Application of Statistics in Physics - Revision history

Jlatimer31 at 00:19, 14 April 2025

2025-04-14T00:19:14Z

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 Claimed by Edwin Solis (April 16th, Spring 2022)
 With the development of Quantum Mechanics and Statistical Mechanics, the subject of <b>Statistics</b> has become quintessential for understanding the foundation of these physical theories.
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 ==History==
 ==References==

Luigi: /* Two State Paramagnet */

2022-11-28T21:18:24Z

Two State Paramagnet

← Older revision		Revision as of 17:18, 28 November 2022
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	<math>		<math>
	S = k\~~log~~ W		S = k\ln W
	</math>		</math>

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	<math>		<math>
	S = k\~~log~~(\frac{N}{N_{\uparrow}!N_{\downarrow}!})		S = k\ln(\frac{N}{N_{\uparrow}!N_{\downarrow}!})
	</math>		</math>

Luigi: /* Two State Paramagnet */

2022-11-28T07:38:06Z

Two State Paramagnet

← Older revision		Revision as of 03:38, 28 November 2022
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	</math>		</math>

	Now, consider an ideal paramagnetic system, where a dipole is either parallel or anti-parallel to the external magnetic field such that dipoles aligned parallel to the field . The energy of an individual magnetic dipole is <math>U=\vec{\mu}\cdot\vec{B}</math> where <math>\mu</math> is the magnetic moment which is constant. Assume that we have an upward external magnetic field, in such case, upward dipoles will have a negative energy (<math>U_{\uparrow}</math>) and downward dipoles will have a positive energy (<math>U_{\downarrow}</math>). Rewriting the original <math>U</math> in the form <math>U=\mu B\cos\theta</math> where theta will be <math>\pi</math>, hence the energy for an individual upward facing dipole is <math>-\mu B</math>. Since there are <math>N_{\uparrow}</math> upward dipoles, the total upward energy is <math>U_{\uparrow}=-\mu BN_{\uparrow}</math>. A similar argument can be made for the total downward dipole energy such that <math>\theta=\0</math> because the downward dipole is anti-parallel resulting in <math>U_{\downarrow}=\mu BN_{\downarrow}</math>. The sum of these quantities gives the total energy of the system <math>U=\mu B(N_{\downarrow}-N_{\uparrow})</math>.		Now, consider an ideal paramagnetic system, where a dipole is either parallel or anti-parallel to the external magnetic field such that dipoles aligned parallel to the field . The energy of an individual magnetic dipole is <math>U=\vec{\mu}\cdot\vec{B}</math> where <math>\mu</math> is the magnetic moment which is constant. Assume that we have an upward external magnetic field, in such case, upward dipoles will have a negative energy (<math>U_{\uparrow}</math>) and downward dipoles will have a positive energy (<math>U_{\downarrow}</math>). Rewriting the original <math>U</math> in the form <math>U=\mu B\cos\theta</math> where theta will be <math>\pi</math>, hence the energy for an individual upward facing dipole is <math>-\mu B</math>. Since there are <math>N_{\uparrow}</math> upward dipoles, the total upward energy is <math>U_{\uparrow}=-\mu BN_{\uparrow}</math>. A similar argument can be made for the total downward dipole energy such that <math>\theta=0</math> because the downward dipole is anti-parallel resulting in <math>U_{\downarrow}=\mu BN_{\downarrow}</math>. The sum of these quantities gives the total energy of the system <math>U=\mu B(N_{\downarrow}-N_{\uparrow})</math>.

	Graph of two state paramagnet of 100 particles where the <math>y</math>-axis represents the entropy divided by <math>k</math> for scale. Notice that the shape is nearly parabolic because as <math>N_{\uparrow}</math>, the number of available microstates increases. However, after a certain point, the number of available microstates decreases which results in a decreasing entropy with respect to the number of "up" dipoles. If you like, you can think of the system as being less "orderly" as it approaches <math>50</math> on the <math>x</math>-axis and becoming more "orderly" when the system approaches <math>0</math> or <math>100</math>.		Graph of two state paramagnet of 100 particles where the <math>y</math>-axis represents the entropy divided by <math>k</math> for scale. Notice that the shape is nearly parabolic because as <math>N_{\uparrow}</math>, the number of available microstates increases. However, after a certain point, the number of available microstates decreases which results in a decreasing entropy with respect to the number of "up" dipoles. If you like, you can think of the system as being less "orderly" as it approaches <math>50</math> on the <math>x</math>-axis and becoming more "orderly" when the system approaches <math>0</math> or <math>100</math>. https://www.glowscript.org/#/user/Luigi/folder/Private/program/EntropyParamag
	[[File:paraomega.jpeg]]		[[File:paraomega.jpeg]]

