# Tag Info

1

Let me elaborate on Ryan's correct comments. The flat background makes all components of the spinors covariantly constant; so the geometry is compatible with all of SUSY. A generic curved 6-real-dimensional manifold has an $O(6)$ holonomy or $SO(6)\sim SU(4)$ if it is orientable. The $SU(3)$ subgroup preserves 1/4 of the original supercharges – it is the ...

2

Say we have a supercharge $Q$ in $\mathbb{R}^{10}$. To turn this into a supercharge on the $\mathbb{R}^4$ effective theory obtained by compactifying on $X$, we need to contract $Q$ with a covariantly constant spinor on $X$. The reason why we want it to be covariantly constant is because we want to take the size of $X$ to zero. Covariant constant spinors are ...

2

I really don’t like that diagram. No, REALLY. I think it conveys a bad intuition that may confuse you. I don’t like what it does with the connexion coefficients (Christoffel symbols). Here’s why. In a general manifold, tangent spaces of course are not comparable at different points. This is in contrast with the situation in Euclidean space, which can be ...

1

I'll try to explain parallel transport first: This image shows how a vector generally does not remain parallel to itself under parallel transport. The vector here is moved along the path A-N-B and is constrained to remain in the tangential plane to the spherical surface at all points of the path because the geometry does not know of the "third" dimension ...

1

This result follows from i) Uniformization theorem and ii) Gauss-Bonnet theorem in 2d. According to the statement of uniformization theorem from this wiki page : every connected Riemann surface X admits a unique complete 2-dimensional real Riemann metric with constant curvature −1, 0 or 1 inducing the same conformal structure On the other hand, ...

1

Why not use explicit construction for such a surface? From The Manifold Atlas: Any hyperbolic metric on a closed, orientable surface $S_g$ of genus $g\ge 2$ is obtained by the following construction: choose a geodesic $4g$-gon in the hyperbolic plane ${\Bbb H}^2$ with area $4(g-1)\pi$. (This implies that the sum of interior angles is $2\pi$.) Then ...

2

Of course, the metric $\eta_{\mu\nu}$ is not a unique solution for Einstein vacuum equations compatible with your given initial data. And yes, we can interpret the alternatives as arising from coordinate functions. Let us take the simplest of such function: redefine time by introducing new 'time' variable $\tau$ through a relation $t=f(\tau)$ (spacial ...

6

The dimension of the string is a special case of the concept of dimension for a much more general class of objects called manifolds. Manifolds are a mathematical abstraction and generalization of the concept of a surface (like the surface of a sphere). The dimension of a (real) manifold is, roughly speaking, the number of coordinates (real numbers) ...

0

I'v got a big surprise about the seemingly extensive uncertainty on the physical interpretation of differential forms, differentials, and infinitesimals. My understanding is this: infinitesimal quantities can take any numbers that are "small without limit" , but they represent always finite quantities (numbers). This is a kind of definition, what made ...

1

Modern mathematicians would use a very rigorous approach to your question but i'll retain the old approach(Euclid's approach) which might be technically wrong but it is how i understand the word one dimensional. i'll mention the informal definition of point and line from the work of Euclid :Euclid's Elements. A point is that of which there is no ...

2

This vector potential can be written in every point on the plane except the origin as: $$A = d\phi$$ where $\phi$ is the polar angle ($\phi = \mathrm{tan}^{-1}\frac{y}{x}$). This does not mean that $A$ is exact, because $\phi$ is singular at the origin. But this means that the magnetic field is zero at every point except the origin. At the origin ...

1

I think you should have a lowered index on the RHS: \begin{align} \frac{dh_{ab}}{dt} &= \frac{d}{dt}\left(h_{ma}h_{nb}h^{mn}\right)\\ &=\frac{dh_{ma}}{dt}\delta_{b}{}^{m} + \frac{dh_{nb}}{dt}\delta_{a}{}^{n} + h_{ma}h_{nb}\frac{dh^{mn}}{dt}\\ \frac{dh_{ab}}{dt}&= 2 \frac{dh_{ab}}{dt} + h_{ma}h_{nb}\frac{dh^{mn}}{dt}\\ \frac{dh_{ab}}{dt} &= ... 0 However, I had difficulty understanding that answer and would like to understand how to do it this way. That is to say, I'd really like to know what property or identity that I'm missing before I can use use the Bianchi identities to show that it is manifestly zero. The other proof uses the first Bianchi identity. That's where the starting assumption ... 4 This is not a complete answer, but fills in some of the missing pieces Trimok asked about in the comments to Stan's answer: Note that I did not verify that Stan's proof actually works. Re 1) \begin{align} \nabla_Z \nabla_Y X &= Z^\lambda(Y^\mu X^\nu_{;\mu})_{;\lambda} \partial_\nu \\&= Z^\lambda(Y^\mu_{;\lambda} X^\nu_{;\mu} + Y^\mu ...

