| Topic: |
Science > Physics |
| User: |
"Gregory L. Hansen" |
| Date: |
08 Jul 2003 03:51:19 PM |
| Object: |
Density of States |
Can someone help me figure this one out? I pretty much know what the
density of states is and what it's good for, and how to calculate it if I
just follow the formulas in the books. But finding those formulas has
always given me some issues. I'm looking specifically at Perkins (3rd),
page 209, if that matters.
How many ways can a particle's momentum go from p to p+dp? That is, look
at the volume in phase space. In phase space expressed in spherical polar
coordinates the momentum part of that element has volume
p^2 sin(theta) d(theta) d(phi)
We assume it's in some spatial volume V that doesn't change (and
ultimately set V=1), partitioned into cubes of size h^3 just because
we're morally beholden to work Planck's constant in there somehow? So
Perkins gives
g V/h^3 d(ANGLE) p^2 dp
where I've added the factor g which would, in other problems, account for
available spin states and other degrees of freedom.
I can't picture why volume is in there when we're changing the momentum,
or why Planck's constant should be there.
--
"Is that plutonium on your gums?"
"Shut up and kiss me!"
-- Marge and Homer Simpson
.
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| User: "Old Man" |
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| Title: Re: Density of States |
09 Jul 2003 03:05:10 PM |
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Gregory L. Hansen <glhansen@steel.ucs.indiana.edu> wrote in message
news:befas7$g42$1@hood.uits.indiana.edu...
Can someone help me figure this one out? I pretty much know what the
density of states is and what it's good for, and how to calculate it if I
just follow the formulas in the books. But finding those formulas has
always given me some issues. I'm looking specifically at Perkins (3rd),
page 209, if that matters.
How many ways can a particle's momentum go from p to p+dp? That is, look
at the volume in phase space. In phase space expressed in spherical polar
coordinates the momentum part of that element has volume
p^2 sin(theta) d(theta) d(phi)
We assume it's in some spatial volume V that doesn't change (and
ultimately set V=1), partitioned into cubes of size h^3 just because
we're morally beholden to work Planck's constant in there somehow? So
Perkins gives
g V/h^3 d(ANGLE) p^2 dp
where I've added the factor g which would, in other problems, account for
available spin states and other degrees of freedom.
I can't picture why volume is in there when we're changing the momentum,
or why Planck's constant should be there.
It's just a statement of the uncertainty principle in three
dimensions. Each state occupies a phase-space volume
of h^3. In one dimension, dp*dx = h, which is just the
uncertainty principle. [Old Man]
.
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| User: "S. Enterprize Company" |
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| Title: Re: Density of States |
09 Jul 2003 05:27:30 PM |
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It's just a statement of the uncertainty principle in three
dimensions. Each state occupies a phase-space volume
of h^3. In one dimension, dp*dx = h, which is just the
uncertainty principle. [Old Man]
There is no uncertainty principle with the Smart Model Theory.
S. Enterprize Co. (Membership)
http://www.s-enterprize.com/
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
.
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| User: "" |
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| Title: Re: Density of States |
08 Jul 2003 04:19:00 PM |
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In article <befas7$g42$1@hood.uits.indiana.edu>, (Gregory L. Hansen) writes:
Can someone help me figure this one out? I pretty much know what the
density of states is and what it's good for, and how to calculate it if I
just follow the formulas in the books. But finding those formulas has
always given me some issues. I'm looking specifically at Perkins (3rd),
page 209, if that matters.
How many ways can a particle's momentum go from p to p+dp? That is, look
at the volume in phase space. In phase space expressed in spherical polar
coordinates the momentum part of that element has volume
p^2 sin(theta) d(theta) d(phi)
We assume it's in some spatial volume V that doesn't change (and
ultimately set V=1), partitioned into cubes of size h^3 just because
we're morally beholden to work Planck's constant in there somehow? So
Perkins gives
g V/h^3 d(ANGLE) p^2 dp
where I've added the factor g which would, in other problems, account for
available spin states and other degrees of freedom.
I can't picture why volume is in there when we're changing the momentum,
or why Planck's constant should be there.
