Welcome to the ToeQuest.
Results 1 to 7 of 7
  1. #1
    Raider of the lost time
    Join Date
    Nov 2003
    Location
    United States
    Posts
    11,784
    Blog Entries
    10
    Thanks Given
    1,106
    Thanked 1,474x in 1,194 Posts
    Rep Power
    158

    questioning the zeroth law

    Would anyone question the zeroth law of thermodynamics? Please visit this link and make your own personal decision. http://en.wikipedia.org/wiki/Zeroth_...thermodynamics
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

  2. The Following User Says Thank You to AntonioLao For This Useful Post:

    Graybeard (01-24-2011)

  3. #2
    9th degree Black Belt
    Join Date
    May 2007
    Location
    United States
    Posts
    1,576
    Blog Entries
    7
    Thanks Given
    2,106
    Thanked 1,355x in 954 Posts
    Rep Power
    36

    Re: questioning the zeroth law

    Quote Originally Posted by AntonioLao View Post
    Would anyone question the zeroth law of thermodynamics? Please visit this link and make your own personal decision. http://en.wikipedia.org/wiki/Zeroth_...thermodynamics
    I would agree with it. What's not to agree with? It seems straight forward to me.

  4. The Following User Says Thank You to Bogie For This Useful Post:

    AntonioLao (01-22-2011)

  5. #3
    Raider of the lost time
    Join Date
    Nov 2003
    Location
    United States
    Posts
    11,784
    Blog Entries
    10
    Thanks Given
    1,106
    Thanked 1,474x in 1,194 Posts
    Rep Power
    158

    Re: questioning the zeroth law

    You still need to clearly state what is it about the zeroth law that you are agreeing to? And it must be supported with your own assertions. It never was straight forward until you understand the second, the third, and the first law, in these order of progressive understanding.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

  6. #4
    Moderator
    Join Date
    Aug 2005
    Location
    Australia
    Posts
    4,896
    Blog Entries
    24
    Thanks Given
    2,967
    Thanked 2,727x in 1,670 Posts
    Rep Power
    92

    Re: questioning the zeroth law

    Quote Originally Posted by AntonioLao View Post
    You still need to clearly state what is it about the zeroth law that you are agreeing to? And it must be supported with your own assertions. It never was straight forward until you understand the second, the third, and the first law, in these order of progressive understanding.
    I agree with it ..... I see it as providing a compelling description of the 'boundaries' between forms that comprise the overall universe.

    cool bananas ... greg
    'Blondie says I must hate all Brunettes. I'll try, but if I can't ... I'll love them both'
    ... graffiti on Tavern wall, Pompeii, circa AD 70.

  7. The Following User Says Thank You to Graybeard For This Useful Post:

    AntonioLao (01-24-2011)

  8. #5
    Raider of the lost time
    Join Date
    Nov 2003
    Location
    United States
    Posts
    11,784
    Blog Entries
    10
    Thanks Given
    1,106
    Thanked 1,474x in 1,194 Posts
    Rep Power
    158

    Re: questioning the zeroth law

    The question is tantamount to asking can thermal equilibrium truly exist in a universe? My argument is that if thermal equality exists then the universe will never evolve to its present state since any form of motion always requires a temperature gradient such that the absolute difference of two given temperatures remains a finite positive quantity even if this quantity is infinitesimally small amount. Although negative temperature is not effectively defined the absolute zero of temperature is not attainable by any thermal experiment.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

  9. The Following User Says Thank You to AntonioLao For This Useful Post:

    Graybeard (01-24-2011)

  10. #6
    Moderator
    Join Date
    Aug 2005
    Location
    Australia
    Posts
    4,896
    Blog Entries
    24
    Thanks Given
    2,967
    Thanked 2,727x in 1,670 Posts
    Rep Power
    92

    Re: questioning the zeroth law

    Quote Originally Posted by AntonioLao View Post
    The question is tantamount to asking can thermal equilibrium truly exist in a universe? My argument is that if thermal equality exists then the universe will never evolve to its present state since any form of motion always requires a temperature gradient such that the absolute difference of two given temperatures remains a finite positive quantity even if this quantity is infinitesimally small amount. Although negative temperature is not effectively defined the absolute zero of temperature is not attainable by any thermal experiment.
    But motion could have always been conserved as chaotic or random. This also is a form of equilibrium or thermal equality.

    cool bananas ... greg
    'Blondie says I must hate all Brunettes. I'll try, but if I can't ... I'll love them both'
    ... graffiti on Tavern wall, Pompeii, circa AD 70.

  11. The Following User Says Thank You to Graybeard For This Useful Post:

    AntonioLao (01-25-2011)

  12. #7
    Raider of the lost time
    Join Date
    Nov 2003
    Location
    United States
    Posts
    11,784
    Blog Entries
    10
    Thanks Given
    1,106
    Thanked 1,474x in 1,194 Posts
    Rep Power
    158

    Re: questioning the zeroth law

    I'm hypothesizing that Local Infinitesimal Motion (LIM) can be conserved as the quantum vacuum fluctuations of the squares of zero-point energies such that the infinitesimal thermal gradient is equal to the product of two temperatures. These are equal only if both are zero.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

 

 

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Back to top