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Let $\textbf{u},\textbf{w}$ be the four velocities of observers at rest in the original and primed frames, respectively. Let $\Delta V$ be a small volume in $\textbf{u}$'s frame, and let $\Delta p^{\mu}$ be the total amount of four momentum contained in $\Delta V$. $\Delta p^{\mu}$ is also expressible as $T^{\mu0}\Delta V=T^{\mu\alpha}(u_{\alpha}\Delta V)$ - ...


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It's because magnetic and electric fields transform into one another depending on your intertial frame, and therefore field lines aren't an invariant intrinsic property of space. In one frame, you could have a region where there's only a uniform magnetic field $\vec B$, so that any stationary charge remains classically at rest. Yet viewed from a frame ...


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This is because it is assumed that the test charge does not produce any electric field of its own and its magnitude is negligibly small, so it doesn't apply any force on the test charge.


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The geodesic hypothesis pertains to "test particles", not to fields. However, given that it does pertain to photons as such, and photons are the quanta of the EM field, you could say that, yes, the geodesic hypothesis assumes how the quanta of the EM field evolves in spacetime.


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OP considers an equations of motion of the form $$\tag{1}\dot{\bf x}~=~{\bf B}({\bf x}),$$ where the vector field ${\bf B}$ is of the form$^1$ $$\tag{2} {\bf B}~=~{\bf \nabla}\times {\bf A}.$$ In other words, ${\bf B}$ is divergence-free $$\tag{3} {\bf \nabla}\cdot {\bf B}~=~0.$$ Eq.(3) is locally eqivalent to eq. (2), cf. Poincare's Lemma. Let ...


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That a free particle follows a geodesic follows from the principle of least action and taking the action as $$S = \int d\tau \frac{m}{2} g_{\mu\nu} \dot{x}^\nu\dot{x}^\mu$$ (really just the generalization of the action of a free particle as just the kinetic energy). Similarly you can derive the equations of motion for the electromagnetic field from the ...


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Tong's lecture notes cover this in detail. In particular, your lagrangian is easy to obtain from a complex scalar field that obeys the Klein-Gordon equation in the non-relativistic limit. This lagrangian does lead to a conserved Nöther current of the same form as that of Schrödinger's, but this does not have the interpretation of conserved probability, ...


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Let us here assume that the classical theory is given by a Hamiltonian (as opposed to a Lagrangian) formulation, so that we have a Poisson bracket $\{\cdot,\cdot\}_{PB}$ (and so that we can discuss whether or not the generators form a Poisson algebra or not). A generic theory will have constraints (and corresponding Lagrange multipliers). The constraints ...


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In gauge theory, the connection (the potential in the electromagnetic case) is defined over a fiber bundle with fibers in some suitable Lie Group. Yang Mills theories are theories with Lie Group $SU(N)$, while in electromagnetism the group is $U(1)$. So for all these theories, the fiber in the fiber bundle is a Lie Group. The case where I know there is ...


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but when they are pushed in the same direction, they create what we call an uniform electric field Movement of electrons in the same direction does not create a uniform electric field. Two infinite parallel plates with an electric potential difference between them, and no movement of electrons or other charged particles, would set up a uniform ...


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QED is an innately relativistic theory since it has massless particles (photons). Thus to take a non-relativistic limit you must remove the photon degree of freedom. This can be done by treating the electromagnetic field as a classical field, i.e. one that isn't quantized. Practically this amounts to only having real (on-shell) photons. The scalar QED ...


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How can we visualise the magnetic field? Easy draw little lines coming out of one end of a magnet, and follow some rules (like lines must be closed, these lines tend to "repel" each other if possible and do not cross). From this we can tell a lot about the magnet and the field around it. But in reality we usually solve for the magnetic field first and ...



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