🚀 go-pugleaf

In article <1992Nov5.190422.1@stsci.edu> zellner@stsci.edu writes:
>
>My goodness!  If we cannot do sophomore physics, how can we settle questions
>about manifolds and Higgs bosons?
>
>  >>  Thus, consider this: If the air surrounding the balloon had exactly 
>  >>  the same density on all sides, then the balloon would not move (I 
>  >>  think).
>  > 
>  >   No, actually it would sink, I think, because there is still the effect
>  >   of gravity.
>
>Double nonsense.  A gas cannot have uniform pressure in a gravitational 

Hey, hey, hey.  As the incarnation of good sense, I take this as an
uncalled-for ad hominem attack.

Someone pointed out earlier that of course balloons float when lighter
than air because of gravitational potential.  Obviously this is true,
and I don't mean to disagree with it, only to reassure the original
poster that his way of seeing things (in terms of forces rather than
potentials) was correct TOO.  In that context, as you go on to say below,
you must conclude from the gravitational potential that the pressure is
NOT the same below and above the balloon.  That's why both methods give
the same answer.  The question, however, was: what if the pressure (actually,
he wrote density) WERE the same?  Though in hydrostatic equil. in a grav.
field this cannot be, one could certainly force the pressure below and
above a balloon to be equal.  (For example, have two different chambers
below and above the balloon, with the balloon constrained to move vertically.
The pressures need to be controlled to a high precision to make them equal
to better than the effect of their own mass.)  In this case, the damned 
thing would obviously sink due to gravity.

(Judging from your first sentence, maybe I should rephrase this and say
that the interaction between the balloon and the vacuum Higgs field causes
the whole mess to exchange gravitons with the earth, and avoid any of this
argument, since we'll be stuck with a Lagrangian formulation once I do so.)


>field, not in hydrostatic equilibrium.  If it did, the gas would move 
>around until it developed pressure differentials such that each volume 
>element would have a pressure gradient between top and bottom equal to its 
>weight.  
>
>In a perfect gas the density is proportional to pressure.  In a liquid like
>water or oil the density is almost independent of pressure, but the above
>requirements of hydrostatic equilibrium are preserved in either case.
>
>Now for balloons.  We can think about an infinitely stretchy balloon, which
>will always maintain the same pressure inside as the air outside, and fill it 
>with helium or hydrogen, which are less dense than air at the same pressure.
>Or we can imagine a perfectly rigid balloon, the volume of which doesn't 
>change with outside pressure, either empty (vacuum) or containing any mass
>(composition irrelevant).  Or you could take an intermediate case. 
>
>In any case it's the outside pressure gradients that count.  If the net 
>pressure gradient is greater than the weight of the balloon, it will rise.  

Exactly.  Since 0 is less than the weight even of a Helium balloon, the
answer to the question is, as I said, yes.  If you want to say a question
is misleading, that's fine, but don't just invert it to what you would
have asked and then call the response "nonsense."  Thank you!
-- 
Aephraim M. Steinberg                    | "WHY must I treat the measuring
UCB Physics                              | device classically??  What will
aephraim@physics.berkeley.edu            | happen to me if I don't??"
                                         |       -- Eugene Wigner