How do we tell which part of kinetic energy gives rise to temperature?Is zero-point energy real?When is temperature not a measure of the average kinetic energy of the particles in a substance?Is kinetic theory part of statistical mechanics?How does temperature relate to the kinetic energy of molecules?How to explain the Venturi effect with Kinetic Theory?What is the relation of particle velocity, temperature, and reaction activation energyIs mean kinetic energy related to temperature of a system of interacting classical particles?Temperature, Kinetic Energy, and Chemical PotentialTemperature as the integral of kinetic energy?Microscopic level - What happen to the particles in a heat transfer through conduction?
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How do we tell which part of kinetic energy gives rise to temperature?
Is zero-point energy real?When is temperature not a measure of the average kinetic energy of the particles in a substance?Is kinetic theory part of statistical mechanics?How does temperature relate to the kinetic energy of molecules?How to explain the Venturi effect with Kinetic Theory?What is the relation of particle velocity, temperature, and reaction activation energyIs mean kinetic energy related to temperature of a system of interacting classical particles?Temperature, Kinetic Energy, and Chemical PotentialTemperature as the integral of kinetic energy?Microscopic level - What happen to the particles in a heat transfer through conduction?
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$begingroup$
I know that macroscopic temperature is a measure of kinetic energy of particles at very low scales (let's call it microscopic kinetic energy).
But how can we derive which part of this microscopic kinetic energy gives rise to temperature, and which part instead gives rise to macroscopic kinetic energy?
thermodynamics energy statistical-mechanics temperature kinetic-theory
edited 1 hour ago
David Z♦
64.7k23 gold badges141 silver badges258 bronze badges
asked 11 hours ago
Federico TosoFederico Toso
544 bronze badges
$endgroup$
add a comment |
$begingroup$
I know that macroscopic temperature is a measure of kinetic energy of particles at very low scales (let's call it microscopic kinetic energy).
But how can we derive which part of this microscopic kinetic energy gives rise to temperature, and which part instead gives rise to macroscopic kinetic energy?
thermodynamics energy statistical-mechanics temperature kinetic-theory
edited 1 hour ago
David Z♦
64.7k23 gold badges141 silver badges258 bronze badges
asked 11 hours ago
Federico TosoFederico Toso
544 bronze badges
$endgroup$
$begingroup$
No, temperature is not proportional to kinetic energy, not when quantum effects play a role. See for example physics.stackexchange.com/questions/413376/…
$endgroup$
– Pieter
3 hours ago
add a comment |
$begingroup$
I know that macroscopic temperature is a measure of kinetic energy of particles at very low scales (let's call it microscopic kinetic energy).
But how can we derive which part of this microscopic kinetic energy gives rise to temperature, and which part instead gives rise to macroscopic kinetic energy?
thermodynamics energy statistical-mechanics temperature kinetic-theory
edited 1 hour ago
David Z♦
64.7k23 gold badges141 silver badges258 bronze badges
asked 11 hours ago
Federico TosoFederico Toso
544 bronze badges
$endgroup$
I know that macroscopic temperature is a measure of kinetic energy of particles at very low scales (let's call it microscopic kinetic energy).
But how can we derive which part of this microscopic kinetic energy gives rise to temperature, and which part instead gives rise to macroscopic kinetic energy?
