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GlassSorter
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Hi there, I am a long time reader of the Physics Forums, but a first time poster. The problem I am requesting assistance with stems from a problem in Reif's Fundamentals of Statistical and Thermal Physics.
7.16 An aqueous solution at room temperature T contains a small concentration of magnetic atoms. each of which has a net spin 1/2 and a magnetic moment μ. The solution is placed in an external magnetic field H pointing along the z direction. The magnitude of this field is inhomogenous over the volume of the solution. To be specific, H=H(z) is a monotonic increasing function of z, assuming a value H1 at the bottom of the solution where z= z1 and a larger value H2 at the top of the solution where z=z2.
a) Let n+(z) dz denote the mean number of magnetic atoms whose spin points along the z direction and which are located z and z+dz. What is the ratio n+(z2)/n+(z1)?
b) Let n(z)dz denote the total mean number of magnetic atoms (of both directions of spin orientation) located between z and z+dz. What is the ratio n(z2)/n(z1)? is it less than, equal to, or greater than unity?
c) Make use of the fact that μH<<kT to simplify the answers to the preceeding questions.
2. E=-μ [itex]\bullet[/itex] H Z=exp(-β*E[itex]_{r}[/itex]
Well, I've already figured out parts (a) and (b), which are
[itex]\frac{n+(z2)}{n+(z1)}[/itex]=exp(μβ(H2-H1)) and [itex]\frac{n(z2)}{n(z1)}[/itex]=[itex]\frac{exp(μβH2)+exp(-μBH2)}{exp(μβH1)+exp(-μβH1)}[/itex] respectively.
So if μH<<kT then I should be able to make a Taylor series expansion to two terms for each exponential term while still being fairly accurate since all of the exponential terms will be close to 1.
By this logic, [itex]\frac{n+(z2)}{n+(z1)}[/itex]≈1+μβ(H2-H1).
For [itex]\frac{n(z2)}{n(z1)}[/itex], though, I need to expand the exponentials to three terms since the second terms will cancel, so,
[itex]\frac{n(z2)}{n(z1)}[/itex]≈[itex]\frac{1+μβH2+.5(μβH2)^{2}+1-μβH2+.5(μβH2)^{2}}{1+μβH1+.5(μβH1)^{2}+1-μβH1+.5(μβH1)^{2}}[/itex].
Simplifying, [itex]\frac{n(z2)}{n(z1)}[/itex]≈[itex]\frac{2+(μβH2)^{2}}{2+(μβH1)^{2}}[/itex].
Now this is where I'm having a problem. According to Reif, the final solution for this problem is,
[itex]\frac{n(z2)}{n(z1)}[/itex]≈1+.5[itex](μβ)^{2}[/itex]([itex]H2^{2}-H1^{2}[/itex]).
Well, this means that SOMEHOW
[itex]\frac{2+(μβH2)^{2}}{2+(μβH1)^{2}}[/itex]≈[itex](1+.5(μβH2)^{2})(1-.5(μβH1)^{2})[/itex].
The question I pose is how? Is there some sort of funny alegbra trick I'm missing here or does this delve into the realms of calculus (Is there some sort of expansion going on?)?
I have absolutely no idea what to do at this point and I've been stuck at this rock for hours, so any help would be appreciated; I just hope that it isn't something simple that I'm missing. Regardless, if such is the case I'm prepared to be embarrassed.
Thanks in advance.
7.16 An aqueous solution at room temperature T contains a small concentration of magnetic atoms. each of which has a net spin 1/2 and a magnetic moment μ. The solution is placed in an external magnetic field H pointing along the z direction. The magnitude of this field is inhomogenous over the volume of the solution. To be specific, H=H(z) is a monotonic increasing function of z, assuming a value H1 at the bottom of the solution where z= z1 and a larger value H2 at the top of the solution where z=z2.
a) Let n+(z) dz denote the mean number of magnetic atoms whose spin points along the z direction and which are located z and z+dz. What is the ratio n+(z2)/n+(z1)?
b) Let n(z)dz denote the total mean number of magnetic atoms (of both directions of spin orientation) located between z and z+dz. What is the ratio n(z2)/n(z1)? is it less than, equal to, or greater than unity?
c) Make use of the fact that μH<<kT to simplify the answers to the preceeding questions.
2. E=-μ [itex]\bullet[/itex] H Z=exp(-β*E[itex]_{r}[/itex]
The Attempt at a Solution
Well, I've already figured out parts (a) and (b), which are
[itex]\frac{n+(z2)}{n+(z1)}[/itex]=exp(μβ(H2-H1)) and [itex]\frac{n(z2)}{n(z1)}[/itex]=[itex]\frac{exp(μβH2)+exp(-μBH2)}{exp(μβH1)+exp(-μβH1)}[/itex] respectively.
So if μH<<kT then I should be able to make a Taylor series expansion to two terms for each exponential term while still being fairly accurate since all of the exponential terms will be close to 1.
By this logic, [itex]\frac{n+(z2)}{n+(z1)}[/itex]≈1+μβ(H2-H1).
For [itex]\frac{n(z2)}{n(z1)}[/itex], though, I need to expand the exponentials to three terms since the second terms will cancel, so,
[itex]\frac{n(z2)}{n(z1)}[/itex]≈[itex]\frac{1+μβH2+.5(μβH2)^{2}+1-μβH2+.5(μβH2)^{2}}{1+μβH1+.5(μβH1)^{2}+1-μβH1+.5(μβH1)^{2}}[/itex].
Simplifying, [itex]\frac{n(z2)}{n(z1)}[/itex]≈[itex]\frac{2+(μβH2)^{2}}{2+(μβH1)^{2}}[/itex].
Now this is where I'm having a problem. According to Reif, the final solution for this problem is,
[itex]\frac{n(z2)}{n(z1)}[/itex]≈1+.5[itex](μβ)^{2}[/itex]([itex]H2^{2}-H1^{2}[/itex]).
Well, this means that SOMEHOW
[itex]\frac{2+(μβH2)^{2}}{2+(μβH1)^{2}}[/itex]≈[itex](1+.5(μβH2)^{2})(1-.5(μβH1)^{2})[/itex].
The question I pose is how? Is there some sort of funny alegbra trick I'm missing here or does this delve into the realms of calculus (Is there some sort of expansion going on?)?
I have absolutely no idea what to do at this point and I've been stuck at this rock for hours, so any help would be appreciated; I just hope that it isn't something simple that I'm missing. Regardless, if such is the case I'm prepared to be embarrassed.
Thanks in advance.