Requiem for Relativity

I think the JPL site is in need of a face lift, it looks dated and it's not that user friendly. There's going to be errors, leaving aside the problem of the year zero and the ten days due to calendar changes. A program that compensates for leap seconds is going to be too slow for the site.

I remember putting up a link here on the problems of historical supernovas. As I recall one of them was of particular interest to South Americans as the nova appeared next to the moon. One European sighting was discounted, because of where the nova remnant is thought to be! It might be an idea to find that article again. I think it's odd that two eleventh century novas should have been reported and them not being where they are supposed to be.

Sloat, A SN next to the moon would only place the event somewhere near the orbit of the moon with no info about the RA of the event. That's a lot of space to search for a remnant.

Hi Jim, the 1006 a.d. supernova is the brightest historical supernova. As i recall, it was seen as a red disk just under the moon by about a half degree, on a certain date. Now this account was rejected, and again from memory, that had something to with the phase of the moon. It simply had to be wrong, because we know where the remnant is, and of course we know where the barycentre of the solar system is.

Well, the link to the paper is somewhere here in this thread. At the time, I mentioned it to Joe as something to look into, Joe wasn't then looking at historical accounts. We've got two supernova that came along like busses. 1006 and 1054, the remnants don't match up to where the historical records mark the positions. This could possibly give us an estimate of whether the barycentre is where we think it is.

Sloat, If the location of the barycenter of the Solar System is the cause of a misplaced SN you can be sure the error is the barycenter. The Barycenter is always misplaced by lazyness of astronomers. They like to use it as a mass center when its easy to look and see its not.

Hi Jim, someone in this thread put up a link to a program that lets you look at the movement of the barycentre for the solar system, and add other planets. It's movement is incredibly complicated. We have to add the fact, that the people who are looking at ancient historical records of astronomical events, tend not to be working astronomers. There are only a handful of people who can read Sanskrit, or eleventh century Korean. Even less can interpret amerindian cave drawings. It's only with those two eleventh century super novae that a team was put together by NASA.

The problem is, if Joe's right, then the barycenre isn't where it's assumed to be. If he can show that the solar system still works as it appears to, then it will be accepted. However, we do have that little problem of the speed of gravity. Let's say that Joe's theory really does put the cat among the pigeons with something like the Mercury problem. Then working astronomers will remain shtum, as they will be seen to be stepping on Einstein's toes.

I thought of another avenue of approach to this yesterday. It's the rather curious orbital behavior of the "Earth's second moon"! It now seems that there's more than one of these, and Mars also has a few. en.wikipedia.org/wiki/3753_Cruithne

On the problem os encounters with the ghost of Einstein, I thought I'd ask a guy who's working on gravito magnetics for his thoughts on the speed of gravity. His reply was, what's your point. Nothing can travel faster than light. End of discussion! Another one, was where I commented on a facebook page of another guy, and said that we can divide mc^2 by the Planck mass and lose the units but I questioned the Planck mass. The post got deleted by the guy. Easy the best one. A guy told me that Newton was confused by the word "instantaneous" because he rode a horse. Wow!

A little update on that Hawking problem.

Power = (16pi G^2 M^2 / c^4) (pi^2k^4 / 60 hbar^3 c^2) (hbar c^3 / 8pi G M k)^4

Just for fun I thought I'd see what it would look like if the Planck particle was a toroid.
area = 4 pi^2 R r

With this, we don't replace the c^4 in the first bracket with b^4 (being the speed of gravity) we have instead (16 pi^2 G^2 M^2 / b^2 c^2)

Just about everything cancels out. We still get that peculiar number in the numerator though. Remember that I divided c^12 by b^4 to get 0.725634065 (100 times the fine structure constant perhaps?)
The same number shows up when I multiply c^12 / b^2 c^2 by hbar.
Anyway I get 16 * 0.725634065 / 60 c^2
This will give a power output of about five times the unit charge. I'm toying with the idea that the square root of 2pi will come into it somewhere.

At the moment I'm just interested in it as a toy but it is interesting as both G and M have dropped out altogether.