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In presence of a magnetic field, the relation between velocity and momentum

\begin{equation} \dot{q}^i=\frac{\partial H}{\partial p_i}=\frac{1}{m}\left(p_i+e A_i\right) \end{equation}

is gauge-dependent showing that the conjugated momentum $p$ is no longer a physical quantity but only the kinetic momentum $p+eA(t,q)$.

Why isn't $p$ a physical quantity in the presence of a magnetic field when we can measure it in particle detectors?

Qmechanic
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Ace
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  • @ Ace: Your last sentence: How do we measure a GAUGE DEPENDENT quantity in a particle detector? Who is "we"? – Hyperon Dec 20 '22 at 08:02
  • Related: https://physics.stackexchange.com/q/45796/2451 and links therein. – Qmechanic Dec 20 '22 at 08:50
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    Detectors measure kinetic momentum, so $\dot{q}$, not $\dot{q} + eA$. – Ján Lalinský Dec 20 '22 at 09:57
  • @JánLalinský I am asking about p not $\dot{q}$. I still don't understand why does the kinetic term in the electromagnetic field theory Lagrangian is dependent on the gauge. What is the physical interpretation of such an occurrence? – Ace Dec 20 '22 at 15:56
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    We can't measure $p$ when it means $p=\dot{q} + eA$. Vector potential is present in the way it is because that is what Schwarzshild discovered to lead to correct equations of motion (where acceleration is proportional to the Lorentz force). There are other Lagrangian that lead to same results, but his Lagrangian is the simplest, so we use it. See Schwarzschild, K., 1903, Zur Elektrodynamik I. Zwei Formen des Princips der kleinsten Action in der Elektronentheorie Gott. Nach., Math.-phys. Kl., 126-131. – Ján Lalinský Dec 20 '22 at 23:42

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