Mixing
Sum to an integral in deriving equipartition theorem
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
I'm reading this derivation of the equipartition theorem for ideal gasses:
http://vallance.chem.ox.ac.uk/pdfs/Equipartition.pdf
On the second page, it is mentioned that the partition function as a simple sum,
$$displaystyle Z=sum _ie^-varepsilon _i/kT$$
is not adequate to describe a classical gas as the distribution of energies is not discrete, but continuous. Instead, an integral is needed where
$$e^-varepsilon/kT$$
is integrated over positions and momenta of which energy is a function of. But why is this conversion from a sum to an integral correct?
I can understand why an integral is needed, as the distribution of energies is continuous. But why is it correct to just integrate the exponential function to get the sum? Doesn't the integral give us the area under the curve of the exponential function, that is (speaking perhaps non-rigorously), the sum of the values of the function at different points, multiplied by the differentials $dx$, rather than the simple sum of the values? I've seen this kind of thing done in other places as well, and it bugs me.
statistical-mechanics integration phase-space partition-function
edited 4 mins ago
Qmechanic♦
98.8k121781089
asked 1 hour ago
S. Rotos
699
add a comment |Â
up vote
3
down vote
favorite
I'm reading this derivation of the equipartition theorem for ideal gasses:
http://vallance.chem.ox.ac.uk/pdfs/Equipartition.pdf
On the second page, it is mentioned that the partition function as a simple sum,
$$displaystyle Z=sum _ie^-varepsilon _i/kT$$
is not adequate to describe a classical gas as the distribution of energies is not discrete, but continuous. Instead, an integral is needed where
$$e^-varepsilon/kT$$
is integrated over positions and momenta of which energy is a function of. But why is this conversion from a sum to an integral correct?
I can understand why an integral is needed, as the distribution of energies is continuous. But why is it correct to just integrate the exponential function to get the sum? Doesn't the integral give us the area under the curve of the exponential function, that is (speaking perhaps non-rigorously), the sum of the values of the function at different points, multiplied by the differentials $dx$, rather than the simple sum of the values? I've seen this kind of thing done in other places as well, and it bugs me.
statistical-mechanics integration phase-space partition-function
edited 4 mins ago
Qmechanic♦
98.8k121781089
asked 1 hour ago
S. Rotos
699
An integral is also the limit of an infinite number of infinitesimal terms
– Wolphram jonny
1 hour ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm reading this derivation of the equipartition theorem for ideal gasses:
http://vallance.chem.ox.ac.uk/pdfs/Equipartition.pdf
On the second page, it is mentioned that the partition function as a simple sum,
$$displaystyle Z=sum _ie^-varepsilon _i/kT$$
is not adequate to describe a classical gas as the distribution of energies is not discrete, but continuous. Instead, an integral is needed where
$$e^-varepsilon/kT$$
is integrated over positions and momenta of which energy is a function of. But why is this conversion from a sum to an integral correct?
I can understand why an integral is needed, as the distribution of energies is continuous. But why is it correct to just integrate the exponential function to get the sum? Doesn't the integral give us the area under the curve of the exponential function, that is (speaking perhaps non-rigorously), the sum of the values of the function at different points, multiplied by the differentials $dx$, rather than the simple sum of the values? I've seen this kind of thing done in other places as well, and it bugs me.
statistical-mechanics integration phase-space partition-function
edited 4 mins ago
Qmechanic♦
98.8k121781089
asked 1 hour ago
S. Rotos
699
I'm reading this derivation of the equipartition theorem for ideal gasses:
http://vallance.chem.ox.ac.uk/pdfs/Equipartition.pdf
On the second page, it is mentioned that the partition function as a simple sum,
$$displaystyle Z=sum _ie^-varepsilon _i/kT$$
is not adequate to describe a classical gas as the distribution of energies is not discrete, but continuous. Instead, an integral is needed where
$$e^-varepsilon/kT$$
is integrated over positions and momenta of which energy is a function of. But why is this conversion from a sum to an integral correct?
