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Quantoken
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We get:
T = 2.7243180 K
The accepted observational value for T is 2.725+-0.005K.
My discrepancy is only 0.00068K, far smaller than the observational uncertainty 0.005K.
See I derived the CMB temperature based on nothing but the fine structure constant. Isn't there new physics here.
But there is more. A more precise result can be derived. Namely the solar constant.
The Big Bang is wrong. The CMB is NOT the remains of the BB, but rather, remains of radiations from stars like our Sun. We know the Hubbard Redshift. What it implies is for whatever reason, the energy radiated from stars do not persist longer than the observational age of the universe.
So we can presume that if we allow stars like the Sun radiate for a total time equals to how long the radiation energy can be preserved before they turn into something else, namely, the age of the universe. Then we average out that energy over the volume of the universe, we should get the CMB energy density or the CMB temperature.
Correspondingly, if we assume the Sun is a rather typical star, which it is, from the known CMB energy density, we can also derive the radiation strength of the Sun, and correspondingly, obtain the solar constant of the Sun, which is defined as radiation power in watts per square meter at the Earth distance.
I will skip the derivation here. Any one can do it based on the above assumption. The final formula, in SI unit, is:
W = (4*alpha/PI^3) * (R*C^2/T^2) * (1/(G*N)) * 1/t0
Here R is the earth-sun distance, 1.496x10^11 m.
T is the Earth orbital period, one year, or 31557825 seconds. G is the Newton gravity constant. N is the big N we talked about. t0 is the natural time unit. And, certainly, alpha is the fine structure constant.
Any one can verify the calculation. The result I get is:
W = 1359.83 W/m^2
The accepted solar constant is 1360W/m^2. My discrepancy is less than 0.01% and far better than the observation error of the solar constant!
My result clearly and indisputably proves facts that are so obvious that simply stating them would seem unnecessary.
Those facts are:
1.The Sun radiates energy, and
2.Other stars radiate energy just as well.
3.The energy radiated by the stars do NOT disappear, at least within 14 billion years. And
4.Ultimately, the ultimate truth that the CMB are star radiations, just like Sr. Eddington said a long time ago.
CMB is NOT the remains of BB. BB theory has never been able to predict the CMB temperature even to a correct order of magnitude. I reached a 0.025% accuracy, after involving huge numbers up to 10^122!
There are new physics here. I have a whole brand new physics theory and I am just beginning to reveal a small portion of it. But before anyone misunderstands: no my universe does not expand and the "age" of the universe is a constant. It's an observation age, not a real age.
QUANTOKEN
T = 2.7243180 K
The accepted observational value for T is 2.725+-0.005K.
My discrepancy is only 0.00068K, far smaller than the observational uncertainty 0.005K.
See I derived the CMB temperature based on nothing but the fine structure constant. Isn't there new physics here.
But there is more. A more precise result can be derived. Namely the solar constant.
The Big Bang is wrong. The CMB is NOT the remains of the BB, but rather, remains of radiations from stars like our Sun. We know the Hubbard Redshift. What it implies is for whatever reason, the energy radiated from stars do not persist longer than the observational age of the universe.
So we can presume that if we allow stars like the Sun radiate for a total time equals to how long the radiation energy can be preserved before they turn into something else, namely, the age of the universe. Then we average out that energy over the volume of the universe, we should get the CMB energy density or the CMB temperature.
Correspondingly, if we assume the Sun is a rather typical star, which it is, from the known CMB energy density, we can also derive the radiation strength of the Sun, and correspondingly, obtain the solar constant of the Sun, which is defined as radiation power in watts per square meter at the Earth distance.
I will skip the derivation here. Any one can do it based on the above assumption. The final formula, in SI unit, is:
W = (4*alpha/PI^3) * (R*C^2/T^2) * (1/(G*N)) * 1/t0
Here R is the earth-sun distance, 1.496x10^11 m.
T is the Earth orbital period, one year, or 31557825 seconds. G is the Newton gravity constant. N is the big N we talked about. t0 is the natural time unit. And, certainly, alpha is the fine structure constant.
Any one can verify the calculation. The result I get is:
W = 1359.83 W/m^2
The accepted solar constant is 1360W/m^2. My discrepancy is less than 0.01% and far better than the observation error of the solar constant!
My result clearly and indisputably proves facts that are so obvious that simply stating them would seem unnecessary.
Those facts are:
1.The Sun radiates energy, and
2.Other stars radiate energy just as well.
3.The energy radiated by the stars do NOT disappear, at least within 14 billion years. And
4.Ultimately, the ultimate truth that the CMB are star radiations, just like Sr. Eddington said a long time ago.
CMB is NOT the remains of BB. BB theory has never been able to predict the CMB temperature even to a correct order of magnitude. I reached a 0.025% accuracy, after involving huge numbers up to 10^122!
There are new physics here. I have a whole brand new physics theory and I am just beginning to reveal a small portion of it. But before anyone misunderstands: no my universe does not expand and the "age" of the universe is a constant. It's an observation age, not a real age.
QUANTOKEN