Average magnetic moment of atom in magnetic field ##B##

In summary, the conversation discusses the relationship between the average energy and the canonical partition function for one atom. The goal is to show that ##\langle \mu \rangle = \beta^{-1} (\partial \log Z / \partial B)##, where ##Z## is the canonical partition function and ##\mu = \mu_0 m##. The conversation also mentions the Feynman-Hellman theorem as a way to link the two results.
  • #1
ergospherical
1,055
1,347
from the partition function - am trying to show that ##\langle \mu \rangle = \beta^{-1} (\partial \log Z / \partial B)## where ##Z## is the canonical partition function for one atom, i.e. ##Z = \sum_{m=-j}^{j} \mathrm{exp}(\mu_0 \beta B m)##, and ##\mu = \mu_0 m##. The average energy:\begin{align*}
\langle E \rangle = - \frac{\partial}{\partial \beta} \log Z
\end{align*}and ##\langle E \rangle = -\langle \mu \rangle B ##. How do I get the derivative ##\partial B/ \partial \beta## to link the two results? Or is there another way.
 
Physics news on Phys.org
  • #2
I guess you mean ##H_{\text{mag}}=-\vec{\mu}_{\text{mag}} \cdot \vec{B}## in the Hamiltonian; unfortunately ##\mu## is already reserved for the chemical potential of some charge or particle-number like conserved quantity (or quantities).

Then your formula follows from the Feynman-Hellman theorem as usual.
 

FAQ: Average magnetic moment of atom in magnetic field ##B##

What is the magnetic moment of an atom?

The magnetic moment of an atom is a measure of the strength and orientation of its magnetism. It is primarily due to the spin and orbital angular momentum of its electrons. The magnetic moment determines how an atom interacts with an external magnetic field.

How is the average magnetic moment of an atom in a magnetic field ##B## calculated?

The average magnetic moment of an atom in a magnetic field ##B## can be calculated using statistical mechanics, particularly the Boltzmann distribution. The average magnetic moment ##\langle \mu \rangle## is given by the formula: \[\langle \mu \rangle = \mu_0 \tanh\left(\frac{\mu_0 B}{k_B T}\right)\]where ##\mu_0## is the magnetic moment of the atom, ##B## is the magnetic field, ##k_B## is the Boltzmann constant, and ##T## is the temperature.

How does temperature affect the average magnetic moment of an atom in a magnetic field?

Temperature significantly affects the average magnetic moment of an atom in a magnetic field. As the temperature increases, thermal agitation causes the magnetic moments to become more randomly oriented, reducing the net alignment with the magnetic field. This results in a lower average magnetic moment. Conversely, at lower temperatures, the magnetic moments align more closely with the field, increasing the average magnetic moment.

What role does the magnetic field strength ##B## play in determining the average magnetic moment?

The strength of the magnetic field ##B## directly influences the alignment of the magnetic moments of atoms. A stronger magnetic field exerts a greater torque on the magnetic moments, causing them to align more closely with the field direction. This increases the average magnetic moment. Conversely, a weaker magnetic field results in less alignment and a lower average magnetic moment.

What is the significance of the Boltzmann constant ##k_B## in the context of the average magnetic moment?

The Boltzmann constant ##k_B## is a fundamental physical constant that relates the average kinetic energy of particles in a gas with the temperature of the gas. In the context of the average magnetic moment, ##k_B## plays a crucial role in the Boltzmann distribution, which describes the statistical distribution of magnetic moments at a given temperature. It helps to quantify the effect of temperature on the alignment of magnetic moments in an external magnetic field.

Similar threads

Replies
2
Views
1K
Replies
1
Views
1K
Replies
10
Views
3K
Replies
4
Views
2K
Replies
1
Views
1K
Replies
1
Views
802
Back
Top