Re: [Phys-L] Textbook Errors re Waterjet Levitation

[John Denker <jsd@av8n.com>]

On 10/20/19 2:09 PM, I wrote:

> A proper analysis of the jet-pack is not easy.
> Nothing involving fluid dynamics is ever easy.

I am not kidding about that.  Fluid /statics/ is not so
bad, e.g. Archimedes principle, or Pascal's principle
as applied to a bottle jack.

Fluid dynamics is something else entirely.  People spend
their lives surrounded by fluids, and they imagine that
they know what's going on, but they don't.  I've been
fooled enough times to be well and truly skittish.  And
I've seen world-famous aerodynamics professors get fooled
enough times.  Feynman once said that fluid dynamics was
more complicated than elementary-particle physics.

======================

For people who are interested in this topic:  I'm not going to
explain things in any detail.  This is more of an outline, not
a pedagogical explanation.  This will perhaps allow you to get
started; you can find the next level of detail by googling the
buzzwords.

*) Nozzles were mentioned.  This is tricky business.

 -- The convergent-divergent ("CD") nozzle aka "de Laval" nozzle
  is a tremendously important invention.  However, it applies
  to rocket engines where the flow is supersonic.  The physics
  is a bit counterintuitive:
    https://www.grc.nasa.gov/WWW/K-12/airplane/nozzled.html

  Same idea with more words, less equations:
    https://blogs.nasa.gov/J2X/tag/convergent-divergent-nozzle/

  The most basic physics of rocket propulsion can be understood
  in terms of momentum.  No nozzle is required, since you can
  imagine using a catapult to throw rocks out the back.
  However, the practical engineering of an /efficient/ rocket
  requires consideration of energy.  You would rather have the
  energy that comes from the rocket fuel exit in the form of
  kinetic energy (related to useful momentum) rather than in
  the form of useless heat.  A well-engineered de Laval nozzle
  is essentially a heat engine, efficiently converting hot
  low-velocity gas into cooler high-velocity gas.

  MEANWHILE... the CD shape is approximately the last thing you
  would want for a garden nozzle, or for a hydraulic jet-pack,
  where the flow is subsonic.  A plain old orifice would work
  better.

  The fire-fighting industry has put a lot of thought into nozzle
  design.  Also, there exists water-jet propulsion for boats large
  and small.  The ducted propulsion in Hunt for Red October was
  unrealistic as to details (probably on purpose) but definitely
  not entirely fanciful.
    https://www.hamiltonjet.com/global/waterjet-overview


*) Tensors were mentioned, in connection with momentum transport.

 As previously mentioned, there are ways of diagramming momentum
 transport, using /double/ arrows:
   https://www.av8n.com/physics/force-intro.htm#sec-momentum-flow

 In spacetime, you probably want the stress-energy tensor, which
 is a 4×4 tensor.  Misner/Thorne/Wheeler is the reference I rely
 on for this.

 For jet-pack purposes the spacelike part of the stress-energy
 tensor suffices.  It is a 3×3 tensor.  Beware:  You might think
 this would be called the stress tensor, but it's not.  It's
 called the momentum-flux tensor.  Actually it's usually just
 called the "flux" tensor, but that is obnoxious, because there
 are lots of different fluxes (electrical flux, mass flux,
 momentum flux, etc.).

 There also exists a stress tensor, but that's something else;
 it is /part/ of the momentum-flux tensor, but not the whole
 story, and definitely not the part we care about for the
 jet-pack, because the 3×3 stress tensor for a moving fluid
 is the /same/ as for a non-moving fluid.

 The diagonal elements of the stress tensor represent the
 /pressure/.

    Again, beware:  The naming conflict between the 4×4
    stress-energy tensor and the 3×3 stress tensor is not
    the least bit obvious, and often goes unexplained.
    Key equation:

	momentum flux = advection - stress

        https://en.wikipedia.org/wiki/Cauchy_momentum_equation#Conservation_form

 In the context of the jet-pack, we especially care about
 the /advection/ of momentum, which can be written as the
 outer product of momentum and velocity:

	A = p ⊗ v		(advection)
	  = m v ⊗ v

 Beware that the notation for outer products can be confusing.
 People who know a little bit of physics can use Dirac bra-ket
 notation to advantage:

	A = |p⟩⟨v|		(outer product = tensor)

 For the basis vector x we can write out the components:

	          [ 1  0  0 ]
	|x⟩⟨x| =  [ 0  0  0 ]
	          [ 0  0  0 ]

 The outer product stands in stark contrast to the dot product:

	2KE = ⟨v|p⟩		(inner product = scalar)
	    = |v⟩ᵀ |p⟩

 where (ᵀ) denotes the transpose, which we use to convert a
 column vector into a row vector.

 In particular, if we rotate the vectors, the inner product
 is unchanged:

	⟨v| Rᵀ R |p⟩ = ⟨v|p⟩	(scalar unaffected by rotation)

 whereas the outer product gets rotated TWICE:

	R |p⟩ ⟨v| Rᵀ		(tensor strongly affected by rotation)

 If we rotate the basis vector x by 90° we get the basis vector
 y, in which case the components are:

	          [ 0  0  0 ]
	|y⟩⟨y| =  [ 0  1  0 ]
	          [ 0  0  0 ]

 For the special case of a 180° rotation, we get

	            [ 1  0  0 ]
	|-x⟩⟨-x| =  [ 0  0  0 ] = |x⟩⟨x|
	            [ 0  0  0 ]

 which is the mathematical version of yesterday's assertion that
 upward transport of upward momentum is equivalent to downward
 transport of downward momentum.


*) Not mentioned yesterday:  A full analysis of the jet-pack would
 include both advection and stress (i.e. pressure) in the feed-tube.

 For a large-diameter feed-tube, pressure is relatively more important.
 Given enough diameter and enough pressure, you could support quite a
 lot of weight on the turgid tube.

 Conversely, for a small-diameter feed-tube, pressure is relatively
 less important and advection is more important.  You're moving the
 same amount of water but with more velocity, therefore more momentum.