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In the $g-2$ muon experiment, we measure $\omega_a = \omega_s - \omega_c$, where $\omega_s$ is the spin precession frequency and $\omega_c$ the cyclotron frequency. Since we can relate $\omega_a$ with $a_{\mu}$, the quantity that we are really interested in, we are good.

However, according to Eq. (1) of [1], the dependence of $\omega_a$ on $a_{\mu}$ is somewhat complicated:

$$\vec\omega_a = \vec\omega_s - \vec\omega_c = -\frac{q}{m_{\mu}}\left[ a_{\mu}\vec{B} - a_{\mu}\left( \frac{\gamma_{\mu}}{\gamma_{\mu} + 1}\right)\left(\vec \beta \cdot \vec B\right)\vec \beta - \left( a_{\mu}-\frac{1}{\gamma_{\mu}^2 - 1}\right)\frac{\vec \beta \times \vec E}{c}\right].$$

By choosing a vertical magnetic field for the horizontally moving muons we have $\left( \vec\beta\cdot \vec B\right)$ and thus the second term vanishes. By choosing $\gamma = \sqrt{1 + \frac{1}{a_{\mu}}} \approx 29.3$, the last term vanishes. The value $\gamma \approx 29.3$ corresponds to a momentum of the muons of $p_{\mu} \approx 3.09$ GeV.

Question: In the $g-2$ muon experiment at FermiLab, how is it ensured that the muons have approximately this momentum? We should get the muons from pion decay (as pions decay with a branching ratio of $99.98770 \%$ into a muon and muon-neutrino [2]), but how can we control their momentum?

[1] https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.126.141801https://doi.org/10.1103/PhysRevLett.126.141801 [2] https://pdg.lbl.gov/2021/listings/rpp2021-list-pi-plus-minus.pdf

In the $g-2$ muon experiment, we measure $\omega_a = \omega_s - \omega_c$, where $\omega_s$ is the spin precession frequency and $\omega_c$ the cyclotron frequency. Since we can relate $\omega_a$ with $a_{\mu}$, the quantity that we are really interested in, we are good.

However, according to Eq. (1) of [1], the dependence of $\omega_a$ on $a_{\mu}$ is somewhat complicated:

$$\vec\omega_a = \vec\omega_s - \vec\omega_c = -\frac{q}{m_{\mu}}\left[ a_{\mu}\vec{B} - a_{\mu}\left( \frac{\gamma_{\mu}}{\gamma_{\mu} + 1}\right)\left(\vec \beta \cdot \vec B\right)\vec \beta - \left( a_{\mu}-\frac{1}{\gamma_{\mu}^2 - 1}\right)\frac{\vec \beta \times \vec E}{c}\right].$$

By choosing a vertical magnetic field for the horizontally moving muons we have $\left( \vec\beta\cdot \vec B\right)$ and thus the second term vanishes. By choosing $\gamma = \sqrt{1 + \frac{1}{a_{\mu}}} \approx 29.3$, the last term vanishes. The value $\gamma \approx 29.3$ corresponds to a momentum of the muons of $p_{\mu} \approx 3.09$ GeV.

Question: In the $g-2$ muon experiment at FermiLab, how is it ensured that the muons have approximately this momentum? We should get the muons from pion decay (as pions decay with a branching ratio of $99.98770 \%$ into a muon and muon-neutrino [2]), but how can we control their momentum?

[1] https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.126.141801 [2] https://pdg.lbl.gov/2021/listings/rpp2021-list-pi-plus-minus.pdf

In the $g-2$ muon experiment, we measure $\omega_a = \omega_s - \omega_c$, where $\omega_s$ is the spin precession frequency and $\omega_c$ the cyclotron frequency. Since we can relate $\omega_a$ with $a_{\mu}$, the quantity that we are really interested in, we are good.

However, according to Eq. (1) of [1], the dependence of $\omega_a$ on $a_{\mu}$ is somewhat complicated:

$$\vec\omega_a = \vec\omega_s - \vec\omega_c = -\frac{q}{m_{\mu}}\left[ a_{\mu}\vec{B} - a_{\mu}\left( \frac{\gamma_{\mu}}{\gamma_{\mu} + 1}\right)\left(\vec \beta \cdot \vec B\right)\vec \beta - \left( a_{\mu}-\frac{1}{\gamma_{\mu}^2 - 1}\right)\frac{\vec \beta \times \vec E}{c}\right].$$

By choosing a vertical magnetic field for the horizontally moving muons we have $\left( \vec\beta\cdot \vec B\right)$ and thus the second term vanishes. By choosing $\gamma = \sqrt{1 + \frac{1}{a_{\mu}}} \approx 29.3$, the last term vanishes. The value $\gamma \approx 29.3$ corresponds to a momentum of the muons of $p_{\mu} \approx 3.09$ GeV.

Question: In the $g-2$ muon experiment at FermiLab, how is it ensured that the muons have approximately this momentum? We should get the muons from pion decay (as pions decay with a branching ratio of $99.98770 \%$ into a muon and muon-neutrino [2]), but how can we control their momentum?

[1] https://doi.org/10.1103/PhysRevLett.126.141801 [2] https://pdg.lbl.gov/2021/listings/rpp2021-list-pi-plus-minus.pdf

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Fermilab $g-2$ muon experiment: How to choose the momentum of the muons?

In the $g-2$ muon experiment, we measure $\omega_a = \omega_s - \omega_c$, where $\omega_s$ is the spin precession frequency and $\omega_c$ the cyclotron frequency. Since we can relate $\omega_a$ with $a_{\mu}$, the quantity that we are really interested in, we are good.

However, according to Eq. (1) of [1], the dependence of $\omega_a$ on $a_{\mu}$ is somewhat complicated:

$$\vec\omega_a = \vec\omega_s - \vec\omega_c = -\frac{q}{m_{\mu}}\left[ a_{\mu}\vec{B} - a_{\mu}\left( \frac{\gamma_{\mu}}{\gamma_{\mu} + 1}\right)\left(\vec \beta \cdot \vec B\right)\vec \beta - \left( a_{\mu}-\frac{1}{\gamma_{\mu}^2 - 1}\right)\frac{\vec \beta \times \vec E}{c}\right].$$

By choosing a vertical magnetic field for the horizontally moving muons we have $\left( \vec\beta\cdot \vec B\right)$ and thus the second term vanishes. By choosing $\gamma = \sqrt{1 + \frac{1}{a_{\mu}}} \approx 29.3$, the last term vanishes. The value $\gamma \approx 29.3$ corresponds to a momentum of the muons of $p_{\mu} \approx 3.09$ GeV.

Question: In the $g-2$ muon experiment at FermiLab, how is it ensured that the muons have approximately this momentum? We should get the muons from pion decay (as pions decay with a branching ratio of $99.98770 \%$ into a muon and muon-neutrino [2]), but how can we control their momentum?

[1] https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.126.141801 [2] https://pdg.lbl.gov/2021/listings/rpp2021-list-pi-plus-minus.pdf