# Is the Explicit Symmetry Breaking of Vorticity Physically Significant in Fixing a Scale?

I have been working on a theoretical framework that draws on concepts from conformal field theory (CFT) to describe fluid dynamics and could potentially be well suited for describing certain turbulent flows. In this approach, I use scalar and vector potentials to capture both conservative and non-conservative aspects of fluids, aiming to simplify complex fluid phenomena with a Maxwell-like formalism.

Overview of the Theory

In this model, fluid flows are described using real scalar $\phi$ and vector potentials $V^\mu = (p, v)$. The key feature is the introduction of a minimal coupling term $(D_\mu \phi)(D^\mu \phi)$, where $D_\mu$ represents a covariant derivative coupling $\phi$ to $V^{\mu}$. Notably, this term is conformally invariant, meaning it respects the symmetry under conformal dilatations (Weyl transformations) $\sigma = e^{\alpha}$. The full list of transformations is:

$\phi \rightarrow \sigma \phi$

$D^{\mu}\phi \rightarrow \sigma D^{\mu}\phi$

$V^{\mu} \rightarrow V^{\mu} + \alpha^{\mu}$

$g_{\mu \nu} \rightarrow \sigma^{-2}g_{\mu \nu}$

To further characterize the fluid dynamics, I define a vorticity tensor $\Omega_{\mu \nu}$, analogous to the electromagnetic field tensor in electrodynamics. Instead of $E$ and $B$ fields, I define the transport field as $U = (v \cdot \nabla)v = -\nabla p - \frac{\partial v}{\partial t}$ and vorticity $W = \nabla \times v$. For the total Lagrangian for the system, $\mathcal{L}_{T}$, I included a kinetic term for the vorticity, $\frac{1}{4} \Omega_{\mu \nu} \Omega^{\mu \nu}$. However, this kinetic term explicitly breaks the conformal symmetry, leading to potentially interesting physical consequences.

$\frac{1}{4} \Omega_{\mu \nu}\Omega^{\mu \nu} \rightarrow \sigma^{-4} \frac{1}{4} \Omega_{\mu \nu}\Omega^{\mu \nu}$

Equations of Motion and Symmetry Breaking

The equations of motion derived from this framework are:

$D_{\mu}D^{\mu} \phi = 0$, which is conformally invariant, and $\partial_\mu \Omega^{\mu \nu} = j^\nu$, where the current $j^\nu$ transforms as $j^\mu \rightarrow \sigma^{2} j^\mu$. This equation is not conformally invariant, indicating a breaking of the symmetry. $\partial_{\mu}\Omega^{\mu \nu}$ and $j^{\mu}$ do not transform the same under $\sigma$.

This explicit symmetry breaking due to the kinetic term seems to introduce a fixed scale into the theory, which in general is crucial for describing fluid phenomena that exhibit characteristic lengths, such as vortex dynamics and turbulence.

Furthermore, the conserved quantity $\frac{Q}{\rho} = \oint U \cdot dS$ acts as an analog to electric charge in electromagnetism, with units corresponding to mass flow rate (kg/s$^2$). This "charge" represents sources or sinks of advective acceleration in the fluid, similar to how electric charge induces an electromagnetic field.

I am particularly interested in understanding whether this explicit symmetry breaking is physically significant. How does this compare to explicit symmetry breaking in other physical theories, and what implications might it have for understanding complex fluid phenomena like vortex shedding and turbulence? Typically the only examples I find are in the context of quantum field theory, not classical field theory.

Any insights or feedback from the community would be greatly appreciated!

Edit:

We want our scalar field $\phi$ to be symmetric under local scaling transformations of the form $\phi \rightarrow \sigma \phi$. To ensure this invariance we introduce the covariant derivative \begin{equation} D^{\mu} = \partial^{\mu} - V^{\mu} \end{equation} where the field $V^{\mu} = (p, v)$ is the solenoidal velocity and it’s associated pressure. The covariant derivative couples $\phi$ to $V^{\mu}$ while still respecting the scaling symmetry

\begin{equation*} D^{\mu}(\sigma \phi) = \sigma D^{\mu} \phi. \end{equation*}We can then modify our Lagrangian for the theory $\mathcal{L}(\phi, \partial_{\mu}\phi) \rightarrow \mathcal{L}(\phi, D_{\mu}\phi)$ so the Lagrangian for this free field theory (minimal coupling) is then \begin{equation*} \mathcal{L}= (D_{\mu}\phi)(D^{\mu}\phi) \end{equation*} which have the equations of motion $D_{\mu}D^{\mu}\phi = 0$ or \begin{equation} \partial_{\mu}\partial^{\mu}\phi - V_{\mu}\partial^{\mu}\phi - V^{\mu}\partial_{\mu}\phi-\partial_{\mu}V^{\mu}\phi + V^{\mu}V_{\mu}\phi = 0. \end{equation}

There are some coupling and potential terms that emerge characterized by the field $V^{\mu}$ and parameter $\alpha$. Perhaps these are the origin of the potentials you were talking about?

