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9

Below follows a handful of excerpts from the book Introduction to the Classical Theory of Particles and Fields (2007) by B. Kosyakov. Controversial/misleading/wrong statements are marked in $\color{Red}{\rm red}$. We agree with OP that the statements marked in $\color{Red}{\rm red}$ are opposite standard terminology/conventions. Some (not all) correct ...


6

I took a quick look at pages 59 and 60 of "Gravitation", section 2.6 "Gradients and Directional Derivatives", to see if there's anything there we can use to clarify this issue. In this section, the gradient of $f$ is $\mathbf df$, the directional derivative along the vector $\mathbf v$ is $\partial_{\mathbf v}f$ and the following relationship holds: ...


5

I believe this is just imprecise use of language by the author - there is nothing mysterious happening, it is just not well stated: As stated in the question, for a hypersurface $\Sigma$ defined by $$ F(x) = c \in \mathbb{R}$$ we find that $$ \mathrm{d}F = 0$$ must hold on $\Sigma$. This is crucial - it means that the 1-form $\mathrm{d}F$ acting upon ...


5

Let's look at your last line: Why not simply postulate a global coordinate system and let the coordinate mapping from $M$ to $\mathbb{R}^n$ and the resulting metric tensor describe the geometry? P.S. By Rn I mean the set of all n-tuples with the usual definitions of sum and scalar multiple, not Euclidean n-space. $\mathbb{R}^n$ has a particular, simple ...


4

OP wants to evaluate $$ {\rm Tr}(-)^F e^{-\beta H}=\int_{PBC}[d\phi][d\psi] e^{-S_E[\phi,\psi]},\tag{2.5} $$ in Ref. 1 with periodic boundary conditions (PBC) for both the boson $\phi\equiv x$ and the fermion $\psi$. One can assume that the corresponding Fourier components are labelled by integers $n\in\mathbb{Z}$. One may argue that (2.5) does not depend ...


4

First, let's go back to flat space. The mentioned SU(2) algebra is just a part of the 10-dimensional Poincare algebra - the algebra of isometries of Minkowski space-time. The generators of isometries are called Killing vector fields, you can easily show that these (they come with the Lie bracket, as usual) obey the Poincare algebra. Because isometries (by ...


3

The identity in question is given as, $$(\nabla_\beta u^\alpha) u_\alpha=(\nabla_\beta u_\alpha)u^\alpha$$ Expanding the left-hand side, we find, $$(\nabla_{\beta}u^\alpha)u_\alpha = (\nabla_\beta g^{\alpha \delta}u_\delta)u_\alpha = (u_\delta \nabla_\beta g^{\alpha\delta} + g^{\alpha\delta}\nabla_\beta u_\delta)u_\alpha$$ The Levi-Civita connection is ...


3

Look more closely at what you write. I'm not going to write it with all the partial derivative and Lagrangians, because the confusions are not dependent on that: The requirement for spatial infinity arises because $\int_\mathcal{M} \mathrm{d}\omega \neq \omega$, but $\int_\mathcal{M} \mathrm{d}\omega = \int_{\partial\mathcal{M}}\omega $ (Stokes' theorem). ...


2

This is not true. There are trivial counter-examples. For example, take $\mathbb{R}$ with the trivial metric. Then, $\gamma :(0,1)\rightarrow \mathbb{R}$ defined by $\gamma (t):=t$ is a geodesic, but it is also clearly extendable. Indeed, it is extended by $\tilde{\gamma}:\mathbb{R}\rightarrow \mathbb{R}$ defined by $\tilde{\gamma}(t):=t$.


2

First of all, the metric tensor is one additional piece of structure one inserts on a smooth manifold to measure lenghts and angles. The metric is indeed not present in all applications of Differential Geometry to Physics (see e.g. Lagrangian Mechanics). In that case, it is important to know also how to deal with manifolds without metric tensors. Now, about ...


2

$$T_{\mu\nu} = \partial_\mu\phi \partial_\nu \phi + g _{\mu\nu} (-1/2) (\partial_\nu\phi \partial^\nu \phi +m^2\phi^2) $$ $$\partial^\mu T_{\mu\nu}=\partial_\mu\partial^\mu\phi \partial_\nu \phi+\partial_\mu\phi\partial_\nu\partial^\mu \phi-\frac{1}{2}\partial_\nu(\partial_\tau\phi \partial^\tau \phi +m^2\phi^2)$$ We have equation of motion ...


