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Requiem for Relativity
14 years 3 months ago #23993
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Thinking about that equation a bit more sqrt(kT/m) We've got T = mc^2/k but I thik we need to multiply that by
1-sqrt(1-v^2/c^2)
We want rid of that root sign so square kT/m and have (kT/m)2 = c^4
C^4 = 8.0784215E 33 and it's reciprocal is going to be 1.2378655E-34 pretty close to barh.
1-sqrt(1-v^2/c^2)
We want rid of that root sign so square kT/m and have (kT/m)2 = c^4
C^4 = 8.0784215E 33 and it's reciprocal is going to be 1.2378655E-34 pretty close to barh.
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14 years 3 months ago #24162
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Joe, you wouldn't happen to know of an "idiots guide to Lie algebra" would you? The only stuff I could find just dives straight in at the deep end. I even went and joined Garrett Lisi's face book page but it's just a fan club. Pretty unhealthy actually, as I think that boat loads of people, without the foggiest idea what he's talking about, want to turn him into the next Einstein.
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14 years 3 months ago #23994
by Stoat
Replied by Stoat on topic Reply from Robert Turner
This might be of interest to you Joe.
www.ras.org.uk/news-and-press/157-news20...bts-on-the-dark-side
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14 years 3 months ago #24293
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Joe, you'll remember that I had a bit of a problem deciding on the speed of gravity being either h = c^2 / b^2 or barh = c^2 / b^2
b being the speed of gravity. All I had to do was divide the mass in E = mb^2 by 2pi but being a bear of little brain this confused me.
Then your post got me thinking about dimensionless numbers again. C^2 / b^2 is dimensionless and I want to work in nothing but those numbers. Remember that for the speed of gravity we get something very close to c^3 Now, I wanted a dimensionless number for particle masses and the obvious choice is the reduced Planck mass. I divide any mass particle by that; or multiply by the reciprocal. Just to see what would happen i multiplied C^2 by the reciprocal and then worked out what the planck mass would have to be, to give me the speed of gravity of about 2.92E 25 The reduced planck mass is approximately 4.34E-9 = Sqrt(barh c / 8pi G )
Just with a pocket calculator and only working to a few decimal places I get a Planck mass of sqrt(barh c / 15.92pi G)
Remember that when we are talking about the planck mass particle, we are talking about a micro black hole. We take the Compton wavelength and make that equal to the Schwarzschild radius,
h / mc = 2Gm / c^2
Its mass is proportionate and not yet settled, it's in the ball park for expressing masses as dimensionless though.
b being the speed of gravity. All I had to do was divide the mass in E = mb^2 by 2pi but being a bear of little brain this confused me.
Then your post got me thinking about dimensionless numbers again. C^2 / b^2 is dimensionless and I want to work in nothing but those numbers. Remember that for the speed of gravity we get something very close to c^3 Now, I wanted a dimensionless number for particle masses and the obvious choice is the reduced Planck mass. I divide any mass particle by that; or multiply by the reciprocal. Just to see what would happen i multiplied C^2 by the reciprocal and then worked out what the planck mass would have to be, to give me the speed of gravity of about 2.92E 25 The reduced planck mass is approximately 4.34E-9 = Sqrt(barh c / 8pi G )
Just with a pocket calculator and only working to a few decimal places I get a Planck mass of sqrt(barh c / 15.92pi G)
Remember that when we are talking about the planck mass particle, we are talking about a micro black hole. We take the Compton wavelength and make that equal to the Schwarzschild radius,
h / mc = 2Gm / c^2
Its mass is proportionate and not yet settled, it's in the ball park for expressing masses as dimensionless though.
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14 years 3 months ago #24163
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Another article that could be very important
news.stanford.edu/news/2010/august/sun-082310.html
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14 years 3 months ago #23995
by Jim
Replied by Jim on topic Reply from
Hi Sloat, You find great links-this one is a gem. I sent an e-mail to the author and will let you know if anything develops.
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