Tag Info

New answers tagged

2

Comments to the question (v4): The question formulation seems to talk about affine parametrizations before applying the principle of stationary action. In the context of Riemannian$^1$ geometry, an affine parametrization of a (not necessarily geodesic) curve means by definition that the arc-length $s$ and the curve parameter $\lambda$ are affinely related ...


0

After thinking about this a bit I realize the question was due to a rather silly misconception,the converse was not mentioned in the usual references because it's trivial from the proof of the usual direction. For future reference I will include a quick version here. $$F = L^2/2$$ $$ EL[F] \equiv \frac{d}{ds}(\frac{\partial}{\partial \dot{x^\mu}}F)- ...


3

Orientation. Let a lagrangian $L$ that is a local function of position and velocity be given. For a parameterized path $x$ on $M$, define a corresponding action $S$ as follows: \begin{align} S[x] = \int_a^b ds\,L(x(s), \dot x(s)). \end{align} Let $\delta x$ be a fixed-endpoint variation, then a standard computation shows that the corresponding variation ...


1

The property you are referring to is called geodesic completeness. It is an important concept in the study of singularities in general relativity. There are somewhat trivial examples of geodesic incompleteness where you are just "missing" part of the spacetime, i.e. you could have Minkowski space with a point removed. In these cases you usually consider ...


1

It is not true. Sometimes the spacetime itself and therefore the geodesics can be extended. Consider for example the manifold described by (t,x,y,z) with $x,y,z> 0$ and Minkowski metric. This is nothing more than Minkowski spacetime truncated to a smaller region, but it is a perfectly valid manifold for all purposes. In the diagram we see schematically ...


1

It's the chain rule: $$ \partial_x f(y(x)) = \partial_y f(y(x)) \cdot \partial_x y(x)$$ Your vector field $A^\mu$ depends on the worldsheet coordinates only through the worldsheet coordinates $x^\mu$. Thus, when how $A^\mu$ behaves under an infinitesimal shift on the world-sheet you need to take into account how $A^\mu$ depends on $z$ - that's $\partial_z ...


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 $\tilde{\gamma}(t):=t$.


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 ...


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 ...


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

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 ...


0

I will do this explicitly. In $AdS_3$, you can see we must have $z=0$ ($x,y,z$ are symmetric so just choose $z$). For general cases, we must have $u,v,x,y \neq 0$, so we must have: $\sin t \neq 0, \cos t \neq 0, \cosh \rho \neq 0, \sinh \rho \neq0, \sin \theta \neq 0, \cos \theta \neq 0, \cos \phi \neq 0$. Thus we have $\sin \phi =0$, i.e, $\phi =0$ ($n ...


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

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 ...


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 ...


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. ...


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

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 ...


0

Coordinates are not physical. They are entirely up to you. In the Eddington Finkelstein coordinates, the lines of constant u theta and phi are null radial lines going out to infinity while r is the coordinate such that the surfaces swept out by the spherical symmetry have an area of 4 pi r^2. (Ie, take a point of the spacetime. Apply a spherical symmetery ...


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 $R_{mnab}$ ...


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 ...


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 ...


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 ...


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 ...


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

(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

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 ...


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 ...



Top 50 recent answers are included