thermodynamics energy statistical-mechanics temperature kinetic-theory
thermodynamics energy statistical-mechanics temperature kinetic-theory
edited 1 hour ago
David Z♦
64.7k23 gold badges141 silver badges258 bronze badges
asked 11 hours ago
Federico TosoFederico Toso
544 bronze badges
edited 1 hour ago
David Z♦
64.7k23 gold badges141 silver badges258 bronze badges
asked 11 hours ago
Federico TosoFederico Toso
544 bronze badges
edited 1 hour ago
David Z♦
64.7k23 gold badges141 silver badges258 bronze badges
edited 1 hour ago
David Z♦
64.7k23 gold badges141 silver badges258 bronze badges
edited 1 hour ago
David Z♦
64.7k23 gold badges141 silver badges258 bronze badges
64.7k23 gold badges141 silver badges258 bronze badges
asked 11 hours ago
Federico TosoFederico Toso
544 bronze badges
asked 11 hours ago
Federico TosoFederico Toso
544 bronze badges
asked 11 hours ago
Federico TosoFederico Toso
544 bronze badges
544 bronze badges
$begingroup$
No, temperature is not proportional to kinetic energy, not when quantum effects play a role. See for example physics.stackexchange.com/questions/413376/…
$endgroup$
– Pieter
3 hours ago
add a comment |
$begingroup$
No, temperature is not proportional to kinetic energy, not when quantum effects play a role. See for example physics.stackexchange.com/questions/413376/…
$endgroup$
– Pieter
3 hours ago
$begingroup$
No, temperature is not proportional to kinetic energy, not when quantum effects play a role. See for example physics.stackexchange.com/questions/413376/…
$endgroup$
– Pieter
3 hours ago
$begingroup$
No, temperature is not proportional to kinetic energy, not when quantum effects play a role. See for example physics.stackexchange.com/questions/413376/…
$endgroup$
– Pieter
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The macroscopic kinetic energy of a system of particles is the kinetic energy due to the velocity of the center of mass of the collection of particles with respect to an external frame of reference.
For example suppose you have a container filled with an ideal gas. The temperature of the gas is a measure of the average kinetic energy of the randomly moving gas particles. That is its internal kinetic energy.
Let the container be moving at constant velocity with respect to an external frame of reference (e.g., the room where the container is located). The external (macroscopic) kinetic energy of the gas in the container is $fracmv^22$ where $m$ is the mass of the gas and $v$ is the velocity of the container with respect to the room. This kinetic energy is independent of the internal kinetic energy, not a part of it.
The total kinetic energy of the gas is the sum of its internal and external kinetic energies.
Hope this helps.
answered 10 hours ago
Bob DBob D
12.1k3 gold badges10 silver badges36 bronze badges
$endgroup$
2
$begingroup$
@Aaron Stevens: It's the defintion of the center of mass. If $bf v_i$ is velocity in center of mass frame, then $sum m_i bf v_i=0$.
$endgroup$
– mike stone
10 hours ago
$begingroup$
@mikestone Ah yes, of course :)
$endgroup$
– Aaron Stevens
10 hours ago
$begingroup$
Rather than "external" and "internal" would it be better to say "center of mass" and "thermodynamic"? It sounds especially weird to talk about the "external" KE of something.
$endgroup$
– abalter
2 hours ago
$begingroup$
@abalter I believe it is customary in thermodynamics to talk about the total energy of a system as the sum of its internal and external kinetic and potential energy. Taking kinetic energy, the internal kinetic energy of the gas (system) is based on the velocities of the particles with respect to an internal frame of reference (in this example, the container). The kinetic energy of the center of mass of the gas is based on the velocity of the COM with respect to an external frame of reference (the room where the gas is located). The term "external" helps to differentiate the two KE components
$endgroup$
– Bob D
2 hours ago
$begingroup$
@abalter The same goes for potential energy. The internal potential energy is based on the intermolecular forces between the molecules, usually ignored for an ideal gas. The external potential energy is based on the position of the COM in the gravitational with respect to an external frame of reference (e.g., the floor of the room).
$endgroup$
– Bob D
2 hours ago
|
show 1 more comment
$begingroup$
I know that macroscopic temperature is a measure of kinetic energy of particales at very low scales (let's call it microscopic kinetic energy).
This is not generally true. The only case where this is true is for an ideal monoatomic gas. For all other materials there are more internal degrees of freedom than merely the kinetic energy.
For the remainder of your question, to determine which portion of the total energy is due to which parts, you have to distinguish between internal and external degrees of freedom. Then the thermal energy is the portion of the total energy contained in all internal degrees of freedom and the kinetic energy is the portion contained in the external rotation and translation degrees of freedom.
answered 10 hours ago
DaleDale
9,0792 gold badges12 silver badges40 bronze badges
$endgroup$
1
$begingroup$
Equipartition is valid for any classical system.
