Integrating over Dirac delta
$begingroup$
TL;DR: How does one show that $(clubsuit)$ holds.
Some context and how I arrived at my problem. In a Thermodynamics problem set I was asked to calculate the partition function of the ideal gas in the microcanonic ensemble using the Hamiltionian
$$mathcal{H} = frac{1}{2m}sum_{i=1}^N p_i^2.$$
The problem seems quite straight forward, but I fail at a certain point when trying to solve the integral:
$$begin{align}Z &= int_{mathbb{R}^{6N}} frac{d^{3N}pd^{3N}q}{h^{3N}N!}deltaleft(E-mathcal{H}right)\
&= frac{1}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}q int_{mathbb{R}^{3N}} d^{3N}p, deltaleft(E-sum_{i=1}^N frac{p_i^2}{2m}right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}u, delta left(E-sum_{i=1}^Nu_i^2right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!} int_0^infty delta (E-r^2)r^{3N-1}dr int d^{3N-1}Omega\
&{=} frac{V^N(2m)^{3N/2}}{h^{3N}N!}frac { 2 pi ^ { 3 N / 2 } } { Gamma left( frac { 3 N } { 2 } right) }int_0^infty delta (E-r^2)r^{3N-1}dr.end{align}$$
And here is where the trouble starts. I'm not really sure on how to proceed from here on. It seems the right thing to do would be a coordinate transformation to get something of the form $delta(x-r)$, but I just can't figure out how to do that. Comparing my current result with the solution I expect something like
$$int_0^infty delta (E-r^2)r^{3N-1}dr=frac { 1 } { 2 } E ^ { 3 N / 2 - 1 }.tag{$clubsuit$}$$
integration definite-integrals physics distribution-theory dirac-delta
edited Jan 12 at 22:19
Qmechanic
4,98711855
asked Jan 12 at 18:40
SitoSito
1706
$endgroup$
add a comment |
$begingroup$
TL;DR: How does one show that $(clubsuit)$ holds.
Some context and how I arrived at my problem. In a Thermodynamics problem set I was asked to calculate the partition function of the ideal gas in the microcanonic ensemble using the Hamiltionian
$$mathcal{H} = frac{1}{2m}sum_{i=1}^N p_i^2.$$
The problem seems quite straight forward, but I fail at a certain point when trying to solve the integral:
$$begin{align}Z &= int_{mathbb{R}^{6N}} frac{d^{3N}pd^{3N}q}{h^{3N}N!}deltaleft(E-mathcal{H}right)\
&= frac{1}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}q int_{mathbb{R}^{3N}} d^{3N}p, deltaleft(E-sum_{i=1}^N frac{p_i^2}{2m}right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}u, delta left(E-sum_{i=1}^Nu_i^2right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!} int_0^infty delta (E-r^2)r^{3N-1}dr int d^{3N-1}Omega\
&{=} frac{V^N(2m)^{3N/2}}{h^{3N}N!}frac { 2 pi ^ { 3 N / 2 } } { Gamma left( frac { 3 N } { 2 } right) }int_0^infty delta (E-r^2)r^{3N-1}dr.end{align}$$
And here is where the trouble starts. I'm not really sure on how to proceed from here on. It seems the right thing to do would be a coordinate transformation to get something of the form $delta(x-r)$, but I just can't figure out how to do that. Comparing my current result with the solution I expect something like
$$int_0^infty delta (E-r^2)r^{3N-1}dr=frac { 1 } { 2 } E ^ { 3 N / 2 - 1 }.tag{$clubsuit$}$$
integration definite-integrals physics distribution-theory dirac-delta
edited Jan 12 at 22:19
Qmechanic
4,98711855
asked Jan 12 at 18:40
SitoSito
1706
$endgroup$
$begingroup$
Isn't it just $dr$? Why would it be $d^{3N}r$?
$endgroup$
– John Doe
Jan 12 at 18:56
$begingroup$
@JohnDoe You‘re right! I will correct it in a couple of minutes..
