Entropy of the cosmological constant and the laws of thermodynamics?
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Holla ^_^
☸ 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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Hey mate ^_^
There are 3 laws of thermodynamics:
=> The First Law of thermodynamics basically says that energy or matter can neither be created nor destroyed.
=> The Second Law of thermodynamics states that entropy of universe is always increasing.
=> Third law of thermodynamics states that the entropy of a perfect crystal at absolute zero temperature is zero.
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And:
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The cosmological constant is most likely a temporary description of a dynamical variable that has been drastically evolving from the early inflationary era to the present.
#Be Brainly❤️
There are 3 laws of thermodynamics:
=> The First Law of thermodynamics basically says that energy or matter can neither be created nor destroyed.
=> The Second Law of thermodynamics states that entropy of universe is always increasing.
=> Third law of thermodynamics states that the entropy of a perfect crystal at absolute zero temperature is zero.
====
And:
====
The cosmological constant is most likely a temporary description of a dynamical variable that has been drastically evolving from the early inflationary era to the present.
#Be Brainly❤️
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