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Why I disagree with static eternal universe
15 years 1 month ago #23715
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Lyndon, Well let's suppose for the moment that the big bang happened but with the proviso that the "speed" of gravity is much greater than the speed of light. Then at one gravitational second the universe would have a radius measured in millions of light years. With my estimate of the "speed" of gravity, one gravitational second will be about a billion light years.
Light then, being much slower, the radius of the electromagnetic universe will be tiny. So let's say that the gravitational mass of this sphere within a sphere is half and half, then the energy density of gravitational space will be close to zero. This also means that the mass is doubled, becuase this gravitational mass cannot be "seen."
So, suppose I wanted to take a sphere, one gravitational second in radius, and move it at the speed of light to someplace else. The gravitational space will Lorentzian contract by a miniscule amount. The electromagnetic space should contract to zero.
Right, so let's now look at the Lorentzian and write it down with some speed of gravity much greater than c. So, sqrt(1 - c^2 / b^2)where b equals the speed of gravity. Tom has the speed of gravity at a minimum value of twenty billion c. So that ratio is about 2.5E-21 We can write that as a refractive index and pop it into the Lorentzian, sqrt(1 - 1 / 4E 20) Newton on the other hand has this as sqrt(1 - 1 / infinity) Newton never saw the Lorentzian but if he had, he would have immediately said that we were looking at an exponential curve.
Now let's usppose that we can have a phase change at the speed of light, all we are doing is allowing for the possibility of negative refractive index. We can then look at teh graph of the cosine, or sine of the natural log of the Lorentzian. We end up with a particle as wave that is a frequency modulated wave. On the x axis, from the value one, the wavelength gets longer and longer along x. From one going leftwards the wavelength gets shorter and shorter. animate it and we have an antiparallel movement of the wavefronts, there's always a peak at the value x equals one. There's also two speeds to the wavefronts.
I take this to mean that such a particle emits a photon when it takes on a graviton, and vis versa.
Back to the Lorentzian again, i think the best bet for the speed of gravity is h = c^2 /b^2 where once again b is the speed of gravity. Note that this a pure dimensionless number, and that it's arrived at from looking at the gravitational/electromagnetic couple.
Well, there has been some disagreement about this on the board, yet I simply cannot see what all the fuss is about. Dirac, Hoyle and Eddington wanted a cosmology that only dealt in dimensionless numbers, so that they could make sense of all the arbitrary constants. We are missing some constant and the best bet, as i see it, is the ratio of the speed of light to the speed of gravity.
Right so let's expand the Lorentzian via the binomial theorem, another of Newton's. We'd get
E = mb^2 where b is the speed of gravity. We get the gravitational energy of any particle, as being numerically equal to the electromagnetic frequency of the particle. But we have E = hf. So what can that mean? An electron is spinning, and its angular momentum is going to be its mass times its angular velocity, c times its radius, the Compton radius. Which gives us the value h. This is its electromagnetic mass, and of course its angular velocity can not exceed light speed. once again let's double the mass of whatever, an electron let's say, has a hidden particle of equal mass ut it cannot be "seen." so we can look at its angular momentum as being h. Then 9.1E-31*b*r = h. Take the angular velocity here to be the speed of gravity, about
1.16E 25 That will give us a radius of 6.25E-29 metres. But if the frequency of a particle were to remain constant then the value of h will approach the value one. Then the radius would be about 94E 3 metres.
Well, that sounds pretty strange but we pays our money and takes our chances here. We can look at gravitational space as being "bigger" or "smaller" than electromagnetic space. Gravitational space is going to be informationally much smaller than e.m. space. We can think of the gravitational mass of an electron as being inside a tiny radius, or we can say that it's anywhere/anywhen within a volume of 3.4E 15 cubic metres. It's gravitational mass due to its ftl spin is pretty huge, as an electromagnetic equivalent, about 1.3 tonnes but it's simply an electron mass in its own space and its spread out very thinly through a huge volume in e.m space.
