Still tentatively feeling my way through this stuff, so please go easy.

The velocity of a fluid at a point P are the components $V^{a}$ of a contravariant vector:$$v^{x},v^{y},v^{z}\equiv\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt},$$ where $x,y,z$ are some arbitrary coordinate system. The metric tensor $g_{ab}$ for that coordinate system can be used to lower the index to give the components of a one-form:$$V_{b}=g_{ab}V^{a}.$$

This one-form will have components$$\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y},\frac{\partial\phi}{\partial z},$$where $\phi$ is a scalar field.

I'm assuming this one-form is just another complementary way of describing the behaviour of the fluid at P. I have some vague notion of a one-form as a set of planes or contours, each one representing different values of $\phi$. So, what is the physical significance of $\phi$? In other words, what is the relationship between $\phi$ and the velocity of the fluid?

Further thoughts – I believe the metric tensor is unitless, which means $V_{b}$ has units of $ms^{-1}$ and $\phi$ has units of $m^{2}s^{-1}$. What on Earth has units of $m^{2}s^{-1}$?

BOUNTY EDIT – On reflection, this is easier for me to visualise in two arbitrary dimensions $(x,y)$, with a tangent vector to a parametric curve having components

$$V^{a}\equiv v^{x},v^{y}\equiv\frac{dx}{dt},\frac{dy}{dt}.$$

Lower the index with the metric to give a one-form with components$$V_{b}\equiv\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y}.$$What is the physical significance of the orientation, value and spacing of the constant $\phi$ contours? Thanks.

  • 2
    $\begingroup$ That the 1-form field $V^{\flat}$ is exact $\mathrm{d}\phi$ is an additional assumption. $\endgroup$
    – Qmechanic
    Apr 6 '19 at 8:16
  • $\begingroup$ Yes, can I keep things simple and assume that. Is it helpful to point out that I also understand (well, sort of) that if the scalar field gave, for example, temperature at every point, the inner product of one-form and vector would give a number – change in temperature per time unit in the direction of the vector? $\endgroup$
    – Peter4075
    Apr 6 '19 at 8:25

So from your definition

$$U_b = g_{ab} V^a,$$

where I have renamed the 1-form $U$ so I don't have to write coordinates to distinguish them, we see that the value of $U$ on a unit vector $W$ is the projection of the velocity $V$ onto $W$, ie. by definition

$$U_bW^b = g_{ab}V^a W^b.$$

Physically we can interpret this as saying that if we have an infinitesimal area element $\hat n dA$, where $\hat n$ is the unit normal vector and $dA$ has units of area, that the value of $U$ on $\hat n$ is the fluid flux per unit time per $dA$ through this area element.

Edits below because I mixed up my curls and divergences.

As Qmechanic mentioned, this 1-form doesn't have to be closed, unless we assume the flow is vorticity-free ie. $\nabla \times V = 0$. This is not a very physical assumption because vorticity tends to just get generated all the time, but in this case we'd be able to define a function $\phi$ such that the velocity field is a gradient: $V = \nabla \phi$.

It's more physical to assume that the flow is divergenceless, ie. that the fluid is incompressible, which could be very physical. So now the cooler thing that I wanted to mention anyway actually becomes necessary to express this condition compactly.

The way to do it is not to speak in terms of $U$ but in terms of a 2-form (or 1-form in 2d, in general it's Poincar\'e dual to $U$)

$$F_{bc} = \epsilon_{abc}V^a$$

whose value on any surface is the flux per unit time through that surface. Here $\epsilon$ is the volume form or Levi-Civita tensor. A useful conservation law for incompressible fluids is $dF = 0$, meaning that in the absence of sources or sinks, the fluid flux through a closed surface is always zero. You can think of this as a Gauss law, which makes fluid flow in 3d look a bit like a gauge theory (with gauge group $\mathbb{R}$), where the flow lines behave like magnetic field lines. Anyway, that's a story for another time. Here's some good reading though.

  • $\begingroup$ I've seen examples where the one-form contours represent height or temperature. What do they represent in this case? $\endgroup$
    – Peter4075
    Apr 6 '19 at 12:13
  • $\begingroup$ @Peter4075 Ah sorry I got mixed up by all the index gymnastics. See my edits. There is not really a $\phi$ that you can define in general, but there is this flux 2-form instead. $\endgroup$ Apr 6 '19 at 13:36
  • $\begingroup$ The thing that still puzzles me is whether the contours of constant $\phi$ (with those weird units of length squared per time) have any physical significance? That's my key question. Can we forget about fluids and just consider the simple case of velocity tangent vectors to a two dimensional parametric curve. We use the metric to lower the index to get a differential one-form. (We could go the other way, start with a differential one-form and raise the index to get velocity tangent vectors.) What do those contours signify? Thanks. $\endgroup$
    – Peter4075
    Apr 7 '19 at 7:00
  • $\begingroup$ @Peter4075 Well $\phi$ only exists if the velocity field is a gradient field. Let's suppose then we have a potential $W(x)$ felt by the particle as a force $dV/dt = -\nabla W$. So we can write $V = - \nabla \int_t W$. There's your $\phi$, it's the time integrated part of the potential, which is one part of the action, the other being the time integrated potential energy. I'm not sure at the moment if there is a better description of it. $\endgroup$ Apr 7 '19 at 7:13
  • $\begingroup$ Sorry for being dim, but what part of your final equation has the units of $\phi$, ie length squared over time? And why isn't there mass in your force equation $dV/dt$? $\endgroup$
    – Peter4075
    Apr 7 '19 at 9:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.