$endgroup$
– Sito
Jan 12 at 18:58
add a comment |
$begingroup$
TL;DR: How does one show that $(clubsuit)$ holds.
Some context and how I arrived at my problem. In a Thermodynamics problem set I was asked to calculate the partition function of the ideal gas in the microcanonic ensemble using the Hamiltionian
$$mathcal{H} = frac{1}{2m}sum_{i=1}^N p_i^2.$$
The problem seems quite straight forward, but I fail at a certain point when trying to solve the integral:
$$begin{align}Z &= int_{mathbb{R}^{6N}} frac{d^{3N}pd^{3N}q}{h^{3N}N!}deltaleft(E-mathcal{H}right)\
&= frac{1}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}q int_{mathbb{R}^{3N}} d^{3N}p, deltaleft(E-sum_{i=1}^N frac{p_i^2}{2m}right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}u, delta left(E-sum_{i=1}^Nu_i^2right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!} int_0^infty delta (E-r^2)r^{3N-1}dr int d^{3N-1}Omega\
&{=} frac{V^N(2m)^{3N/2}}{h^{3N}N!}frac { 2 pi ^ { 3 N / 2 } } { Gamma left( frac { 3 N } { 2 } right) }int_0^infty delta (E-r^2)r^{3N-1}dr.end{align}$$
And here is where the trouble starts. I'm not really sure on how to proceed from here on. It seems the right thing to do would be a coordinate transformation to get something of the form $delta(x-r)$, but I just can't figure out how to do that. Comparing my current result with the solution I expect something like
$$int_0^infty delta (E-r^2)r^{3N-1}dr=frac { 1 } { 2 } E ^ { 3 N / 2 - 1 }.tag{$clubsuit$}$$
integration definite-integrals physics distribution-theory dirac-delta
edited Jan 12 at 22:19
Qmechanic
4,98711855
asked Jan 12 at 18:40
SitoSito
1706
$endgroup$
TL;DR: How does one show that $(clubsuit)$ holds.
Some context and how I arrived at my problem. In a Thermodynamics problem set I was asked to calculate the partition function of the ideal gas in the microcanonic ensemble using the Hamiltionian
$$mathcal{H} = frac{1}{2m}sum_{i=1}^N p_i^2.$$
The problem seems quite straight forward, but I fail at a certain point when trying to solve the integral:
$$begin{align}Z &= int_{mathbb{R}^{6N}} frac{d^{3N}pd^{3N}q}{h^{3N}N!}deltaleft(E-mathcal{H}right)\
&= frac{1}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}q int_{mathbb{R}^{3N}} d^{3N}p, deltaleft(E-sum_{i=1}^N frac{p_i^2}{2m}right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}u, delta left(E-sum_{i=1}^Nu_i^2right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!} int_0^infty delta (E-r^2)r^{3N-1}dr int d^{3N-1}Omega\
&{=} frac{V^N(2m)^{3N/2}}{h^{3N}N!}frac { 2 pi ^ { 3 N / 2 } } { Gamma left( frac { 3 N } { 2 } right) }int_0^infty delta (E-r^2)r^{3N-1}dr.end{align}$$
And here is where the trouble starts. I'm not really sure on how to proceed from here on. It seems the right thing to do would be a coordinate transformation to get something of the form $delta(x-r)$, but I just can't figure out how to do that. Comparing my current result with the solution I expect something like
$$int_0^infty delta (E-r^2)r^{3N-1}dr=frac { 1 } { 2 } E ^ { 3 N / 2 - 1 }.tag{$clubsuit$}$$
integration definite-integrals physics distribution-theory dirac-delta
integration definite-integrals physics distribution-theory dirac-delta
edited Jan 12 at 22:19
Qmechanic
4,98711855
asked Jan 12 at 18:40
SitoSito
1706
edited Jan 12 at 22:19
Qmechanic
4,98711855
asked Jan 12 at 18:40
SitoSito
1706
edited Jan 12 at 22:19
Qmechanic
4,98711855
edited Jan 12 at 22:19
Qmechanic
4,98711855
edited Jan 12 at 22:19
Qmechanic
4,98711855
4,98711855
asked Jan 12 at 18:40
SitoSito
1706
asked Jan 12 at 18:40
SitoSito
1706
asked Jan 12 at 18:40
SitoSito
1706
1706
$begingroup$
Isn't it just $dr$? Why would it be $d^{3N}r$?
