David Levy

- How could it be that we see galaxies at a range of 46 Billion light years away while their light has been traveling for only 13.8 Billion years? - Is it 46 billion years for any direction from us? so can we assume that the total diameter of the visible universe should be at the range of 92 Billion years?

The science has an explanation for those galaxies which are located at almost 14 billion light years away from us. However, if one day we will discover a galaxy at 50 (or 500) Billion light years away (or at infinite distance) – is it going to set any change in our understanding about the age of the universe?

The atmosphere set the green house effect and keeps the heat of the Earth. Why the Earth is considered as Habitable Planet for more than 3.5 By? Don't you think that it is due to the atmosphere? Hence, the emissivity of the atmosphere is the most relevant issue. Without the atmosphere - the Earth will be considered as a black body. Without the atmosphere – there is no possibility to set any form of life on Earth. (So, instead of black body it is a dead body). Stefan Boltzmann law uses specifically the ambient temperature. This temperature is outside the atmosphere. So, how can we ignore the atmosphere? Therefore, if the atmosphere "has a negligible effect" let's verify what might be the impact without it. We don't have to go far away. Just look at the Moon. Day temperature - about 250K, Night temperature – less than 100K. The moon represents a black body. The earth should be considered as gray body. It also seems that the Earth is considered as black body or gray body just in order meet the current theories. As we all know, the science assume that in the Earth had been cool down to 300K in just 30My. Please see again that calculation: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime2.html#c1 The assumption is that the emissivity of the Earth is one, therefore, it is a black body. During this period of time, the science didn't calculate the effect of the Sun radiation. Why? It just used as to proof that in a very short time the Earth had lost significant amount of temperature (from over 6000K to about 300K). However, in our days: -Suddenly there is an impact of the sun radiation -Suddenly "The Stefan Boltzmann law works well in the correct applications." -Suddenly we need to take in account the "Surface area, convection rates as the crust develops greater depth, this influences emissivity rates" -Suddenly "your also dealing with different materials with different conductivity values" Where were all of those impacts and factors during the first 30 My? Therefore, Stefan Boltzmann law was perfectly O.K. for the first 30My, but unfortunately, it is not applicable for our days Hence, when ever is convenient to the science – The emissivity of the atmosphere is irrelevant. Its value could be 1 or 0.61. The Earth can be considered as black body or gray body with or without atmospher .

Sorry It seems that you have no real information about the emissivity of the Earth/atmosphere. Do you really believe that the emissivity of the Atmosphere should be the same as silver, iron or silicon? How can you use those elements on the surface of the Earth as a represented objects for the emissivity of the Earth itself? There is something between the surface of the Earth and the space. This something is called atmosphere. I have got similar results to that article by using different approach of calculation. They have calculated the effect of the atmosphere and the influence of carbon dioxide, while I extract it directly from Stefan Boltzmann law. You didn't provide any article which try to set the calculation of the emissivity of the atmosphere. It's easy to claim that you don't agree with that article. However, it is expected that you should offer other alternative for measuring the real effect of the atmosphere. Please, if you reject that article, you should offer other real alternative for the emissivity of the atmosphere.

It's not me. I just expect to get some basic appreciation.

Is it? Please see the following: http://jennifermarohasy.com/2011/03/total-emissivity-of-the-earth-and-atmospheric-carbon-dioxide/ "Introducing 7700 meters as the average altitude of the troposphere and the real partial pressure of the atmospheric carbon dioxide (0.00038 atm-m), the resulting total emissivity of the carbon dioxide is 0.0017 (0.002, rounding up the number). Evidently, the carbon dioxide is not a blackbody, but a very inefficient emitter (a gray-body). In this note I have calculated the real total emissivity of the atmospheric carbon dioxide at its current partial pressure and instantaneous temperature to be 0.002." It is quite close to my calculation.

It seems that you prefer to avoid a direct answer to my question. O.K. - no problem. With regards to emissivity I would expect the emissivity of the Earth should be quite different from any object or matter on the planet. What is your advice?

