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Classical theory fails to explain quantization of motions?
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I understand everything written here.
But the last point, I cannot get, at all.
How does it point towards Quantization of the two motions, since the energy change is not sudden, but gradual?
And if anything is wrong with the given image, please tell what it is.
quantum-mechanics thermodynamics molecules degrees-of-freedom
asked Aug 10 at 13:23
Aditya Agarwal
839
add a comment |Â
up vote
7
down vote
favorite
I understand everything written here.
But the last point, I cannot get, at all.
How does it point towards Quantization of the two motions, since the energy change is not sudden, but gradual?
And if anything is wrong with the given image, please tell what it is.
quantum-mechanics thermodynamics molecules degrees-of-freedom
asked Aug 10 at 13:23
Aditya Agarwal
839
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I understand everything written here.
But the last point, I cannot get, at all.
How does it point towards Quantization of the two motions, since the energy change is not sudden, but gradual?
And if anything is wrong with the given image, please tell what it is.
quantum-mechanics thermodynamics molecules degrees-of-freedom
asked Aug 10 at 13:23
Aditya Agarwal
839
I understand everything written here.
But the last point, I cannot get, at all.
How does it point towards Quantization of the two motions, since the energy change is not sudden, but gradual?
And if anything is wrong with the given image, please tell what it is.
quantum-mechanics thermodynamics molecules degrees-of-freedom
asked Aug 10 at 13:23
Aditya Agarwal
839
edited Aug 10 at 13:55
asked Aug 10 at 13:23
Aditya Agarwal
839
asked Aug 10 at 13:23
Aditya Agarwal
839
asked Aug 10 at 13:23
Aditya Agarwal
839
839
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
The "wrong" thing in this picture is an illusion of "horizonality" of some parts of this curve. According to the Maxwell distribution $propto textexp(-mv^2/2kT)$, in the volume there are always high velocity molecules capable to get rotational and vibrational excitations, thus the curve has always a slope as a function of $T$.
For one molecule you have clear thresholds (a step-wise curve), but for a volume of molecules the thresholds are smeared due to statistics applied to calculate/measure the heat capacity of the volume.
Still, one can see a "thershold-like" behavior of the gas heat capacity indicating quantization of rotational and vibrational energies. Without quantization the curve would not be a step-wise at all.
By the way, with $T$ growing, excitations of electron levels come into play. Finally one can finish with fully ionized plasma ;-).
answered Aug 10 at 14:14
Vladimir Kalitvianski
10.2k11232
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up vote
5
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Because the degrees of freedom look like they're 'frozen' at low T. Statistically, we know there's going to be an average energy of $RT/2$ per mole for each degree of freedom. If you have translation, rotation and vibration that makes for a total of 7 degrees of freedom. By classical mechanics, there should be no lower limit to how much energy goes into them, so it should be $C_v = 7R/2$ from the very beginning. Instead, because of QM, there's an energy gap between ground and first excited state for each of these motions, and that means they don't contribute up to the point when $kT sim hbar omega$ for each of them. That's what causes the 'steps' to appear in the heat capacity, and the very existence of those steps is only possible because of quantum effects. That the steps are smoothed out is merely a statistical effect due to the fact that not all modes will activate instantly across the gas.
answered Aug 10 at 13:55
Okarin
31318
Average energy of $RmathbbT/2$ for a mole?
– Aditya Agarwal
Aug 10 at 13:58
Yes, sorry. Fixed that, now it's clearer.
– Okarin
Aug 10 at 14:01
The molar heat capacity should be $C_V=7R/2$
– probably_someone
Aug 10 at 14:05
Yes, sorry, I changed everything in a sweep again and made another stupid mistake. Fixed.
– Okarin
Aug 10 at 14:07
1
Because for classical mechanics, energy is a continuous quantity. That energy comes in discrete packets was one of the key discoveries that led to developing quantum mechanics. You can have any tiny fraction of energy 'stored' in a specific degree of freedom. In quantum mechanics instead you need to have at least a certain minimum amount for each mode.