2

Eddington–Finkelstein coordinates use the same position coordinates as Schwarzschild coordinates, only the time coordinate is transformed, so first consider how to define Schwarzschild coordinates in a physical way. This pdf explains a way of defining the position coordinates in section 9.1.1: • We may assign a practical definition to the radial ...


2

It seems you want to calculate the laplacian in the spacial surface of $t$ constant. If the global metric is defined as: $$ ds^2 = (1-\frac{r_s}{r}) \, dt^2 + (1-\frac{r_s}{r})^{-1} \, dr^2 + r^2 \, d\theta^2 + r^2 \, \sin^2(\theta) \, d\phi^2 $$ The metric on the surface $t$ constant is $$ ds^2 = (1-\frac{r_s}{r})^{-1} \, dr^2 + r^2 \, d\theta^2 + r^2 \, ...


2

I'll answer my own question and hope this information is useful to someone. I'll take $\hbar = 1$ and will deal with systems of one degree of freedom (the generalisation should be obvious). The normalisation factors are very confusing, so I'll omit most of the reasons for them to be what they are. First, let's take a look at the good old continuous Wigner ...


1

I believe you are confused because you are mixing up related but slightly different quantities. Yes, a partial derivative is a vector and yes, a vector is an object with an upper index. The above statement may seem contradictory, but in fact it is not for the following reason. A vector is an abstract quantity that is an element of a "vector space". In ...


1

(I am dropping the bothersome factors of $\mathrm{i}$ in this answer, they contribute nothing to understanding what is going on) The gauge covariant derivative exists for all forms on the spacetime manifold $\mathcal{M}$ taking value in a representation of the gauge group. (Formally, these are sections of associated vector bundles to the gauge principal ...


1

Actually, they are one and the same thing. Before I delve into the question you asked, let me quickly describe a closely related analogy - the covariant derivative in GR. This is a quantity $\nabla_\mu$ that acts differently on different objects. In particular $$ \nabla_\mu \phi = \partial_\mu \phi,~~~ (\nabla_\mu V)_\nu = \partial_\mu V_\nu - ...


1

In Riemann Normal Coordinates; you can take advantage of the special symmetry \begin{equation} g_{ab, cd} = g_{cd, ab} \end{equation} which allows you to express the curvature (very simply) as \begin{equation} R_{\mu \nu \alpha \beta} = g_{\mu \beta, \alpha \nu} - g_{\mu \alpha, \nu \beta} \end{equation} If you transpose the first two indices of $R_{mnab}$ ...


1

A covariant derivative of a tensor is itself a tensor. Actually, when we say something is covariant (or invariant under coordinate transformation), we mean that thing is a tensor. So, in this case $\nabla_\mu V^\nu\equiv T_\mu{}^\nu$. Now calculate $\nabla_\alpha T_\mu{}^\nu$ easily. \begin{equation} \nabla_\alpha T_\mu{}^\nu=\partial_\alpha ...


1

The covariant derivative for a general tensor of the form $T^{a_1\dots a_n}_{b_1 \dots b_n}$ is given by, $$\nabla_c T^{a_1\dots a_n}_{b_1 \dots b_n} = \partial_c T^{a_1\dots a_n}_{b_1 \dots b_n} + \Gamma^{a_1}_{cd}T^{d\dots a_n}_{b_1 \dots b_n} + \dots - \Gamma^d_{c b_1}T^{a_1\dots a_n}_{d \dots b_n} - \dots$$ Taking the covariant derivative of a ...


1

Yes, the proper time along a timelike curve in relativity is very much analogous to the length of a curve. Just as the length of a curve is invariant under rotations, the proper time along a curve is invariant under Lorentz transformations. One difference with conventional length is that although a straight line is the shortest length between two points, a ...


1

All curves can be parameterised by an affine parameter (commonly written as $\lambda$ in the GR books I have). However only for timelike curves does the parameter $\lambda$ have a physical meaning i.e. it's the elapsed time $\tau$ shown on a clock by the observer following the curve. So there's no mathematical difference between parameterising a timelike ...


1

The first point to consider is that the Riemann tensor can be expressed in terms of the Weyl tensor and the Ricci Tensor: $$R_{abcd}=C_{abcd}-g_{a[d}R_{c]b}-g_{b[c}R_{d]a}-\frac{1}{3}Rg_{a[c}g_{d]b}$$ The Ricci tensor is given by Einstein's equation: $$R_{ab}-\frac{1}{2}g_{ab}R+\Lambda g_{ab}=8\pi T_{ab}$$ Now the Weyl tensor is not specified by the EFE. ...



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