	When discussing paramagnets, it is useful to consider the ''magnetization''. Magnetization (<math>M</math>), in this context, can be defined as the net magnetic moment: <math>M=\mu (N_{\uparrow}-N_{\downarrow})=-\frac{U}{B}</math>. Using magnetization results in figure below.		When discussing paramagnets, it is useful to consider the ''magnetization''. Magnetization (<math>M</math>), in this context, can be defined as the net magnetic moment: <math>M=\mu (N_{\uparrow}-N_{\downarrow})=-\frac{U}{B}</math>. Using magnetization results in figure below.https://www.glowscript.org/#/user/Luigi/folder/MyPrograms/program/paramagenergy
	[[File:energypara.png]]		[[File:energypara.png]]

Luigi: /* Two State Paramagnet */

2022-11-28T07:35:16Z

Two State Paramagnet

← Older revision		Revision as of 03:35, 28 November 2022
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	</math>		</math>

	Now, consider an ideal paramagnetic system, where a dipole is either parallel or anti-parallel to the external magnetic field such that dipoles aligned parallel to the field . The energy of an individual magnetic dipole is <math>U=\vec{\mu}\cdot\vec{B}</math> where <math>\mu</math> is the magnetic moment which is constant. Assume that we have an upward external magnetic field, in such case, upward dipoles will have a ~~positive~~ energy (<math>U_{\uparrow}</math>) and downward dipoles will have a ~~negative~~ energy (<math>U_{\downarrow}</math>). Rewriting the original <math>U</math> in the form <math>U=\mu B\cos\theta</math> where theta will be <math>0</math>, hence the energy for an individual upward facing dipole is <math>\mu B</math>. Since there are <math>N_{\uparrow}</math> upward dipoles, the total upward energy is <math>U_{\uparrow}=\mu BN_{\uparrow}</math>. A similar argument can be made for the total downward dipole energy such that <math>\theta=\pi</math> because the downward dipole is anti-parallel resulting in <math>U_{\downarrow}=-\mu BN_{\downarrow}</math>. The sum of these quantities gives the total energy of the system <math>U=\mu B(N_{\~~uparrow~~}-N_{\~~downarrow~~})</math>.		Now, consider an ideal paramagnetic system, where a dipole is either parallel or anti-parallel to the external magnetic field such that dipoles aligned parallel to the field . The energy of an individual magnetic dipole is <math>U=\vec{\mu}\cdot\vec{B}</math> where <math>\mu</math> is the magnetic moment which is constant. Assume that we have an upward external magnetic field, in such case, upward dipoles will have a negative energy (<math>U_{\uparrow}</math>) and downward dipoles will have a positive energy (<math>U_{\downarrow}</math>). Rewriting the original <math>U</math> in the form <math>U=\mu B\cos\theta</math> where theta will be <math>\pi</math>, hence the energy for an individual upward facing dipole is <math>-\mu B</math>. Since there are <math>N_{\uparrow}</math> upward dipoles, the total upward energy is <math>U_{\uparrow}=-\mu BN_{\uparrow}</math>. A similar argument can be made for the total downward dipole energy such that <math>\theta=\0</math> because the downward dipole is anti-parallel resulting in <math>U_{\downarrow}=\mu BN_{\downarrow}</math>. The sum of these quantities gives the total energy of the system <math>U=\mu B(N_{\downarrow}-N_{\uparrow})</math>.

	Graph of two state paramagnet of 100 particles where the <math>y</math>-axis represents the entropy divided by <math>k</math> for scale. Notice that the shape is nearly parabolic because as <math>N_{\uparrow}</math>, the number of available microstates increases. However, after a certain point, the number of available microstates decreases which results in a decreasing entropy with respect to the number of "up" dipoles. If you like, you can think of the system as being less "orderly" as it approaches <math>50</math> on the <math>x</math>-axis and becoming more "orderly" when the system approaches <math>0</math> or <math>100</math>.		Graph of two state paramagnet of 100 particles where the <math>y</math>-axis represents the entropy divided by <math>k</math> for scale. Notice that the shape is nearly parabolic because as <math>N_{\uparrow}</math>, the number of available microstates increases. However, after a certain point, the number of available microstates decreases which results in a decreasing entropy with respect to the number of "up" dipoles. If you like, you can think of the system as being less "orderly" as it approaches <math>50</math> on the <math>x</math>-axis and becoming more "orderly" when the system approaches <math>0</math> or <math>100</math>.

