# Tag Info

2

Let's do this explicitly for both cases. For these examples, the classical formula for the geodesic curvature $k_g$ suffices. Let $\gamma(t)$ be a curve in a surface $S \subset \mathbb{R}^3$, and let $n(t)$ be the unit normal to $S$ at the point $\gamma(t)$. Then $$k_g = \frac{\ddot{\gamma}(t) .(n(t) \times \dot{\gamma}(t))}{|\dot{\gamma}(t)|^3}$$ First ...

1

You don't need to use the metric of the hemisphere. This is because the pullback of arbitrary forms onto the submanifold is the trivial pullback operator. All you need to do is apply the projection operator. Therefore, the extrinsic curvature tensor is just $K_{ab} = - \gamma_{a}{}^{c}\gamma_{b}{}^{d}\nabla_{c}n_{d}$, where $\gamma_{ab}$ is the metric of ...

0

There is a real Lie group $\tilde{Diff}(S^1)$ which is a $U(1)$ central extension of the real Lie group $Diff(S^1)$, and the Virasoro algebra is the Lie algebra of this Lie group. The central extension $\tilde{Diff}(S^1)$ can be realized geometrically in two ways. The first is via a Hilbert space embedding (as in the book of Pressley-Segal), and the second ...

1

I got the answer reading a book from E. Poisson, what I was doing was indeed wrong, you have to start with the induced metric given by $$h_{ab}= g_{\mu\nu}e^{\mu}_a e^{\nu}_b$$ where $$e^{\mu}_a=\frac{\partial x^{\mu}}{\partial y^a}$$ are the tangent vectors to curves of the hypersurface. Then, you just replace $g$ by $h$ in the usual relation ...

0

It took me a couple of days to figure it out, but I can now answer my own question: The problem with directly substituting the expression for $\delta g_{\mu\nu}$ into that for ${\delta \Gamma^\alpha}_{\beta\mu}$ is that there is no obvious source for the term $\xi^\sigma \partial_\sigma { \Gamma^\alpha}_{\beta\mu}$ that has to be the leading term in the ...

0

This question is subtler than one might think at a first glance. Three points may be worth noting. It will make a difference whether the manifold $M$ is compact or not. The nice answer by Michael Seifert implicitly assumes the non-compact case. Otherwise the infinite Gaussian integrals (over $\mathbb{R}^n$) will not make sense. In the compact situation ...

4

You can get a nice expression for the leading-order correction to the flat-manifold result via the use of Riemann normal coordinates. Basically, imagine expanding the metric in a power series at the point $x_0$: $$g_{\mu \nu}(x) = g_{\mu \nu} (x_0) + \partial_\rho g_{\mu \nu} (x^\rho - x_0^\rho) + \frac{1}{2} \partial_\rho \partial_\sigma g_{\mu \nu} ... 0 Indeed as was commented before usually in physics these derivatives are 'derived' by postulating how the object transforms. Obviously there are more rigorous definitions, as is usually the case. In this particular case you could try to find the structure which gives you these 'covariant derivatives' in the first place. In books they are often referred to ... 1 Well, this is linked to what the cotetrad e_\mu^I is. It is customary to present the cotetrad as a diagonalization of the metric and indeed, we have:$$g_{\mu\nu} = e_\mu^I e_\nu^J \eta_{IJ}$$Note here that the cotetrad has two kind of indices. The greek type corresponds to spacetime coordinates but the latin type indices are indices in the tangent ... 2 you can find the detail in the manuscript 3 R_{\alpha\beta}=R{^{\mu}}_{\alpha\mu\beta} is not the Einstein tensor G_{\alpha\beta}, but the Ricci tensor. You get the Einstein tensor via$$ G_{\alpha\beta} = R_{\alpha\beta} - \frac{1}{2} g_{\alpha\beta} \mathcal{R} \textrm{,} $$with \mathcal{R} = R^{\alpha}_{\textrm{ }\alpha} being the Ricci scalar. In your last equation for R_{\alpha\beta}, ... 0 Why do we need coordinate-free description in the first place? In physics, we have coordinate-free description in the first place. Consider your example: When I am walking from school to my house The description of this consist of you and the school (building) departing from each other, you and your house meeting each other, and, filling in ... 2 The formula you have for the metric is not quite right. The curvature tensor is skew in the first pair of indices and the last pair but your metric has it being symmetric. Noneltheless, I agree with you calculation of the Christoffel symbol. I note that our formula for {\Gamma^\lambda}_{\mu\nu} is symmetric in the last two indices as befits a ... 0 When I'm confused about all the functional derivatives, I'm doing that:$$ S(\phi_1,\phi_2,\phi_3,g_1,g_2,g_3) =\sqrt{g_1}\,L(\phi_1,g_1) + \sqrt{g_2}\,L(\phi_2,g_2) + \sqrt{g_3}\,L(\phi_3,g_3)$$I hope you'll recognize your \int\sqrt{g}\, L[\phi,g] in the expression above. The functional derivatives then become more familiar gradients:$$\frac{\partial ...