– Okarin
Aug 10 at 14:14
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
The "wrong" thing in this picture is an illusion of "horizonality" of some parts of this curve. According to the Maxwell distribution $propto textexp(-mv^2/2kT)$, in the volume there are always high velocity molecules capable to get rotational and vibrational excitations, thus the curve has always a slope as a function of $T$.
For one molecule you have clear thresholds (a step-wise curve), but for a volume of molecules the thresholds are smeared due to statistics applied to calculate/measure the heat capacity of the volume.
Still, one can see a "thershold-like" behavior of the gas heat capacity indicating quantization of rotational and vibrational energies. Without quantization the curve would not be a step-wise at all.
By the way, with $T$ growing, excitations of electron levels come into play. Finally one can finish with fully ionized plasma ;-).
answered Aug 10 at 14:14
Vladimir Kalitvianski
10.2k11232
add a comment |Â
up vote
5
down vote
accepted
The "wrong" thing in this picture is an illusion of "horizonality" of some parts of this curve. According to the Maxwell distribution $propto textexp(-mv^2/2kT)$, in the volume there are always high velocity molecules capable to get rotational and vibrational excitations, thus the curve has always a slope as a function of $T$.
For one molecule you have clear thresholds (a step-wise curve), but for a volume of molecules the thresholds are smeared due to statistics applied to calculate/measure the heat capacity of the volume.
Still, one can see a "thershold-like" behavior of the gas heat capacity indicating quantization of rotational and vibrational energies. Without quantization the curve would not be a step-wise at all.
By the way, with $T$ growing, excitations of electron levels come into play. Finally one can finish with fully ionized plasma ;-).
answered Aug 10 at 14:14
Vladimir Kalitvianski
10.2k11232
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
The "wrong" thing in this picture is an illusion of "horizonality" of some parts of this curve. According to the Maxwell distribution $propto textexp(-mv^2/2kT)$, in the volume there are always high velocity molecules capable to get rotational and vibrational excitations, thus the curve has always a slope as a function of $T$.
For one molecule you have clear thresholds (a step-wise curve), but for a volume of molecules the thresholds are smeared due to statistics applied to calculate/measure the heat capacity of the volume.
Still, one can see a "thershold-like" behavior of the gas heat capacity indicating quantization of rotational and vibrational energies. Without quantization the curve would not be a step-wise at all.
By the way, with $T$ growing, excitations of electron levels come into play. Finally one can finish with fully ionized plasma ;-).
answered Aug 10 at 14:14
Vladimir Kalitvianski
10.2k11232
The "wrong" thing in this picture is an illusion of "horizonality" of some parts of this curve. According to the Maxwell distribution $propto textexp(-mv^2/2kT)$, in the volume there are always high velocity molecules capable to get rotational and vibrational excitations, thus the curve has always a slope as a function of $T$.
For one molecule you have clear thresholds (a step-wise curve), but for a volume of molecules the thresholds are smeared due to statistics applied to calculate/measure the heat capacity of the volume.
Still, one can see a "thershold-like" behavior of the gas heat capacity indicating quantization of rotational and vibrational energies. Without quantization the curve would not be a step-wise at all.
By the way, with $T$ growing, excitations of electron levels come into play. Finally one can finish with fully ionized plasma ;-).
answered Aug 10 at 14:14
Vladimir Kalitvianski
10.2k11232
edited Aug 10 at 14:21
answered Aug 10 at 14:14
Vladimir Kalitvianski
10.2k11232
answered Aug 10 at 14:14
Vladimir Kalitvianski
10.2k11232
answered Aug 10 at 14:14
Vladimir Kalitvianski
10.2k11232
10.2k11232
add a comment |Â
add a comment |Â
up vote
5
down vote
Because the degrees of freedom look like they're 'frozen' at low T. Statistically, we know there's going to be an average energy of $RT/2$ per mole for each degree of freedom. If you have translation, rotation and vibration that makes for a total of 7 degrees of freedom. By classical mechanics, there should be no lower limit to how much energy goes into them, so it should be $C_v = 7R/2$ from the very beginning. Instead, because of QM, there's an energy gap between ground and first excited state for each of these motions, and that means they don't contribute up to the point when $kT sim hbar omega$ for each of them. That's what causes the 'steps' to appear in the heat capacity, and the very existence of those steps is only possible because of quantum effects. That the steps are smoothed out is merely a statistical effect due to the fact that not all modes will activate instantly across the gas.
answered Aug 10 at 13:55
Okarin
31318
Average energy of $RmathbbT/2$ for a mole?
