Entropy of the cosmological constant and the laws of thermodynamics?
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In the Laws of Thermodynamics, we have still more evidence for a transcendent origin of the cosmos. The First Law of Thermodynamics is often called the Law of Conservation of Energy.This law suggests that energy can be transferred from one system to another in many forms, but it cannot be created ordestroyed. Matter can be converted into energy, as Albert Einstein observed when he offered his grand equation E=MC2. However, the total amount of energy available in the universe since it came into existence is constant.
The Second Law of Thermodynamics is commonly known as the Law of Increased Entropy. While the quantity of matter/energy remains the same (First Law), the quality of matter/energy deteriorates gradually over time. How so? Usable energy is inevitably used for productivity, growth, and repair. In the process, usable energy is converted into unusable energy. Thus, usable energy is irretrievably “lost” in the form of unusable energy.
The Second Law of Thermodynamics is commonly known as the Law of Increased Entropy. While the quantity of matter/energy remains the same (First Law), the quality of matter/energy deteriorates gradually over time. How so? Usable energy is inevitably used for productivity, growth, and repair. In the process, usable energy is converted into unusable energy. Thus, usable energy is irretrievably “lost” in the form of unusable energy.
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☸ Required Answer is ⬇⬇ ⬇⬇ ⬇
⭕ Entropy of the cosmological constant -
=> Let the number of micro-states of the particles created (not virtual particles) be Ω=N!∏njΩ=N!∏nj. Where njnj is the particle in the jj'th state and energy (ϵc)j(ϵc)j
Hence,
ln(Ω)=ln(N!)−∑jln(nj!)ln(Ω)=ln(N!)−∑jln(nj!)
=>ln(Ω)=NlnN−N−(∑jnjlnnj−nj)=>ln(Ω)=NlnN−N−(∑jnjlnnj−nj)
=>ln(Ω)=NlnN−(∑jnjlnnj)=>ln(Ω)=NlnN−(∑jnjlnnj)
=>d(ln(Ω))=(lnN)dN−∑jlnnjd(nj)=>d(ln(Ω))=(lnN)dN−∑jlnnjd(nj)
We note for Λ=0=>dN=0Λ=0=>dN=0 :
=>d(ln(Ω)=−∑jlnnjd(nj)=d(Sckb)=dUc−PcdVcT=>d(ln(Ω)=−∑jlnnjd(nj)=d(Sckb)=dUc−PcdVcT
Going back to Λ≠0Λ≠0
=>d(ln(Ω)=lnNdN+dUc−PcdVc=>d(ln(Ω)=lnNdN+dUc−PcdVc
Rewriting in terms of SS
=>dS=kblnNdN dSΛ+d(Sc)=>dS=kblnNdN dSΛ+d(Sc)
Defining kbln NdN kbln NdN as the entropy of the cosmological constant SΛSΛ
=>dS=dSΛ+dSc=>dS=dSΛ+dSc
Writing everything explicitly:
TdS=kbTlnNcdNc+dUc−PcdVcTdS=kbTlnNcdNc+dUc−PcdVc
⭕ Law of thermodynamics states that the energy can not be created or can't be destroyed in any isolated system. This laws is also known as the law of conservation of energy. (first law of thermodynamics) .
⭕ 2nd Law of Thermodynamics states that the entropy of any isolated system always increases never decreased.
Vielen Dank ♥
☸ Required Answer is ⬇⬇ ⬇⬇ ⬇
⭕ Entropy of the cosmological constant -
=> Let the number of micro-states of the particles created (not virtual particles) be Ω=N!∏njΩ=N!∏nj. Where njnj is the particle in the jj'th state and energy (ϵc)j(ϵc)j
Hence,
ln(Ω)=ln(N!)−∑jln(nj!)ln(Ω)=ln(N!)−∑jln(nj!)
=>ln(Ω)=NlnN−N−(∑jnjlnnj−nj)=>ln(Ω)=NlnN−N−(∑jnjlnnj−nj)
=>ln(Ω)=NlnN−(∑jnjlnnj)=>ln(Ω)=NlnN−(∑jnjlnnj)
=>d(ln(Ω))=(lnN)dN−∑jlnnjd(nj)=>d(ln(Ω))=(lnN)dN−∑jlnnjd(nj)
We note for Λ=0=>dN=0Λ=0=>dN=0 :
=>d(ln(Ω)=−∑jlnnjd(nj)=d(Sckb)=dUc−PcdVcT=>d(ln(Ω)=−∑jlnnjd(nj)=d(Sckb)=dUc−PcdVcT
Going back to Λ≠0Λ≠0
=>d(ln(Ω)=lnNdN+dUc−PcdVc=>d(ln(Ω)=lnNdN+dUc−PcdVc
Rewriting in terms of SS
=>dS=kblnNdN dSΛ+d(Sc)=>dS=kblnNdN dSΛ+d(Sc)
Defining kbln NdN kbln NdN as the entropy of the cosmological constant SΛSΛ
=>dS=dSΛ+dSc=>dS=dSΛ+dSc
Writing everything explicitly:
TdS=kbTlnNcdNc+dUc−PcdVcTdS=kbTlnNcdNc+dUc−PcdVc
⭕ Law of thermodynamics states that the energy can not be created or can't be destroyed in any isolated system. This laws is also known as the law of conservation of energy. (first law of thermodynamics) .
⭕ 2nd Law of Thermodynamics states that the entropy of any isolated system always increases never decreased.
Vielen Dank ♥
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