$endgroup$
– John Doe
Jan 12 at 18:56
$begingroup$
@JohnDoe You‘re right! I will correct it in a couple of minutes..
$endgroup$
– Sito
Jan 12 at 18:58
add a comment |
$begingroup$
Isn't it just $dr$? Why would it be $d^{3N}r$?
$endgroup$
– John Doe
Jan 12 at 18:56
$begingroup$
@JohnDoe You‘re right! I will correct it in a couple of minutes..
$endgroup$
– Sito
Jan 12 at 18:58
$begingroup$
Isn't it just $dr$? Why would it be $d^{3N}r$?
$endgroup$
– John Doe
Jan 12 at 18:56
$begingroup$
Isn't it just $dr$? Why would it be $d^{3N}r$?
$endgroup$
– John Doe
Jan 12 at 18:56
$begingroup$
@JohnDoe You‘re right! I will correct it in a couple of minutes..
$endgroup$
– Sito
Jan 12 at 18:58
$begingroup$
@JohnDoe You‘re right! I will correct it in a couple of minutes..
$endgroup$
– Sito
Jan 12 at 18:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have the integral$$int_0^infty delta (E-r^2)r^{3N-1}drtag1$$
Then let $g(r)=-r^2+E$. Of course, $g(r)$ is only zero when $r=sqrt E$ ($E>0$ since it is an energy, and we don't care about $r=-sqrt E$ since this is not within out integration bounds), so we use the definition for the composition of the Dirac delta function with a different function given on Wikipedia to get the delta function into a more managable form.
$$delta(g(x))=sum_ifrac{delta(x-x_i)}{|g'(x_i)|}$$ where the sum is over all the roots of $g(x)$.
For our function $g(r)$, this gives $$g(-r^2+E)=frac{delta(r-sqrt E)}{2sqrt E}$$
Plugging this into $(1)$, $$int_0^infty delta (E-r^2)r^{3N-1}dr=frac1{2sqrt E}int_0^inftydelta(r-sqrt E)r^{3N-1}dr=frac12 E^{-1/2}E^{3N/2} E^{-1/2}=frac12E^{{3N}/2-1}$$ as required.
answered Jan 12 at 18:57
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
Beautiful! Didn't know about the composition, so thank you for that!
$endgroup$
– Sito
Jan 12 at 19:14
add a comment |
$begingroup$
User John Doe has already given a good answer. Here is a slightly different calculation:
$$ int_{mathbb{R}_+}! mathrm{d}r~r^{3N-1}delta(r^2-E)
~stackrel{u=r^2}{=}~
frac{1}{2}int_{mathbb{R}_+}! mathrm{d}u~u^{3N/2-1}delta(u-E)
~=~frac{1}{2}theta(E)E^{3N/2-1},tag{$clubsuit$}$$
where $theta$ denotes the Heaviside step function.
answered Jan 12 at 22:17
QmechanicQmechanic
4,98711855
$endgroup$
add a comment |
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2 Answers
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oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
You have the integral$$int_0^infty delta (E-r^2)r^{3N-1}drtag1$$
Then let $g(r)=-r^2+E$. Of course, $g(r)$ is only zero when $r=sqrt E$ ($E>0$ since it is an energy, and we don't care about $r=-sqrt E$ since this is not within out integration bounds), so we use the definition for the composition of the Dirac delta function with a different function given on Wikipedia to get the delta function into a more managable form.
$$delta(g(x))=sum_ifrac{delta(x-x_i)}{|g'(x_i)|}$$ where the sum is over all the roots of $g(x)$.
