munion Posted September 15, 2009 Share Posted September 15, 2009 I have another question about relativity combined with thermodynamics: Let assume that we have a Box of Volume Vo. Inside the box there is photons (photon gas). The gas has internal energy Eo, pressure Po and the box is accelerated to a velocity near to the speed of light. What will happen to the internal energy and the pressure of the gas? Link to comment Share on other sites More sharing options...
swansont Posted September 15, 2009 Share Posted September 15, 2009 In which frame? Link to comment Share on other sites More sharing options...
munion Posted September 16, 2009 Author Share Posted September 16, 2009 Whatever frame you want except the rest frame. Link to comment Share on other sites More sharing options...
swansont Posted September 16, 2009 Share Posted September 16, 2009 Now that I think about it, I don't think the frame matters. For an isotropic gas, every photon that gets redshifted will have another that is blueshifted by an equal amount. Based on that, I'd say the energy remains constant. Link to comment Share on other sites More sharing options...
munion Posted September 16, 2009 Author Share Posted September 16, 2009 If remain the same then because ΔQ=0 (we assume that we have isolated box) and take notice that V'=1/γV0 this means from the 1rst law of thermodynamics that: ΔU=-ΔW+ΔQ => ΔW=0 => P=0 ??? Link to comment Share on other sites More sharing options...
ydoaPs Posted September 16, 2009 Share Posted September 16, 2009 Now that I think about it, I don't think the frame matters. For an isotropic gas, every photon that gets redshifted will have another that is blueshifted by an equal amount. Based on that, I'd say the energy remains constant. Assuming the walls of the box are perfect reflectors. Link to comment Share on other sites More sharing options...
munion Posted September 16, 2009 Author Share Posted September 16, 2009 Assuming the walls of the box are perfect reflectors. Yeap thats the word Link to comment Share on other sites More sharing options...
swansont Posted September 16, 2009 Share Posted September 16, 2009 Assuming the walls of the box are perfect reflectors. And mounted on an elephant whose mass may be ignored, which is standing on a frictionless surface. Link to comment Share on other sites More sharing options...
Sisyphus Posted September 16, 2009 Share Posted September 16, 2009 And mounted on a spherical elephant whose mass may be ignored, which is standing on a frictionless surface. Fixed. Link to comment Share on other sites More sharing options...
swansont Posted September 16, 2009 Share Posted September 16, 2009 If remain the same then because ΔQ=0 (we assume that we have isolated box) and take notice that V'=1/γV0 this means from the 1rst law of thermodynamics that: ΔU=-ΔW+ΔQ => ΔW=0 => P=0 ??? Why would P=0? One thing to consider: is volume constant? Link to comment Share on other sites More sharing options...
ydoaPs Posted September 16, 2009 Share Posted September 16, 2009 Fixed. In a perfect vacuum. Link to comment Share on other sites More sharing options...
munion Posted September 16, 2009 Author Share Posted September 16, 2009 (edited) Why would P=0? One thing to consider: is volume constant? The Volume is not constant ΔV not 0. The P=0 is derived by 1rst thermodyn law because as you say ΔU=0 the box *mounted on an elephant whose ΔQ=0 so ΔW=0 => P=0.... Merged post follows: Consecutive posts mergedW = \Delta (p\,V) I think would be helpful http://en.wikipedia.org/wiki/First_law_of_thermodynamics Edited September 16, 2009 by munion Consecutive posts merged. Link to comment Share on other sites More sharing options...
swansont Posted September 16, 2009 Share Posted September 16, 2009 The Volume is not constant ΔV not 0. The P=0 is derived by 1rst thermodyn law because as you say ΔU=0 the box *mounted on an elephant whose ΔQ=0 so ΔW=0 => P=0.... Merged post follows: Consecutive posts mergedW = \Delta (p\,V) I think would be helpful http://en.wikipedia.org/wiki/First_law_of_thermodynamics I think it would be helpful for you to take a course in thermodynamics and learn about thermodynamic potentials. Link to comment Share on other sites More sharing options...
munion Posted September 17, 2009 Author Share Posted September 17, 2009 (edited) I think it would be helpful for you to take a course in thermodynamics and learn about thermodynamic potentials. Yes you think.. at least i m not doing the smart guy here not the physics "expert" i have a lesson also for you for the ridiculous answer that you gave me... http://puhep1.princeton.edu/~mcdonald/examples/uoveromega.pdf By the way with the link in Wikipedia i would not intent to insult you i put it there as a reference and only that... You guys really what is your job here to humiliate the question of the physics amateur? If you have an objection or you find a mistake please say it with no more comments... Merged post follows: Consecutive posts mergedI would not ask something again i will leave you to discuss the physics with your superior intelligence and as one last thing i would like to ask sorry for my silly questions. Edited September 17, 2009 by munion Link to comment Share on other sites More sharing options...
