Stellar Splitting and pairing NEW Black holes foun

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16 years 1 month ago #15468 by Jim
Replied by Jim on topic Reply from
Sloat, You first said h is the angular momentum and now you say h is the angular velosity and the angular momentum is mvr. I used the mvr definition and solved for mass using h equals mvr. Thats how the strange units of mass got in here. (Anyway, all his got me thinking about nuclear fission notation which maybe better left out of this thread.) If electrons have different radii how to keep constant angular momentum and velosity gets tricky even if you tweek the orbits because c and h are constant. And getting back to the vacuum energy detail the electrons still would fill that space wouldn't they? What is the frequency of Compton wavelenght-we can plug that in h and see what happens.

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16 years 1 month ago #20386 by Stoat
Replied by Stoat on topic Reply from Robert Turner
Sorry Jim, you're right, I did write h is the angular velocity. That's simply a mistake it should read, Taking h as the angular momentum. h = mvr for an electron 9.1093897E-31 * 2.99792458E 08 * r divide h by mv to get r.

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16 years 1 month ago #15469 by Jim
Replied by Jim on topic Reply from
Sloat, It would be just as a big mess either way because h,c,v&m are not without attached units. Writing h=mvr has a balance issue that makes one part of mvr very odd. Writing h=massxc^2xtime=mvr where mass cancels out, leaves c^2=vr/time. We need to know what the time unit is so I guess you can use that to make the units balance. Sweep the mess into time. Does that help at all? I guess not because that defines time in terms of c.

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16 years 1 month ago #15470 by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Jim, what we have is hf = m c^2 which we rewrite as
mvrf = mc^2 Cancel out and we get f = c / r

Divide c by the Compton wavelength which is the radius of an electron and we get a frequency of 1.2355897798E 20 cycles per second. Thats the frequency of an electron.

What has to be of interest though, is that for my proposed speed of gravity we have E = m b^2 where b is the speed of gravity.
E = 9.1093897E-31 * 1.35639139448E 50 = 1.2355897798E 20 Joules.

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16 years 1 month ago #15471 by Jim
Replied by Jim on topic Reply from
Sloat, You have a worse messs with this result because you cancel time out along with the mass. Its ok to do thiswith the mass because it is clear the mass is equal on both sides but time only shows up on one side because of how h is constructed. Everyone just does the logical(simplest from a math perspective)thing and sweeps the time out of the constant. Its just not right though because you always get the wrong answer.

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16 years 1 month ago #15513 by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Jim, no, thats a perfectly good equation. The dimensions are just length and time, time does not get lost. Let's do it the long way round. E = m c^2 for an electron
E = 9.1093897E-31 * 8.98755178737E 16
E = 8.18711116801E-14J
hf = 8.18711116801E-14 divide by h to get the frequency, 1.2355897798E 20 cps

Where people could go ballistic is with my proposed speed of gravity. E = mb^2 where b is the speed of gravity.
E = 9.1093897E-31 * 1.35639139448E 50
E = 1.2355897798E 20J
hf = E
h * 1.2355897798E 20 = 1.2355897798E 20

I think that that lends weight (excuse the pun) to Newton's concept of an infinite speed of gravity. I do believe he was working on it, his idea that light was particulate in nature strongly suggests that. Where I would differ with him, is that I would say that rather than e^x we have e^nx

A bec cosine energy profile particle will be more complicated by this stonking great exponential. Its core will be densely packed waves but they'll have a wave superposed that would give us shells separated by the wave troughs, the casimer effect would have to be taken into account. Any gaps in the shells are the true vacuum.

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