$endgroup$
– Pieter
3 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The macroscopic kinetic energy of a system of particles is the kinetic energy due to the velocity of the center of mass of the collection of particles with respect to an external frame of reference.
For example suppose you have a container filled with an ideal gas. The temperature of the gas is a measure of the average kinetic energy of the randomly moving gas particles. That is its internal kinetic energy.
Let the container be moving at constant velocity with respect to an external frame of reference (e.g., the room where the container is located). The external (macroscopic) kinetic energy of the gas in the container is $fracmv^22$ where $m$ is the mass of the gas and $v$ is the velocity of the container with respect to the room. This kinetic energy is independent of the internal kinetic energy, not a part of it.
The total kinetic energy of the gas is the sum of its internal and external kinetic energies.
Hope this helps.
answered 10 hours ago
Bob DBob D
12.1k3 gold badges10 silver badges36 bronze badges
$endgroup$
2
$begingroup$
@Aaron Stevens: It's the defintion of the center of mass. If $bf v_i$ is velocity in center of mass frame, then $sum m_i bf v_i=0$.
$endgroup$
– mike stone
10 hours ago
$begingroup$
@mikestone Ah yes, of course :)
$endgroup$
– Aaron Stevens
10 hours ago
$begingroup$
Rather than "external" and "internal" would it be better to say "center of mass" and "thermodynamic"? It sounds especially weird to talk about the "external" KE of something.
$endgroup$
– abalter
2 hours ago
$begingroup$
@abalter I believe it is customary in thermodynamics to talk about the total energy of a system as the sum of its internal and external kinetic and potential energy. Taking kinetic energy, the internal kinetic energy of the gas (system) is based on the velocities of the particles with respect to an internal frame of reference (in this example, the container). The kinetic energy of the center of mass of the gas is based on the velocity of the COM with respect to an external frame of reference (the room where the gas is located). The term "external" helps to differentiate the two KE components
$endgroup$
– Bob D
2 hours ago
$begingroup$
@abalter The same goes for potential energy. The internal potential energy is based on the intermolecular forces between the molecules, usually ignored for an ideal gas. The external potential energy is based on the position of the COM in the gravitational with respect to an external frame of reference (e.g., the floor of the room).
$endgroup$
– Bob D
2 hours ago
|
show 1 more comment
$begingroup$
The macroscopic kinetic energy of a system of particles is the kinetic energy due to the velocity of the center of mass of the collection of particles with respect to an external frame of reference.
For example suppose you have a container filled with an ideal gas. The temperature of the gas is a measure of the average kinetic energy of the randomly moving gas particles. That is its internal kinetic energy.
Let the container be moving at constant velocity with respect to an external frame of reference (e.g., the room where the container is located). The external (macroscopic) kinetic energy of the gas in the container is $fracmv^22$ where $m$ is the mass of the gas and $v$ is the velocity of the container with respect to the room. This kinetic energy is independent of the internal kinetic energy, not a part of it.
The total kinetic energy of the gas is the sum of its internal and external kinetic energies.
Hope this helps.
answered 10 hours ago
Bob DBob D
12.1k3 gold badges10 silver badges36 bronze badges
$endgroup$
2
$begingroup$
@Aaron Stevens: It's the defintion of the center of mass. If $bf v_i$ is velocity in center of mass frame, then $sum m_i bf v_i=0$.
$endgroup$
– mike stone
10 hours ago
$begingroup$
@mikestone Ah yes, of course :)
$endgroup$
– Aaron Stevens
10 hours ago
$begingroup$
Rather than "external" and "internal" would it be better to say "center of mass" and "thermodynamic"? It sounds especially weird to talk about the "external" KE of something.