Thanks Mordred Somehow, I'm not sure that I get a consistent feedback with regards to Stefan Boltzmann law. It one had you claim that Stefan Boltzmann law works well in the correct applications: Also, in the article which you have pointed it is stated: " The cooling time from an initial temperatureTi to a final temperature Tf is given approximately by t = (1/T 3− 1/T 3 )mc/3σA where Ti is the temperature at time t , Tf the ambient temperature (both expressed in kelvins), m is the mass, c the specific heat capacity and A the area of the object (see Block A2). Applying this formula to the small iron sphere, we find that the cooling time is about 47 minutes. For the time of cooling of a globe of the size of the Earth that is made entirely of iron, we obtain about 45 000 years (see Block A2)." Hence, the science is using Stefan Boltzmann law to calculate the cooling time of the Earth. However, in the other hand you claim: The only difference between my calculation and the above one is the emissivity. They assume that the emissivity is 1 and I tried to find the real value. As you can see they do not care about - "continous heat source such as radiation, convection and conduction" They do not care about - "Surface area, convection rates as the crust develops greater depth, this influences emmisivity rates, your also dealing with different materials with different conductivity values" If they can use it, then I can also use it. If you wish to discuss about the emissivity - you are more than welcome. So please - try to take a decision. Can we use the above formola for cooling time? Yes or no (without but..) With regards to emissivity Based on my calculation : ε = 0.0013 You claim that it is 0.7. Can you please explain how did you get this value?

It's very interesting article. However, it is focusing on the emissivity of volcano/lava activity on the surface of the Earth. However, I have asked specifically about the emissivity of the Earth itself. Yes, I assume that the Earth could be considered as a good reflector. Therefore its emissivity should be low. Hence, what is value of the emissivity which we can apply to Stefan Boltzmann law in order to calculate the real cooling time of the Earth?

Thanks You claim that Stefan Boltzmann law covers radiation only. So how can we add the impact of the solar heat to this formula? What is the real meaning of the emissivity in Stefan Boltzmann law? What kind of factors it represent? What is the estimated emissivity value of the Earth?

Thanks Now, we all agree that Stefan Boltzmann law is valid and we can use it. However, we need to verify that we are using it correctly. So, let's look again on Stefan Boltzmann law formula: P = dE/dt = ε·σ·A·(T(hot)4 –T(ambient)4) In this formula there is no requirement for: Surface area, convection rates as the crust develops greater depth or even solar radiation There is only one small sign which covers all of those factors. This is the emissivity - ε As you have stated: "Surface area, convection rates as the crust develops greater depth, this influences emissivity rates" Hence, if we can find the correct values of the emissivity, there is no need to calculate the green house effect, the radiation, the curst of Iron/silicon conductivity or solar radiation. All included in one small sign - emissivity – ε. However, based on this formula, the cooling time is as follow: t(cooling) = Nk / (2·ε·σ·A) · (1/T(final) 3 –1/T(hot)3) Please advice if you agree with that?

Is it real? Stefan Boltzmann is not just one more formula. It isn't a hypothetical idea or theory It's a Law. http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law "The Stefan–Boltzmann law, also known as Stefan's law, describes the power radiated from a black body in terms of its temperature. Specifically, the Stefan–Boltzmann law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black-body radiant exitance or emissive power)," You are the master. How can we claim that Stefan–Boltzmann law is approximation? Shall we also claim that Newton law is approximation? Please – let's set a constructive discussion. Can we use this law? Yes or no

Let me explain you clearly. There were two clever scientists that developed a formula for heat radiation. This formula is called - Stefan-Boltzmann Law http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2 In this formula they have added the emissivity factor. If e = 1 then it is consider as ideal radiator or black body object. The science had fully accepted their formula. This formula had been used to develop the frame calculation for Radiative Cooling Time http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html#c1 The science used that frame to calculate the requested time for Cooling the Earth to 300K. http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime2.html#c1 Unfortunately, with or without attention, it was assumed that the Earth is an ideal radiator (or blackbody). The science ignores completely the yellow star in the sky, the greenhouse effect and the atmosphere. Based on that assumption – it was proved that the Earth can be cool down from any temp to 300K in just 30 My. So far the science was quite proud with this formula as it meets the main stream concept. Unfortunately, (for some scientists) it is clear now that the Earth shouldn't be considered as a black body. Based on a very simple calculation it is proved that the age of the Earth is much higher than the current expectation. In the past it was believed that the Earth is the center of the Universe. Later on it was discovered that this theory was incorrect. So what? Now we understand that the age of the Earth is longer than our current expectation. So what? With regards to Metallicity or any other unproved theory. The science should be proud with this discovery. It doesn't contradict with any real evidence or observation. In the contrary, it opens a new perspective on the Universe. Please be happy. We are facing a new dawn for the science.

There is no contradiction with any evidence. None of the above is considered as a proved theory. Therefore, this discovery could potentially contradict with some unproved cosmology/astronomy concepts.

