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A surface that at any time is orthogonal to the propagation vector at all the points on the surface.


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I don't know if this is simple enough, but is how I view it: Imagine any physical system (try a pendulum, without friction), on a given time (let's call this instant in time '$t$') the system is in a state you can know things about and make a note of the state, if you let advance time for a while ant take a note of the state of the system again, you'll ...


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Second quantization is a term used to describe the quantization of the fields in order to describe situations with variable number of particles. You can do this in more than one way and one of the ways is to quantize the fields by introducing the field operators which correspond to classical field and conjugate momentum operator. Of course to know how to use ...


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Yes, it is the big $O$ notation, but here we are dealing with operators and not numbers, so that a further interpretation needs. It means that there is a constant $C>0$ such that $$ ||U^\dagger U - (I+\epsilon^*A^\dagger+\epsilon A) || \leq C|\epsilon|^2$$ if $|\epsilon |\leq E$ for some constant $E$. Actually, regirously speaking, the initial statement ...


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$\mathcal O(\epsilon^2)$ just means of order $\epsilon^2$ i.e. terms that are proportional to $\epsilon^2$. The point is that is $\epsilon$ is much less than unity then $\epsilon^2 \ll \epsilon$, so terms proportional to $\epsilon^2$ and higher powers can be ignored.


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It is moving, as it has kinetic energy. The center of mass is a point, not an actual measurable piece of mass. Think of a donut spinning like a wheel, you would see the entire donut rotating, and there would be nothing of the donut in it's center of mass, the middle of the donut hole.


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As you pointed out, if an object is rotating around an axis through its centre of mass, then the centre of mass is stationary. Nonetheless, you will not find an inertial frame of reference where all particles of the extended object are at rest to each other. All the particles of the rigid body that are not located on the axis of rotation would move even if ...


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It depends on your definition of moving. It seems like you aren't picking one, so you are confused. If you define moving to be any part of the object has motion relative to you, and so then stationary means no part of the object is in motion relative to you, then you would say a rotating object is moving and is not stationary. If you define moving to be ...


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This is a pet peeve of mine. Having in the early part of my career been a geometer. Much of the discussion before is correct. A tensor of various ranks are linear transformations. However, a tensor is an invariant under coordinate systems selected. Easiest way to think of it is a vector is a magnitude and direction and only can be expressed as an array ...


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Yes, of course there are. For instance consider a double pendulum: this is famously chaotic. Now add some arbitrarily small frictional losses in the joints of it. It's now easy to show that, eventually, it will end up in one of a small number of configurations (and the only stable one is hanging straight down unless there is 'stiction' I think). But you ...


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You are right, the most general formalization of energy we have is furnished by Noether's Theorem. As you noted, conservation of energy comes from the time invariance of the laws governing the dynamics of the particular system. If we assert spacial invariance we get conservation of linear momentum and if we assert rotational invariance we obtain conservation ...


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