swansont Posted September 17, 2009 Share Posted September 17, 2009 Yes you think.. at least i m not doing the smart guy here not the physics "expert" i have a lesson also for you for the ridiculous answer that you gave me... http://puhep1.princeton.edu/~mcdonald/examples/uoveromega.pdf By the way with the link in Wikipedia i would not intent to insult you i put it there as a reference and only that... You guys really what is your job here to humiliate the question of the physics amateur? If you have an objection or you find a mistake please say it with no more comments... Merged post follows: Consecutive posts mergedI would not ask something again i will leave you to discuss the physics with your superior intelligence and as one last thing i would like to ask sorry for my silly questions. What was ridiculous about my answer? My suggestion was not meant to humiliate you. Relativistic thermodynamics is still sort of an open area of physics. http://www.aip.org/pnu/2007/split/843-1.html Hoping to have an understanding of it without a good background in thermodynamics and statistical mechanics is going to be problematic. There are many different thermodynamic potentials, each applying to different situations. The problem here is that the statement of conservation of energy doesn't account for all of the energy terms. I don't know what happens to entropy in the new frame, but I'm not convinced that it would be the same, so the Gibbs Free Energy is probably the term you should be looking at here. G = U + pV - TS If you're intimidated by all of this, consider not blaming the messenger. Merged post follows: Consecutive posts merged"And mounted on a spherical elephant whose mass may be ignored, which is standing on a frictionless surface." Why wouldn't the point-elephant approximation work here? Link to comment Share on other sites More sharing options...
munion Posted September 17, 2009 Author Share Posted September 17, 2009 What was ridiculous about my answer? My suggestion was not meant to humiliate you. Relativistic thermodynamics is still sort of an open area of physics. http://www.aip.org/pnu/2007/split/843-1.html Hoping to have an understanding of it without a good background in thermodynamics and statistical mechanics is going to be problematic. There are many different thermodynamic potentials, each applying to different situations. The problem here is that the statement of conservation of energy doesn't account for all of the energy terms. I don't know what happens to entropy in the new frame, but I'm not convinced that it would be the same, so the Gibbs free energy.Free Energy is probably the term you should be looking at here. G = U + pV - TS If you're intimidated by all of this, consider not blaming the messenger. Merged post follows: Consecutive posts merged"And mounted on a spherical elephant whose mass may be ignored, which is standing on a frictionless surface." Why wouldn't the point-elephant approximation work here? If you meant that way i m own you an apology... i misunderstood. But you could wrote to me immediately i didn't knew about the gibs free energy Link to comment Share on other sites More sharing options...
insane_alien Posted September 17, 2009 Share Posted September 17, 2009 it should probably be noted that swansont did not give himself the physics expert title but it was the admins. they gave him the title because (appart from being a proffesional physicist) he has repeatedly shown himself to have expert knowledge in one or more fields of physics. so he does actually know what he's talking about. i've taken a few classes in (non-relativistic) thermodynamics and its clear to me that you don't have sufficient knowledge of it to start introducing complicating factors such as relativity. Link to comment Share on other sites More sharing options...
munion Posted September 17, 2009 Author Share Posted September 17, 2009 (edited) it should probably be noted that swansont did not give himself the physics expert title but it was the admins. they gave him the title because (appart from being a proffesional physicist) he has repeatedly shown himself to have expert knowledge in one or more fields of physics. so he does actually know what he's talking about. i've taken a few classes in (non-relativistic) thermodynamics and its clear to me that you don't have sufficient knowledge of it to start introducing complicating factors such as relativity. Ok then i wont learn anything about thermodynamics thank you for your support here... For the god shake it is only a question. Ps i have taken also some lessons in thermodynamics in university but that doesn't mean anything.Could you help me out with this? Edited September 17, 2009 by munion Link to comment Share on other sites More sharing options...
insane_alien Posted September 17, 2009 Share Posted September 17, 2009 umm, how did you get 'don't learn thermodynamics' from a suggestion to get a stronger foundation in the basics before adding in extra factors? Link to comment Share on other sites More sharing options...
munion Posted September 17, 2009 Author Share Posted September 17, 2009 (edited) umm, how did you get 'don't learn thermodynamics' from a suggestion to get a stronger foundation in the basics before adding in extra factors? Ok could you provide me some please?? Where should i start? I want to know what influences has the relativity to thermodynamics.... Could you explain me how my question will be answered? The papers of plank-einstein that i read there are in controversy with the paper of Ott and really baffled me And if you know what will happen especially with the entropy of the photon gas please explain it to me... Merged post follows: Consecutive posts mergedImagine that you have a box with steam if that box is moving then according with einstein - plank papers the temperature is falling (According to the Ott happens the opposite!!). this mean that one observer will observer steam another with greater speed water and another with much more speed ice? except if the relativity does not affect the entropy of the system. Merged post follows: Consecutive posts mergedhttp://www.aip.org/pnu/2007/split/843-1.html Now there is a third approach T'=T0.... Edited September 17, 2009 by munion Link to comment Share on other sites More sharing options...
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