--
Start with one dimension. Just a particle in a well of length L. Ask
Shroedinger what are the possible momentum states in this well (you'll
find that they're integer multiples of h/L, thus their density is
proportional to L/h. Then generalize to 3D.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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| User: "Gregory L. Hansen" |
|
| Title: Re: Density of States |
09 Jul 2003 10:04:04 AM |
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In article <8zGOa.110$Y4.39316@news.uchicago.edu>,
<meron@cars3.uchicago.edu> wrote:
In article <befas7$g42$1@hood.uits.indiana.edu>,
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) writes:
Can someone help me figure this one out? I pretty much know what the
density of states is and what it's good for, and how to calculate it if I
just follow the formulas in the books. But finding those formulas has
always given me some issues. I'm looking specifically at Perkins (3rd),
page 209, if that matters.
How many ways can a particle's momentum go from p to p+dp? That is, look
at the volume in phase space. In phase space expressed in spherical polar
coordinates the momentum part of that element has volume
p^2 sin(theta) d(theta) d(phi)
We assume it's in some spatial volume V that doesn't change (and
ultimately set V=1), partitioned into cubes of size h^3 just because
we're morally beholden to work Planck's constant in there somehow? So
Perkins gives
g V/h^3 d(ANGLE) p^2 dp
where I've added the factor g which would, in other problems, account for
available spin states and other degrees of freedom.
I can't picture why volume is in there when we're changing the momentum,
or why Planck's constant should be there.
--
Start with one dimension. Just a particle in a well of length L. Ask
Shroedinger what are the possible momentum states in this well (you'll
find that they're integer multiples of h/L, thus their density is
proportional to L/h. Then generalize to 3D.
I have to think about this. Momentum eigenstates would be
exp i(wt - n pi x/L)
P psi = nh/2L psi
If you're in the ground state, you have to hop by h/2L to get to the first
excited state. If you're in the 100th state, you have to hop by the same
amount to get to the 101st state. That gives you a linear density of
momentum states, and (h/2L)*dp is the number you sweep through when
momentum changes by dp. And v/h^3 d(ANGLE) p^2 dp is the number of
momentum states contained in that volume. Add the factor g when momentum
states aren't the only degree of freedom for the final state.
--
"Is that plutonium on your gums?"
"Shut up and kiss me!"
-- Marge and Homer Simpson
.
|
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| User: "S. Enterprize Company" |
|
| Title: Re: Density of States |
09 Jul 2003 05:25:27 PM |
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|
In article <8zGOa.110$Y4.39316@news.uchicago.edu>,
<meron@cars3.uchicago.edu> wrote:
In article <befas7$g42$1@hood.uits.indiana.edu>,
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) writes:
Can someone help me figure this one out? I pretty much know what the
density of states is and what it's good for, and how to calculate it if I
just follow the formulas in the books. But finding those formulas has
always given me some issues. I'm looking specifically at Perkins (3rd),
page 209, if that matters.
How many ways can a particle's momentum go from p to p+dp? That is, look
at the volume in phase space. In phase space expressed in spherical polar
coordinates the momentum part of that element has volume
p^2 sin(theta) d(theta) d(phi)
We assume it's in some spatial volume V that doesn't change (and
ultimately set V=1), partitioned into cubes of size h^3 just because
we're morally beholden to work Planck's constant in there somehow? So
Perkins gives
g V/h^3 d(ANGLE) p^2 dp
where I've added the factor g which would, in other problems, account for
available spin states and other degrees of freedom.
I can't picture why volume is in there when we're changing the momentum,
or why Planck's constant should be there.
--
Start with one dimension. Just a particle in a well of length L. Ask
Shroedinger what are the possible momentum states in this well (you'll
find that they're integer multiples of h/L, thus their density is
proportional to L/h. Then generalize to 3D.
I have to think about this. Momentum eigenstates would be
exp i(wt - n pi x/L)
P psi = nh/2L psi
If you're in the ground state, you have to hop by h/2L to get to the first
excited state. If you're in the 100th state, you have to hop by the same
amount to get to the 101st state. That gives you a linear density of
momentum states, and (h/2L)*dp is the number you sweep through when
momentum changes by dp. And v/h^3 d(ANGLE) p^2 dp is the number of
momentum states contained in that volume. Add the factor g when momentum
states aren't the only degree of freedom for the final state.
