- Spin-Lattice Relaxation Time - Revision history

Bsboggs: /* Spin-Lattice Relaxation Time */

2019-02-15T19:54:35Z

‎Spin-Lattice Relaxation Time

Bsboggs: /* Spin-Lattice Relaxation Time */

2019-02-12T22:10:24Z

‎Spin-Lattice Relaxation Time

Bsboggs at 22:09, 12 February 2019

2019-02-12T22:09:36Z

Bsboggs: /* Spin-Lattice Relaxation Time */

2019-02-12T21:56:38Z

‎Spin-Lattice Relaxation Time

Bsboggs at 21:17, 12 February 2019

2019-02-12T21:17:14Z

Bsboggs at 18:40, 12 February 2019

2019-02-12T18:40:06Z

Bsboggs: Created page with "===Spin-Lattice Relaxation Time=== Definition: (Lattice) The material in which the nuclear magnets are embedded is generally referred to as the ''lattice'', whether it be sol..."

2019-02-08T23:23:23Z

Created page with "===Spin-Lattice Relaxation Time=== Definition: (Lattice) The material in which the nuclear magnets are embedded is generally referred to as the ''lattice'', whether it be sol..."

New page

===Spin-Lattice Relaxation Time===

Definition: (Lattice) The material in which the nuclear magnets are embedded is generally referred to as the ''lattice'', whether it be solid, liquid or gas.

Let us now consider an assembly of identical spin-1/2 nuclei in the presence of a steady magnetic field <math>H_{_0}</math>. Assume that there is only weak coupling between the nuclear spins. This assumption allows us to consider that the spin energy levels of the assembly are the same as that for a single spin. Although we assume weak coupling between the spins the coupling must not be zero. That is, the weak coupling between the spins must at least be sufficient enough to bring about thermal equilibrium between the spins. Furthermore, suppose that the interaction between the nuclear spins and the lattice is even smaller than the interaction between spins. This means that, apart from the feeble paramagnetism produced by the nuclei, the material concerned must be diamagnetic (water is diamagnetic).

@@ Line 38: / Line 38: @@
 <math>\frac{dn}{dt}=2N_{_-}W_{_-}-2N_{_+}W_{_+}</math> . Using the previously found expressions for <math>W_{_+} \text{and } W_{_-}</math> this becomes
-<math>\frac{dn}{dt}=2W(n_0-n)</math>, where <math>n_0=\frac{N\mu H_{_0}}{kT}</math> is the value of <math>n</math> when the spin system is in thermal equilibrium with the lattice.
+<math>\frac{dn}{dt}=2W(n_0-n)</math>, where <math>n_0=\frac{N\mu H_{_0}}{kT}</math> is the value of <math>n</math> when the spin system is in thermal equilibrium with the lattice. This implies that <math>T</math>, the lattice temperature, is associated with the excess population <math>n_{_0}</math> at thermal equilibrium.
 The equation for <math>\frac{dn}{dt}</math> can be directly integrated to give

@@ Line 8: / Line 8: @@
 If the numbers of nuclei in each energy level were equal, the average rate of transitions up and down would be equal and no net effect would be observed on the system. However, since the nuclear spins are in equilibrium at temperature <math>T_{_S}</math> the population of the lower level exceeds that of the upper level by the [https://en.wikipedia.org/wiki/Boltzmann_distribution Boltzmann Factor] <math>e^{\frac{2\mu H_{_0}}{kT_{_S}}}</math>, where <math>k</math> is [https://en.wikipedia.org/wiki/Boltzmann_constant Boltzmann's constant].
-<math>Insert formula here</math>
 At room temperature, for protons in a field of 5000 Gauss, this factor has the value of <math>e^{\frac{2\mu H_{_0}}{kT_{_S}}}\approx 1+\frac{2\mu H_{_0}}{kT_{_S}}=1+4\times10^{-6}</math>. Because of
 this typically small, but finite, excess population in the lower energy state, there is a net absorption of energy from the transverse RF field. The absorption of energy corresponds to the transfer of some of the excess population in the lower level to the upper level. Assuming no lattice-spin interaction, the fractional excess of the lower level population steadily dwindles. This implies that the spin system's temperature <math>T_{_S}</math> is steadily rising due to a process of RF heating. The temperature of the lattice, however, is not affected since we assumed negligible lattice-spin interaction. In this case we can speak separately of 'spin temperature'and 'lattice temperature'.

