The general belief on the conditions of the deep space, beyond the terrestrial exosphere, is about a completely empty place without temperature.
However, highly accurate measurements made by satellites, like the Wilkinson Microwave Anisotropy Probe (WMAP) , have corroborated that the deep space has a temperature and, additionally, that it is not an absolutely empty space.
WMAP has revealed a deep space temperature of 2.7251 K and a density of 1 particle/cm^3 (density based on protons in the outer space) .
The theoretical temperature was confirmed by WMAP measurements. The theoretical basis related to the temperature of the deep space is given by the correlation between the temperature and the kinetic energy of the particle. On this case, the root mean square (rms) speed vrms of protons in deep space is 260 m/s.
The purpose of this essay is to know the amount of energy emitted by the Earth towards the outer space and the concept of microstates.
The Earth in the Cold Space
The formula to calculate the temperature of deep space is as follows:
T = (m*v^2rms) / (3*k)
Where m is the mass of particles, vrms is the root mean square velocity of those particles in that medium –because protons speed is highly variable, and k is for Boltzmann constant.
Given that ionized Hydrogen is the main constituent in the outer space, we consider the mass and the root mean square velocity of a proton in deep space to make our calculations.
m = 1.67 x 10^-27 kg (mass of a proton)
vrms = 260 m/s (root mean square velocity of protons in the outer space).
k = Boltzmann’s constant = 1.38 x 10^-23 J/K
By introducing magnitudes into the formula T = (mv^2rms)/( 3*k), the theoretical temperature T of deep space, taking into account the kinetic energy of protons in deep space gives a result of 2.72686. The rms error is 0.00176 K, which is quite insignificant (0.06%), therefore, the theoretical value is in conformity with direct measurements.
The outer space is the environment of the Earth. The question is:
How much power the Earth radiates per unit area toward the deep space? To answer this question, let us resort to the Stephan-Boltzmann Equation:
P = e (A) (σ) (TEarth^4 – Tds^4)
Here, a problem arises with respect to the emissivity of the Earth. However, careful examinations and calculations of the Earth’s emissivity give a mean correlation factor of 0.82.  Introducing this correlation factor, the power emitted by the Earth, per square meter, during one second, is 329.51 W.
To correct this apparent incongruence with respect to the supposed amount of the incident solar IR radiation on Earth’s surface, some authors resort to iterate the quantity until the resulting power equals to the supposed incident solar IR radiation.
However, we only are allowed to take into account the sphericity of the Earth, so the value changes to esph = e / (4(π)) = 0.644.
The result after introducing the new correlation factor of 0.644 is 258.8 W.
Our last option to get the emissivity is to invent it by means of introducing a flawed value of the emissions from the Earth:
e = (249 W) / [A (σ) (TEarth^4 – Tds^4)] = 0.62
This way, we make the hypothesis matches with the Earth’s energy budget model.
However, this is not a valid procedure in science because the scientific methodology starts with observations and after it proceeds to produce hypotheses, which must be proven by means of experimentation, or more observations.
If we consider the correlation coefficient 0.82 as the total emissivity of the Earth, the absorbed energy by the Earth would be 601.92 J. The latter magnitude represents 44% of the solar constant (1368 W/m^2 x 0.44 = 601.92 W/m^2). NASA assigns a theoretical absorption of solar energy by the Earth of 48%.
Here, the power emitted by an ideal Earth should be 401 W. However, the measurements of the Earth’s emissivity reveal a correlation factor of 0.82.
Consequently, the observations of the real world reveal that the value of 0.62 assigned a priori to the emissivity of the Earth is not real and rise serious doubts about the total amount of solar power absorbed and the amount of power emitted by our planet.
Microstates and the Outer Space
To properly talk about microstates, we need that any amount of matter is present in a given medium. We cannot talk about microstates if we have not, at least, one Hydron (H+) in a given medium.
A microstate refers to any initial of final configuration of the energy in a given system.
The Second Law of Thermodynamics, although initially was derived from the observation of thermal processes, has been proven to be acting on every level of energy exchange between two or more systems.
