# Re: GPT4 + Mathematica

```Following up on all that. Sounds Chat_GPT4 sounds friendly to me. But I, can
easily be fooled, for a time.
Hey, here's a thingy from The Mormon Transhumanist Association. Here's a link.
It ain't mathematical, but it is logical. Go ahead and kick the tires, and take
it for a spin, all! You may hate it, and that's ok by me, because I am
suspecting the Universe doesn't mind happy atheists as it's challengers as long
as they do the science? The religious folk as well. The agnostics are yummy,
good too.
I haven't gotten deep in it yet, seems a bit, plausible.
I tend to be open to Sci guys like Prisco, or Moravec, and Tipler, as well as
Tim Anderson too.
Should you be? Meh!  Totally up to you. Still, I like philosophy too. ```
```
https://new-god-argument.com/god-conclusion.html

-----Original Message-----
From: John Clark <johnkcl...@gmail.com>
To: everything-list@googlegroups.com
Sent: Sun, Apr 9, 2023 1:56 pm
Subject: Re: GPT4 + Mathematica

On Sun, Apr 9, 2023 at 12:53 PM smitra <smi...@zonnet.nl> wrote:

> For some simple physics or math result consider different
ways of getting to that result where one of these ways is not widely
published and is likely not in GPT's database. Take e.g. different ways
of computing the moment of inertia of a ball of uniform density of
radius R and mass M (relative to an axis through the center). There are
many ways to do this, but I've not seen my favorite way of doing this on
any webpage, which is to restore spherical symmetry by adding up the
three identical moments of inertia relative to 3 orthogonal axes.

I just send this to GPT:
Me: Derive the formula for the moment of inertia (relative to an axis through
the center) of a ball of uniform density of radius R and mass M by adding up
the three identical moments of inertia relative to 3 orthogonal axes and
restoring spherical symmetry.

And in less than three seconds it gave me this:
GPT: To derive the formula for the moment of inertia of a ball of uniform
density, we will start by finding the moment of inertia relative to three
orthogonal axes passing through the center of the ball. Let's call these axes
x, y, and z.

The moment of inertia of a point mass m relative to an axis at a distance r
from the mass is given by:

I = m * r^2

For a solid sphere of uniform density, we can divide it into small volume
elements, each with mass dm. The volume element located at a distance r from
the center of the sphere has a moment of inertia relative to the x-axis given
by:

dIx = dm * y^2

Similarly, the moment of inertia of this volume element relative to the y-axis
is:

dIy = dm * x^2

The moment of inertia relative to the z-axis is:

dIz = dm * r^2

where r is the distance from the center of the sphere to the volume element.

We can express the differential mass element dm in terms of the density ρ and
the volume element dV:

dm = ρ * dV

For a sphere of radius R, the volume element is:

dV = 4/3 * π * r^2 * dr

Substituting dm and dV into the expressions for dIx, dIy, and dIz, and
integrating over the volume of the sphere, we get:

Ix = ∫∫∫ y^2 * ρ * (4/3 * π * r^2 * dr) dV

Iy = ∫∫∫ x^2 * ρ * (4/3 * π * r^2 * dr) dV

Iz = ∫∫∫ r^2 * ρ * (4/3 * π * r^2 * dr) dV

where the limits of integration are:

0 ≤ r ≤ R, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π

To simplify these integrals, we can use spherical coordinates, where x = r sin
θ cos φ, y = r sin θ sin φ, and z = r cos θ. The volume element in spherical
coordinates is:

dV = r^2 sin θ dr dθ dφ

Substituting this into the expressions for Ix, Iy, and Iz, we get:

Ix = 2/5 * M * R^2

Iy = 2/5 * M * R^2

Iz = 2/5 * M * R^2

where M is the mass of the sphere, given by:

M = ρ * V = 4/3 * π * R^3 * ρ

where V is the volume of the sphere.

Note that Ix = Iy, since the sphere has spherical symmetry, and the moments of
inertia relative to any two orthogonal axes in a plane perpendicular to the
third axis are equal. Therefore, the moment of inertia of a sphere of uniform
density relative to an axis through the center is:

I = 2/5 * M * R^2

which is the desired formula.
John K Clark    See what's on my new list at  Extropolis
grxq

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```