$endgroup$
– abalter
2 hours ago
$begingroup$
@abalter I believe it is customary in thermodynamics to talk about the total energy of a system as the sum of its internal and external kinetic and potential energy. Taking kinetic energy, the internal kinetic energy of the gas (system) is based on the velocities of the particles with respect to an internal frame of reference (in this example, the container). The kinetic energy of the center of mass of the gas is based on the velocity of the COM with respect to an external frame of reference (the room where the gas is located). The term "external" helps to differentiate the two KE components
$endgroup$
– Bob D
2 hours ago
$begingroup$
@abalter The same goes for potential energy. The internal potential energy is based on the intermolecular forces between the molecules, usually ignored for an ideal gas. The external potential energy is based on the position of the COM in the gravitational with respect to an external frame of reference (e.g., the floor of the room).
$endgroup$
– Bob D
2 hours ago
|
show 1 more comment
$begingroup$
The macroscopic kinetic energy of a system of particles is the kinetic energy due to the velocity of the center of mass of the collection of particles with respect to an external frame of reference.
For example suppose you have a container filled with an ideal gas. The temperature of the gas is a measure of the average kinetic energy of the randomly moving gas particles. That is its internal kinetic energy.
Let the container be moving at constant velocity with respect to an external frame of reference (e.g., the room where the container is located). The external (macroscopic) kinetic energy of the gas in the container is $fracmv^22$ where $m$ is the mass of the gas and $v$ is the velocity of the container with respect to the room. This kinetic energy is independent of the internal kinetic energy, not a part of it.
The total kinetic energy of the gas is the sum of its internal and external kinetic energies.
Hope this helps.
answered 10 hours ago
Bob DBob D
12.1k3 gold badges10 silver badges36 bronze badges
$endgroup$
The macroscopic kinetic energy of a system of particles is the kinetic energy due to the velocity of the center of mass of the collection of particles with respect to an external frame of reference.
For example suppose you have a container filled with an ideal gas. The temperature of the gas is a measure of the average kinetic energy of the randomly moving gas particles. That is its internal kinetic energy.
Let the container be moving at constant velocity with respect to an external frame of reference (e.g., the room where the container is located). The external (macroscopic) kinetic energy of the gas in the container is $fracmv^22$ where $m$ is the mass of the gas and $v$ is the velocity of the container with respect to the room. This kinetic energy is independent of the internal kinetic energy, not a part of it.
The total kinetic energy of the gas is the sum of its internal and external kinetic energies.
Hope this helps.
answered 10 hours ago
Bob DBob D
12.1k3 gold badges10 silver badges36 bronze badges
edited 7 hours ago
answered 10 hours ago
Bob DBob D
12.1k3 gold badges10 silver badges36 bronze badges
answered 10 hours ago
Bob DBob D
12.1k3 gold badges10 silver badges36 bronze badges
answered 10 hours ago
Bob DBob D
12.1k3 gold badges10 silver badges36 bronze badges
12.1k3 gold badges10 silver badges36 bronze badges
2
$begingroup$
@Aaron Stevens: It's the defintion of the center of mass. If $bf v_i$ is velocity in center of mass frame, then $sum m_i bf v_i=0$.
$endgroup$
– mike stone
10 hours ago
$begingroup$
@mikestone Ah yes, of course :)
$endgroup$
– Aaron Stevens
10 hours ago
$begingroup$
Rather than "external" and "internal" would it be better to say "center of mass" and "thermodynamic"? It sounds especially weird to talk about the "external" KE of something.
$endgroup$
– abalter
2 hours ago
$begingroup$
@abalter I believe it is customary in thermodynamics to talk about the total energy of a system as the sum of its internal and external kinetic and potential energy. Taking kinetic energy, the internal kinetic energy of the gas (system) is based on the velocities of the particles with respect to an internal frame of reference (in this example, the container). The kinetic energy of the center of mass of the gas is based on the velocity of the COM with respect to an external frame of reference (the room where the gas is located). The term "external" helps to differentiate the two KE components
$endgroup$
– Bob D
2 hours ago
$begingroup$
@abalter The same goes for potential energy. The internal potential energy is based on the intermolecular forces between the molecules, usually ignored for an ideal gas. The external potential energy is based on the position of the COM in the gravitational with respect to an external frame of reference (e.g., the floor of the room).