1

Comments to the question (v3): The vielbein $e^A{}_{\mu}$ in the Cartan formalism is an intertwiner $$\tag{1} g_{\mu\nu}~=~e^A{}_{\mu} ~\eta_{AB} ~e^B{}_{\nu}$$ between the curved (pseudo) Riemannian metric $g_{\mu\nu}$ and the corresponding flat metric $\eta_{AB}$. Here Greek indices $\mu,\nu,\lambda, \ldots,$ are so-called curved indices, while ...

0

Your post has many details, but overall you were a little vague about the signature (++++, +++-, ++--, +---, ----) of the metric. In fact it only came up when you said the metric is close to $\eta$ when two points are close. And that could be interpreted ina way that isn't true. For any point there exists a coordinate system where the metric is very close ...

2

Comments to the question (v4): First a disclaimer. Note that even for a smooth local functional, the existence of a functional/variational derivatives is not guaranteed, but depends on appropriately chosen boundary conditions. Calculus of variations, functional/variational derivatives, Frechet derivatives, Gateaux derivatives, etc, is a huge mathematical ...

0

A physical derivation of that result can be provided in the framework of general relativity using the equivalence principle and the principle of general covariance. The principle of general covariance tells us that if a physical reult is valid in a frame of reference, then it must be valid in every other frame of reference. On the other hand, the equivalence ...

9

This is an axiom, not a result. When defining the covariant derivative, we choose for it to obey a number of properties. Carroll's Spacetime and Geometry summarizes these well -- look at the preprint for Chapter 3 here. We of course want $\nabla$ to act linearly on its argument and to obey the product rule. We also demand that it commute with contractions ...

