User:Diamond

From SkepticWiki

1 Personal Details
2 Religion and Philosophy
3 Special Interests
4 Comments
5 Question I'd like an answer to
6 My test pages

[edit] Personal Details

Age: 40+
Sex: Yes, please (serious answer: male)
Star Sign: Skepticus the Challenger of Fakery
Family: One wife and one daughter

[edit] Religion and Philosophy

I am a philosophical naturalist and rationalist.

I was brought up in a not-terribly strict Roman Catholic family and became a fundamentalist evangelical Christian in my early 20s. Through reading science books, a passion from my youth, I became less and less interested in religion finally finding religion to be unnecessary to any explanation of the Universe. I am now an agnostic.

[edit] Special Interests

Science, especially astronomy. I did do some courses at college in Earth sciences so I am now also interested in climate science and the current fascination of "global warming"

[edit] Comments

It was my idea to set this wiki up. Whether it lives or dies is up to its contributors, not me.

I am not a bureaucrat or SysOp and cannot help with administration of this wiki. For adminsitrative assistance, please contant Fowlsound. For the real technical stuff about wiki, ask geni...

I am an occasional contributor.

[edit] Question I'd like an answer to

The Schwarzschild Metric on a spacial plane passing through the center of a spherically symmetric (non-spinning) center of gravitational attraction is:

$d \tau^2 = (1- \frac{2M}{r})dt^2 - \frac{dr^2}{(1- \frac{2M}{r})} - r^2d \phi^2$

If there are two spaceships at a distance $r 1$ from a gravitating body, and one moves radially ( $d φ = 0$ ) to a radius $r 2$ , in a time $τ 1$ as measured by its own clock, and then immediately returns from $r 2$ to $r 1$ in a time $τ 1$ , how much time will have elapsed on the clock on the spaceship which remains at $r 1$ ?

$\tau = - \left( -\sqrt {r \left( {E}^{2}r-r+2\,M \right) }\sqrt {{E}^{2}-1}+M \ln \left( {\frac {M+{E}^{2}r-r+\sqrt {r \left( {E}^{2}r-r+2\,M \right) }\sqrt {{E}^{2}-1}}{\sqrt {{E}^{2}-1}}} \right) \right) \sqrt {{\frac {{E}^{2}r-r+2\,M}{r}}}r{\frac {1}{\sqrt {r \left( {E}^{2 }r-r+2\,M \right) }}} \left( {E}^{2}-1 \right) ^{-3/2}$