Dear Mordred You are a best teacher. I consider you as my partner. You have significant part of this discovery. I have learned many aspects of modern science from you. Therefore, I have some sort of disappointment from your feedback. I would expect that you would verify the calculation, look at the formulas and ask me about blackbody or graybody. Instead, it seems that you are unhappy as this discovery contradicts some of the current unproved theories. So let's try to distinguish between theory, reality and none relevant issue: Theories It isn't preception.It isn't approximation. It is a real verification. Please read Stefan-Boltzmann Law. I have used the formula exactly as it should be! http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2 Let's assume that you have some zircon and we estimate that the age is 4.5 My. What does it mean? It just give us an indication for the min age. We have no guarantee that the minerals that we have are the oldest ever made on earth or in the universe. I have High appriciation for all the scientists. But in this case, it is expected to verify the formula and the calculation. .

Our current assumption about the Nebula couldn't consider as an evidence! It is a theory. Please read again my following reply: The assumption that the Earth could be considered as a blackbody is a severe mistake. However, the willing of the science to fit each evidence to the current concept is a fatal error. Therefore, we all must verify every observation and get a conclusion of what we see. Just after getting all the real evidences/observations/calculations we can set down and figure out the source of the Universe enigma. I have a simple questions: Do you agree with my calculation based on Stefan-Boltzmann law? Do you agree that the Earth is a graybody, and it should never ever consider as blackbody?

Meteorites are not part of the Earth itself. There is no need to explain the current dating of those Meteorites. In the future, we might find older dating meteorites. So what? The assumption that the Earth could be considered as a blackbody is a severe mistake. However, the willing of the science to fit each evidence to the current concept is a fatal error. Therefore, we all must verify every observation and get a conclusion of what we see. Just after getting all the real evidences/observations/calculations we can set down and figure out the source of the Universe enigma. We have to ask ourselves the following questions – Do we agree that Stefan-Boltzmann law is correct formula? Do we agree that the Earth is a graybody, and it should never ever consider as blackbody? If the answer for both questions is yes, then there is no need for further evidences.

There is no contradiction between this discovery and the geological evidences. Those evidences give us the min age of the Earth. There is good chance that in the future we will discover new evidence which will confirms that the age of the earth is significantly higher than our current expectation. Don't forget that in the past people believed that the age of the Earth was only 5,000 years

My calculation is 100% correct. I have already explained that it was a severe mistake to assume that the Earth is a black body. In all the articles it is stated that the Earth is a gray body due to the solar radiation and greenhouse effect. Therefore, it was great mistake to assume that in the first 30 My the Earth will be considered as black body while in the rest of 4.4 By it will become a gray body. So, there is no error in Stefan-boltzmann law. There is also no error in the formulas which the science is using. All of them are 100% correct. The error was in the assumption that the thermal emissivity for the earth could be same as for a theoretical blackbody ϵ = 1.0. The Earth was never ever a blackbody! I have just used the Stefan-Boltzmann formula correctly. The science must verify this critical issue. It is expected to forward this breakthrough discovery to all the relevant scientists. I'm quite sure that they will confirm it. We all must be proud of this discovery and in this forum. Well, the title of this tread is: "Earth - What is the real age?". I have proved that by using correctly the formulas we can get the real answer, without digging in solar radiation and find its impact on the Earth at any specific temperature. However, if you wish, I will try to help with your request, although - it might take me some time to set it.

Thanks for the remarks about the T and t. I have fixed it. I'm using Stefan-Boltzmann law correctly. There is no error in my calculation. Stefan-Boltzmann can be proud of using their formula as it should be. If there is any error in this calculation, please do not hesitate to inform. Stefan-Boltzmann law specifically covers a gray body star. Unfortunately, the science had decided to ignore completely that yellow star in the sky. Therefore, the emissivity had been set to one. That is the biggest mistake of the modern science. As I have already stated, a black body planet means – that the planet had been set outside the solar system, in full darkness, without any green house effect or atmosphere protection. This is a great mistake. There is no possibility to assume that in the first 30 My of the Earth life, it had been set at this unrealistic conditions – black body. But just after cooling down to the current temperature, the Earth had been set back to the solar system and starts to work as gray body. Please look again at this calculation. You also should be proud of it. This discovery is a breakthrough. The science should consider a Novel price. Not for me, but for this forum and all the members that participated in this thread.

