# Tag Info

## Hot answers tagged metric-tensor

9

In the MTW book Gravitation as well as most relativity literature, index raising and lowering is taken as self-evident and/or implicit. This means that when we refer to four-velocity, we may mean the object with components $u_\alpha$ or $u^\alpha$. These are related by index lowering/raising: $$u_\alpha = u^\beta g_{\beta \alpha}$$ Since the metric has $g_{\... 8 This cannot work in general since there are manifolds that admit no Killing vectors at all. It also cannot work in general because the Killing vectors are not unique - if$K$is a Killing vector, then so is$\alpha K$for$\alpha\in\mathbb{R}$and if$K$and$G$are Killing vectors, then so is$K + \alpha G$. The formula$g^{\mu\nu} = K_i^\mu K_i^\nu$is not ... 6 The net gravitational force on any mass inside a spherical shell is zero. This is a consequence of Newtons' shell theorem. As per this link it states "In classical mechanics, the shell theorem gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetrical body. This theorem has particular application to ... 5 Adjoint refresher If you consider the Lie algebra$\mathfrak{g}$as a vector space, then the Lie algebra has$\mathfrak{g}$a natural action on the vector space$\mathfrak{g}$. This is called the adjoint representation$\mathrm{ad}_\mathfrak{g}$. It acts as, for$X, Y \in \mathfrak{g}$, \begin{equation} \mathrm{ad}_X Y = [X, Y]. \end{equation} This is a ... 3 You're going to need some sort of extra assumption in order to make this idea work, because not every spacetime even has a single global Killing vector, much less enough Killing vectors to span the spacetime. 3 Actually if you look through Tong's notes, the derivation he is presenting does not consider the metric to be the fundamental variable (it sort of doesn't make sense to talk about isometries of a dynamical metric), but rather the worldline of a particle moving on a fixed background metric. In particular, after Eq 4.32, Tong writes down the action \begin{... 3 The simplest possible concrete example demonstrates this to be wrong. Take$[\eta]=\text{diag}(-1,+1,+1,+1)$, then: $$\eta^{00}\eta_{11}=-1\cdot +1=-1\tag{1},$$ on the other hand: $$\delta^0_1\delta^0_1=0. \tag{2}$$ 2 All FLRW spacetimes are conformally flat, and they are not maximally symmetric unless$ρ=-p=\text{constant}$(in which case they are Minkowski or (anti) de Sitter). 2 Yes, they do commute. However, your last equation has a mistake: your$A^l_i$tensor should contract with$\phi_{,l}$to give$\phi_{,i}$. Your final answer should be $$\frac{\partial \phi}{\partial x^j} g_{ik}-\frac{\partial \phi}{\partial x^i} g_{jk}$$ You can check the indices to see if your result is consistent. 2 This is how the dot product can be defined for covariant and contravariant vectors (without explicitly inserting the metric) i.e, with the metric, the dot product would look like $$\beta \cdot \beta = \beta_i \beta^i = \gamma_{ij} \beta^j \beta^i$$ Note that $$\beta_i = \gamma_{ij} \beta^j$$ and $$\beta^i = \gamma^{ij}\beta_{j}$$ The metric$\gamma_{ij}$... 2 The dot product on a Minkowski manifold is defined to have indefinite signature, so the second option is correct. P.S. uses$\eta_{\mu\nu} = \text{diag}(1, -1, -1, -1)$. 2 There are a couple of notations that are sometimes seen in tensor calculus:$u_{\alpha,\beta}$and$u_{\alpha;\beta}$.$u_{\alpha,\beta}$(with a comma) stands for the partial derivative of$u_\alpha$with respect to$x^\beta.$This is often also written$\partial_\beta u_\alpha.u_{\alpha;\beta}$(with a semicolon) stands for the covariant derivative of$...

2

A small comment: "If $\eta$ was non-degenerate..." What you meant to say is "positive-definite"; note that the term "non-degenerate" has a specific technical meaning in linear algebra (according to which $\eta$ is non-degenerate) so we shouldn't mix up the terminology. We can actually generalize the result much further, though ...

