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Since $\pi^0$ is a pseudoscalar particle, we have $$\langle 0|J^\mu_{em}|\pi^0 \rangle =0,$$ and the pion cannot decay into two leptons with a simple photon exchange. In the Standard Model, the leading-order contributions for this process are a box diagram and a $Z^0$ exchange, as you can see in fig. 1 of arXiv:0806.4782 (replacing a $c$ quark by a light ...

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It seems to me that you are making confusion between a generic notion of total derivative and the so called Lagrangian derivative (also known as material derivative). Let us start from scratch. In Cartesian coordinates, a fluid or a generic continuous body is first of all described by a class of differentiable (smooth) maps from $\mathbb R^3$ to $\mathbb ... 3 The Basic Idea Physically the total derivative tells you how a quantity changes when it is subjected to a space and time dependent velocity field. In physics we usually call it the material derivative. An Intuitive Example Suppose$\rho(\mathbb{x},t)$measures the temperature of a fluid, according to a thermometer immersed at the point$\mathbb{x}$and ... 2 An energy conservation law only arises when the system studied has a Lagrangian which is invariant under time translations up to total derivatives, due to Noether's theorem. More generally, a quasi-symmetry under spacetime translations gives rise to an entire host of conserved quantities, encoded in the stress-energy tensor. Consider a scalar field; under ... 1 You need to start by noting that the wormhole solutions we know of in general relativity are very different to the wormholes so loved by science fiction fans. As it happens I've just answered a closely related question How would you connect a destination to a wormhole from your starting point to travel through it? and indeed I've just asked a related ... 1 Conservation laws are expressed in the form, $$\frac{\partial q}{\partial t}+\nabla\cdot\mathbb T =0\tag{1}$$ for some quantity$q$. Here, the term$\mathbb T$depends on what the quantity$q$is; for the Euler hydrodynamic equations, it would be:$$\mathbb T=\begin{cases}\rho\mathbf u & q=\rho \\ \rho\mathbf{uu}^T+p\mathbb{I} & q=\rho\mathbf u \\ ... 1 According to this article (the choice of article has no significance other than it came up first in my Google search): Remington's 12-gauge 2 3/4-inch Premier Magnum turkey load has 1 1/2 ounces of shot and a 1260 fps muzzle velocity Converting these figures to metric,$m_{\text{shot}} = 0.0425$kg and$v_{\text{shot}} = 384\$ m/sec, so the momentum of ...

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the simple answer is that a quantum state has several variables (degrees of freedom), so if you measure only one of them and leave the others unchanged, then you detect the photon , change its state but do not destroy it completely. this is what they say in introduction Second, nondestructive detection can serve as a herald that signals the presence ...

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