I think an example might help elude what the parameter $\sigma$ could represent physically. The most tractable example I have is the simple case of acoustic waves, as they are easily represented with a scalar field $\phi$ in one spatial dimension $x$.

First I want to clarify that we are fixing the background field $V^{\mu}$ with $V^{\mu} \approx \delta V^{\mu} = \alpha^{\mu}$ were $\alpha^1$ is a constant which effectively suppresses the vorticity.

The equations of motion in $(t, x)$ reduce to

\begin{equation*} \frac{\partial^2 \phi}{\partial t^2} - \frac{\partial^2 \phi}{\partial x^2} - 2\alpha \frac{\partial \phi}{\partial x} + \alpha^{2}\phi = 0 \end{equation*}which is the Webster horn equation but with an extra term $\alpha^2 \phi$ which has the effect of phase shifting the solutions. We can recognize the symmetry operator here as $\sigma = e^{\alpha x}$ representing the geometry (curvature in this case) of the exponential horn. This is reflected in the canonical Webster Horn equation, where the equations of motion are invariant under any choice of initial cross-sectional area due to the term $\sigma^{-1} \frac{d \sigma}{d x}$. If the cross sectional area can be represented by $S(x) = S_0 \sigma = S_0 e^{\alpha x}$ then the term will always reduce to $\alpha$ no matter the choice of $S_0$. This type of self-similarity is expected in a theory that is scale invariant.

As far as why do all this when Helmholtz-decomposition is perfectly adequate? Why not? Recasting theories in different frameworks can add new perspectives and potentially insights into known phenomenon. I have gained some weird deep intuition about certain principles while engaging in this representation and have found myself baffled many times. I also intend to use this framework as the foundation for future work in emergent fluid-like structures, but I am not nearly there yet.

## 1 answer

There is so much to unpack here. I am going to start this post and will keep adding to it as I get more thoughts.

First, let me reiterate the most important thing - I do not understand the physics behind this derivation at all. I keep trying to come up with some explanation that would make any sense (in particular why the chosen Lagrangian is what it is).

Second, in the mean time we have to do the most obvious thing: test the derivation by reducing it to as many standard solutions as possible. Eowyn has already gave a great start - the derivation of the Webster equation. Personally, I had never seen it in my life. But since he managed to arrive at something that had been already established, we can be reasonably confident the derivation is correct and we should keep trying getting other existent solutions.

My bigger goal currently is to do Reynold's decomposition in Eowyn's framework and see how MKE and TKE work out, see how turbulence energy dissipation term works out for standard flows (wake, jet, shear flow), see how Taylor scale works out, see if we can derive convective scales (for vortex dynamics) and see if we can derive Kolmogorov scales.

**The first baby step**, though, is to do the simplest thing possible: let's try to derive an equation for an infinite steady inviscid uniform flow in the absence of the gravitational force in a three dimensional "flat" space.

Assume, we do not experience any coordinate transformation ever. In this case, covariant derivative in flat space (which is our case) is equal to a regular derivative.

Since the flow is inviscid, the vector potential must be zero $\vec V (\vec x, t) = 0$. Note, I do not understand why Eowyn expresses the vector potential in the phase space. Helmholtz decomposition is performed on the given velocity field $\vec v (\vec x, t)$. Therefore, both flow potential and vector potential must be the functions of $(\vec x, t)$. Before Lagrangian is "applied" to the vector potential, it must be written as a function of regular Decart (i.e., Cartesian) coordinates, I think. And that is what I did.

At this stage we have $D^{\mu}(D^{\mu}(\phi)) = \frac{\partial}{\partial \mu}(\frac{\partial}{\partial \mu}(\phi)) = 0$, where $\mu$ can be either $x$, or $y$, or $z$, or $t$.

Since the flow is assumed steady, $\mu$ reduces to $\mu = \vec x = (x,y,z)$ and our expression reduces to

$$D^{\mu}(D^{\mu}(\phi)) = \frac{\partial}{\partial \mu}\left(\frac{\partial}{\partial \mu}(\phi)\right) = \frac{\partial}{\partial \vec x}\left(\frac{\partial}{\partial \vec x}(\phi)\right) = grad (grad (\phi)) = \nabla(\nabla(\phi)) = \nabla (\vec v)$$The last equality came from Helmholtz decomposition where the irrotational part of the velocity field is given by $v^{\phi} = \nabla \phi$.

And we end up with the following equation for our uniform flow derived in the Eowyn's framework:

$$\nabla (\vec v) = 0$$Let's see if we get the same from the Navier-Stokes equation:

$$\rho \left(\frac{\partial\vec v}{\partial t} + \vec v \cdot \nabla\vec v\right) = -\nabla P + \rho\vec g + \nabla \cdot \tau$$Here: $\tau=0$ because the flow is inviscid, $\rho\vec g = 0$ because the absence of gravity is assumed, $\nabla P =0$ because we need curved surface for a flow to have pressure gradient, which is not the case in our problem, $\partial\vec v / \partial t = 0$ because the flow is assumed steady. And we are left with the equation for our uniform flow

$$\nabla \vec v = 0$$Which is exactly the same as derived in Eowyn's framework.

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