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pressure, symmetry, relativity, and conservation



At 02:36 PM 11/10/00 -0500, I wrote:
Static pressure is independent of the observer's frame of reference.

There are two ways of understanding this statement:

1) One can note that pressure is _defined_ in the rest frame of the fluid,
and therefore it is practically a tautology to say that the observer's
frame of reference doesn't matter. This tautology by itself doesn't tell
us anything about physics.

2) But there is much more to it than that:

At 08:24 PM 11/8/00 -0500, I wrote that "static pressure tells the story"
as to whether fluid will flow horizontally under the influence of a
pressure difference. This is a nontrivial nontautological statement about
physics.

I also said this was "the only answer consistent with relativity and
symmetry and other basic principles".

It has come to my attention that some people would like more details on how
to apply those principles to this problem, so I'll spell it out now.

Nitpickers note: This discussion is restricted to speeds
small compared to the speed of light. The foregoing
mention of "relativity" denotes Galileo's principle of
relativity. The concepts can be generalized to higher
speeds, but that would be a needless distraction.

So, as before, consider two huge airmasses "A" and "B" sliding past each
other. One airmass is northbound at 20 knots. The other airmass is
southbound at 10 knots. They meet at a sharp frontal boundary which lies
in a vertical north/south plane. All this happens at the equator, so we
don't need to worry about Coriolis effects. The sea-level pressure of each
airmass is 1013 millibars.

We will analyze this situation using the principles of
-- symmetry
-- Galilean relativity
-- conservation of momentum

We will need the following lemma about relativity: If we have two
observers whose relative velocity is purely in the north/south direction,
they will disagree about the north/south component of momentum a given
observed object, but they will _agree_ about any components of momentum in
orthogonal directions (such as the east/west component, which is the one we
care about).

We now set up an appeal to symmetry: Consider the point of view of
observer "C" who is moving at the average velocity of the two
airmasses; in this example that means northbound at 5 knots. To this
observer, everything is symmetric. To this observer, there is no reason
why the pressure from airmass A should not exactly balance the pressure
from airmass B. There should be no net flow of east/west momentum from one
airmass to the other.

Technical detail for those who are interested: Note that
the quantity I call "east/west momentum" is one component
of the momentum vector. The question of which _component_
that quantity represents is separate from the question of
which direction that quantity is flowing. There is actually
a tensor involved, in which one of the matrix elements is
the x-flow of the x-component of momentum.

Applying the lemma: All observers moving in purely north/south directions
will agree that there is no net flow of east/west momentum from one airmass
to the other.

As always, a pressure P acting normal to an area A for a time t will
transfer momentum equal to PAt. When we say there is no _net_ flow of
momentum from one airmass to the other, that is because the PAt eastward
from one airmass is just balanced by the PAt westward from the other.

So now consider observer "A" (comoving with airmass A) and observer "B"
(comoving with airmass B) and all other observers moving in purely
north/south directions. They agree about the area of the frontal
boundary. They agree about the time. The lemma says they agree about the
momentum. The momentum is PAt. Therefore (drumroll, please ...) they must
agree that P is the physically-relevant pressure in this situation and that
it is independent of the velocity of the observer.

I hope this helps de-mystify the question that Robert A Cohen posted at
06:26 PM 11/8/00 -0500.

BTW, note the limits to the validity of this result: We have assumed that
the fluid is flowing parallel to a flat boundary. If that is not true, the
result doesn't hold in general. In the more-complicated situation, a full
analysis would need to include dynamic effects, not just the static pressure.

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

We have the following useful "take home messages":
-- It is often helpful to think about pressure in terms of momentum flow.
-- Look for opportunities to apply symmetry principles.
-- Look for opportunities to apply relativity principles.
-- Look for opportunities to apply conservation principles.

The problem RAC teed up for us is an unusually juicy opportunity to apply
symmetry, relativity, and conservation in rapid succession: bing, bang, boom.

Physics can be viewed as a two-step process: The _second_ step involves
writing down a formula, plugging in numbers, and turning the crank. But
before that, the first step is to figure out what is the right formula, and
what the symbols in it mean. In Bernoulli's formula, P means pressure and
V means velocity, but pressure of what? velocity of what? in whose
reference frame? and what is that "constant" on the RHS? You need the
grand principles of physics (symmetry, relativity, conservation, etc.) to
get you through this first step.

A related take-home message: As John Archibald Wheeler is fond of saying,
if a complicated calculation gives a simple answer, step back and see if
there is a more-elegant way of analyzing things. Yesterday I explained the
gory details of why the "static port" on an altimeter is sensitive to
P+Q-Q. But as I said way back on Wednesday, and again today, the grand
principles (symmetry, relativity, conservation) tell us directly that it
_must_ be sensitive to P. Sometimes having a good feel for the grand
principles is very practical.