Re: The antics of thermodynamics, the depravity of relativity, the bunkum of quantum

Liste des GroupesRevenir à col advocacy 
Sujet : Re: The antics of thermodynamics, the depravity of relativity, the bunkum of quantum
De : peter (at) *nospam* pmoylan.org (Peter Moylan)
Groupes : alt.usage.english sci.physics
Date : 11. Mar 2025, 01:29:28
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vqo05a$1kf0u$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9
User-Agent : Mozilla/5.0 (OS/2; Warp 4.5; rv:38.0) Gecko/20100101 Thunderbird/38.8.0
On 11/03/25 06:54, jerryfriedman wrote:
On Sun, 9 Mar 2025 9:50:06 +0000, Peter Moylan wrote:
>
On 09/03/25 09:58, Phil wrote:
On 08/03/2025 22:46, Bertietaylor wrote:
>
It's just as implausible as the suggestion (easily
disproved) that the pressure is zero at the centre of the
earth.
>
The pressure is most certainly zero at the centre of the
stars and planets. Read a first year book on physics.
>
Which will say that within an enclosed surface with mass the
net gravitational force or pressure is zero.
>
Read that first year book yourself. Did you find the words "or
pressure"? No, I didn't think so. You've tried to conclude
something about the pressure from the gravitational force. That
doesn't work, because they are different quantities.
>
Gravitational force, like all forces, is a vector quantity. It has
a magnitude and a direction. That makes it possible that a number
of nonzero vectors can sum to zero; and, indeed, that is what
happens inside a spherical shell.
>
Pressure is a scalar. If you add two pressures, you get a higher
pressure. There's no such thing as a negative pressure to cancel
out the first pressure.
>
Think of a cone, or similar shape, whose point is at the centre of
the earth. You can separate out a section with thickness dr, and
write down the force balance equation for that slab. (This, too, is
first year physics.) From that you get a differential equation for
the pressure as a function of radius. No matter what
simplifications you make, you will get the same conclusion: the
deeper you go, the higher the pressure. Which is something that
ocean divers can confirm from their own experience.
>
Even swimming pool divers.
>
At the centre of the earth, the gravitational force is zero but
the pressure is at a maximum.
>
Presumably, by an analogous argument, the pressure at the centre
of a balloon is also zero?
>
Actually, the gravitational force at the centre of a balloon is
zero, if you count only the force due to the balloon itself. But of
course, you do have to count the attraction from the earth as
well.
>
Either way, what you conclude about the gravitational force says
nothing about the pressure. They're different quantities.
>
They're different, but they are connected.  Since the gravitational
force at the center of the Earth is 0, you can conclude that the
pressure /gradient/ there is 0.  Hint to Arindam.
Re-reading this thread, I have suddenly realised what has misled
Arindam. He keeps using "force" and "gravitational force" as if they
were interchangeable. Not all forces are gravitational forces; and to
calculate pressure you have to add up all the forces, not just the
gravitational component. At the centre of the earth, you also have to
taken into account the radially directed non-gravitational  force that
comes from the weight of all the rocks (and so on) above your head.
--
Peter Moylan       peter@pmoylan.org    http://www.pmoylan.org
Newcastle, NSW

Date Sujet#  Auteur
16 Apr 25 o 

Haut de la page

Les messages affichés proviennent d'usenet.

NewsPortal