3

As one way, consider the parallel transport of two vectors $p^\mu$ and $q^\nu$ along the same curve $C$ such that the same angle between the two, $$\cos\phi=\frac{p^\alpha q_\alpha}{\sqrt{p^\beta q_\beta}\sqrt{p^\gamma q_\gamma}}$$ is preserved. Parallel transport tells us that for a general vector $f^\alpha$, $$... 5 This is a great question, with three great answers, and here I am, a bit late to the party. There are two crucial things which the above answers don't seem to address, so I am going to try to give a really simple explanation at those levels. Locality / Manifolds I'm going to come close to giving you one technical definition of a manifold, following ... 0 On the very theoretical side of solid state physics is the holographic AdS/CFT correspondence which links strongly coupled condensed matter systems to gravitational theories. Recent work has been done on describing things like phase transitions in this theory. For example models of superconductivity in the gravity dual are promising in describing difficult ... 8 Why do we need coordinate free in the first place? Let me tell you about a related experience I have had teaching students. When I ask them to define the scalar product then the vast majority will write down something along the lines of$$ \vec{v}\cdot \vec{w} = v_x w_x + v_y w_y + v_z w_z \qquad (1) $$i.e. a coordinate-based description. However, ... 5 Let me first tell you that what you read is very vague. In GR the laws of physics are assumed to be independent of the observer. An observer is represented by the a reference frame the observer uses to measure physical phenomena. There are a set of coordinate transformations that relates observables for different observers. Say for example the speed of a ... 0 There is theory of dispersion in crystals. One can say that the differential geometry is used there. I think it is Group theory + differential geometry. 19 That's a very good question. While it may seem "natural" that the world is ordered like a vector space (it is the order that we are accustomed to!), it's indeed a completely unnatural requirement for physics that is supposed to be built on local laws only. Why should there be a perfect long range order of space, at all? Why would space extend from here to ... 2 If you computed arc length you'd chop your curve into pieces, find the distance between the end points of each piece, add them up, and then take the limit as the pieces get individually smaller and hence more numerous. Now image that instead you took the square of the distance between the end points of each piece, that wouldn't work because you would get ... 3 Schutz's A First Course in General Relativity explains it pretty well, I think. The pertinent page is here, but if international copyright law won't let you read it, the basic gist of it is the following: If the gravitational acceleration was the same everywhere in space, then we could shift into a freely falling reference frame, and freely falling ... 1 It is not clear what OP is asking, but consider the following chain of reasoning: Scalar curvature R is an invariant independent of choice of coordinates. In GR, regions of matter, e.g., a star, is modelled with non-zero curvature. The Minkowski space has zero curvature. Hence, if a spacetime (M,g) has non-zero scalar curvature R at a point p\in M, ... 0 as far as i know these two expressions are used synonymically. and your coordinates h_{\mu \nu}=g_{\mu \nu}-\sigma n_{\mu}n_{\nu} arent gaussian coordinates in general. gaussian are for example flrw metrics like ds^2=-dt^2+h_{ij}(t)dx^idx^j with spacelike i,j where there is no mixed time/space basis term in the second part of the line element, only a ... 1 I am doing my Bachelors on something like that and we even chose to include double boundary terms in the variation. In addition to this paper, Variational principle and 1-point functions in 3-dimensional flat space Einstein gravity by Stephane Detournay et al., we get one further term:$$-\frac{(1-\alpha)}{16\pi G}\int_{\mathcal{\partial ...

-1

Assuming that g is constant, the only way your equations "hold," is if $F[\phi] = e^\phi$. I hope this is helpful.

4

A null geodesic is a geodesic (that is: with respect to length extremal line in a manifold), whose tangent vector is a light-like vector everywhere on the geodesic (that is $x(s)$ is a geodesic and $g_{\mu\nu} \frac{dx^\mu}{ds}\frac{dx^\nu}{ds} = 0$ for all $s$, where $s$ is an affine parameter along the curve). The null geodesics are exactly the paths that ...

7

A null geodesic is the path that a massless particle, such as a photon, follows. That's why it's called null, it's interval (it's "distance" in 4 D spacetime) is equal to zero and it does not have a proper time associated with it. When they are drawn on a spacetime diagram, they are the edges of the light cones, as in the picture below, the lines at 45 ...

0

The calculation is quite easy and you can find it here: http://www.physicspages.com/2014/01/02/covariant-derivative-of-the-metric-tensor/

4

If you stick to one convention out of many other conventions, you should have same results regardless of signature of metric. Here I follow Carroll's conventions: http://amzn.com/0805387323. For Christoffel symbol, we have \Gamma^\lambda_{\mu\nu}=\frac{1}{2}g^{\lambda\rho}(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho ...

1

The signature is one convention (both in special relativity and in general relativity). But in general relativity there are many different conventions besides just the signature. The front inside cover of Misner Thorne and Wheeler lists conventions for signature, for the Riemann Tensor, for the Einstein Tensor, and for the use of Greek and Latin indices and ...

1

Our rigorous definition of "dimension" comes from linear algebra. This is going to be a quick run-through of the mathematical way of describing dimensions, and then its physical significance. The first concept needed is a vector space. A vector space can effectively be thought of as a collection of points that satisfy a few particular (and very useful) ...