O.K. Let's start by extracting the emissivity during the habitable planet phase of time. This value should cover the effects of the solar radiation and atmosphere factors. During this phase of time the planet had been cooled down to the current temperature – 300K. It is known that life on Earth had started about 3.5 By ago. Unfortunately, there is no clear evidence about the temperature on Earth at that time. Therefore, let's assume that in the worst case scenario the temperature at that time was 318K (which is 45 c). Hence: t(cooling) = 3.5 By, T(final) = 300K, T(hot) = 318 Based on that info, we will use Stefan-Boltzmann law to extract the Emissivity of the Earth (for a gray body). The t(cooling) time formula is as follow: t(cooling) = Nk · (1/T(final)3 – 1/T(hot)3)/(2·ε·σ·A) Hence: t(cooling) · ε = Nk·(1/T(final)3 – 1/T(hot)3)/(2·σ·A) We also know that for black body (ε=1) at T(final) of 300K the t(cooling) is 30 My. Therefore, by the following extrapolation we can find the emissivity for the habitable planet phase. So, 3.5 By · ε = 30 My · 3003 · (1/3003 – 1/3183) 3.5 · 109 · ε = 30 · 106 · 27 · 106 · 5.94· 10-9 3.5 · 109 · ε = 4,811 · 103 = 4.811 · 106 ε = 4.811 · 106 · / (3.5 · 109) = 1.375 · 10-3 Let's assume that this emissivity is applicable also for the wet planet phase. In this phase of time the planet had been cooled down from 373K to 318K. t(cooling) · ε = 30 My · 3003 · (1/3183 – 1/3733) t(cooling) = 30 My · 27 · 106 · 1.1827 · 10-8 / ε t(cooling) = 30 My · 27 · 106 · 1.1827 · 10-8 / (1.375 · 10-3) t(cooling) = 6.967 By Hence, base on our worst case scenario the minimum requested cooling time for the wet planet phase time is 6.967 By. Therefore, the total cooling time from 373K (100 c) to current time is: t(cooling from 100 c till today) = t(cooling from 100 c till 45 c) + t(cooling from 45 c till today) t(cooling from 100 c till today) = 3.5 + 6.967 = 10.467 By. Please be aware that if 3.5 By ago, the temperature was lower than 45 c then it should decrease the value of the emissivity and therefore it could increase the requested time for the wet planet phase. Let's set a brief calculation for 308K, which is 35 c 3.5 By · ε = 30 My · 3003 · (1/3003 – 1/3083) 3.5 · 109 · ε = 30 · 106 · 27 · 106 · 2.812 · 10-9 3.5 · 109 · ε = 2.812 · 106 ε = 2.812 · 10-6 · 10-9 / 3.5 = 0.803 · 10-3 t(cooling) · ε = 30 My · 3003 · (1/3083 – 1/3733) t(cooling) = 30 My · 27 · 106 · 1.495 · 10-8 / ε t(cooling) = 30 My · 27 · 106 · 1.1827 · 10-8 / (0.803 · 10-3) t(cooling) = 11.930 Gy Therefore, in this scenario, the total cooling time from 373K (100 c) to current time is: 3.5 + 11.93 By = 15.43 By. Now we need to add the requested cooling time for phase one and two.

Well it's not straight forward. Hence, let me explain how we should solve this issue: -Earth phases: Verify the four main phases of the planet. -Emissivity: Verify emissivity at each phase. -Solar Radiation: Verify the sun radiation at each phase -Formula: Use the correct formulas for each phase as needed. -Cooling time: Extract the requested cooling time for each phase of the Earth. Earth phases Phase one – Full melting Planet, from its first day till setting the crust all over the surface. During this phase, the Expected temperature on the surface should decrease from about 6,000K (or above) to about 1,000K (lava curst temp) Phase two – Hot Planet, from getting the crust till sustaining the first drop of water. During this phase, the Expected temperature on the surface should decrease from about 1,000K to about 373K (100 c) Phase three – Wet Planet, from getting the first drop of water till having the first sign of life. During this phase, the expected temperature should decrease from 373K to 318K (45 c) Phase four – Habitable Planet, from having the first sign of life till today. Emissivity The thermal emissivity range of an object is (0 ≤ ϵ ≤ 1 ). A theoretical blackbody has ϵ = 1.0. However, due to the greenhouse effect the Earth is considered as a gray body. We need to verify the emissivity of Earth for each phase. Solar Radiation http://nptel.ac.in/courses/112108149/pdf/M9/Student_Slides_M9.pdf "The magnitude of the energy leaving the Sun varies with time and is closely associated with such factors as solar flares and sunspots. Nevertheless, we often choose to work with an average value. The energy leaving the sun is emitted outward in all directions so that at any particular distance from the Sun we may imagine the energy being dispersed over an imaginary spherical area. Because this area increases with the distance squared, the solar flux also decreases with the distance squared. At the average distance between Earth and Sun this heat flux is 1353 W/m2." However, we need to verify the emissivity effect "A body that does not absorb all incident radiation (sometimes known as a grey body) emits less total energy than a black body and is characterized by an emissivity ϵ <1". Therefore, it is needed to verify the real impact of this sun radiation on Earth. We also need to take in account that the sun is drifting outwards from the Earth (by about 1.5 cm per year). Hence, it could potentially increase the impact on Earth as we go back on time. More to come.

