On 11-06-16 08:57 AM, David Jonsson wrote:
Good to hear. I have been thinking since March last year. First step
is to determine if Coreolis or centrifugal acceleration is the case.
OK, I think I've got it.
I'm assuming your reasoning is "purely Newtonian" -- you are using a
pure Newtonian mechanics model here. As I already said, reasoning
within that model doesn't prove anything about the real world; it can
only prove what will happen within the model. That, in turn, is a
*prediction* about the real world, which may be tested if one feels so
moved. However, as long as we can stick within the model, we can assume
certain basic things, like conservation of linear (and angular
momentum). The conservation laws must hold in the model, unless some
fundamental piece of basic mathematics is logically inconsistent.
My conclusion is that, working within the Newtonian model, the assertion
that hot and cold air of equal density differ in buoyancy when in a
rotating frame is false.
The reasoning consists of a simple gedanken experiment, which uses
conservation of momentum to arrive at its result.
*******
Assume we have a rigid, hollow sphere of negligible mass, and that it's
filled with gas. Its total mass is M (sphere + gas), and its velocity is V.
The total linear momentum of the sphere and its enclosed gas is the
average velocity of all particles in the gas, times the total mass of
the system (which is just the mass of the gas, as the shell is assumed
to be very light). The average velocity of the gas particles is (almost
exactly) the same as the velocity of the containing sphere, so the
linear momentum of the system is
P = M*V
That does *not* depend on the temperature of the gas.
Now, let's let the sphere follow a circular path around a planet (as it
would if it were sitting on the surface of the Earth). Your assertion is:
The sphere will "weigh" less if the gas is hot than it will if the
gas is cold.
Newtonian gravity is just another force -- it isn't special. So we can
replace the gravitational field with a rope tied to the sphere and
attached to a stationary post, and it won't affect the results.
Let's spin the sphere around the post, at the end of its rope. By your
assertion, the tension on the rope will be less when the gas is hot than
it will be when the gas is cold.
Let's have the whole thing oriented vertically, in the plane of the
paper, with the sphere going around a central point counter-clockwise.
At the top it's going to the left, at the bottom it's going to the right.
Let's take the "positive" direction to be to the right.
At the top of the path, the sphere is moving to the left, and its
momentum is -M*V.
At the bottom of its path, the sphere's momentum is +M*V.
The difference is 2*M*V.
As the sphere travels half way around the circle, the net momentum
transfered from the post to the sphere (through the rope) is
Delta-P = 2*M*V
If momentum is to be conserved, the net momentum transfered from the
sphere to the post (also through the rope) must be exactly
- Delta-P = -2*M*V
That value does not depend on the temperature of the gas in the sphere.
The tension on the rope, which is felt by the post, is the "weight" of
the sphere. That doesn't depend on the heat of the gas. Therefore, the
weight of the sphere can't depend on the heat of the gas.
The buoyancy of the sphere is the weight of the air it displaces, minus
its weight; so, the buoyancy can't depend on the temperature of the gas,
either.
QED.
David
On Jun 15, 2011 10:42 PM, "Stephen A. Lawrence" <sa...@pobox.com
<mailto:sa...@pobox.com>> wrote:
>
>
> On 11-06-15 09:03 AM, David Jonsson wrote:
>> On Sun, Jun 12, 2011 at 10:50 PM, Stephen A. Lawrence
<sa...@pobox.com <mailto:sa...@pobox.com>
>> <mailto:sa...@pobox.com <mailto:sa...@pobox.com>>> wrote:
>>
>>
>>
>> But using the Newtonian mechanics model itself, if you arrive at
>> the conclusion that the box is lighter when the ball is bouncing,
>> you can safely conclude that you did something wrong. That's not
>> a conclusion you can ever get to from the Newtonian model.
>>
>>
>> OK, sorry, but I also later came with a correction.
>>
>> Lets change the setup so that the ball bounces sideways. Do you agree
>> that it now becomes lighter? This is because the centrifugal forces.
>> The increase and decrease does not balance to zero.
>>
>> Do you also agree that with the sideways bouncing ball there is also a
>> small torque on the box, due to the same differences in centrifugal
>> acceleration?
>
> Dunno -- I'm going to have to think about that one, and I haven't had
> the time to really understand it. It seemed wrong when a similar
> assertion was first posted (months ago) and still seems wrong to me but
> I haven't got a proof that it's wrong, so I could be the one who's
wrong.
>