– Aditya Agarwal
Aug 10 at 13:58
Yes, sorry. Fixed that, now it's clearer.
– Okarin
Aug 10 at 14:01
The molar heat capacity should be $C_V=7R/2$
– probably_someone
Aug 10 at 14:05
Yes, sorry, I changed everything in a sweep again and made another stupid mistake. Fixed.
– Okarin
Aug 10 at 14:07
1
Because for classical mechanics, energy is a continuous quantity. That energy comes in discrete packets was one of the key discoveries that led to developing quantum mechanics. You can have any tiny fraction of energy 'stored' in a specific degree of freedom. In quantum mechanics instead you need to have at least a certain minimum amount for each mode.
– Okarin
Aug 10 at 14:14
 |Â
show 2 more comments
up vote
5
down vote
Because the degrees of freedom look like they're 'frozen' at low T. Statistically, we know there's going to be an average energy of $RT/2$ per mole for each degree of freedom. If you have translation, rotation and vibration that makes for a total of 7 degrees of freedom. By classical mechanics, there should be no lower limit to how much energy goes into them, so it should be $C_v = 7R/2$ from the very beginning. Instead, because of QM, there's an energy gap between ground and first excited state for each of these motions, and that means they don't contribute up to the point when $kT sim hbar omega$ for each of them. That's what causes the 'steps' to appear in the heat capacity, and the very existence of those steps is only possible because of quantum effects. That the steps are smoothed out is merely a statistical effect due to the fact that not all modes will activate instantly across the gas.
answered Aug 10 at 13:55
Okarin
31318
Average energy of $RmathbbT/2$ for a mole?
– Aditya Agarwal
Aug 10 at 13:58
Yes, sorry. Fixed that, now it's clearer.
– Okarin
Aug 10 at 14:01
The molar heat capacity should be $C_V=7R/2$
– probably_someone
Aug 10 at 14:05
Yes, sorry, I changed everything in a sweep again and made another stupid mistake. Fixed.
– Okarin
Aug 10 at 14:07
1
Because for classical mechanics, energy is a continuous quantity. That energy comes in discrete packets was one of the key discoveries that led to developing quantum mechanics. You can have any tiny fraction of energy 'stored' in a specific degree of freedom. In quantum mechanics instead you need to have at least a certain minimum amount for each mode.
– Okarin
Aug 10 at 14:14
 |Â
show 2 more comments
up vote
5
down vote
up vote
5
down vote
Because the degrees of freedom look like they're 'frozen' at low T. Statistically, we know there's going to be an average energy of $RT/2$ per mole for each degree of freedom. If you have translation, rotation and vibration that makes for a total of 7 degrees of freedom. By classical mechanics, there should be no lower limit to how much energy goes into them, so it should be $C_v = 7R/2$ from the very beginning. Instead, because of QM, there's an energy gap between ground and first excited state for each of these motions, and that means they don't contribute up to the point when $kT sim hbar omega$ for each of them. That's what causes the 'steps' to appear in the heat capacity, and the very existence of those steps is only possible because of quantum effects. That the steps are smoothed out is merely a statistical effect due to the fact that not all modes will activate instantly across the gas.