$endgroup$
– Bob D
2 hours ago
|
show 1 more comment
2
$begingroup$
@Aaron Stevens: It's the defintion of the center of mass. If $bf v_i$ is velocity in center of mass frame, then $sum m_i bf v_i=0$.
$endgroup$
– mike stone
10 hours ago
$begingroup$
@mikestone Ah yes, of course :)
$endgroup$
– Aaron Stevens
10 hours ago
$begingroup$
Rather than "external" and "internal" would it be better to say "center of mass" and "thermodynamic"? It sounds especially weird to talk about the "external" KE of something.
$endgroup$
– abalter
2 hours ago
$begingroup$
@abalter I believe it is customary in thermodynamics to talk about the total energy of a system as the sum of its internal and external kinetic and potential energy. Taking kinetic energy, the internal kinetic energy of the gas (system) is based on the velocities of the particles with respect to an internal frame of reference (in this example, the container). The kinetic energy of the center of mass of the gas is based on the velocity of the COM with respect to an external frame of reference (the room where the gas is located). The term "external" helps to differentiate the two KE components
$endgroup$
– Bob D
2 hours ago
$begingroup$
@abalter The same goes for potential energy. The internal potential energy is based on the intermolecular forces between the molecules, usually ignored for an ideal gas. The external potential energy is based on the position of the COM in the gravitational with respect to an external frame of reference (e.g., the floor of the room).
$endgroup$
– Bob D
2 hours ago
2
2
$begingroup$
@Aaron Stevens: It's the defintion of the center of mass. If $bf v_i$ is velocity in center of mass frame, then $sum m_i bf v_i=0$.
$endgroup$
– mike stone
10 hours ago
$begingroup$
@Aaron Stevens: It's the defintion of the center of mass. If $bf v_i$ is velocity in center of mass frame, then $sum m_i bf v_i=0$.
$endgroup$
– mike stone
10 hours ago
$begingroup$
@mikestone Ah yes, of course :)
$endgroup$
– Aaron Stevens
10 hours ago
$begingroup$
@mikestone Ah yes, of course :)
$endgroup$
– Aaron Stevens
10 hours ago
$begingroup$
Rather than "external" and "internal" would it be better to say "center of mass" and "thermodynamic"? It sounds especially weird to talk about the "external" KE of something.
$endgroup$
– abalter
2 hours ago
$begingroup$
Rather than "external" and "internal" would it be better to say "center of mass" and "thermodynamic"? It sounds especially weird to talk about the "external" KE of something.
$endgroup$
– abalter
2 hours ago
$begingroup$
@abalter I believe it is customary in thermodynamics to talk about the total energy of a system as the sum of its internal and external kinetic and potential energy. Taking kinetic energy, the internal kinetic energy of the gas (system) is based on the velocities of the particles with respect to an internal frame of reference (in this example, the container). The kinetic energy of the center of mass of the gas is based on the velocity of the COM with respect to an external frame of reference (the room where the gas is located). The term "external" helps to differentiate the two KE components
$endgroup$
– Bob D
2 hours ago
$begingroup$
@abalter I believe it is customary in thermodynamics to talk about the total energy of a system as the sum of its internal and external kinetic and potential energy. Taking kinetic energy, the internal kinetic energy of the gas (system) is based on the velocities of the particles with respect to an internal frame of reference (in this example, the container). The kinetic energy of the center of mass of the gas is based on the velocity of the COM with respect to an external frame of reference (the room where the gas is located). The term "external" helps to differentiate the two KE components
$endgroup$
– Bob D
2 hours ago
$begingroup$
@abalter The same goes for potential energy. The internal potential energy is based on the intermolecular forces between the molecules, usually ignored for an ideal gas. The external potential energy is based on the position of the COM in the gravitational with respect to an external frame of reference (e.g., the floor of the room).
$endgroup$
– Bob D
2 hours ago
$begingroup$
@abalter The same goes for potential energy. The internal potential energy is based on the intermolecular forces between the molecules, usually ignored for an ideal gas. The external potential energy is based on the position of the COM in the gravitational with respect to an external frame of reference (e.g., the floor of the room).