I can understand why an integral is needed, as the distribution of energies is continuous. But why is it correct to just integrate the exponential function to get the sum? Doesn't the integral give us the area under the curve of the exponential function, that is (speaking perhaps non-rigorously), the sum of the values of the function at different points, multiplied by the differentials $dx$, rather than the simple sum of the values? I've seen this kind of thing done in other places as well, and it bugs me.
statistical-mechanics integration phase-space partition-function
statistical-mechanics integration phase-space partition-function
edited 4 mins ago
Qmechanic♦
98.8k121781089
asked 1 hour ago
S. Rotos
699
edited 4 mins ago
Qmechanic♦
98.8k121781089
asked 1 hour ago
S. Rotos
699
edited 4 mins ago
Qmechanic♦
98.8k121781089
edited 4 mins ago
Qmechanic♦
98.8k121781089
edited 4 mins ago
Qmechanic♦
98.8k121781089
98.8k121781089
asked 1 hour ago
S. Rotos
699
asked 1 hour ago
S. Rotos
699
asked 1 hour ago
S. Rotos
699
699
An integral is also the limit of an infinite number of infinitesimal terms
– Wolphram jonny
1 hour ago
add a comment |Â
An integral is also the limit of an infinite number of infinitesimal terms
– Wolphram jonny
1 hour ago
An integral is also the limit of an infinite number of infinitesimal terms
– Wolphram jonny
1 hour ago
An integral is also the limit of an infinite number of infinitesimal terms
– Wolphram jonny
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
Whenever a sum goes to an integral in the continuous limit, you need to take account of the issue raised by this question. When in doubt, proceed in two steps, as follows.
- First write the sum with an explicit 'delta something' to signify the change in the index being summed over. In the present example:
beginequation
Z = sum_i e^-epsilon_i/kT delta i
endequation
The value of this $delta i$ is 1 in this sum.
- Next, replace $delta i$ by the product of a density of something and a change in that something. In the present example, you should ask, "how many states are there per unit range of $x,y,z$ and $p_x, p_y, p_z$? The answer is one state per volume $h^3$ of phase space. The value $delta i = 1$ represents the count in $Z$ increasing by 1 state, so the relationship is
beginequation
delta i = delta x delta y delta z delta p_x delta p_y delta p_z / h^3
endequation
which gives us
beginequation
Z = sum_i e^-epsilon_i/kT delta x delta y delta z delta p_x delta p_y delta p_z / h^3
endequation
Now the continuous limit can be taken:
beginequation
Z = int e^-epsilon_i/kT dx dy dz, dp_x dp_y dp_z / h^3
endequation
where the integral sign is a shorthand for six integrals in this example.
That's it. To repeat: this issue comes up whenever a sum goes to an integral; it is not a special feature of kinetic theory or statistical mechanics. I guess some textbook writers simply assume it as a known aspect of this area of mathematics.
answered 38 mins ago
Andrew Steane
3396
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Whenever a sum goes to an integral in the continuous limit, you need to take account of the issue raised by this question. When in doubt, proceed in two steps, as follows.
- First write the sum with an explicit 'delta something' to signify the change in the index being summed over. In the present example:
beginequation
Z = sum_i e^-epsilon_i/kT delta i
endequation
The value of this $delta i$ is 1 in this sum.
- Next, replace $delta i$ by the product of a density of something and a change in that something. In the present example, you should ask, "how many states are there per unit range of $x,y,z$ and $p_x, p_y, p_z$? The answer is one state per volume $h^3$ of phase space. The value $delta i = 1$ represents the count in $Z$ increasing by 1 state, so the relationship is
beginequation
delta i = delta x delta y delta z delta p_x delta p_y delta p_z / h^3
endequation
which gives us
beginequation
Z = sum_i e^-epsilon_i/kT delta x delta y delta z delta p_x delta p_y delta p_z / h^3
endequation
Now the continuous limit can be taken:
beginequation
Z = int e^-epsilon_i/kT dx dy dz, dp_x dp_y dp_z / h^3
endequation
where the integral sign is a shorthand for six integrals in this example.
That's it. To repeat: this issue comes up whenever a sum goes to an integral; it is not a special feature of kinetic theory or statistical mechanics. I guess some textbook writers simply assume it as a known aspect of this area of mathematics.
answered 38 mins ago
Andrew Steane
3396
add a comment |Â
up vote
4
down vote
Whenever a sum goes to an integral in the continuous limit, you need to take account of the issue raised by this question. When in doubt, proceed in two steps, as follows.
- First write the sum with an explicit 'delta something' to signify the change in the index being summed over. In the present example:
beginequation
Z = sum_i e^-epsilon_i/kT delta i
endequation
The value of this $delta i$ is 1 in this sum.
