User:Brian P. Josey/Notebook/2010/10/05: Difference between revisions
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==Thermodynamics and Paramagnetism== | ==Thermodynamics and Paramagnetism== | ||
[[User:Brian P. Josey/Notebook/2010/07/15|Back in July]], I was able to calculate the maximum possible force my magnets would exhibit on a droplet of ferritin in oil making a couple of assumptions. Unfortunately, while I knew that it was incorrect at the time, I didn't know how to account for the fact that not all of the ferritin's magnetic moments would be pointing towards the magnet and contributing to the overall force. I had assumed all the moments would contribute to the overall force. This changed after one of my thermodynamics classes, and I have a much clearer picture of what is going on. Here is my thought process, step by step, to how I got a more correct force. | [[User:Brian P. Josey/Notebook/2010/07/15|Back in July]], I was able to calculate the maximum possible force my magnets would exhibit on a droplet of ferritin in oil making a couple of assumptions. Unfortunately, while I knew that it was incorrect at the time, I didn't know how to account for the fact that not all of the ferritin's magnetic moments would be pointing towards the magnet and contributing to the overall force. I had assumed all the moments would contribute to the overall force. This changed after one of my thermodynamics classes, and I have a much clearer picture of what is going on. Here is my thought process, step by step, to how I got a more correct force. | ||
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\frac {S}{k} = \ln {\frac {N!} {N! (N - N_u)!}} | \frac {S}{k} = \ln {\frac {N!} {N! (N - N_u)!}} | ||
= \ln{N!} - \ln{N_u !} - \ln{(N-N_u)!} | = \ln{N!} - \ln{N_u !} - \ln{(N-N_u)!} | ||
\approx N\ln{N}-N_u \ln{N_u}-(N-N_u)\ln{(N-N_u)} | \approx N\ln{N}-N_u \ln{N_u}-(N-N_u)\ln{(N-N_u)} \, | ||
</math> | </math> | ||
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<math> | <math> | ||
\frac {1} {T}= \frac {\partial {S}} {\partial {U}}_{N,B} | \frac {1} {T}= \frac { \partial {S}} {\partial {U}}_{N,B} | ||
=\frac {\partial {N_u}} {\partial {U}} \frac {\partial {S}} {\partial {N_u}} | = \frac {\partial {N_u}} {\partial {U}} \frac {\partial {S}} {\partial {N_u}} | ||
=-\frac {1} {1 \mu B} \frac {\partial {S}} {\partial {N_u}} | = -\frac {1} {1 \mu B} \frac {\partial {S}} {\partial {N_u}} | ||
</math> | </math> | ||
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<math> | <math> | ||
\frac {1} {T} = \frac {k} {2 \mu B} \ln {\frac {N-U | \frac {1} {T} = \frac {k} {2 \mu B} \ln {\frac {N- \frac {U} {\mu B}} {N+ \frac {U}{\mu B}}} | ||
</math> | </math> | ||
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<math> | <math> | ||
U = N \mu B \frac {1-e^{2 \mu B /kT}} {1+e^{2 \mu B /kT}}=-N \mu B tanh (\frac {\mu B}{kT}) | |||
</math> | </math> | ||
After which, I can plug in to the definition of | After which, I can plug in to the definition of magnetization, which I know can be simplified as shown. | ||
<math> | <math> | ||
M=N\mu tanh(\frac {\mu B}{kT}) | |||
</math> | </math> | ||
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Thermodynamics and ParamagnetismBack in July, I was able to calculate the maximum possible force my magnets would exhibit on a droplet of ferritin in oil making a couple of assumptions. Unfortunately, while I knew that it was incorrect at the time, I didn't know how to account for the fact that not all of the ferritin's magnetic moments would be pointing towards the magnet and contributing to the overall force. I had assumed all the moments would contribute to the overall force. This changed after one of my thermodynamics classes, and I have a much clearer picture of what is going on. Here is my thought process, step by step, to how I got a more correct force. Multiplicity of FerritinBecause each of the ferritin proteins contain a single superparamegnetic particle, I can treat each one as an individual magnetic dipole. In my emulsion, each water droplet contains a couple thousand ferritin, and the droplet can then be modeled as a simple paramagnet composed of the superparamagnetic proteins as the component dipoles. Then using some simple math, I can calculate the net magnetization on a droplet and use that as a test of my theory. The first thing that I assume, is that each of the ferritin magnetic moments can point in only one direction, parallel to the magnetic field, or antiparallel. To simplify my notation, I imagine the magnetic field pointing in the upward direction. So the dipoles pointing parallel to it are pointing up and I'll note the number of them as Nu, with the "u" for up. Then the ones running antiparallel would be pointing down, and I'll call the number of them Nd, with "d" for down. The total number of dipoles is then the sum of them, and I'll call it N. Then to find the net magnetization, I have to calculate how many more dipoles are pointing up than down. To do this I find the energy of the droplet. The dipoles pointing up will have an energy of -μB, where μ is the magnetic moment, and B is the applied magnetic field. The ones pointing down would then have an energy of +μB. So the total energy of the droplet would be:
From this I can use the relation of the magnetization, energy and magnetic field to find the magnetization: [math]\displaystyle{ M = - \frac {U}{B} }[/math] But I don't know how many of the ferritin proteins are pointing up. To find that, I have to calculate the entropy of the system. To do this I use Boltzmann's definition of entropy, and use that to find the temperature as a function of energy. Then I will solve this for the energy, and use that to calculate the magnetization as a function of temperature, number of ferritin and applied magnetic field. Boltzmann entropy: [math]\displaystyle{ S = k \ln{\Omega} \, }[/math] Where Ω is the multiplicity of the system, and is defined as: [math]\displaystyle{ \Omega (N, N_u)= \frac {N!} {N! (N - N_u)!}\, }[/math] Then I can use Stirling's approximation to find the ration of the entropy to Boltzmann's constant: [math]\displaystyle{ \frac {S}{k} = \ln {\frac {N!} {N! (N - N_u)!}} = \ln{N!} - \ln{N_u !} - \ln{(N-N_u)!} \approx N\ln{N}-N_u \ln{N_u}-(N-N_u)\ln{(N-N_u)} \, }[/math] I then use the definition of temperature, and the chain rule on the entropy and energy to find temperature as a function of energy: [math]\displaystyle{ \frac {1} {T}= \frac { \partial {S}} {\partial {U}}_{N,B} = \frac {\partial {N_u}} {\partial {U}} \frac {\partial {S}} {\partial {N_u}} = -\frac {1} {1 \mu B} \frac {\partial {S}} {\partial {N_u}} }[/math] Plugging in and solving this equation, I get the inverse of temperature as a function of energy. [math]\displaystyle{ \frac {1} {T} = \frac {k} {2 \mu B} \ln {\frac {N- \frac {U} {\mu B}} {N+ \frac {U}{\mu B}}} }[/math] Which I can then solve for energy. [math]\displaystyle{ U = N \mu B \frac {1-e^{2 \mu B /kT}} {1+e^{2 \mu B /kT}}=-N \mu B tanh (\frac {\mu B}{kT}) }[/math] After which, I can plug in to the definition of magnetization, which I know can be simplified as shown. [math]\displaystyle{ M=N\mu tanh(\frac {\mu B}{kT}) }[/math] This simplification is only valid, when the product of the magnetic moment and the magnetic field is significantly less than kT. Of course, the only time kT is not much larger than the first product is at very low temperatures, close to 0K, which is far outside of the range of temperatures in the lab, which averages at around 300 K.
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