f*******d
发帖数: 339 | 1 I learned about negative tempertature on class before, which is basically
an inversion of the occupation number, and laser for example, is produced
this way. At that time, I thought I understand this point. However, later
on I became puzzled.
Suppose you have some energy levels with different occupation numbers.
Now in order to use negative temperature, you have to have your distribution
such that f_n = g_n exp(-E_n/kT), with a negative T. Now although for a pair
of inverted energy levels it is | s*a
发帖数: 33 | 2 For such a multi-level system, maybe one could consider each energy level
(or each pair of levels) as a sub-system which has a particular negative
temperature easily determined by your formula. Without fully relaxation,
there are not enough thermal exchange between different sub-systems so they
may have different temperatures. (or we can say they are not in thermal
equilibrium). Only after the "fine-tuning" you called, it is possible they
reach a uniform temperature and satisfy the Boltzmann dis
【在 f*******d 的大作中提到】
: I learned about negative tempertature on class before, which is basically
: an inversion of the occupation number, and laser for example, is produced
: this way. At that time, I thought I understand this point. However, later
: on I became puzzled.
: Suppose you have some energy levels with different occupation numbers.
: Now in order to use negative temperature, you have to have your distribution
: such that f_n = g_n exp(-E_n/kT), with a negative T. Now although for a pair
: of inverted energy levels it is
| c****u
发帖数: 584 | 3
1/T=dS/dE. that is the definition of the temperature. When the system is the
energy bounded system which means the system's energy can't goes to infinite,
such as magnet system, when you keep heating the system, you increase the
entropy by increasing the energy, however when you increase the temperature to
infinite,the system can't gain more entropy, since all particles tends to go to
the highest state. At that time temperature will goes to negative and goes to
negative zero, when all particles
【在 f*******d 的大作中提到】
: I learned about negative tempertature on class before, which is basically
: an inversion of the occupation number, and laser for example, is produced
: this way. At that time, I thought I understand this point. However, later
: on I became puzzled.
: Suppose you have some energy levels with different occupation numbers.
: Now in order to use negative temperature, you have to have your distribution
: such that f_n = g_n exp(-E_n/kT), with a negative T. Now although for a pair
: of inverted energy levels it is
| f*******d
发帖数: 339 | 4
I think you are wrong. First of all, I know that there are systems with unbounded
energy levels which can have population inversion, the key point there is that
one or more of the higher energy levels receive a large inbound transition
rate. For example, this is what happens in laser. Also, in reality, all systems
have unbounded states.
Second, you did not answer my question. Even for bounded systems, there is no
reason that adding more energy would lead to a Bolzmann distribution with
T<0, unl
【在 c****u 的大作中提到】
:
: 1/T=dS/dE. that is the definition of the temperature. When the system is the
: energy bounded system which means the system's energy can't goes to infinite,
: such as magnet system, when you keep heating the system, you increase the
: entropy by increasing the energy, however when you increase the temperature to
: infinite,the system can't gain more entropy, since all particles tends to go to
: the highest state. At that time temperature will goes to negative and goes to
: negative zero, when all particles
| s*a
发帖数: 33 | 5
Second, you did not answer my question. Even for bounded systems, there is no
reason that adding more energy would lead to a Bolzmann distribution with
T<0, unless there is some physical principle requires this.
As for the suggestion of treating level pairs as subsystems, it is possible, but
I can hardly see the advantage of introducing a negative T in that case.
=========================
That is the problem of your question. You introduced an artificial system
which is not thermally relaxed an
【在 f*******d 的大作中提到】
:
: I think you are wrong. First of all, I know that there are systems with unbounded
: energy levels which can have population inversion, the key point there is that
: one or more of the higher energy levels receive a large inbound transition
: rate. For example, this is what happens in laser. Also, in reality, all systems
: have unbounded states.
: Second, you did not answer my question. Even for bounded systems, there is no
: reason that adding more energy would lead to a Bolzmann distribution with
: T<0, unl
| c****u
发帖数: 584 | 6
Can you show me some examples? As I know most laser systems are 4 level or
three level systems.
1/T=dS/dE, U add energy however entropy decrease, so temperature should be
negative. That is thermodynamics result. We got that not just for convient.
【在 f*******d 的大作中提到】
:
: I think you are wrong. First of all, I know that there are systems with unbounded
: energy levels which can have population inversion, the key point there is that
: one or more of the higher energy levels receive a large inbound transition
: rate. For example, this is what happens in laser. Also, in reality, all systems
: have unbounded states.
: Second, you did not answer my question. Even for bounded systems, there is no
: reason that adding more energy would lead to a Bolzmann distribution with
: T<0, unl
| f*******d
发帖数: 339 | 7
That is when their higher energy states are not excited. If you just add energy,
but does not specify how this energy is added, then they could actually excite
higher states and make the system unbound. In laser, it is because this energy
is added as quanta, so that they can not be used for transition to higher states.
That is what make me puzzle. For this formula to work, you have to assume there
is already a temperature well defined for your system. Also, you have to specify
the condition or
【在 c****u 的大作中提到】
:
: Can you show me some examples? As I know most laser systems are 4 level or
: three level systems.
: 1/T=dS/dE, U add energy however entropy decrease, so temperature should be
: negative. That is thermodynamics result. We got that not just for convient.
| f*******d
发帖数: 339 | 8
but the question is, for thermally relaxed system, can there be negative
temperature?
【在 s*a 的大作中提到】
:
: Second, you did not answer my question. Even for bounded systems, there is no
: reason that adding more energy would lead to a Bolzmann distribution with
: T<0, unless there is some physical principle requires this.
: As for the suggestion of treating level pairs as subsystems, it is possible, but
: I can hardly see the advantage of introducing a negative T in that case.
: =========================
: That is the problem of your question. You introduced an artificial system
: which is not thermally relaxed an
| c****u
发帖数: 584 | 9
OK that is the same thing as bounded. So we use 4 or 3 levels model to
study laser.
Of course there is a certain temperature if the system is in thermo equilibrium
That is not assumption. It is true there are many ways to do in experiment,
in NMR they inverse the magnet field. In laser they pump the media or using
Q switch to build up the inverse...I don't care. As long as you prepare a
negative
temperature system, that system should obey the thermodynamic laws. And
1/T=dS/dE is the definiti
【在 f*******d 的大作中提到】
:
: but the question is, for thermally relaxed system, can there be negative
: temperature?
| s*a
发帖数: 33 | 10 I think it is possible, one trivial example is the two level system in
which the sponetaneous transition between the two levels is forbidden.
For a multi level system, I guess there must not be any sponetaneous
transtions between any levels involved, which means, the system can't
exchange energies or reach thermal equilibrium by itself, some outside/other
mechanism is needed to adjust the population of each level (just like
in the 2 level system). If the system is designed well, it is quite
poss
【在 f*******d 的大作中提到】
:
: but the question is, for thermally relaxed system, can there be negative
: temperature?
| l**n
发帖数: 67 | 11
I cannot remember clearly, but i think if the relaxation time is long
enough, then you can also define the temperature. And most negative
temperature states belong to this case.
【在 f*******d 的大作中提到】
:
: but the question is, for thermally relaxed system, can there be negative
: temperature?
| G********t
发帖数: 334 | 12
If a system if in thermal equilibrium,
negative temperature is not possible.
But any laser system is a example of negative
temperature system.
【在 l**n 的大作中提到】
:
: I cannot remember clearly, but i think if the relaxation time is long
: enough, then you can also define the temperature. And most negative
: temperature states belong to this case.
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