The oscillators of a perfect blackbody are considered as non-interacting entities. Thus, Bose-Einstein Condensation is possible for these entities. The Bose-Einstein Condensate (BEC) temperature of a perfect blackbody is calculated from the Planck’s theory of blackbody radiation and de Broggle’s wave-particle duality relation. It is observed that the BEC temperature of an ideal blackbody is 4.0*K*. Thus, bellow 4.0*K* temperature the energy density vs wavelength plot of a blackbody would look like a delta function. In this region, a blackbody would absorb or emit radiation of unique frequency depending upon its temperature. It is also possible to calculate the rest mass and the ground state vibrational energy of the oscillators of a blackbody using present formalism.

## I. INTRODUCTION

Bose-Einstein condensation (BEC) was predicted in 1924.^{1–4} But, it took nearly seventy years to obtain such state through experiment.^{5} Observation of BEC in a dilute atomic vapor at 170 nanokelvin temperature by Anderson *et. al.*^{5} in 1995 is immediately followed by several groups of researchers.^{6–9} Presently, BEC is achieved for different systems and its properties are well studied. Non-interacting Bose-Einstein condensation is also reported for quasi-periodic lattice.^{10} Thus, it is possible to obtain a state of unique features similar to BEC for the oscillators of a blackbody which are considered as non-interacting entities.

BEC requires cooling of a sample so that its thermal de Brogglie wavelength, *λ*_{db}, becomes larger than the mean spacing between two particles of the system. Exploiting this condition it is possible to calculate the condensation temperature of a blackbody. Here it should be mentioned that, condensation of the oscillators of a blackbody implies a state where all oscillators of it exist in the ground state of vibration. This study is very important because the cosmic background temperature is very near to zero Kelvin.^{11} Thus, the nature of the interaction of matter with radiation at very low temperature has immense importance in astrophysics and space science research. Near 0*K* temperature, the thermal momentum of constituent particles, i.e. atoms or molecules, is near to 0. Thus, according to the uncertainty principle, their position certainty is very high and at the same time, their de Broggle’s wavelength^{12} is very large. So, the existence of BEC is possible for the astroparticles at this temperature. Due to the very longer wavelength of each individual particle and nearly zero interaction between two particles, the system acquires a unique energy state with an infinite number of occupancy.

On the other hand, the study of lattice vibration at very low temperature is also very important due to its application in various fields of scientific research. Heat transfer, i.e. absorption or emission of heat of a body depends on its atomic level energy partition. A gas particle has rotational, translational and vibrational energy partition. But, with a decrease in temperature different degrees of freedom of a gas particle decreases. Thus, it is very important to know at a very low temperature which degree of freedom of a gas molecule exists and which degree of freedom diminishes. We also want to know at what temperature all degrees of freedom of a gas molecule become zero. Heat exchange of a solid material changes its lattice vibration which is commonly known as the phonon. From the third law of thermodynamics, we know that the entropy of a pure crystalline substance is zero at 0*K* and for other substances, it is minimum at 0*K*. Here it could be questioned - is it possible to make the thermal entropy of a system zero at any temperature higher than 0*K*? We are curious to find a proper answer to this question as at the Bose-Einstein Condensate, a system has only one microstate and its entropy is zero. Thus, zero entropy of a system at a finite temperature is possible. Of course, this temperature is the critical temperature of the system for Bose-Einstein Condensation. From the gas law of thermodynamics, we could not derive any relation to find out the critical temperature of a material at which its entropy is zero. Thus we use the wave-particle duality relation and Planck’s theory of blackbody radiation to find out the critical temperature of a perfect blackbody at which its thermal entropy becomes zero. It is obvious that for a blackbody, oscillators are considered as non-interacting and the number of oscillators are infinite. But, a real system has a finite number of interacting particles. Thus, the use of Planck’s law of blackbody radiation to a real system is not trivial. Here an ideal system is considered for this study. The study of a real system would be followed. In a recent article,^{13} critical temperature of matter-dark matter^{14–18} conversion is reported. Using the critical temperature, the mass of the resonator of a perfect blackbody is calculated which is very close to the value of the mass of a resonator reported earlier.

## II. THEORY

According to Max Planck,^{19–21} a blackbody is a collection of numbers of oscillators. An oscillator emits or absorbs energy proportional to its oscillation frequency. Thus, emission or absorption of energy of a blackbody is discrete rather continuous which is termed as quantization of energy. This concept of quantization of electromagnetic radiation is used to explain the energy density distribution nature of a blackbody.