5

the article in wikipedia says that in string theory the particles at lower level are broken down into one dimensional strings, but I understand that only a straight line can be one dimensional, how are these loop like strings still said to be one dimensional ? Maybe this will help: In mathematics, the dimension of an object is an intrinsic property ...

4

Edit: Note: I have posted another proof of this in another question, here. Those who prefer coordinates may find it slightly more palatable. I gather from your comments that you can do this if you have $\mathcal{L}_X\text{Ric} = 0$. Thus I will outline a somewhat more general result, assuming a certain identity connecting Killing vectors and Riemann ...

3

In the language of differential forms, the Maxwell-Lorentz equations are simply $$\begin{eqnarray*}\mathrm{d}\!\star\!F = J/\lambda_0 &\text{,}\quad&\mathrm{d}F = 0\text{,}\end{eqnarray*}$$ where $1/\lambda_0$ is the characteristic impedance of free space, and can be fixed to $1$ in appropriate units. From that point of view, all you need to "move" ...

3

The calculus of forms is already well defined on curved manifolds, so you can use $d$ right off the bat.

1

Is teleparallelism an alternative to the introduction of a metric? Teleparallel gravity still comes with a metric - just take the tetrad field as orthonormal basis and there it is. The main difference between GR and teleparallelism is that the former uses curvature, the latter torsion to model gravity. According to Kleinert, there's actually a type of ...

1

There is a sense in which metric theories of spacetime are "general". I simplify to four dimensions, but the argument generalizes to higher dimensions. Consider a particle whose path is parameterized by four coordinates $x^{a} = (t(s),x(s),y(s),z(s))$. We wish to describe the motion of the particle, given that at s=0, each of these functions has a known ...

3

If you look at the first law of thermodynamics, $$dU=\delta Q-\delta W=TdS - pdV$$ then consider a reversible processes ($dU=0$), then we get $$TdS=pdV$$ Then using the ideal gas law, $pV=nRT$, we find $$dS \sim \frac{dV}{V}$$ The volume considered would be the volume of the system (e.g., a gas), with its infinitesimal increase(decrease) signified by ...

0

I would like to add that the geometric picture and relationship between Nambu-Goto and Polyakov actions are only hints and heuristics. Specifically, string scattering amplitudes are computed in a Lorentzian space, but the worldsheets are Euclidean. One way to see it is that topology changes don't respect causality, so branching worldsheets are problematic ...

3

First equation refers to the passive view of coordinate transformations while the second is the active view. Let $M$ be a manifold with metric $g$ which in local coordinates $x$ is written as $g_{\alpha\beta}(x)dx^{\alpha}\otimes dx^{\beta}$. Let $\phi:M\rightarrow M$ be a differentiable function and let $g'=\phi^*g$ be the pullback of the metric $g$, ...

3

Let's look at an example. Let's consider $0+1$ dimensions. Our manifold will be $M=\mathbb{R}$ and the coordinate system we will use will map the coordinate $x \in \mathbb{R}$ to the point $p \in M$ according to the rule $p(x^a) = x^0$. Now suppose the metric in this coordinate system has coordinates $g_{00}=x^0$ in this coordinate system. Now we would ...

4

The books are correct. The statement is a definite relation that is being imposed between the 'old' metric structure and the transformed one, for the transformation to be conformal. The equation you're unhappy about, $$g_{\mu \nu}'(x') = \Omega(x) g_{\mu \nu}(x)$$ states that the transformed metric $g_{\mu \nu}'$ at the transformed point $x'$ can be ...

0

The unit normal to the 3-surfaces is proportional to the Killing field. The coefficient being the norm of the Killing filed. You can compute the derivative using the Killing equations a couple of times to get the result.