Light then, being much slower, the radius of the electromagnetic universe will be tiny. So let's say that the gravitational mass of this sphere within a sphere is half and half, then the energy density of gravitational space will be close to zero. This also means that the mass is doubled, becuase this gravitational mass cannot be "seen."
So, suppose I wanted to take a sphere, one gravitational second in radius, and move it at the speed of light to someplace else. The gravitational space will Lorentzian contract by a miniscule amount. The electromagnetic space should contract to zero.
Right, so let's now look at the Lorentzian and write it down with some speed of gravity much greater than c. So, sqrt(1 - c^2 / b^2)where b equals the speed of gravity. Tom has the speed of gravity at a minimum value of twenty billion c. So that ratio is about 2.5E-21 We can write that as a refractive index and pop it into the Lorentzian, sqrt(1 - 1 / 4E 20) Newton on the other hand has this as sqrt(1 - 1 / infinity) Newton never saw the Lorentzian but if he had, he would have immediately said that we were looking at an exponential curve.
Now let's usppose that we can have a phase change at the speed of light, all we are doing is allowing for the possibility of negative refractive index. We can then look at teh graph of the cosine, or sine of the natural log of the Lorentzian. We end up with a particle as wave that is a frequency modulated wave. On the x axis, from the value one, the wavelength gets longer and longer along x. From one going leftwards the wavelength gets shorter and shorter. animate it and we have an antiparallel movement of the wavefronts, there's always a peak at the value x equals one. There's also two speeds to the wavefronts.
I take this to mean that such a particle emits a photon when it takes on a graviton, and vis versa.
Back to the Lorentzian again, i think the best bet for the speed of gravity is h = c^2 /b^2 where once again b is the speed of gravity. Note that this a pure dimensionless number, and that it's arrived at from looking at the gravitational/electromagnetic couple.
Well, there has been some disagreement about this on the board, yet I simply cannot see what all the fuss is about. Dirac, Hoyle and Eddington wanted a cosmology that only dealt in dimensionless numbers, so that they could make sense of all the arbitrary constants. We are missing some constant and the best bet, as i see it, is the ratio of the speed of light to the speed of gravity.
Right so let's expand the Lorentzian via the binomial theorem, another of Newton's. We'd get
E = mb^2 where b is the speed of gravity. We get the gravitational energy of any particle, as being numerically equal to the electromagnetic frequency of the particle. But we have E = hf. So what can that mean? An electron is spinning, and its angular momentum is going to be its mass times its angular velocity, c times its radius, the Compton radius. Which gives us the value h. This is its electromagnetic mass, and of course its angular velocity can not exceed light speed. once again let's double the mass of whatever, an electron let's say, has a hidden particle of equal mass ut it cannot be "seen." so we can look at its angular momentum as being h. Then 9.1E-31*b*r = h. Take the angular velocity here to be the speed of gravity, about
1.16E 25 That will give us a radius of 6.25E-29 metres. But if the frequency of a particle were to remain constant then the value of h will approach the value one. Then the radius would be about 94E 3 metres.
Well, that sounds pretty strange but we pays our money and takes our chances here. We can look at gravitational space as being "bigger" or "smaller" than electromagnetic space. Gravitational space is going to be informationally much smaller than e.m. space. We can think of the gravitational mass of an electron as being inside a tiny radius, or we can say that it's anywhere/anywhen within a volume of 3.4E 15 cubic metres. It's gravitational mass due to its ftl spin is pretty huge, as an electromagnetic equivalent, about 1.3 tonnes but it's simply an electron mass in its own space and its spread out very thinly through a huge volume in e.m space.
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15 years 1 month ago #23020
by lyndonashmore
Replied by lyndonashmore on topic Reply from lyndon ashmore
Hi Stoat,
I will need to think about this. One thing I need claryfying;
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Stoat</i>
Back to the Lorentzian again, i think the best bet for the speed of gravity is h = c^2 /b^2 where once again b is the speed of gravity. Note that this a pure dimensionless number, and that it's arrived at from looking at the gravitational/electromagnetic couple.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
In your equation what is 'h'? is it the Planck constant - since you use it later in E = hf; and I am assuming that c = speed of light in a vacuum).
