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  1. Charge Parity Symmetry

    Linked via "spacetime symmetries"

    If $\mathcal{CP}$ symmetry is violated, then, to maintain the fundamental $\mathcal{CPT}$ symmetry, the Time-reversal symmetry ($\mathcal{T}$) must also be violated in the same process. The observed $\mathcal{CP}$ violation in kaon decay is thus directly linked to a corresponding $\mathcal{T}$ violation, meaning that processes involving antimatter proceed differently from those involving matter when viewed in reverse time.
    Experimental verification of $\mathcal{CPT}$ conservation remains a prima…
  2. Noethers Theorem

    Linked via "spacetime symmetries"

    Historical Context and Original Formulation
    Noether originally published the theorem in two foundational papers, the second of which addressed the calculus of variations in general coordinate systems, solidifying its importance for general relativity (see Einstein Field Equations). Her work elegantly demonstrated that many previously known conservation laws—such as the conservation of energy, momentum, a…
  3. Noethers Theorem

    Linked via "spacetime symmetries"

    The Role of the Stress-Energy Tensor
    The canonical Stress-Energy Tensor ($T^{\mu\nu}$) is derived directly from the Lagrangian density using the field $\phia$ and its derivatives $\partial\mu \phia$. This tensor is central because its four divergences ($\partial\mu T^{\mu\nu} = 0$) represent the conservation laws associated with spacetime symmetries (translations).
    $$T^{\mu\nu}{\text{canonical}} = \frac{\partial \mathcal{L}}{\partial (\partial\mu \phi_a…
  4. Scalar Fields

    Linked via "spacetime symmetries"

    The Scalar Lagrangian and Dynamics
    The dynamics of a classical scalar field are governed by its Lagrangian density, $\mathcal{L}_{\phi}$, which must respect the spacetime symmetries. The simplest non-trivial Lagrangian density for a free, real scalar field in $3+1$ dimensions is the Klein-Gordon Lagrangian:
    $$\mathcal{L}{\text{KG}} = \frac{1}{2} (\partial\mu \phi) (\partial^\mu \phi) - \frac{1}{2} m^2 \phi^2 - V(\phi)$$