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Ref : Padmanabhan, Gravitation, Cambridge, p $531-534$ In ADM formalism, we may write : $ds^2=g_{mn}dx^mdx^n= -N^2dt^2+h_{\alpha\beta}(dx^\alpha+N^\alpha dt)(dx^\beta+N^\beta dt)$ $N$ is called the lapse function, and $N^\alpha$ are called shift functions. Here the latin letters are for $4$-metrics, while greek letters are for the induced $3$- metrics. ...

1

Look at the ADM equations. Apply the staticity condition to them.

3

The Weyl tensor is the trace-free part of the Riemann tensor. The latter describes the curvature of spacetime. In the absence of sources, the trace part of the Riemann tensor will vanish due to the Einstein equations, but the Weyl tensor can still be non-zero. This is the case for gravitational waves propagating in vacuum. The physical reason is that even ...

0

I found the answer: The apparent horizon $\mathcal{H}$ is defined as the outer boundary of the region of $\Sigma$ (a hypersurface of spacetime with induced metric $h_{ab}$ and extrinsic curvature $K_{ab}$) which contains trapped or marginally trapped surfaces. $\mathcal{H}$ itself must be a marginally trapped surface, and thus it satisfies ...

4

The integral you wrote down would simply be computed as follows: \begin{align} \int_\Sigma f\,d\theta\wedge d\phi = \int_0^{2\pi}d\phi\int_0^\pi d\theta f(\theta, \phi) \end{align} You just "erase the wedge." The extra factor of $\sin\theta$ is included if you are integrating a 2-form $\omega$ that is proportional to the volume form; \begin{align} ...

1

$[[D_{\mu}, D_{\nu}],D_{\lambda}]A^{\rho} = [D_{\mu}, D_{\nu}]D_{\lambda}A^{\rho}-D_{\lambda}[D_{\mu}, D_{\nu}]A^{\rho}$ $=-R^{\tau}_{{\lambda}\mu \nu}D_{\tau}A^{\rho}+R^{\rho}_{\sigma \mu \nu}D_{\lambda}A^{\sigma}- D_{\lambda}(R^{\rho}_{\sigma \mu \nu}A^{\sigma})$ $=-R^{\tau}_{{\lambda}\mu \nu}D_{\tau}A^{\rho}+ R^{\rho}_{\sigma \mu \nu ; ... 1 The Weyl tensor contains the information necessary to describe solutions of the Einstein equations in vacuum, given by $$R_{\mu\nu}=0.$$ From this we can deduce that the trace part of the Riemann tensor vanishes, but not its traceless part, which is given by the Weyl tensor. The latter therefore describes curvature phenomena in the absence of matter, like ... 3 The Riemann tensor encapsulates all information about the 4-dimensional space-time. This information can generally divided into two sectors: Information about the curvature of space-time due to the existence of matter. This is given by the Ricci tensor according to the Einstein equation $$R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = 8 \pi G T_{\mu\nu}$$ ... 3 In addition to the other answers here, I would like to add that your intuition would have been right on target in lower dimensions ($2+1$and$1+1$). There,$T_{ab}=0$and$\Lambda=0$does imply (locally) flat space (although nontrivial topological effects are still possible). In$3+1$(and higher) dimensions the geometry is not uniquely defined by Ricci ... 3 A point is that the stress-energy tensor$T_{ab}$is not conserved (see Wiki). This is because gravitation has its own stress-energy (pseudo) tensor$t_{ab}$, even if we may always choose a frame such as$t_{ab}=0$for some particular space-time point. Only the sum of the two, (up to a factor$g$if we take the Landau–Lifshitz pseudotensor) is conserved : ... 3 If you consider first the Schwarzschild metric describing the space outside some sphere on non-zero radius, the Ricci curvature is indeed zero outside the sphere and indeed the stress-energy tensor is also zero there. The black hole is the limit of taking the radius of the sphere to zero. In that case both the Ricci curvature and stress-energy tensor are ... 0 Since I hate (for reasons I can not make objective I confess) the$\star$symbol, I propose below to reconstruct the Maxwell's equations from the definition of the 2-form $$F=E_{i}dx^{i}\wedge dt+B_{\left\vert i\right\vert }dx^{\left\vert i+1\right\vert }\wedge dx^{\left\vert i+2\right\vert }$$ where I introduce the modulus notation$\left\vert i\right\vert ...

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