$endgroup$
– Bob D
2 hours ago
|
show 1 more comment
$begingroup$
I know that macroscopic temperature is a measure of kinetic energy of particales at very low scales (let's call it microscopic kinetic energy).
This is not generally true. The only case where this is true is for an ideal monoatomic gas. For all other materials there are more internal degrees of freedom than merely the kinetic energy.
For the remainder of your question, to determine which portion of the total energy is due to which parts, you have to distinguish between internal and external degrees of freedom. Then the thermal energy is the portion of the total energy contained in all internal degrees of freedom and the kinetic energy is the portion contained in the external rotation and translation degrees of freedom.
answered 10 hours ago
DaleDale
9,0792 gold badges12 silver badges40 bronze badges
$endgroup$
1
$begingroup$
Equipartition is valid for any classical system.
$endgroup$
– Pieter
3 hours ago
add a comment |
$begingroup$
I know that macroscopic temperature is a measure of kinetic energy of particales at very low scales (let's call it microscopic kinetic energy).
This is not generally true. The only case where this is true is for an ideal monoatomic gas. For all other materials there are more internal degrees of freedom than merely the kinetic energy.
For the remainder of your question, to determine which portion of the total energy is due to which parts, you have to distinguish between internal and external degrees of freedom. Then the thermal energy is the portion of the total energy contained in all internal degrees of freedom and the kinetic energy is the portion contained in the external rotation and translation degrees of freedom.
answered 10 hours ago
DaleDale
9,0792 gold badges12 silver badges40 bronze badges
$endgroup$
1
$begingroup$
Equipartition is valid for any classical system.
$endgroup$
– Pieter
3 hours ago
add a comment |
$begingroup$
I know that macroscopic temperature is a measure of kinetic energy of particales at very low scales (let's call it microscopic kinetic energy).
This is not generally true. The only case where this is true is for an ideal monoatomic gas. For all other materials there are more internal degrees of freedom than merely the kinetic energy.
For the remainder of your question, to determine which portion of the total energy is due to which parts, you have to distinguish between internal and external degrees of freedom. Then the thermal energy is the portion of the total energy contained in all internal degrees of freedom and the kinetic energy is the portion contained in the external rotation and translation degrees of freedom.
answered 10 hours ago
DaleDale
9,0792 gold badges12 silver badges40 bronze badges
$endgroup$
I know that macroscopic temperature is a measure of kinetic energy of particales at very low scales (let's call it microscopic kinetic energy).
This is not generally true. The only case where this is true is for an ideal monoatomic gas. For all other materials there are more internal degrees of freedom than merely the kinetic energy.
For the remainder of your question, to determine which portion of the total energy is due to which parts, you have to distinguish between internal and external degrees of freedom. Then the thermal energy is the portion of the total energy contained in all internal degrees of freedom and the kinetic energy is the portion contained in the external rotation and translation degrees of freedom.
answered 10 hours ago
DaleDale
9,0792 gold badges12 silver badges40 bronze badges
answered 10 hours ago
DaleDale
9,0792 gold badges12 silver badges40 bronze badges
answered 10 hours ago
DaleDale
9,0792 gold badges12 silver badges40 bronze badges
answered 10 hours ago
DaleDale
9,0792 gold badges12 silver badges40 bronze badges
9,0792 gold badges12 silver badges40 bronze badges
1
$begingroup$
Equipartition is valid for any classical system.
$endgroup$
– Pieter
3 hours ago
add a comment |
1
$begingroup$
Equipartition is valid for any classical system.
$endgroup$
– Pieter
3 hours ago
1
1
$begingroup$
Equipartition is valid for any classical system.
$endgroup$
– Pieter
3 hours ago
$begingroup$
Equipartition is valid for any classical system.
$endgroup$
– Pieter
3 hours ago
add a comment |
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$begingroup$
No, temperature is not proportional to kinetic energy, not when quantum effects play a role. See for example physics.stackexchange.com/questions/413376/…
$endgroup$
– Pieter
3 hours ago