Luigi: /* Two State Paramagnet */

2022-11-28T07:31:39Z

Two State Paramagnet

← Older revision		Revision as of 03:31, 28 November 2022
Line 219:		Line 219:

	Now, consider an ideal paramagnetic system, where a dipole is either parallel or anti-parallel to the external magnetic field such that dipoles aligned parallel to the field . The energy of an individual magnetic dipole is <math>U=\vec{\mu}\cdot\vec{B}</math> where <math>\mu</math> is the magnetic moment which is constant. Assume that we have an upward external magnetic field, in such case, upward dipoles will have a positive energy (<math>U_{\uparrow}</math>) and downward dipoles will have a negative energy (<math>U_{\downarrow}</math>). Rewriting the original <math>U</math> in the form <math>U=\mu B\cos\theta</math> where theta will be <math>0</math>, hence the energy for an individual upward facing dipole is <math>\mu B</math>. Since there are <math>N_{\uparrow}</math> upward dipoles, the total upward energy is <math>U_{\uparrow}=\mu BN_{\uparrow}</math>. A similar argument can be made for the total downward dipole energy such that <math>\theta=\pi</math> because the downward dipole is anti-parallel resulting in <math>U_{\downarrow}=-\mu BN_{\downarrow}</math>. The sum of these quantities gives the total energy of the system <math>U=\mu B(N_{\uparrow}-N_{\downarrow})</math>.		Now, consider an ideal paramagnetic system, where a dipole is either parallel or anti-parallel to the external magnetic field such that dipoles aligned parallel to the field . The energy of an individual magnetic dipole is <math>U=\vec{\mu}\cdot\vec{B}</math> where <math>\mu</math> is the magnetic moment which is constant. Assume that we have an upward external magnetic field, in such case, upward dipoles will have a positive energy (<math>U_{\uparrow}</math>) and downward dipoles will have a negative energy (<math>U_{\downarrow}</math>). Rewriting the original <math>U</math> in the form <math>U=\mu B\cos\theta</math> where theta will be <math>0</math>, hence the energy for an individual upward facing dipole is <math>\mu B</math>. Since there are <math>N_{\uparrow}</math> upward dipoles, the total upward energy is <math>U_{\uparrow}=\mu BN_{\uparrow}</math>. A similar argument can be made for the total downward dipole energy such that <math>\theta=\pi</math> because the downward dipole is anti-parallel resulting in <math>U_{\downarrow}=-\mu BN_{\downarrow}</math>. The sum of these quantities gives the total energy of the system <math>U=\mu B(N_{\uparrow}-N_{\downarrow})</math>.
	[[File:paraomega.jpeg \|Graph of two state paramagnet of <math>100</math> particles where the <math>y</math>-axis represents the entropy divided by <math>k</math> for scale. Notice that the shape is nearly parabolic because as <math>N_{\uparrow}</math>, the number of available microstates increases. However, after a certain point, the number of available microstates decreases which results in a decreasing entropy with respect to the number of "up" dipoles. If you like, you can think of the system as being less "orderly" as it approaches <math>50</math> on the <math>x</math>-axis and becoming more "orderly" when the system approaches <math>0</math> or <math>100</math>.]]

	When discussing paramagnets, it is useful to consider the ''magnetization''. Magnetization (<math>M</math>), in this context, can be defined as the net magnetic moment: <math>M=\mu (N_{\uparrow}-N_{\downarrow})=-\frac{U}{B}</math>. Using magnetization results in figure [].		Graph of two state paramagnet of 100 particles where the <math>y</math>-axis represents the entropy divided by <math>k</math> for scale. Notice that the shape is nearly parabolic because as <math>N_{\uparrow}</math>, the number of available microstates increases. However, after a certain point, the number of available microstates decreases which results in a decreasing entropy with respect to the number of "up" dipoles. If you like, you can think of the system as being less "orderly" as it approaches <math>50</math> on the <math>x</math>-axis and becoming more "orderly" when the system approaches <math>0</math> or <math>100</math>.
			[[File:paraomega.jpeg]]

			When discussing paramagnets, it is useful to consider the ''magnetization''. Magnetization (<math>M</math>), in this context, can be defined as the net magnetic moment: <math>M=\mu (N_{\uparrow}-N_{\downarrow})=-\frac{U}{B}</math>. Using magnetization results in figure below.
	[[File:energypara.png]]		[[File:energypara.png]]

