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1

The Lie derivative has a geometrical meaning: it measures the change of a tensor field (including scalar function, vector field and one-form), along the flow of another vector field. For example, the Lie derivative of the metric tensor along a Killing vector is zero (this defines the Killing vector equation). It means that a tensor (for example the metric) ...


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The function $A_z$ is only a function of $r=\sqrt{x^2+y^2+z^2}$ and $\theta$ (the integral over $z'$ is not a function of $z$, but does depend on $\theta$). I.e., $$ A_z=\frac{\mu e^{-iBr}}{4\pi r}{\tt C(\theta)} $$ Since only (apparently) $A_z$ is non-zero we only need to be able to take the derivative w.r.t. $z$ to get $\nabla \cdot A_z$. $$ \nabla \cdot ...


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I suppose the right function to consider for $R$ is $R=\Vert\mathbf x_i-\mathbf x_j\Vert$ and the usual differentiation operators for the $\nabla$s. So, if $\mathbf x_i = (x_i,y_i,z_i)$, then $$R = \sqrt{(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2}$$ and $$\nabla_i = \left(\frac\partial{\partial x_i},\frac\partial{\partial y_i},\frac\partial{\partial z_i}\right)$$ ...


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We have the frame $\{e_\mu\}_{\mu=0,\dotsc,3}$ in terms of which the velocity vector is $v=v^\mu e_\mu$. There are a few properties of the affine connection which I would like to summarize: $$\nabla_{fX}Y=f\nabla_XY$$ $$\nabla_X(fY)=f\nabla_XY+X(f)Y$$ $$\nabla_{e_\mu}e_\nu=\Gamma^\lambda{}_{\mu\nu}e_\lambda$$ Using this, let's get to work. We have ...


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Yes, there is a subtlety there. Also notice that in quantum mechanics we have finely resolved energy eigenstates, and in a typical complicated system all degeneracies are broken so that at any given energy there is at most one state ($\Omega = 1$) but most likely there is no state at exactly that energy ($\Omega = 0$). So it would seem that $\log \Omega$ is ...


2

When contracting with the metric $\eta_{\mu\nu}$ one has explicitly, $$\eta_{\mu\nu} \frac{\partial \mathcal L}{\partial (\partial_\mu \phi)} \partial^\nu \phi = \frac{\partial \mathcal L}{\partial (\partial_0 \phi)} \partial^0 \phi - \sum_{i=1}^3\frac{\partial \mathcal L}{\partial (\partial_i \phi)} \partial^i \phi$$ For the other term, one has, ...


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The number of microstates is usually so large that we may approximate using a derivative. After all, look at $10^{23}$ molecules in a jar. It is quite easy to think that you are looking at a continuum.


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In a sense, parallel transport, covariant derivative and connection are all synonym for you can recover one from the other. So given a manifold one usually starts by giving one notion (e.g. how a vector field is transported parallel to itself along a family of curves) and then, if needed, the other objects are derived. In physics, when dealing with a ...


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$\nabla^2\vec A = (\nabla \cdot \nabla)\vec A$, not $\nabla(\nabla \cdot \vec A)$. The former is usually written $\nabla^2$, whereas the latter is usually written out explicitly.


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There is nothing wrong with having a divergence-less vector potential, as you can always arrange for this situation (cf. Coulomb gauge) without changing the fields, which are the physically relevant objects. Observe that the curl of a radial field is always zero.


3

Basically you are missing a term. Think of Lagrangian mechanics. If $q(t)$ is your generalized coordinate, and you make the change $q(t) \to q(t) + \delta q(t)$, then remember that the change in the action is $\delta S = \int_{t_i}^{t_f} \left(\partial_q L - \partial_t \partial_\dot{q} L \right) \delta q$. Thus you can basically think of the change in the ...


1

The group velocity depends on the relationship between $\omega$ and $k$ which is not, in general, constant. Thus, the group velocity will be a function of $\omega$. But for a given value of $\omega$ there is a corresponding value of $k$. If you permit the initial wave packet to contain a wide range of frequencies, you get a phenomenon called dispersion in ...


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The important point is that both $t^a$ and $v^b$ are vectoror fields defined (at least) on $C$. The idea is that if $t^a\nabla_av^b=0$ and you take $v^b$ at a particular point $p$ on $C$ then $v^b$ at all other points on $C$ is the parallel transported vector from $p$. Another way of thinking about it is that the covariant derivative in a given direction is ...


3

The inverse operators are Green's functions, this is rather common notation for them. For example, $\square^{-1}$ is the Green's function for the Klein-Gordon operator $\square$. Of course, boundary conditions (or a pole prescription) must be imposed in order to find a unique inverse (Green's function), so such a choice is implicit in this notation. Often a ...



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