For our function $g(r)$, this gives $$g(-r^2+E)=frac{delta(r-sqrt E)}{2sqrt E}$$
Plugging this into $(1)$, $$int_0^infty delta (E-r^2)r^{3N-1}dr=frac1{2sqrt E}int_0^inftydelta(r-sqrt E)r^{3N-1}dr=frac12 E^{-1/2}E^{3N/2} E^{-1/2}=frac12E^{{3N}/2-1}$$ as required.
answered Jan 12 at 18:57
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
Beautiful! Didn't know about the composition, so thank you for that!
$endgroup$
– Sito
Jan 12 at 19:14
add a comment |
$begingroup$
You have the integral$$int_0^infty delta (E-r^2)r^{3N-1}drtag1$$
Then let $g(r)=-r^2+E$. Of course, $g(r)$ is only zero when $r=sqrt E$ ($E>0$ since it is an energy, and we don't care about $r=-sqrt E$ since this is not within out integration bounds), so we use the definition for the composition of the Dirac delta function with a different function given on Wikipedia to get the delta function into a more managable form.
$$delta(g(x))=sum_ifrac{delta(x-x_i)}{|g'(x_i)|}$$ where the sum is over all the roots of $g(x)$.
For our function $g(r)$, this gives $$g(-r^2+E)=frac{delta(r-sqrt E)}{2sqrt E}$$
Plugging this into $(1)$, $$int_0^infty delta (E-r^2)r^{3N-1}dr=frac1{2sqrt E}int_0^inftydelta(r-sqrt E)r^{3N-1}dr=frac12 E^{-1/2}E^{3N/2} E^{-1/2}=frac12E^{{3N}/2-1}$$ as required.
answered Jan 12 at 18:57
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
Beautiful! Didn't know about the composition, so thank you for that!
$endgroup$
– Sito
Jan 12 at 19:14
add a comment |
$begingroup$
You have the integral$$int_0^infty delta (E-r^2)r^{3N-1}drtag1$$
Then let $g(r)=-r^2+E$. Of course, $g(r)$ is only zero when $r=sqrt E$ ($E>0$ since it is an energy, and we don't care about $r=-sqrt E$ since this is not within out integration bounds), so we use the definition for the composition of the Dirac delta function with a different function given on Wikipedia to get the delta function into a more managable form.
$$delta(g(x))=sum_ifrac{delta(x-x_i)}{|g'(x_i)|}$$ where the sum is over all the roots of $g(x)$.
For our function $g(r)$, this gives $$g(-r^2+E)=frac{delta(r-sqrt E)}{2sqrt E}$$
Plugging this into $(1)$, $$int_0^infty delta (E-r^2)r^{3N-1}dr=frac1{2sqrt E}int_0^inftydelta(r-sqrt E)r^{3N-1}dr=frac12 E^{-1/2}E^{3N/2} E^{-1/2}=frac12E^{{3N}/2-1}$$ as required.
answered Jan 12 at 18:57
John DoeJohn Doe
11.1k11238
$endgroup$
You have the integral$$int_0^infty delta (E-r^2)r^{3N-1}drtag1$$
Then let $g(r)=-r^2+E$. Of course, $g(r)$ is only zero when $r=sqrt E$ ($E>0$ since it is an energy, and we don't care about $r=-sqrt E$ since this is not within out integration bounds), so we use the definition for the composition of the Dirac delta function with a different function given on Wikipedia to get the delta function into a more managable form.
$$delta(g(x))=sum_ifrac{delta(x-x_i)}{|g'(x_i)|}$$ where the sum is over all the roots of $g(x)$.