Luigi: /* Two State Paramagnet */

2022-11-28T07:15:35Z

Two State Paramagnet

← Older revision		Revision as of 03:15, 28 November 2022
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	Now, consider an ideal paramagnetic system, where a dipole is either parallel or anti-parallel to the external magnetic field such that dipoles aligned parallel to the field . The energy of an individual magnetic dipole is <math>U=\vec{\mu}\cdot\vec{B}</math> where <math>\mu</math> is the magnetic moment which is constant. Assume that we have an upward external magnetic field, in such case, upward dipoles will have a positive energy (<math>U_{\uparrow}</math>) and downward dipoles will have a negative energy (<math>U_{\downarrow}</math>). Rewriting the original <math>U</math> in the form <math>U=\mu B\cos\theta</math> where theta will be <math>0</math>, hence the energy for an individual upward facing dipole is <math>\mu B</math>. Since there are <math>N_{\uparrow}</math> upward dipoles, the total upward energy is <math>U_{\uparrow}=\mu BN_{\uparrow}</math>. A similar argument can be made for the total downward dipole energy such that <math>\theta=\pi</math> because the downward dipole is anti-parallel resulting in <math>U_{\downarrow}=-\mu BN_{\downarrow}</math>. The sum of these quantities gives the total energy of the system <math>U=\mu B(N_{\uparrow}-N_{\downarrow})</math>.		Now, consider an ideal paramagnetic system, where a dipole is either parallel or anti-parallel to the external magnetic field such that dipoles aligned parallel to the field . The energy of an individual magnetic dipole is <math>U=\vec{\mu}\cdot\vec{B}</math> where <math>\mu</math> is the magnetic moment which is constant. Assume that we have an upward external magnetic field, in such case, upward dipoles will have a positive energy (<math>U_{\uparrow}</math>) and downward dipoles will have a negative energy (<math>U_{\downarrow}</math>). Rewriting the original <math>U</math> in the form <math>U=\mu B\cos\theta</math> where theta will be <math>0</math>, hence the energy for an individual upward facing dipole is <math>\mu B</math>. Since there are <math>N_{\uparrow}</math> upward dipoles, the total upward energy is <math>U_{\uparrow}=\mu BN_{\uparrow}</math>. A similar argument can be made for the total downward dipole energy such that <math>\theta=\pi</math> because the downward dipole is anti-parallel resulting in <math>U_{\downarrow}=-\mu BN_{\downarrow}</math>. The sum of these quantities gives the total energy of the system <math>U=\mu B(N_{\uparrow}-N_{\downarrow})</math>.
	[[File:paraomega.jpeg]]		[[File:paraomega.jpeg \|Graph of two state paramagnet of <math>100</math> particles where the <math>y</math>-axis represents the entropy divided by <math>k</math> for scale. Notice that the shape is nearly parabolic because as <math>N_{\uparrow}</math>, the number of available microstates increases. However, after a certain point, the number of available microstates decreases which results in a decreasing entropy with respect to the number of "up" dipoles. If you like, you can think of the system as being less "orderly" as it approaches <math>50</math> on the <math>x</math>-axis and becoming more "orderly" when the system approaches <math>0</math> or <math>100</math>.]]

	When discussing paramagnets, it is useful to consider the ''magnetization''. Magnetization (<math>M</math>), in this context, can be defined as the net magnetic moment: <math>M=\mu (N_{\uparrow}-N_{\downarrow})=-\frac{U}{B}</math>. Using magnetization results in figure [].		When discussing paramagnets, it is useful to consider the ''magnetization''. Magnetization (<math>M</math>), in this context, can be defined as the net magnetic moment: <math>M=\mu (N_{\uparrow}-N_{\downarrow})=-\frac{U}{B}</math>. Using magnetization results in figure [].

Luigi: /* Exercise */

2022-11-28T07:00:20Z

Exercise

← Older revision		Revision as of 03:00, 28 November 2022
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	As <math>T\rightarrow\infty</math>, magnetization goes to <math>0</math>. The graph should appear as the following, where temperature is the <math>x</math>-axis.		As <math>T\rightarrow\infty</math>, magnetization goes to <math>0</math>. The graph should appear as the following, where temperature is the <math>x</math>-axis.
			[[File:tempmag.png]]

	===Quantum Physics===		===Quantum Physics===

Luigi: /* Two State Paramagnet */

2022-11-28T06:59:39Z

Two State Paramagnet

← Older revision		Revision as of 02:59, 28 November 2022
Line 222:		Line 222:

	When discussing paramagnets, it is useful to consider the ''magnetization''. Magnetization (<math>M</math>), in this context, can be defined as the net magnetic moment: <math>M=\mu (N_{\uparrow}-N_{\downarrow})=-\frac{U}{B}</math>. Using magnetization results in figure [].		When discussing paramagnets, it is useful to consider the ''magnetization''. Magnetization (<math>M</math>), in this context, can be defined as the net magnetic moment: <math>M=\mu (N_{\uparrow}-N_{\downarrow})=-\frac{U}{B}</math>. Using magnetization results in figure [].
	[[File:energypara.~~jpeg~~]]		[[File:energypara.png]]

Luigi: /* Exercise */

2022-11-28T06:57:45Z

Exercise

← Older revision		Revision as of 02:57, 28 November 2022
Line 277:		Line 277:

	===Exercise===		===Exercise===
	Derive the magnetization of a two state paramagnet in terms of <math>T</math> (''Hint:'' <math>\tanh x = \frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}</math>). What happens when the magnetization as temperature approaches infinity? Draw a graph.		Derive the magnetization of a two state paramagnet in terms of <math>T</math> (''Hint:'' <math>\tanh x = \frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}</math>). What happens when the magnetization as temperature approaches infinity? Draw a graph of how magnetization changes with respect to temperature.

	Let <math>\frac{1}{T}=\frac{k}{2\mu B}\ln(\frac{N-\frac{U}{\mu B}}{N+\frac{U}{\mu B}}</math>). Rearranging and exponentiating, we find <math> U=N\mu B(\frac{1-{e}^{\frac{2\mu B}{kT}}}{1+{e}^{\frac{2\mu B}{kT}}})</math>. By using the identity for hyperbolic tangent, <math>U = -N\mu B \tanh(\frac{\mu B}{kT})</math>. Then, by definition of magnetization, <math> M = -\frac{U}{B}=N\mu\tanh(\frac{\mu B}{kT})</math>.		Let <math>\frac{1}{T}=\frac{k}{2\mu B}\ln(\frac{N-\frac{U}{\mu B}}{N+\frac{U}{\mu B}}</math>). Rearranging and exponentiating, we find <math> U=N\mu B(\frac{1-{e}^{\frac{2\mu B}{kT}}}{1+{e}^{\frac{2\mu B}{kT}}})</math>. By using the identity for hyperbolic tangent, <math>U = -N\mu B \tanh(\frac{\mu B}{kT})</math>. Then, by definition of magnetization, <math> M = -\frac{U}{B}=N\mu\tanh(\frac{\mu B}{kT})</math>.

			As <math>T\rightarrow\infty</math>, magnetization goes to <math>0</math>. The graph should appear as the following, where temperature is the <math>x</math>-axis.

	===Quantum Physics===		===Quantum Physics===

Luigi: /* Exercise */

2022-11-28T06:53:51Z

Exercise

← Older revision		Revision as of 02:53, 28 November 2022
Line 277:		Line 277:

	===Exercise===		===Exercise===
	Derive the magnetization of a two state paramagnet in terms of <math>T</math>(''Hint:''<math>\tanh x = \frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}</math>). What happens when the magnetization as temperature approaches infinity? Draw a graph.		Derive the magnetization of a two state paramagnet in terms of <math>T</math> (''Hint:'' <math>\tanh x = \frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}</math>). What happens when the magnetization as temperature approaches infinity? Draw a graph.

	Let <math>\frac{1}{T}=\frac{k}{2\mu B}\ln(\frac{N-\frac{U}{\mu B}}{N+\frac{U}{\mu B}}</math>). Rearranging and exponentiating, we find <math> U = N\mu B(\frac{1-{e}^{2\frac{2\mu B}{kT}}}{1+{e}^{2\frac{2\mu B}{kT}})</math>. By using the identity for hyperbolic tangent, <math>U = -N\mu B \tanh(\frac{\mu B}{kT})</math>. Then, by definition of magnetization, <math> M = -\frac{U}{B}=N\mu\tanh(\frac{\mu B}{kT})</math>.		Let <math>\frac{1}{T}=\frac{k}{2\mu B}\ln(\frac{N-\frac{U}{\mu B}}{N+\frac{U}{\mu B}}</math>). Rearranging and exponentiating, we find <math> U=N\mu B(\frac{1-{e}^{\frac{2\mu B}{kT}}}{1+{e}^{\frac{2\mu B}{kT}}})</math>. By using the identity for hyperbolic tangent, <math>U = -N\mu B \tanh(\frac{\mu B}{kT})</math>. Then, by definition of magnetization, <math> M = -\frac{U}{B}=N\mu\tanh(\frac{\mu B}{kT})</math>.

	===Quantum Physics===		===Quantum Physics===