1

You just need to plug your expression for $dT^2$ back into the original metric, with the substitution $\psi' = -C/A$. This gets you \begin{align*} ds^2 &= -A(r) \left[ dT^2 - {\psi'}^2 dr^2 + 2 \frac{C}{A} dr dt \right] + B(r) dr^2 + 2 C(r) \, dr \, dt + D(r) r^2 \, d\Omega^2 \\ &= -A(r) dT^2 + \left[ B(r) + A(r) \left( \psi'(r) \right)^2 ...

3

In a geometrical context, the dimension (at a point) roughly speaking is the number of coordinates you need to identify any point in a fixed neighbourhood. Note that in a geometrical context, the least you need is a topology, in order to be able to speak of continuity, neigbourhoods, etc. This intuitive definition used to work quite well, but at a certain ...

0

Consider the metric in comoving conformal coordinates: $ds^2 = a^2(\eta)[-d\eta^2 + \delta_{ij}dx^i dx^j]$. It is easy to show that $\partial_{\eta}$ solves $\mathcal{L}_{\partial_{\eta}}g_{\mu\nu} = \frac{2}{a(\eta)}g_{\mu\nu}$; furthermore $\partial_{\eta}$ is clearly time-like.

6

I think this depends a lot on what you are doing and how you look at whatever you are looking at. Speaking of which, how many dimensions does the content displayed by your computer monitor have? Two, I guess, could be one answer. It's not three dimensional and it certainly is not one strip of pixels. Let me quote from Carl's great answer that I like to ...

28

In this context, I usually explain it (non-mathematically) by saying that the number of dimensions is the number of values you need to specify where an event occurs. For most people this involves space and time (but for particle physicists it might involve more values ;). Anyway, certainly even people before Einstein would need to specify the time as well ...

1

A normal Killing vector is obtanied by solving the usual Killing equation $\nabla_{(\mu} \xi_{\nu)} = 0$ A conformal Killing vector is obtanied by solving a slightly different equation, the conformal Killing equation: $\nabla_\mu \xi_\nu + \nabla_\nu \xi_\mu = 2 \alpha g_{\mu \nu}$ where $\alpha$ is obtanied by taking the trace of the equation above. ...

1

Your first line is incorrect, gravity is real, just jump up and you will confirm that. Frames of reference are used to find how gravity affects our measurements of spacetime. Your second line is correct. Objects, from photons to planets, moving in general relativity, obey the geodesic equation, so a curved line can be a "straight line", that is, if you ...

35

Coming from a math perspective, I would define a dimension as "any property which is orthogonal to all other properties." "Orthogonal" here means you cannot get to one property by applying scalar operations on another. For example, the x-axis dimension can never become a y-axis value, and similarly for time vs. spatial dimensions. For that matter, ...

4

Nice question. One way to think about it is that given a metric $g$, the statement $\mathcal L_Xg = 0$ says something about the metric, whereas $\nabla_Xg = 0$ says something about the connection. Now what $\mathcal L_Xg = 0$ says, is that the flow of $X$, where defined, is an isometry for the metric, while $\nabla_Xg = 0$ says that $\nabla$ transports a ...

2

To decide if two metrics are related by a change of frame and/or coordinate transformation is called the equivalence problem. It can be solved using the Cartan-Karlhede algorithm. Given a metric $g$ expressed in some coordinates $x_i$, the algorithm computes a set of invariantly defined curvature invariants expressed as functions of $x_i$. For example, the ...

0

It makes no sense to demand that $dx' = a(t)\ dx$. Suppose that your coordinate transformation was of the form $x' = x'(x)$. Then you would have $dx' = \frac{dx'}{dx}\ dx$, but $\frac{dx'}{dx}$ would have to be a function of $x$ only, and so it couldn't be $a(t)$. Now suppose we tried to fix that by doing a transformation $x' = x'(x,t)$. Now \$dx' = ...

2

Your introductory claims are not correct. The usual wavefunction is not a function of spacetime, for two reasons: It is not Lorentz invariant (time is singled out, so it is not a function of a point on a spacetime manifold, and it is not even time dependent at all in the Heisenberg picture) and it can depend on more than one "set" of positions - for a system ...

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