phi(x) \to \phi(x) + a.$$ Both of these seem to be some sort of translational symmetry. The conserved quantity resulting from the first symmetry is what we usually call 'the momentum', and it is $$P = -\int \pi \partial_x \phi \, dx.$$ The conserved quantity resulting from the second symmetry is the total canonical momentum $$\Pi = \int \pi\, dx.$$ So we started with one conserved momentum, took the continuum limit, and now we have two! They're totally different quantities, not even the same order in the fields. What is going on?

• What is the physical difference between these two symmetries? Since this is a longitudinal wave, they both seem to be the very same translational symmetry.
• Exactly what part of the continuum limit 'doubles the symmetry'? Why didn't we see this in the original system?
• What is the physical interpretation of $\Pi$?

Some more calculations about $P$ and $\Pi$ follow. Let's impose the reasonable boundary conditions that $\phi(x) \to 0$ for $x \to \infty$, in some frame. There are two reasonable-looking boost transformations. The first one is $$\phi(x) \to \phi(x) + vt, \quad \pi(x) \to \pi(x) + v.$$ The quantity $\Pi$ changes by $vL$, where $L$ is the length of the line. But the quantity $P$ changes to $$P \to P - v \int \frac{d\phi}{dx} dx = P - v(vt - vt) = P$$ where I applied the boundary conditions. Then $P$ is boost invariant.

The second reasonable-looking boost transformation is $$\phi(x) \to \phi(x) + v t \frac{d\phi}{dx}$$ which is equivalent to $x \to x + vt$. By similar calculations, $\Pi$ is invariant under this boost, but $P$ is not.

Physically, I think the second boost is "boosting to a new coordinate frame", while the first boost is "boosting the medium with respect to the wave". That means that $P$ should be interpreted "the total momentum" and $\Pi$ should be interpreted as "the momentum of the wave in the medium". But this all sounds like nonsense to me because there is no 'wave' and 'medium'. There are just masses on springs. What's the difference?

• Have you looked at the discrete translation symmetry that becomes the continuum symmetry in the limit, $\phi_i \rightarrow \phi_{i+k}$? That is where the symmetry $\phi(x) \rightarrow \phi(x + a)$ comes from. – Sean E. Lake Dec 17 '16 at 5:10
• For the longitudinal wave, how do $\pi(x)$ and $P$ transform under boosts by velocity $v$? Does boosting to a frame where $\Pi = 0$ alter $P$ (relevant to this question: what are the boundary conditions on $\phi$)? – Sean E. Lake Dec 17 '16 at 11:13
• @SeanLake Well, there are two reasonable-looking boost transformations one can do, and $P$ is invariant under one while $\Pi$ is invariant under the other. But this reduces the problem to asking, what's the physical interpretation of the 'other' boost-like transformation, which is just as opaque to me. – knzhou Dec 18 '16 at 19:57
• If the wave is truly longitudinal, neither should be invariant under boosts. Starting with the Lagrangian, the boost for a longitudinal wave in this notation is: $$x \rightarrow x \pm vt,$$ $$\phi \rightarrow \phi \pm vt,$$ $$\dot{\phi}\rightarrow \dot{\phi} \pm v.$$ In a problem like this $\phi$ has a preferred frame of reference, so I don't think invariance under Lorentz boosts is inherent. Galilean boosts should still work, though. – Sean E. Lake Dec 18 '16 at 20:07
• I think that in the discrete case you could have some problem defining your simmetry $\phi_i \to \phi _i +a$. Since $i$ is allowed to range from $-\infty$ to $+\infty$, to have a well posed problem you should require some condition, maybe $\phi (t)\in \ell ^2$ for all $t$, in which case $\phi _i \to \phi _i +a$ is not admissible (this wave would have infinite energy). – pppqqq Dec 18 '16 at 20:34

Consider $\phi:x\rightarrow R$, to be a time dependent map from the position space labeled by $x$ or $i$ to a 1 dimensional real line that I'll call the 'target space'. Since you are talking about periodic boundary conditions $\phi$ traces out a loop in the target space as you vary $x$ (it is a loop embedded in 1 dimension so it retraces its path).

This loop is localized in the target space. You can define a `center of mass' at a mean position $\phi_0$. There are then two translation symmetries. You can shift the $\phi\rightarrow \phi+a$ which shifts the position in the target space. The conjugate momentum $\Pi$ is the momentum of the center of mass in the target space, and as you saw you can define a boost operation on $\Pi$.

The other symmetry is translating in the $x$ or $i$ space. As people have pointed out in the comments this symmetry also exists in the discrete model and has nothing to do with the continuum limit. The conjugate momentum $P$ describes the flow of energy around the loop for fixed center of mass in the target space.

There is another hidden conserved quantity if $\phi$ is considered to map to a compact target space like a circle instead of the real line (i.e. $\phi$ is an angular coordinate). Then the loop can fully wind around the target space before it comes back to its starting point and the number of windings $m$ is conserved with time along with $\Pi$ and $P$.

As you might have guessed this has a connection with string theory. Fields like $\phi$ describe coordinates in the target space and the two coordinates $x$ and $t$ that label points along the loop of string are called coordinates of the 'world sheet'.

