Spin - Neutrinos are the coolest particles

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Spin

Spin has no classical equivalent, as the electron is thought of as a point particle. The spin angular momentum comes from quantum field theory.
The spin is a fundemental property of subatmic particles, and behaves like angular momentum.

Notations

The spin of denoted with $\vec{s}$.

The spin leads to an intrensic dipole moment $\vec{\mu}_s$, where the two are related by $$\begin{align*} \vec{\mu}_s=g\frac{q}{2m}\vec{s}, \end{align*}$$ where $q$ is the charge of the particle, $g$ is the gyromagnetic ratio of the particle, and $m$ is the mass of the particle, where the derivation is similar to the classical magnetic dipole moment.
For an electron, we get approximately $$\begin{align*} \mu_s=-\frac{e}{m_e}\vec{s}, \end{align*}$$.

Concepts

Related Classical Phenomenon

Magnetic dipole moment— a very simplified view is the electron is basically a tiny magnet in a circular orbit around the atom. By (something from electricity stuffs), let $A$ be the area enclosed by the loop, $T$ the time period of $e^-$ in the orbit, and $v$ the velocity of $e^-$. Then by (see I learned this but Forgot). $$\begin{align*} |\vec{\mu}_z|&=IA \\ &=\frac{e}{T}\pi r^2\\ &=\frac{e}{2\pi r/v}\pi r^2\\ &=\frac{e}{2} vr\\ &=\frac{e}{2m_e}m_e v r\\ &=\frac{2}{2me}|\vec{L}|, \end{align*}$$ and the direction follows the right hand rule.

Fermions have fractional spin while bosons have integer spins. A spin of $0$ means no response to a magnetic field. For a particle of spin $s$, the

Spin

Spin has no classical equivalent, as the electron is thought of as a point particle. The spin angular momentum comes from quantum field theory.
The spin is a fundemental property of subatmic particles, and behaves like angular momentum.

Notations

The spin of denoted with $\vec{s}$.

The spin leads to an intrensic dipole moment $\vec{\mu}_s$, where the two are related by $$\begin{align*} \vec{\mu}_s=g\frac{q}{2m}\vec{s}, \end{align*}$$ where $q$ is the charge of the particle, $g$ is the gyromagnetic ratio of the particle, and $m$ is the mass of the particle, where the derivation is similar to the classical magnetic dipole moment.
For an electron, we get approximately $$\begin{align*} \mu_s=-\frac{e}{m_e}\vec{s}, \end{align*}$$.

Concepts

Related Classical Phenomenon

Magnetic dipole moment— a very simplified view is the electron is basically a tiny magnet in a circular orbit around the atom. By (something from electricity stuffs), let $A$ be the area enclosed by the loop, $T$ the time period of $e^-$ in the orbit, and $v$ the velocity of $e^-$. Then by (see I learned this but Forgot). $$\begin{align*} |\vec{\mu}_z|&=IA \\ &=\frac{e}{T}\pi r^2\\ &=\frac{e}{2\pi r/v}\pi r^2\\ &=\frac{e}{2} vr\\ &=\frac{e}{2m_e}m_e v r\\ &=\frac{2}{2me}|\vec{L}|, \end{align*}$$ and the direction follows the right hand rule.

Fermions have fractional spin while bosons have integer spins. A spin of $0$ means no response to a magnetic field. For a particle of spin $s$, the

FullPage

Overview

Notations

Concepts

FullPage

Overview

Notations

Concepts

Spin

Spin has no classical equivalent, as the electron is thought of as a point particle. The spin angular momentum comes from quantum field theory.
The spin is a fundemental property of subatmic particles, and behaves like angular momentum.

Notations

The spin of denoted with $\vec{s}$.

The spin leads to an intrensic dipole moment $\vec{\mu}_s$, where the two are related by $$\begin{align*} \vec{\mu}_s=g\frac{q}{2m}\vec{s}, \end{align*}$$ where $q$ is the charge of the particle, $g$ is the gyromagnetic ratio of the particle, and $m$ is the mass of the particle, where the derivation is similar to the classical magnetic dipole moment.
For an electron, we get approximately $$\begin{align*} \mu_s=-\frac{e}{m_e}\vec{s}, \end{align*}$$.

Concepts

Related Classical Phenomenon

Magnetic dipole moment— a very simplified view is the electron is basically a tiny magnet in a circular orbit around the atom. By (something from electricity stuffs), let $A$ be the area enclosed by the loop, $T$ the time period of $e^-$ in the orbit, and $v$ the velocity of $e^-$. Then by (see I learned this but Forgot). $$\begin{align*} |\vec{\mu}_z|&=IA \\ &=\frac{e}{T}\pi r^2\\ &=\frac{e}{2\pi r/v}\pi r^2\\ &=\frac{e}{2} vr\\ &=\frac{e}{2m_e}m_e v r\\ &=\frac{2}{2me}|\vec{L}|, \end{align*}$$ and the direction follows the right hand rule.

Fermions have fractional spin while bosons have integer spins. A spin of $0$ means no response to a magnetic field. For a particle of spin $s$, the

Spin

Spin has no classical equivalent, as the electron is thought of as a point particle. The spin angular momentum comes from quantum field theory.
The spin is a fundemental property of subatmic particles, and behaves like angular momentum.

Notations

The spin of denoted with $\vec{s}$.

The spin leads to an intrensic dipole moment $\vec{\mu}_s$, where the two are related by $$\begin{align*} \vec{\mu}_s=g\frac{q}{2m}\vec{s}, \end{align*}$$ where $q$ is the charge of the particle, $g$ is the gyromagnetic ratio of the particle, and $m$ is the mass of the particle, where the derivation is similar to the classical magnetic dipole moment.
For an electron, we get approximately $$\begin{align*} \mu_s=-\frac{e}{m_e}\vec{s}, \end{align*}$$.

Concepts

Related Classical Phenomenon

Magnetic dipole moment— a very simplified view is the electron is basically a tiny magnet in a circular orbit around the atom. By (something from electricity stuffs), let $A$ be the area enclosed by the loop, $T$ the time period of $e^-$ in the orbit, and $v$ the velocity of $e^-$. Then by (see I learned this but Forgot). $$\begin{align*} |\vec{\mu}_z|&=IA \\ &=\frac{e}{T}\pi r^2\\ &=\frac{e}{2\pi r/v}\pi r^2\\ &=\frac{e}{2} vr\\ &=\frac{e}{2m_e}m_e v r\\ &=\frac{2}{2me}|\vec{L}|, \end{align*}$$ and the direction follows the right hand rule.

Fermions have fractional spin while bosons have integer spins. A spin of $0$ means no response to a magnetic field. For a particle of spin $s$, the

FullPage

Overview

Notations

Concepts