Does string theory state as vibrational entropy increases, mass increases?
Related: What is a D-brane?
Reference: Cambridge Relativity
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1$\begingroup$ Can you make your question more specific? $\endgroup$– Ernesto UlloaCommented Jun 19, 2012 at 19:26
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$\begingroup$ @ErnestoUlloa: I am not sure how much more specific I can make this question for fear of being over specific. $\endgroup$– ArgusCommented Jun 19, 2012 at 23:34
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$\begingroup$ @ErnestoUlloa: Maybe, is a string with less entropy less massive relative a string with higher entropy. $\endgroup$– ArgusCommented Jun 19, 2012 at 23:37
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$\begingroup$ Usually entropy is used for systems of many string states. Maybe a stringy black hole, or black-branes. $\endgroup$– Ernesto UlloaCommented Jun 20, 2012 at 20:28
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$\begingroup$ @ErnestoUlloa: I was thinking of a single string as its vibrational energy decreases (ie. Entropy increases) does its mass increase? $\endgroup$– ArgusCommented Jun 20, 2012 at 21:18
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1 Answer
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I think that your question is about string thermodynamics in the microcanonical ensemble. The entropy obtained by calculating the partition function of a relativistic quantum string is just the Boltzmann constant multiplied by the number of vibrational microstates, calculated by using the Hardy-Ramanujan asymptotic expansion. The entropy energy relation in relativistic open and closed strings is worked in detail in chapter 16 of Barton Zwiebach book on string theory. It turns out that entropy is directly proportional to energy. This means that increasing vibrational energy (and also mass) the entropy increases linearly. So mass increased with vibrational entropy. The curious thing is that the derivative of energy with respect to entropy is the temperature of the strings thermodynamic system. This temperature is then constant; no matter how large is the energy! So the answer to your question, according to this reference, is yes, in the high energy regime, mass increases with entropy. I hope this is the answer you were looking for.
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$\begingroup$ That is exactly the answer I was looking for. $\endgroup$– ArgusCommented Jun 21, 2012 at 12:23