Each state you hop to is a function of a Smart Model Sub-atomic Oscillator.
Each transformed energy state corresponds to a group of energy levels that make
a line on the spectra which froms from harmonic energy oscillations.
S. Enterprize Co. (Membership)
http://www.s-enterprize.com/
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
.
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| User: "" |
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| Title: Re: Density of States |
09 Jul 2003 02:27:09 PM |
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In article <behat4$41v$2@hood.uits.indiana.edu>, (Gregory L. Hansen) writes:
In article <8zGOa.110$Y4.39316@news.uchicago.edu>,
<meron@cars3.uchicago.edu> wrote:
In article <befas7$g42$1@hood.uits.indiana.edu>,
(Gregory L. Hansen) writes:
Can someone help me figure this one out? I pretty much know what the
density of states is and what it's good for, and how to calculate it if I
just follow the formulas in the books. But finding those formulas has
always given me some issues. I'm looking specifically at Perkins (3rd),
page 209, if that matters.
How many ways can a particle's momentum go from p to p+dp? That is, look
at the volume in phase space. In phase space expressed in spherical polar
coordinates the momentum part of that element has volume
p^2 sin(theta) d(theta) d(phi)
We assume it's in some spatial volume V that doesn't change (and
ultimately set V=1), partitioned into cubes of size h^3 just because
we're morally beholden to work Planck's constant in there somehow? So
Perkins gives
g V/h^3 d(ANGLE) p^2 dp
where I've added the factor g which would, in other problems, account for
available spin states and other degrees of freedom.
I can't picture why volume is in there when we're changing the momentum,
or why Planck's constant should be there.
--
Start with one dimension. Just a particle in a well of length L. Ask
Shroedinger what are the possible momentum states in this well (you'll
find that they're integer multiples of h/L, thus their density is
proportional to L/h. Then generalize to 3D.
I have to think about this. Momentum eigenstates would be
exp i(wt - n pi x/L)
P psi = nh/2L psi
If you're in the ground state, you have to hop by h/2L to get to the first
excited state. If you're in the 100th state, you have to hop by the same
amount to get to the 101st state. That gives you a linear density of
momentum states, and (h/2L)*dp is the number you sweep through when
momentum changes by dp. And v/h^3 d(ANGLE) p^2 dp is the number of
momentum states contained in that volume. Add the factor g when momentum
states aren't the only degree of freedom for the final state.
Yep, here you go. A decent textbook should present this derivation,
IMO, but few do. The first place I saw it, as I recall, was in Reif's
"Statistical Physics".
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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| User: "Gregory L. Hansen" |
|
| Title: Re: Density of States |
09 Jul 2003 04:16:38 PM |
|
|
In article <h0_Oa.10$_4.2148@news.uchicago.edu>,
<meron@cars3.uchicago.edu> wrote:
In article <behat4$41v$2@hood.uits.indiana.edu>,
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) writes:
In article <8zGOa.110$Y4.39316@news.uchicago.edu>,
<meron@cars3.uchicago.edu> wrote:
In article <befas7$g42$1@hood.uits.indiana.edu>,
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) writes:
Can someone help me figure this one out? I pretty much know what the
density of states is and what it's good for, and how to calculate it if I
just follow the formulas in the books. But finding those formulas has
always given me some issues. I'm looking specifically at Perkins (3rd),
page 209, if that matters.
How many ways can a particle's momentum go from p to p+dp? That is, look
at the volume in phase space. In phase space expressed in spherical polar
coordinates the momentum part of that element has volume
p^2 sin(theta) d(theta) d(phi)
We assume it's in some spatial volume V that doesn't change (and
ultimately set V=1), partitioned into cubes of size h^3 just because
we're morally beholden to work Planck's constant in there somehow? So
Perkins gives
g V/h^3 d(ANGLE) p^2 dp
where I've added the factor g which would, in other problems, account for
available spin states and other degrees of freedom.
I can't picture why volume is in there when we're changing the momentum,
or why Planck's constant should be there.
--
Start with one dimension. Just a particle in a well of length L. Ask
Shroedinger what are the possible momentum states in this well (you'll
find that they're integer multiples of h/L, thus their density is
proportional to L/h. Then generalize to 3D.