If so there is a problem.
Cheers,
lyndon
lyndon ashmore - bringing cosmology back down to earth.
I will need to think about this. One thing I need claryfying;
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Stoat</i>
Back to the Lorentzian again, i think the best bet for the speed of gravity is h = c^2 /b^2 where once again b is the speed of gravity. Note that this a pure dimensionless number, and that it's arrived at from looking at the gravitational/electromagnetic couple.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
In your equation what is 'h'? is it the Planck constant - since you use it later in E = hf; and I am assuming that c = speed of light in a vacuum).
If so there is a problem.
Cheers,
lyndon
lyndon ashmore - bringing cosmology back down to earth.
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15 years 1 month ago #23021
by Stoat
Replied by Stoat on topic Reply from Robert Turner
It's the numerical part of h, without any dimensions. As you say c is the speed of light.
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15 years 1 month ago #23023
by lyndonashmore
Replied by lyndonashmore on topic Reply from lyndon ashmore
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Stoat</i>
<br />It's the numerical part of h, without any dimensions. As you say c is the speed of light.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
In that case we have a problem!
Quantities have both a numerical part and a unit and these cannot be separated. I am a mainstream physicist (its only the bit about the universe expanding that I disagree with!)
You have:
h = c^2 /b^2
as the ratio of the squares of the speeds.
well values of 'h' vary depending upon which system of units one uses.
In SI it is:
6.62x10^-34 Js
In electron Volt (eV) it is
4.13x10^-15 eVs
Whilst in CGS it is:
6.62x10^-27 ergs
Which numerical part are you using as the ratio of the speeds.
Surely there can only be one ratio?
Any equation must be homogeneous - that is have the same units both sides.
Cheers,
lyndon
lyndon ashmore - bringing cosmology back down to earth.
<br />It's the numerical part of h, without any dimensions. As you say c is the speed of light.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
In that case we have a problem!
Quantities have both a numerical part and a unit and these cannot be separated. I am a mainstream physicist (its only the bit about the universe expanding that I disagree with!)
You have:
h = c^2 /b^2
as the ratio of the squares of the speeds.
well values of 'h' vary depending upon which system of units one uses.
In SI it is:
6.62x10^-34 Js
In electron Volt (eV) it is
4.13x10^-15 eVs
Whilst in CGS it is:
6.62x10^-27 ergs
Which numerical part are you using as the ratio of the speeds.
Surely there can only be one ratio?
Any equation must be homogeneous - that is have the same units both sides.
Cheers,
lyndon
lyndon ashmore - bringing cosmology back down to earth.
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15 years 1 month ago #23651
by Stoat
Replied by Stoat on topic Reply from Robert Turner
6.626E-34 Now this is dimensionless because the speed of light and the speed of gravity are in the same units.
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15 years 1 month ago #23024
by lyndonashmore
Replied by lyndonashmore on topic Reply from lyndon ashmore
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Stoat</i>
<br />6.626E-34 Now this is dimensionless because the speed of light and the speed of gravity are in the same units.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Its true that your ratio of speeds is dimensionless and has one value only.
But I have a friend who is old and uses eVs for h. He wants to know why his value for the speed of gravity compared to 'c' is only one tenth of yours? Is 'b' variable?
To put it another way, from all the possible numerical values for h out of all the values for 'h' in the world(depending upon the units) why do you choose 6.626E-34?
Cheers,
lyndon
lyndon ashmore - bringing cosmology back down to earth.
<br />6.626E-34 Now this is dimensionless because the speed of light and the speed of gravity are in the same units.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Its true that your ratio of speeds is dimensionless and has one value only.
But I have a friend who is old and uses eVs for h. He wants to know why his value for the speed of gravity compared to 'c' is only one tenth of yours? Is 'b' variable?
To put it another way, from all the possible numerical values for h out of all the values for 'h' in the world(depending upon the units) why do you choose 6.626E-34?
Cheers,
lyndon
lyndon ashmore - bringing cosmology back down to earth.
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