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#1
Dec2412, 02:17 PM

P: 4

Is there any maximum temperature that could be reached ( since temp is directly prop. To average kinetic energy , knowing that no particle can exceed the speed of light .



#2
Dec2412, 03:13 PM

Sci Advisor
P: 2,193

You might define a maximum temperature to be the Planck temperature, similar to how planck lengths and times are sometimes quoted to be the shortest units of distance/time. But this is more a statement that it's the highest temperature up to which we can, in principle with our current theories, explain physics as opposed to a real largest temperature scale. If you want to define temperature as [itex] k T \sim E [/itex] then putting enough energy in a small enough volume will just result in a black hole, which is likely to have a smaller thermodynamic temperature than before collapse. This is really the only decent 'maximum temperature' that might exist. 


#3
Dec2412, 06:15 PM

P: 741

The Planck temperature can be defined as the temperature at which the thermal radiation in a small region has to collapse into a black hole. Another poster defined the upper limit of temperature this way. However, he didn’t quite use the phrase “Planck temperature.” Here is a link on the Planck temperature. http://en.wikipedia.org/wiki/Planck_temperature “As for most of Planck units, a Planck temperature of 1 (unity) is a fundamental limit of quantum theory, in combination with gravitation, as presently understood. At temperatures greater than or equal to TP, current physical theory breaks down because we lack a theory of quantum gravity.” 


#4
Dec2412, 06:59 PM

Mentor
P: 16,316

Maximum temperature;
First, the Planck units are not where "general relativity and quantum mechanics contradict each other". They are usually a rough estimate on where general relativistic effects become important, but this could be off by a factor of 10 or 1/10. Second, the exception is temperature, where it is solely based on dimensional analysis: T = E/k, rather than any actual properties of temperature. Wikipedia is not right here. It's just adding confusion. If one uses the Freshman physics definition of temperature, there is no limit on T because there is no limit of KE. If one uses a more advanced definition of temperature, 1/T = dS/dE, the same thing happens, in a more complex fashion  but it's clear that it's not as simple as T=E/k. 


#5
Dec2612, 01:46 PM

P: 40

The Planck temperature (or any higher temperature) is no problem for systems with finite number of energy states. In these systems, any temperature can be attained according to the known laws of physics, including negative temperatures.



#6
Dec2612, 02:17 PM

P: 741

The Third Law of Thermodynamics is also a known law of physics. One definition of the third law is that all systems at stable equilibrium have an absolute minimum of entropy at a certain universal temperature, referred to as absolute zero. This would apply even to systems that have a finite number of energy states. What you said is probably true for systems that are not in thermal equilibrium. Systems that are not in stable equilibrium can have an unstable equilibrium state which is not an absolute minimum. The temperature of this unstable equilibrium state can be negative. One example would be the gain medium in a laser. The gain medium of a laser has a temperature that is negative. However, the gain medium is not in stable equilibrium. Although it is in a stable state, one photon can initial a chain reaction where all the available energy in the gain medium is released as photons. However, the gain medium can have a large number o energy states. One way to formalize this is to define temperature as the derivative of internal energy with respect to entropy. Or rather, T=dU/dS where U is the internal energy, S is the entropy and T is the temperature. For a system that is in a stable equilbrium,, either T=0 or T>0. If T<0, then the system is not in stable equilibrium. If T<0, then the dU/dS<0. The system can then be in an unstable energy state. A system with a finite number of energy states can have a T<0. In fact, a system with just a few interacting states can be made into an unstable equilibrium state easily. The gain medium of a laser is a good example. Many lasers work using the electronic transitions of just a few energy states. The population inversion of a laser is by definition an unstable equilibrium state, also called the metastable state. The ruby laser has about 3 relevant energy states. In Dye lasers often work with three broad bands of energy states. A free electron laser has a high number of energy states. In all these cases, the lasing medium reaches a "temperature" where T<0. I think what you intended to say is that systems that have a ‘small number of high energy states’ can have a ‘negative temperature’. A very ‘small number of low energy states’ has to have a ‘positive temperature.’ This is the case for all ‘classical physics’ systems. http://en.wikipedia.org/wiki/Negative_temperature “Most familiar systems cannot achieve negative temperatures, because adding energy always increases their entropy. The possibility of decreasing in entropy with increasing energy requires the system to "saturate" in entropy, with the number of high energy states being small. These kinds of systems, bounded by a maximum amount of energy, are generally forbidden classically. Thus, negative temperature is a strictly quantum phenomenon. Some systems, however (see the examples below), have a maximum amount of energy that they can hold, and as they approach that maximum energy their entropy actually begins to decrease.” Here is a general definition of "the Thrid Law of Thermodynamics." http://en.wikipedia.org/wiki/Third_l...thermodynamics “A more general form of the third law applies to systems such as glasses that may have more than one minimum energy state: The entropy of a system approaches a constant value as the temperature approaches zero. The constant value (not necessarily zero) is called the residual entropy of the system.” Note that there is no upper limit to temperature implied by any of these laws. Planck's temperature is not an upper limit to temperature in terms of any of these laws. I meant in a previous post that Planck's temperature is considered a sort of boundary to known laws of physics. This was a very general statement. General relativity is considered a fairly good theory for cosmology except in those first moments after the Big Bang when the temperature was above the Planck temperature. Before that, general relativity runs into problems. 


#7
Dec3012, 05:13 PM

P: 40

The example I have in mind is a nuclear spin system in a magnetic field. Your derivative is not well defined for these systems, since two values of U correspond to a given entropy. But even if you write 1/T=dS/dU, I don't see what the problem is. 


#8
Dec3012, 07:44 PM

P: 159

I understand that if they really are isolated degrees of freedom then you can consider them in thermal equilibrium at a negative temperature. 


#9
Dec3012, 09:29 PM

P: 40

I don't think anything in nature is truly isolated. 


#10
Dec3112, 04:51 PM

P: 741

“In these systems, any temperature can be attained according to the known laws of physics, including negative temperatures.” One expression for the Third Law of Thermodynamics is that “absolute zero can not be achieved with a finite number of steps.” Unless one can apply an infinite number of steps to a physical process, then absolute zero temperature can not be attained according to the Third Law. I understand that there is more than one way to express the third law. Furthermore, I know that the finite number of steps definition may not be considered a truly fundamental expression of the third law. However, the third law does imply that absolute zero of temperature is an unphysical limit. This doesn’t mean the Third Law is correct! Here is a link to the Wiki article on the third law. Note the quotation concerning “finite number of steps”. http://en.wikipedia.org/wiki/Third_l...thermodynamics “Physically, the law implies that it is impossible for any procedure to bring a system to the absolute zero of temperature in a finite number of steps.” I like your definition of temperature, i.e., 1/T=dS/dU Note that the derivative is not well defined when "T=0". Thus, one can not achieve absolute zero ... 


#11
Jan113, 04:48 PM

P: 40




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