I think that I start to understand the solution for this enigma. So far we only discuss about "The rate of radiative energy emission from a hot surface is given by the Stefan-Boltzmann law ." Hence, if we take the Earth outside the solar system in a dark place without any protection (Atmosphere as an example), then yes – after 30 My the earth should decrease its temperature from any starting temp to 300K. This is fully correct. Never the less, this isn't applicable for real Earth. We must take in account the effect of the Sun radiation, Atmoshere and several other factors. So, let's try to understand the impact of a real surface and the effect of sun radiation by the following article: RADIATION HEAT TRANSFER http://nptel.ac.in/courses/112108149/pdf/M9/Student_Slides_M9.pdf "Real Surfaces Thus far we have spoken of ideal surfaces, i.e. those that emit energy according to the Stefan-Boltzman law: Eb = σ·Tabs4 Real surfaces have emissive powers, E, which are somewhat less than that obtained theoretically by Boltzman. To account for this reduction, we introduce the emissivity, ε. so that the emissive power from any real surface is given by: E = ε·σ·Tabs4 However: Radiosity We have developed the concept of intensity, I, which let to the concept of the view factor. We have discussed various methods of finding view factors. There remains one additional concept to introduce before we can consider the solution of radiation problems. Radiosity, J, is defined as the total energy leaving a surface per unit area and per unit time. This may initially sound much like the definition of emissive power, but the sketch below will help to clarify the concept. J ≡ ε·Eb + ρ·G Receiving Properties: Targets receive radiation in one of three ways; they absorption, reflection or transmission. To account for these characteristics, we introduce three additional properties: Absorptivity, α, the fraction of incident radiation absorbed. Reflectivity, ρ, the fraction of incident radiation reflected. Transmissivity, τ, the fraction of incident radiation transmitted. We see, from Conservation of Energy, that: α + ρ + τ = 1 So, let's go back to Radiative Cooling Time: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html#c1 P = dE/dt = ε·σ·A·(T(hot)4 –T(ambient)4) Here P is the power emitted from the area, and E is the energy contained by the object. However, we must add to this formula the contribution of the Sun radiation as well as the effect of real surface. Therefore: E(updated total energy) = E(energy contained by the object) + E(energy from the sun) Hence: P(updated) = dE(updated total energy)/dt =dE(energy contained by the object)/dt + dE(energy from the sun)/dt If we do it correctly, we will be able to extract the real cooling time!

There is one more explanation for this paradox: Let's assume that we don't know the temperature on Earth. So, we already know that for T(final) = 300K the T(cooling) time is 30 My. Let's try to verify the requested time for T(final) = 200K As : (300/200)3 = 3.375 Then, the T(cooling) time for T(final) at 200K is: 30 x 3.375 = 101.25 My. However, we know that after at least 4.4 By the temperature on Earth is still 300K. Therefore, we have to look again on Stefan-Boltzmann law . http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html#c1 There is no error in this formula! We just need to understand it and use it correctly. The answer is already integrated in this formula. It is called - Emissivity. The biggest error of the modern science is that they have used an ideal case for the Emissivity and set it to 1. This is a severe error! The value shouldn't be one ever and never. This sign represents the effects of the Atmosphere and Crust heat conductance and other factors which I have started to highlight at thread no. 54 For 4.4 By, this emissivity is in full balance with the temperature of the Earth. However, it is expected that the emissivity of the earth is changing over time. Therefore, the science must understand what might be the emissivity of the Earth from its first day. By finding the correct value of the emissivity we can get more realistic cooling time verification.

Thanks Based on the Cooling time formula, there is virtually no meaning for the starting temperature. (As long as it's high enough with relative to the final temp). Hence, the requested cooling time for T(final) at 1000K is: 30 My x (300/1000)3 = 30 x 0.027 = 81,000 Year I still estimate that there is a paradox in this calculation. Therefore, I would like to verify what might be the outcome of the Integral without neglecting the T ambient (although its contribution is less than 1%).