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Stellar Splitting and pairing NEW Black holes foun
16 years 1 month ago #15477
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Jim, Ill get back to you on those points.
I still think were making some progress. Heres something that should be of interest to people on this board. Weve got a lorentzian of sqrt(1 - v^2 / b^2) Where b is the speed of gravity. When v equals the speed of light we have h. Now we can put anything in for v, with no problems until we get to v = b
So, lets put in 269.126797287E 09 times c. This will give us
4.79921566384E-11 = v^2 / b^2 thats h / k = v^2 / b^2
Now lets write the lorentzian in terms of the refractive index:- sqrt(1 - 1 / eta) When v equals the speed of light we get a refractive index of the reciprocal of h, 1.5091889611E 33. For the other value we get a refractive index of 20.8367381265E 09
A little Newton pops up and says, okay when v equals one, you get a refractive index of 1.35639139448E 50 and when its less than one? When its zero, dont we have one minus minus infinity, which is just infinity. The speed of gravity is infinite but locally you can have steps in your graph. Locally is huge but.
I still think were making some progress. Heres something that should be of interest to people on this board. Weve got a lorentzian of sqrt(1 - v^2 / b^2) Where b is the speed of gravity. When v equals the speed of light we have h. Now we can put anything in for v, with no problems until we get to v = b
So, lets put in 269.126797287E 09 times c. This will give us
4.79921566384E-11 = v^2 / b^2 thats h / k = v^2 / b^2
Now lets write the lorentzian in terms of the refractive index:- sqrt(1 - 1 / eta) When v equals the speed of light we get a refractive index of the reciprocal of h, 1.5091889611E 33. For the other value we get a refractive index of 20.8367381265E 09
A little Newton pops up and says, okay when v equals one, you get a refractive index of 1.35639139448E 50 and when its less than one? When its zero, dont we have one minus minus infinity, which is just infinity. The speed of gravity is infinite but locally you can have steps in your graph. Locally is huge but.
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16 years 1 month ago #20388
by Jim
Replied by Jim on topic Reply from
Sloat, I suppose you use k as the Boltzmann constant. With that in mind writing h/k is equal to writing time.temperature. Its just a trick math guys use to make nonsense out of data in order to spend more time on figuring it all out. The facts about k only apply to mass and do not work on energy. The facts about h apply to huge bundles of energy and these facts are misused in all science. This is a problem that dates to the begining of modern physics and really needs to be fixed in order to address the other stuff in a logical process.
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16 years 1 month ago #15515
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Jim, I think theres a possible test open to us. I believe that any particle moving at any velocity will not change its frequency but will change its wavelength. So gravitational and electromagnetic frequencies are the same but we have different grav and elec wavelengths.
I also say that there is a bog standard photon, which is h of the mass of an electron. That means that there is a distinct cut off to long wave radiation.
Weve got lambda = h / m c Lambda giving us the radius of a particle. For my mass of a photon we have 6.03595039111E-64 kgs So we get a radius of the photon as being 3.66176117373E 21 metres. That is just ginormous! What can it be? Lets say that its the gravitational wavelength, so divide that number into the speed of gravity to give us its frequency. My speed of gravity is 1.16464217444E 25 metres per second This gives us a gravitational an electromagnetic frequency of 3.18055197809E 03 cps. What we then do, because we believe that nothing travels faster than light, is to divide this into c to give us a wavelength, that comes out at 9.42579967456E 04 metres.
The test is, can we build a tank circuit to transmit and receive a longer wavelength signal?
(Edited) I suppose this would mean that there are 3.18055197881E 03 gravitons to the photon.
I also say that there is a bog standard photon, which is h of the mass of an electron. That means that there is a distinct cut off to long wave radiation.
Weve got lambda = h / m c Lambda giving us the radius of a particle. For my mass of a photon we have 6.03595039111E-64 kgs So we get a radius of the photon as being 3.66176117373E 21 metres. That is just ginormous! What can it be? Lets say that its the gravitational wavelength, so divide that number into the speed of gravity to give us its frequency. My speed of gravity is 1.16464217444E 25 metres per second This gives us a gravitational an electromagnetic frequency of 3.18055197809E 03 cps. What we then do, because we believe that nothing travels faster than light, is to divide this into c to give us a wavelength, that comes out at 9.42579967456E 04 metres.
The test is, can we build a tank circuit to transmit and receive a longer wavelength signal?
(Edited) I suppose this would mean that there are 3.18055197881E 03 gravitons to the photon.
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16 years 1 month ago #15480
by Jim
Replied by Jim on topic Reply from
Sloat, If lambea=h/mc it is equal to c times time with time being any unit from 10E-1111s to 10E1111 or so. Basically, its nothing and makes no sense. The misuse of h is still poison.
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16 years 1 month ago #20151
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Jim, I think I see where youre coming from but there really is no problem here. If we have h = c^2 / b^2 we have to rewrite it as x = v^2 / b^2
Forget for the moment that h has any significance, its just a number. So, lets peg the speed of gravity, nothing can travel faster than it. Things can travel faster or slower that the speed of light. All we have is a ratio and infinities lurk in all ratios.
When v^2 equals zero we have minus infinity. Thats just saying that where McCrimon sits is head of the table. An ultimate preferred frame of reference. This means that we can write the lorentzian as (1 + 1 / eta) where eta is the gravitational refractive index, infinity is this case. If we say that the lorentzian takes an exponential form, then we can sum the binomial to the value e.
A bec particle, as wave packet, can have an infinity of waves but its actually just a frequency modulated object.
Theres absolutely nothing wrong with Newtons concept of an infinite speed of gravity but I think we have a natural log curve that has little quantum steps in it. Why? At the moment Im thinking along the lines of Maxwells daemon gatekeeper. The gate being at the Shwarztchild radius.
Forget for the moment that h has any significance, its just a number. So, lets peg the speed of gravity, nothing can travel faster than it. Things can travel faster or slower that the speed of light. All we have is a ratio and infinities lurk in all ratios.
When v^2 equals zero we have minus infinity. Thats just saying that where McCrimon sits is head of the table. An ultimate preferred frame of reference. This means that we can write the lorentzian as (1 + 1 / eta) where eta is the gravitational refractive index, infinity is this case. If we say that the lorentzian takes an exponential form, then we can sum the binomial to the value e.
A bec particle, as wave packet, can have an infinity of waves but its actually just a frequency modulated object.
Theres absolutely nothing wrong with Newtons concept of an infinite speed of gravity but I think we have a natural log curve that has little quantum steps in it. Why? At the moment Im thinking along the lines of Maxwells daemon gatekeeper. The gate being at the Shwarztchild radius.
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16 years 1 month ago #20157
by Jim
Replied by Jim on topic Reply from
Sloat, Ok taking h out makes sense to me but h is not just a number- its a constant that has been misused. As for gravity, I don't understand why it has to have a speed so why you are looking for it mystifies me. Why not just think of gravity as a force that never moves at all. Force is a mystery that could be explored in its own right having nothing to do with any of particles or effects.
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