Initially, the Second Law was described in terms of the directionality in the flow of the energy in transit (a process function), which depends on the states of the systems involved in the exchange of such energy in transit. The Second Law clearly specified that the work only can be done by a higher energy density system on a lower energy density system and not the opposite.
However, with the advent of Quantum Physics, the scientists wondered whether this Law was valid at the quantum level or not. The answer to this question was given heuristically through the calculations of Maxwell, Boltzmann and Gibbs. The heuristic character of the calculations vanished when those hypotheses were later confirmed by experimentation.
In consequence, the definition of the Second Law was amplified to include its influence on the quantum level and not only on those process functions where heat and work were implied.
This shift was important because it defined the real concept of entropy and detached it from contextual derivations. For example, now we know that the fundamental concept of entropy has nothing to do with disorder, movement, complexity or heat “content”, but with the configurations that the energy adopts in a given system and the directionality of the energy exchange.
Entropy is now defined as the natural trend of the energy to flow towards the system or systems with a higher number of available microstates.
Let us say that two systems permit six configurations of the energy. One of them, let us say the system A, has four “occupied” configurations and only two available configurations. The other system, or system B, has only one “occupied” configuration and five available configurations. According to the Second Law of Thermodynamics, the energy will flow spontaneously from the system A to the system B and never the opposite.
Perhaps, you are wondering if the energy could flow from B to A during the process. The answer is no because two systems implied in an energy exchange process cannot adopt the same configuration at once, although any system could adopt any configuration.
To calculate the number of microstates that a system can adopt, we resort to the following formula:
Nms = N! / (n1! * n2! * n3! …)
Where Nms is the number of available microstates (Maxwell-Boltzmann Number), N is the number of particles, and n is the number of particles in a determined occupied microstate. For example, we have a system A that have six particles from which four are in the microstate 0E, one is in the microstate 3E and one is in the microstate 5E. The number of available microstates for system A is:
Nms = 6! / (4! * 1! * 1!) = 720 / 24 = 30
Then, 30 is the number of available microstates for this system.
Let us consider a system B with the same number of particles (six) and the same number of levels of energy, i.e. six, but where each particle is occupying a level of energy, i.e. one particle at level 0E, one particle at level 1E, one particle at level 2E, etc. The solution is as follows:
Nms = 6! / (1! * 1! * 1! *1! * 1! * 1!) = 720 / 1 = 720
This system offers more available microstates, that is, more configurations to be adopted by the energy in a radiation process. Therefore, the radiation will flow from system A, with 30 available microstates, towards system B, with 720 available microstates.
What about the outer space, where there is only one particle per cubic meter? Is it possible that it has more available microstates than the massive Earth?
All the particles in the deep space are in their basic configuration, that is, there are no particles occupying any level of energy, but only high speed protons, therefore:
Nms = 6! / (0!) = 720 / 1 = 720
Consequently, the radiation trajectory will be always from the Earth towards the deep space, the most efficient sink of radiation of any kind. Notice that it has nothing to do with temperature, disorder, complexity, etc.
Kevan Hashemi asked Cohenite if a particle at 300 K will or will not emit photons. Any particle at 300 K is at its fundamental energy state, i.e. its available microstates will be higher than those of any particle with a temperature above 300 K. Such particle won’t radiate, but it will absorb energy.
By Nasif S. Nahle, Scientific Research Director at Biology Cabinet Mexico,
Nasif’s website http://www.biocab.org
More from Nasif at this blog http://jennifermarohasy.com/blog/author/nasif-s-nahle/
This topic has been magisterially touched by Alan Siddons in Chapters 2 and 3 of the book “Slaying the Sky Dragon-Death of the Greenhouse Theory”.
Alan Siddons explains how the models on the Earth’s energy budget cannot depict the real exchange of energy through radiation between the Sun and the Earth and the Earth and the deep space .
This essay deals with the exchange of energy between the Earth and the deep space, already explained by Alan Siddons in his articles, and with the quantum concept of microstates, which agrees with the explanation given by Alan Siddons on his articles in the book Slaying the Sky Dragon-Death of the Greenhouse Theory.
- John D. Cutnell and Kenneth W. Johnson. Physics, 3rd Edition. John Wiley and Sons, Inc. 1995. New York. Page 434.