On the other hand, Louis de Broggle proposed that every moving object has a wave-particle dual character.^{12} According to the de Broggle, wavelength (*λ*) of a particle having momentum *p*, is $hp$, where *h* is the Planck’s constant. Since oscillators are in motion and have momentum, they should have de Broggle wavelength.

Following Planck’s postulate,^{22} let us consider that the number of identical resonators per unit volume of a blackbody, having oscillation frequency *ν*, is *N* at the equilibrium temperature (*T*). If *U* is the vibrational energy of a resonator, the total energy per unit volume is *NU*. It is known that the rate of change of vibrational entropy, *S*, with respect to the change of vibrational energy, *U*, at a temperature *T* is -

If *N*_{i} number of particles occupy the *i*^{th} vibrational state, the total energy of these particles are

The number of modes per unit volume for frequency *ν*_{i} is $8\pi \nu i2c3$. Thus, considering the equipartition principle (classical approach), the total energy per unit volume for this frequency is

Comparing Equation (2) and (3) we get an expression for the number of particles at *i*^{th} vibrational state which is

where *ν*_{0} is the frequency for the ground state vibrational energy. Total internal energy (*U*) of a system could be calculated using Equation (4). If the number of particles per unit volume of a system is *N*, the total energy per unit (*U*) volume of this system is

From Equation (5) we get

or,

where,

Putting the value of *dU* in Equation (1) we get

On the integration of Equation (9), we get

because, for an ideal system, entropy is zero at 0*K* according to the third law of thermodynamics. It is also known that the thermal entropy is zero at the Bose-Einstein Condensate temperature.^{23,24}

Following the Planck’s derivation^{22} of the entropy of an oscillator, we get

where *ν* is the frequency of the oscillator. Equation (11) may be rearranged as

Equation (13) is also valid for *ν*_{0}. If *T*_{0}*K* is the Bose-Einstein condensate temperature of the system, then at *T*_{0}*K* temperature all the oscillators should exist in the ground vibrational state. Thus, from Equation (13) we get

From equation (5) we get

Let,

where,

Putting this value of $Uh\nu 0$ in equation (14) we get the Bose-Einstein condensate temperature of the system of interest in terms of *Z* and *B*, which is

It is impossible to find the value of *T*_{0} unless *Z* and *B* are known. But, we can study the variation of *T*_{0} with the changes of *Z* and *B*.

## III. DISCUSSIONS

To understand the variational behavior of *T*_{0} with respect to the independent change of *Z* and *B*, a numerical simulation is performed. Computed results are presented in graphical form in Figure 1. Both *Z* and *B* are varied without any constraint from 1.0 to 10.0. We obtain the area of condensation which looks like a feather (blue colored area in Figure 1(a)). Outside of this area, condensation is not possible irrespective of $ZB$. It is observed that *T*_{0} not only depends on $ZB$ but also depends on individual values of *Z* and *B*. We get different values of *T*_{0} for the same value of $ZB$ coming from different values of *Z* and *B*. Interestingly, *T*_{0} decreases for the same value of $ZB$ of larger values of *Z* and *B*. Though the decrease in *T*_{0} is very very small compared to the increase of *Z* and *B*. Thus, as a whole, we observe that for small values of $ZB$, *T*_{0} is small. Different systems of different $ZB$ values may condense at the same temperature. If, *Z* and *B* varied conditionally, the condensation temperature increases with the increase of $ZB$. But, the rate of increase depends on the initial value of $ZB$.

To understand the physical significance of this behavior of *T*_{0} we have to search for the meaning of *Z* and *B*. From Equation (16) we know that *Z* is the ratio of the total energy per unit volume of the system to the ground state energy of a single oscillator. Thus, it is related to the number of the oscillators of the system. On the other hand, *B* is the total mode of vibrations available for a particular frequency. Thus, $ZB$ implies the crowding of particles or oscillators of the system.

If we take a few small selective values of *Z* and *B* and calculate $ZB$ and *T*_{0} we should be able to understand the physical relation between *Z*, *B*, and *T*_{0}. Let us consider four values of *Z* as (*Z* = 1, 2, 2, 1) and *B* as (*B* = 1, 1, 2, 2). Now the corresponding values of $ZB$ and *T*_{0} are $ZB=1,2,1,0.5$ and *T*_{0} = 4.0, 6.75, 2.6, 2.0. It is observed that when *B* > *Z*, particles are free to move to give rise more numbers of microstates which produces higher values of the entropy of the system. Thus, a lower temperature is required for condensation. But, when *B* < *Z*, particles are forced to occupy the same energy state though it may not be quantum mechanically allowed, due to the unavailability of the energy states. This type of condensation may be considered as conditional condensation. It is obvious that such condensate is very unstable. Physical and mechanical properties of this type of condensation must be different from the other type. Forced pairing or clustering may exist in this system. Thus, condensation occurs at the higher temperature.