2

I you read it carefully you will understand that Schwartz only does this for ease of notation. He specifically says that index position is important when you plug in explicit vectors or matrices. I.e. he basically says that the reader is smart enough to do the contractions correctly. But it is indeed rather sloppy and unnecessary if you ask me.

2

Much of the (endless) confusion about this subject can be attributed to the fact that differential geometry can be formulated in several different ways. One approach goes as follows. We consider a vector space $V$ with an inner product provided by a metric tensor $g:V\times V \rightarrow \mathbb R$. Given a basis $\{\hat e_\mu\}$ for $V$, we can expand ...

1

The correction term is, by definition, $$h_{\mu\nu} \equiv g_{\mu\nu} - g_{\mu\nu}^{\rm background}$$ and you have $$g_{tt}^{\rm background} = -1 + \frac{GM}{r}.$$ Thus the equation comes about by a mathematical choice or definition: we choose to explore the difference between the actual metric and the Schwarzschild metric. It is not clear in the ...

1

Correct is the second 'geometrical' representation. Note the properties : (1) the vectors of the dual basis are parallel to the heights of the parallelogram formed by the vectors of the original basis with magnitudes inversely proportional to these heights and (2) increasing the magnitude of a vector of the original basis the corresponding component is ...

1

Using the product rule on an arbitrary $(r,s)$ tensor $$T = T^{i_1 \ldots i_r}_{\qquad\, j_1 \ldots j_s} \mathbf{e}_{i_1} \otimes \ldots \otimes \mathbf{e}_{i_r} \otimes \mathbf{e}^{j_1} \otimes \ldots \otimes \mathbf{e}^{j_s}$$ we get $$\mathcal{L}_{\mathbf{v}} T^{i_1 \ldots i_r}_{\qquad\, j_1 \ldots j_s} = v^k \partial_k T^{i_1 \ldots i_r}_{\qquad\, j_1 \... 1 You do not need distance. What is considered are curves on a manifold. The curve on a manifold (M) is map form real numbers into manifold, i.e. a map that takes real number and assigns a point in the manifold:$$P(\lambda):\mathbb{R}\rightarrow M.$$\lambda is simply parameter of a curve. The tangent vector is then considered to be$$\frac{dP}{d\lambda}=\...

1

The sub and superscripts represent contra and covariant vectors. We can (sometimes) think of these vectors as column vectors (contravariant) and row vectors (covariant or covectors). An intuitive way to understand the difference between the two is to consider how each transforms under a change of basis. A covector transforms in the same way as its basis ...

1

My first question is how can I see that the (𝑟,𝜃,𝜙) in the FLRW metric are comoving coordinates? Comoving distances remain constant with time. The metric in the form you wrote it places all the time dependence into the function $R(t)$, and the coordinates $r,\theta,\phi$ are independent of time. So any distance expressed in terms of $r,\theta,\phi$ is a ...

1

By the chain rule $$\frac{d}{d\tau}(x(b\tau)) = b \cdot\left(\frac{dx}{d\tau}\right)(b\tau)$$ (indices suppressed for brevity). The rest is simple algebra. Proper time is a special case of a world line parameter. It serves the purpose of giving you the velocity seen from the locally comoving observer at any point of the worldline, by just deriving w.r.t. ...

1

First of all I would recommend having a copy of "A First Course in String Theory" by Barton Zwiebach while working through Polchinski. Now lets return to your question. The Nambu-Goto action is given by $$S_{NG}=-\frac{1}{2\pi\alpha'}\int d\tau\ d\sigma\ \sqrt{(\dot{X}\cdot X')^2-(\dot{X})^2(X')^2}.$$ The reasoning behind is quite simple. We want ...

1

It seems that be that you are automatically using a fixed set of coordinates, which is not how you should be thinking about the metric. Mathematically by construction, the metric, $g$, is a geometrical object, specifically a 2 rank tensor. As such we are free to choose which local coordinates we want to use to represent it. In the context of general ...

Only top voted, non community-wiki answers of a minimum length are eligible