- Next, replace $delta i$ by the product of a density of something and a change in that something. In the present example, you should ask, "how many states are there per unit range of $x,y,z$ and $p_x, p_y, p_z$? The answer is one state per volume $h^3$ of phase space. The value $delta i = 1$ represents the count in $Z$ increasing by 1 state, so the relationship is
beginequation
delta i = delta x delta y delta z delta p_x delta p_y delta p_z / h^3
endequation
which gives us
beginequation
Z = sum_i e^-epsilon_i/kT delta x delta y delta z delta p_x delta p_y delta p_z / h^3
endequation
Now the continuous limit can be taken:
beginequation
Z = int e^-epsilon_i/kT dx dy dz, dp_x dp_y dp_z / h^3
endequation
where the integral sign is a shorthand for six integrals in this example.
That's it. To repeat: this issue comes up whenever a sum goes to an integral; it is not a special feature of kinetic theory or statistical mechanics. I guess some textbook writers simply assume it as a known aspect of this area of mathematics.
answered 38 mins ago
Andrew Steane
3396
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Whenever a sum goes to an integral in the continuous limit, you need to take account of the issue raised by this question. When in doubt, proceed in two steps, as follows.
- First write the sum with an explicit 'delta something' to signify the change in the index being summed over. In the present example:
beginequation
Z = sum_i e^-epsilon_i/kT delta i
endequation
The value of this $delta i$ is 1 in this sum.
- Next, replace $delta i$ by the product of a density of something and a change in that something. In the present example, you should ask, "how many states are there per unit range of $x,y,z$ and $p_x, p_y, p_z$? The answer is one state per volume $h^3$ of phase space. The value $delta i = 1$ represents the count in $Z$ increasing by 1 state, so the relationship is
beginequation
delta i = delta x delta y delta z delta p_x delta p_y delta p_z / h^3
endequation
which gives us
beginequation
Z = sum_i e^-epsilon_i/kT delta x delta y delta z delta p_x delta p_y delta p_z / h^3
endequation
Now the continuous limit can be taken:
beginequation
Z = int e^-epsilon_i/kT dx dy dz, dp_x dp_y dp_z / h^3
endequation
where the integral sign is a shorthand for six integrals in this example.
That's it. To repeat: this issue comes up whenever a sum goes to an integral; it is not a special feature of kinetic theory or statistical mechanics. I guess some textbook writers simply assume it as a known aspect of this area of mathematics.
answered 38 mins ago
Andrew Steane
3396
Whenever a sum goes to an integral in the continuous limit, you need to take account of the issue raised by this question. When in doubt, proceed in two steps, as follows.
- First write the sum with an explicit 'delta something' to signify the change in the index being summed over. In the present example:
beginequation
Z = sum_i e^-epsilon_i/kT delta i
endequation
The value of this $delta i$ is 1 in this sum.
- Next, replace $delta i$ by the product of a density of something and a change in that something. In the present example, you should ask, "how many states are there per unit range of $x,y,z$ and $p_x, p_y, p_z$? The answer is one state per volume $h^3$ of phase space. The value $delta i = 1$ represents the count in $Z$ increasing by 1 state, so the relationship is
beginequation
delta i = delta x delta y delta z delta p_x delta p_y delta p_z / h^3
endequation
which gives us
beginequation
Z = sum_i e^-epsilon_i/kT delta x delta y delta z delta p_x delta p_y delta p_z / h^3
endequation
Now the continuous limit can be taken:
beginequation
Z = int e^-epsilon_i/kT dx dy dz, dp_x dp_y dp_z / h^3
endequation
where the integral sign is a shorthand for six integrals in this example.
That's it. To repeat: this issue comes up whenever a sum goes to an integral; it is not a special feature of kinetic theory or statistical mechanics. I guess some textbook writers simply assume it as a known aspect of this area of mathematics.
answered 38 mins ago
Andrew Steane
3396
edited 26 mins ago
answered 38 mins ago
Andrew Steane
3396
answered 38 mins ago
Andrew Steane
3396
answered 38 mins ago
Andrew Steane
3396
3396
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f438889%2fsum-to-an-integral-in-deriving-equipartition-theorem%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
An integral is also the limit of an infinite number of infinitesimal terms
– Wolphram jonny
1 hour ago