It is observed that at $ZB=1$,

Thus BEC temperature is proportional to the ground state vibrational frequency of the oscillators and also proportional to the square root of its oscillator density. But, we have to know either of them to calculate the other.

Since the oscillators are harmonic oscillators, energy of the *n*^{th} vibrational state is

where *ω* is the angular frequency of the oscillator. *ω* = 2*πν* where *ν* is linear frequency. At temperature *T*, the energy of the oscillator is *kT*. Thus, from Equation (20), we get

For ground state (i. e. *n* = 0) vibrational energy (*E*_{0}) we get

For an ideal harmonic oscillator, at the equilibrium position its total energy is equal to its kinetic energy at this position as potential energy is zero. If *p* is the momentum of the oscillator at its equilibrium position, then its kinetic energy is $p22m$ where *m* is the mass of the oscillator. According to the de Broggle wave-particle duality,

For the ground state vibrational energy

Simplifying Equation (25) we get

Equation (27) is not a general one, but this is the only way we can calculate the mass of the oscillator of a blackbody or say phonon of any crystalline solid subject to the constraint mentioned above. For a perfect blackbody, $ZB$ should be unity following the condition that both *Z* and *B* are individually unity. Thus, *T*_{0} of a perfect blackbody is 4.0*K*. This is a very important temperature for the universe, because, bellow this temperature a perfect blackbody would not follow the Planck’s energy distribution law. Bellow 4.0*K*, a blackbody absorbs or emits radiation of a particular frequency only. But, a non-ideal system may behave differently. In previous sections, we observe that for *B* > *Z* BEC temperature is less than 4.0*K*. Thus, we may conclude that 4.0*K* is the BEC temperature of an ideal system only. A non-ideal system has a BEC temperature either greater or less than 4.0*K*. But, their nature of condensation must be different from each other.

Considering *T*_{0} = 4.0*K*, the rest mass of the oscillators of an ideal blackbody is calculated which is 6.1 × 10^{−40}*kg*. This value is nearly 10^{11} times larger than the previously reported value of the mass of an oscillator of a blackbody.^{13} But, the reported value is the lower limit of it. Thus, this result doesn’t contradict with the earlier result. Using this value of the rest mass of the oscillator, the ground state vibrational energy is calculated which is 13.53 × 10^{−24}*J*.

## IV. CONCLUSIONS

In the present study, a theoretical model is reported which relates the Bose-Einstein condensate (BEC)^{3,4} temperature of a system with its lattice phenomenon with the help of Planck’s theory of blackbody radiation and de Broggle’s wave-particle duality relation. This formulated theory is very handy to study the nature of a material near its BEC temperature. It is observed that the BEC temperature of an ideal system is 4.0*K*. BEC temperature of any non-ideal system is either greater or less than 4.0*K*. The deviation of BEC temperature of a system from 4.0*K* is due to different types of non-ideal nature of the system which depends upon the particle density and the number of available vibrational modes. Considering the BEC temperature of a blackbody is 4.0*K*, the rest mass and the ground state vibrational energy of the oscillators are calculated. This is the first report in this field of research in this direction. There is no other theoretical or experimental result available to compare this result. Thus, we have to wait for the advancement in the experimental section to test its validity.

Here the classical approach is used to calculate the total internal energy of a system. But one may use the quantum mechanical approach for it. Present derivation is done considering non-interacting particle. Thus, it may not be applicable for bosons and fermions as it is. Modifications must be done before its use for such particles. Here we observe two different types of deviation from the ideal system. These two types of deviation may be due to the differences in the fundamental nature of the particles. Thus, we may conclude that, BEC type condensation is possible for both bosons as well as fermions. This prediction is highly optimistic at this moment. Further studies considering the interacting particle in the quantum mechanical constraint, is necessary to test this prediction. At this moment it could be concluded that if it is possible to condense fermions it should occur at either extreme end of the temperature scale.

## ACKNOWLEDGMENTS

I like to thank Dr. Pradip Ghorai, IISER Kolkata, India, for providing research facilities and Tanmoy Mandal for potential academic discussions.