Back to longitudinal springs

It occured to me you might want some more down to earth explanation in terms of longitudinal springs. There is a peculiarity here in the sense that the target space position $\phi$ appears to be existing in the same space as the internal word sheet position $x$. But note that you are using periodic boundary conditions in order to say it is translationally invariant in x. There is in effect a spring connecting the right and left endpoints in our $x$ label so these two endpoints want to be close to each other and the whole system forms a loop just as I described above. So $P$ still describes an internal energy flow along the chain of springs.

And note that your derivation of the Lagrangian is still valid in a boosted frame where all springs are moving with the same forward velocity, so $\Pi$ still describes the momentum of the center of mass of the entire system as opposed to the internal energy flow around the loop.

If you compactify your overall $\phi$ position with periodic boundary conditions then you can also talk about the conserved number of windings $m$ as I mentioned above. But note these periodic boundary conditions are distinct from the periodic boundary conditions in $x$ which amounts to adding an extra spring between the endpoints.

Actually, there was an equivalent for the translation symmetry $\phi(x)\rightarrow \phi(x+a)$ in the discrete case, but it was a discrete symmetry $\phi_i\rightarrow \phi_{i+n}$, which for that reason did not have a Noether current.

There is an ambiguity in the description in the theory: both $x$ and $\phi(x)$ seem to represent position along the string. This is the most obvious origin for the ambiguity in the two symmetries. This ambiguity is not so serious in the discrete case, but it becomes problematic in the continuous limit, as you discovered. For the sake of distinguishing this ambiguity, I'm going to assume that $\phi(x)$ represents transverse displacement as opposed to longitudinal displacements.

The other symmetry $\phi(x)\rightarrow \phi(x)+a$ is actually then not a translation symmetry, but an overall constant, almost like a gauge symmetry, except that it is a global symmetry and not a local symmetry. In the context of vibration of a string, this transformation corresponds to a constant transverse displacement of the string, as opposed to a longitudinal translation. (OK maybe one can interpret is as a translation symmetry in the transverse direction.) It comes about because the fields are massless; there is no mass term. If you would add a mass term this symmetry would disappear. Such a mass term would penalize transverse displacements.

If I can try to interpret what $\Pi$ means in the context of the model that you investigate (but with transverse displacements), I'd say that it looks like the net transverse velocity. The conjugate variable $\pi=\dot{\phi}$ seems to be the transverse velocity. Integrated over the whole system this then becomes the net transverse velocity. In the case of a massless string, this net transverse velocity becomes a conserved quantity. Interesting. Makes sense if one sees this transformation as a transverse translation.

• I agree that the physical picture is much clearer in the transverse case, but that's why I specified the longitudinal one. There, the interpretation of $\Pi$ is trickier, and it's harder to disentangle the two similar-looking symmetries. – knzhou Dec 17 '16 at 7:57
• If @knzhou explicitly added the mass density of the string (perhaps even as a function of position) it would be clear that $\pi(x)$ isn't just a velocity, but a momentum density, with $\Pi$ being the weighted net transverse momentum of the string. – Sean E. Lake Dec 17 '16 at 11:05
• As a fun exercise, you can look at $\Pi$ for the retarded Green's function of the wave equation in any dimensionality and find that $\Pi$ increases linearly with time after the delta function. A little harder: if you do the same for the Klein-Gordon equation $\Pi$ undergoes simple harmonic motion after the delta function. – Sean E. Lake Dec 17 '16 at 11:16
• If there is some way one can distinguish between $x$ and $\phi(x)$ for the longitudinal case, then I would argue that the interpretation would be the same or at least similar to the case for the transverse case. First one must find a way to distinguish $x$ and $\phi(x)$ for the longitudinal case though. Any suggestions? – flippiefanus Dec 17 '16 at 15:34

flippiefanus's answer is exactly correct. To add to it, a point that's often glossed over in the discussion of field theory is that it's crucial to carefully distinguish between internal transformations/symmetries and spatial transformations/symmetries. Strictly speaking, a field is a map from a spacetime manifold M to a field space F, and the two spaces $M$ and $F$ are completely unrelated. This distinction is crucial for understanding the Coleman-Mandula theorem and its various loopholes. See here for another instance in which the distinction is important.

For a system of springs moving longitudinally, the spacetime manifold $M$ is the discrete set of points where the ends of the springs lie when they are all relaxed. The field space is the small interval $[-d\varphi, d\varphi]$ over which they oscillate about their rest positions. As long as the oscillation amplitude is much less than the spring length, there is no ambiguity about which "rest site" each mass lives at. But when they become comparable, the coordinate $x$ and the field value $\varphi(x)$ become ambiguously interrelated, so the continuum limit is not well-defined. Strictly speaking, you need to take the proper order of continuum limits, where $d\varphi \to 0$ much faster than the spring lengths do. In other words, $d \varphi \ll dx$, or $d \varphi/dx$ is "small," in the appropriate units. This is the usual statement that continuum fields only make sense when they are very slowly varying.

TLDR: in general, you can't take the naive continuum limit for a system of coupled longitudinally oscillating springs, although everything's fine for transversely oscillating springs. (So for intuition, it's best to always visualize transverse oscillations so that this subtlety doesn't trip you up.) To get a well-defined field theory, you need to unambiguously differentiate the "internal/field" and "spatial" degrees of freedom at the microscopic level.