I have to think about this. Momentum eigenstates would be
exp i(wt - n pi x/L)
P psi = nh/2L psi
If you're in the ground state, you have to hop by h/2L to get to the first
excited state. If you're in the 100th state, you have to hop by the same
amount to get to the 101st state. That gives you a linear density of
momentum states, and (h/2L)*dp is the number you sweep through when
momentum changes by dp. And v/h^3 d(ANGLE) p^2 dp is the number of
momentum states contained in that volume. Add the factor g when momentum
states aren't the only degree of freedom for the final state.
Yep, here you go. A decent textbook should present this derivation,
IMO, but few do. The first place I saw it, as I recall, was in Reif's
"Statistical Physics".
So it occurs to me that you couldn't be so cavalier about defining a V=1
for your density of states if the final state were to be in a meaningful
potential, like a harmonic oscillator or an atom. But the typical
particle physics assumption is a free particle that flies off into the
vacuuum.
--
"Is that plutonium on your gums?"
"Shut up and kiss me!"
-- Marge and Homer Simpson
.
|
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| User: "" |
|
| Title: Re: Density of States |
09 Jul 2003 05:00:21 PM |
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|
In article <bei0nm$cc2$1@hood.uits.indiana.edu>, (Gregory L. Hansen) writes:
In article <h0_Oa.10$_4.2148@news.uchicago.edu>,
<meron@cars3.uchicago.edu> wrote:
In article <behat4$41v$2@hood.uits.indiana.edu>,
(Gregory L. Hansen) writes:
In article <8zGOa.110$Y4.39316@news.uchicago.edu>,
<meron@cars3.uchicago.edu> wrote:
In article <befas7$g42$1@hood.uits.indiana.edu>,
(Gregory L. Hansen) writes:
Can someone help me figure this one out? I pretty much know what the
density of states is and what it's good for, and how to calculate it if I
just follow the formulas in the books. But finding those formulas has
always given me some issues. I'm looking specifically at Perkins (3rd),
page 209, if that matters.
How many ways can a particle's momentum go from p to p+dp? That is, look
at the volume in phase space. In phase space expressed in spherical polar
coordinates the momentum part of that element has volume
p^2 sin(theta) d(theta) d(phi)
We assume it's in some spatial volume V that doesn't change (and
ultimately set V=1), partitioned into cubes of size h^3 just because
we're morally beholden to work Planck's constant in there somehow? So
Perkins gives
g V/h^3 d(ANGLE) p^2 dp
where I've added the factor g which would, in other problems, account for
available spin states and other degrees of freedom.
I can't picture why volume is in there when we're changing the momentum,
or why Planck's constant should be there.
--
Start with one dimension. Just a particle in a well of length L. Ask
Shroedinger what are the possible momentum states in this well (you'll
find that they're integer multiples of h/L, thus their density is
proportional to L/h. Then generalize to 3D.
I have to think about this. Momentum eigenstates would be
exp i(wt - n pi x/L)
P psi = nh/2L psi
If you're in the ground state, you have to hop by h/2L to get to the first
excited state. If you're in the 100th state, you have to hop by the same
amount to get to the 101st state. That gives you a linear density of
momentum states, and (h/2L)*dp is the number you sweep through when
momentum changes by dp. And v/h^3 d(ANGLE) p^2 dp is the number of
momentum states contained in that volume. Add the factor g when momentum
states aren't the only degree of freedom for the final state.
Yep, here you go. A decent textbook should present this derivation,
IMO, but few do. The first place I saw it, as I recall, was in Reif's
"Statistical Physics".
So it occurs to me that you couldn't be so cavalier about defining a V=1
for your density of states if the final state were to be in a meaningful
potential, like a harmonic oscillator or an atom.
Indeed, true.
But the typical particle physics assumption is a free particle that flies
off into the vacuuum.
Yes. So, in principle, this is a case of V -> infinity. But, it ends
cancelling out of the calculation anyway, as I recall, so you can take
it as 1 or whatever. In case of doubt it is probably better to keep
it in the calculation (instead of setting it to 1) and verifying
explicitly that it cancels at the end.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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