For our function $g(r)$, this gives $$g(-r^2+E)=frac{delta(r-sqrt E)}{2sqrt E}$$
Plugging this into $(1)$, $$int_0^infty delta (E-r^2)r^{3N-1}dr=frac1{2sqrt E}int_0^inftydelta(r-sqrt E)r^{3N-1}dr=frac12 E^{-1/2}E^{3N/2} E^{-1/2}=frac12E^{{3N}/2-1}$$ as required.
answered Jan 12 at 18:57
John DoeJohn Doe
11.1k11238
edited Jan 12 at 19:34
answered Jan 12 at 18:57
John DoeJohn Doe
11.1k11238
answered Jan 12 at 18:57
John DoeJohn Doe
11.1k11238
answered Jan 12 at 18:57
John DoeJohn Doe
11.1k11238
11.1k11238
$begingroup$
Beautiful! Didn't know about the composition, so thank you for that!
$endgroup$
– Sito
Jan 12 at 19:14
add a comment |
$begingroup$
Beautiful! Didn't know about the composition, so thank you for that!
$endgroup$
– Sito
Jan 12 at 19:14
$begingroup$
Beautiful! Didn't know about the composition, so thank you for that!
$endgroup$
– Sito
Jan 12 at 19:14
$begingroup$
Beautiful! Didn't know about the composition, so thank you for that!
$endgroup$
– Sito
Jan 12 at 19:14
add a comment |
$begingroup$
User John Doe has already given a good answer. Here is a slightly different calculation:
$$ int_{mathbb{R}_+}! mathrm{d}r~r^{3N-1}delta(r^2-E)
~stackrel{u=r^2}{=}~
frac{1}{2}int_{mathbb{R}_+}! mathrm{d}u~u^{3N/2-1}delta(u-E)
~=~frac{1}{2}theta(E)E^{3N/2-1},tag{$clubsuit$}$$
where $theta$ denotes the Heaviside step function.
answered Jan 12 at 22:17
QmechanicQmechanic
4,98711855
$endgroup$
add a comment |
$begingroup$
User John Doe has already given a good answer. Here is a slightly different calculation:
$$ int_{mathbb{R}_+}! mathrm{d}r~r^{3N-1}delta(r^2-E)
~stackrel{u=r^2}{=}~
frac{1}{2}int_{mathbb{R}_+}! mathrm{d}u~u^{3N/2-1}delta(u-E)
~=~frac{1}{2}theta(E)E^{3N/2-1},tag{$clubsuit$}$$
where $theta$ denotes the Heaviside step function.
answered Jan 12 at 22:17
QmechanicQmechanic
4,98711855
$endgroup$
add a comment |
$begingroup$
User John Doe has already given a good answer. Here is a slightly different calculation:
$$ int_{mathbb{R}_+}! mathrm{d}r~r^{3N-1}delta(r^2-E)
~stackrel{u=r^2}{=}~
frac{1}{2}int_{mathbb{R}_+}! mathrm{d}u~u^{3N/2-1}delta(u-E)
~=~frac{1}{2}theta(E)E^{3N/2-1},tag{$clubsuit$}$$
where $theta$ denotes the Heaviside step function.
answered Jan 12 at 22:17
QmechanicQmechanic
4,98711855
$endgroup$
User John Doe has already given a good answer. Here is a slightly different calculation:
$$ int_{mathbb{R}_+}! mathrm{d}r~r^{3N-1}delta(r^2-E)
~stackrel{u=r^2}{=}~
frac{1}{2}int_{mathbb{R}_+}! mathrm{d}u~u^{3N/2-1}delta(u-E)
~=~frac{1}{2}theta(E)E^{3N/2-1},tag{$clubsuit$}$$
where $theta$ denotes the Heaviside step function.
answered Jan 12 at 22:17
QmechanicQmechanic
4,98711855
answered Jan 12 at 22:17
QmechanicQmechanic
4,98711855
answered Jan 12 at 22:17
QmechanicQmechanic
4,98711855
answered Jan 12 at 22:17
QmechanicQmechanic
4,98711855
4,98711855
add a comment |
add a comment |
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$begingroup$
Isn't it just $dr$? Why would it be $d^{3N}r$?
$endgroup$
– John Doe
Jan 12 at 18:56
$begingroup$
@JohnDoe You‘re right! I will correct it in a couple of minutes..
$endgroup$
– Sito
Jan 12 at 18:58