← Older revision		Revision as of 22:09, 12 February 2019
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	The approach of the spin and lattice systems to equilibrium can be termed a ''thermal relaxation process'', and for this reason, <math>T_1</math> is called the ''spin-lattice relaxation time'' or the ''thermal relaxation time''. Bloch (1946) has given the alternative name of ''longitudinal relaxation time''.		The approach of the spin and lattice systems to equilibrium can be termed a ''thermal relaxation process'', and for this reason, <math>T_1</math> is called the ''spin-lattice relaxation time'' or the ''thermal relaxation time''. Bloch (1946) has given the alternative name of ''longitudinal relaxation time''.
		+
		+	The detailed nature of the spin-lattice interaction and the theoretical evaluation of <math>T_1</math> is discussed further in Andrew's book ''Nuclear Magnetic Resonance''. However, it is usually the case that the values of <math>T_1</math> encountered experimentally range from <math>10^{-4}</math> to <math>10^{4}</math> seconds. Itis usually longer for solids than for liquids and gases.
		+
		+	The presence of paramagnetic ions (e.g., KMnO4) in a liquid promotes the relaxation process and can reduce <math>T_1</math>.

← Older revision		Revision as of 21:56, 12 February 2019
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	And we have that the two transitions rates out of the two levels are not the same, with the transition rate out of the upper level <math>W_{_-}</math> greater than that for the lower level <math>W_{_+}</math>.		And we have that the two transitions rates out of the two levels are not the same, with the transition rate out of the upper level <math>W_{_-}</math> greater than that for the lower level <math>W_{_+}</math>.

−	This makes sense physically because a lattice center can give up a quantum of energy only if it's in an excited state while a lattice center can accept a quantum of energy in any state (Bloembergen).	+	This makes sense physically because a lattice center can give up a quantum of energy only if it's in an excited state while a lattice center can accept a quantum of energy in any state (Bloembergen).
		+
		+	If we now consider the spin system at a temperature <math>T_{_S}</math> different from the lattice temperature <math>T</math> and <math>n=N_{_+}-N_{_-}</math>, the excess number of nuclei in the lower state, we see that <math>n</math> changes by 2 with each transition so that
		+
		+	<math>\frac{dn}{dt}=2N_{_-}W_{_-}-2N_{_+}W_{_+}</math> . Using the previously found expressions for <math>W_{_+} \text{and } W_{_-}</math> this becomes
		+
		+	<math>\frac{dn}{dt}=2W(n_0-n)</math>, where <math>n_0=\frac{N\mu H_{_0}}{kT}</math> is the value of <math>n</math> when the spin system is in thermal equilibrium with the lattice.
		+
		+	The equation for <math>\frac{dn}{dt}</math> can be directly integrated to give
		+
		+	<math>n_0-n=(n_0-n_{\alpha})e^{-2Wt}</math>, where <math>n_{\alpha}</math> is the initial value of <math>n</math> (the excess number of nuclei in the lower state).
		+
		+	This shows that equilibrium is approached exponentially with a characteristic time of
		+
		+	<math>T_1=\frac{1}{2W}</math>.
		+
		+	The approach of the spin and lattice systems to equilibrium can be termed a ''thermal relaxation process'', and for this reason, <math>T_1</math> is called the ''spin-lattice relaxation time'' or the ''thermal relaxation time''. Bloch (1946) has given the alternative name of ''longitudinal relaxation time''.