answered Aug 10 at 13:55
Okarin
31318
Because the degrees of freedom look like they're 'frozen' at low T. Statistically, we know there's going to be an average energy of $RT/2$ per mole for each degree of freedom. If you have translation, rotation and vibration that makes for a total of 7 degrees of freedom. By classical mechanics, there should be no lower limit to how much energy goes into them, so it should be $C_v = 7R/2$ from the very beginning. Instead, because of QM, there's an energy gap between ground and first excited state for each of these motions, and that means they don't contribute up to the point when $kT sim hbar omega$ for each of them. That's what causes the 'steps' to appear in the heat capacity, and the very existence of those steps is only possible because of quantum effects. That the steps are smoothed out is merely a statistical effect due to the fact that not all modes will activate instantly across the gas.
answered Aug 10 at 13:55
Okarin
31318
edited Aug 10 at 14:07
answered Aug 10 at 13:55
Okarin
31318
answered Aug 10 at 13:55
Okarin
31318
answered Aug 10 at 13:55
Okarin
31318
31318
Average energy of $RmathbbT/2$ for a mole?
– Aditya Agarwal
Aug 10 at 13:58
Yes, sorry. Fixed that, now it's clearer.
– Okarin
Aug 10 at 14:01
The molar heat capacity should be $C_V=7R/2$
– probably_someone
Aug 10 at 14:05
Yes, sorry, I changed everything in a sweep again and made another stupid mistake. Fixed.
– Okarin
Aug 10 at 14:07
1
Because for classical mechanics, energy is a continuous quantity. That energy comes in discrete packets was one of the key discoveries that led to developing quantum mechanics. You can have any tiny fraction of energy 'stored' in a specific degree of freedom. In quantum mechanics instead you need to have at least a certain minimum amount for each mode.
– Okarin
Aug 10 at 14:14
 |Â
show 2 more comments
Average energy of $RmathbbT/2$ for a mole?
– Aditya Agarwal
Aug 10 at 13:58
Yes, sorry. Fixed that, now it's clearer.
– Okarin
Aug 10 at 14:01
The molar heat capacity should be $C_V=7R/2$
– probably_someone
Aug 10 at 14:05
Yes, sorry, I changed everything in a sweep again and made another stupid mistake. Fixed.
– Okarin
Aug 10 at 14:07
1
Because for classical mechanics, energy is a continuous quantity. That energy comes in discrete packets was one of the key discoveries that led to developing quantum mechanics. You can have any tiny fraction of energy 'stored' in a specific degree of freedom. In quantum mechanics instead you need to have at least a certain minimum amount for each mode.
– Okarin
Aug 10 at 14:14
Average energy of $RmathbbT/2$ for a mole?
– Aditya Agarwal
Aug 10 at 13:58
Average energy of $RmathbbT/2$ for a mole?
– Aditya Agarwal
Aug 10 at 13:58
Yes, sorry. Fixed that, now it's clearer.
– Okarin
Aug 10 at 14:01
Yes, sorry. Fixed that, now it's clearer.
– Okarin
Aug 10 at 14:01
The molar heat capacity should be $C_V=7R/2$
– probably_someone
Aug 10 at 14:05
The molar heat capacity should be $C_V=7R/2$
– probably_someone
Aug 10 at 14:05
Yes, sorry, I changed everything in a sweep again and made another stupid mistake. Fixed.
– Okarin
Aug 10 at 14:07
Yes, sorry, I changed everything in a sweep again and made another stupid mistake. Fixed.
– Okarin
Aug 10 at 14:07
1
1
Because for classical mechanics, energy is a continuous quantity. That energy comes in discrete packets was one of the key discoveries that led to developing quantum mechanics. You can have any tiny fraction of energy 'stored' in a specific degree of freedom. In quantum mechanics instead you need to have at least a certain minimum amount for each mode.
– Okarin
Aug 10 at 14:14
Because for classical mechanics, energy is a continuous quantity. That energy comes in discrete packets was one of the key discoveries that led to developing quantum mechanics. You can have any tiny fraction of energy 'stored' in a specific degree of freedom. In quantum mechanics instead you need to have at least a certain minimum amount for each mode.
– Okarin
Aug 10 at 14:14
 |Â
show 2 more comments
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