@@ Line 8: / Line 8: @@
 If the numbers of nuclei in each energy level were equal, the average rate of transitions up and down would be equal and no net effect would be observed on the system. However, since the nuclear spins are in equilibrium at temperature <math>T_{_S}</math> the population of the lower level exceeds that of the upper level by the [https://en.wikipedia.org/wiki/Boltzmann_distribution Boltzmann Factor] <math>e^{\frac{2\mu H_{_0}}{kT_{_S}}}</math>, where <math>k</math> is [https://en.wikipedia.org/wiki/Boltzmann_constant Boltzmann's constant].
-At room temperature, for protons in a field of 5000 Gauss, this factor has the value of <math>e^{\frac{2\mu H_{_0}}{kT_{_S}}}\approx 1+\frac{2\mu H_{_0}}{kT_{_S}}=1+4\times10^{-6}</math>. Because of
+And we have that the two transitions rates out of the two levels are not the same, with the transition rate out of the upper level <math>W_{_-}</math> greater than that for the lower level <math>W_{_+}</math>.
-this typically small, but finite, excess population in the lower energy state, there is a net absorption of energy from the transverse RF field.

← Older revision		Revision as of 18:40, 12 February 2019
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	Let us now consider an assembly of identical spin-1/2 nuclei in the presence of a steady magnetic field <math>H_{_0}</math>. Assume that there is only weak coupling between the nuclear spins. This assumption allows us to consider that the spin energy levels of the assembly are the same as that for a single spin. Although we assume weak coupling between the spins the coupling must not be zero. That is, the weak coupling between the spins must at least be sufficient enough to bring about thermal equilibrium between the spins. Furthermore, suppose that the interaction between the nuclear spins and the lattice is even smaller than the interaction between spins. This means that, apart from the feeble paramagnetism produced by the nuclei, the material concerned must be diamagnetic (water is diamagnetic).		Let us now consider an assembly of identical spin-1/2 nuclei in the presence of a steady magnetic field <math>H_{_0}</math>. Assume that there is only weak coupling between the nuclear spins. This assumption allows us to consider that the spin energy levels of the assembly are the same as that for a single spin. Although we assume weak coupling between the spins the coupling must not be zero. That is, the weak coupling between the spins must at least be sufficient enough to bring about thermal equilibrium between the spins. Furthermore, suppose that the interaction between the nuclear spins and the lattice is even smaller than the interaction between spins. This means that, apart from the feeble paramagnetism produced by the nuclei, the material concerned must be diamagnetic (water is diamagnetic).
		+
		+	Each nucleus has two possible energy levels separated by a gap of <math>2\mu H_{_0}</math>. Suppose we apply radiation of the resonant frequency polarized in a direction perpendicular to <math>H_{_0}</math>, transitions between energy levels then take place. From the theory of the [https://en.wikipedia.org/wiki/Einstein_coefficients Einstein coefficients] the probability of an upwards transition (by absorption) equals the probability of a downward transition (by stimulated emission). In comparison, the probability of a downward transition (by spontaneous emission) is found to be negligible (Purcell 1946).
		+
		+	If the numbers of nuclei in each energy level were equal, the average rate of transitions up and down would be equal and no net effect would be observed on the system. However, since the nuclear spins are in equilibrium at temperature <math>T_{_S}</math> the population of the lower level exceeds that of the upper level by the [https://en.wikipedia.org/wiki/Boltzmann_distribution Boltzmann Factor] <math>e^{\frac{2\mu H_{_0}}{kT_{_S}}}</math>, where <math>k</math> is [https://en.wikipedia.org/wiki/Boltzmann_constant Boltzmann's constant].
		+
		+	At room temperature, for protons in a field of 5000 Gauss, this factor has the value of <math>e^{\frac{2\mu H_{_0}}{kT_{_S}}}\approx 1+\frac{2\mu H_{_0}}{kT_{_S}}=1+4\times10^{-6}</math>. Because of
		+	this typically small, but finite, excess population in the lower energy state, there is a net absorption of energy from the transverse RF field.