New answers tagged

2

Notation $C(M)$ represents smooth functions $f:M\rightarrow \mathbb{R}$. $\tau$ represents a tensor field, and $\mathcal{T}^r_s(M)$ represents the set of all $(r,s)$ tensor fields on $M$. The function $\langle \cdot,\cdot \rangle$ represents the operation of the metric $g(\cdot,\cdot)$. And denote $g=\sum_{i,j} g_{i,j} dx^i\otimes dx^j$ where $\partial_i$ ...


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In general relativity, we don't equip spacetime itself with a vector space structure, which means that the "four-position" is not a four-vector. As a result, the object $x_\mu$ whose components are given by $x_{\mu}=g_{\mu \nu} x^\nu$ is not a covector. One can show this by its transformation properties. If we change coordinate systems from $x$ to $y$, ...


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In the general case, 4-position is not a vector, so it does not have a covariant derivative. Describing it as a vector seems to assume Minkowski metric, in which case the covariant derivative is simply the partial derivative We do have $$x^i_{,j} = {\partial x^i \over \partial x^j} = g^i_j $$ $$x_{i,j} = g_{ij} $$


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If someone give me a metric tensor and ask: "what are the components of 4-velocity?" how can I calculate the components? You can’t. A massive particle at any point in any metric can have any time-like four-velocity, and a massless particle can have any lightlike four-velocity. All the metric lets you do is compute the covariant components of the four-...


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The source of your problem seems to be C). The Christoffel symbols satisfy $$\Gamma_{abc} + \Gamma_{bac} = g_{ab,c} $$ So (assuming no silly typos) $$g_{ab;i} = g_{ab,i} - \Gamma^c_{bi}g_{ac} - \Gamma^c_{ai}g_{cb}= g_{ab,i} - \Gamma_{abi} - \Gamma_{bai} =0 $$ Thus, the metric behaves as a constant with respect to covariant differentiation; this simply states ...


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In general coordinates, the metric does not even have to be diagonal. However, for any metric you can always find a change of coordinates such that at a particular point in spacetime the metric is diagonal. It can be proven that the signs of the coefficients on the diagonal are the same (up to reordering, i.e. there will always be same number of positive and ...


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I'll base my answer on the first question: your title. Could Ellis Wormholes be used for Interstellar Travel? The short answer is no, due to the physical content of standard (well-stablished) General Relativity. But let's see more deeply about this (boring) fact. I) General Relativity in a (planck-lenght) nutshell: So, gravity is a natural interaction ...


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A recent (2019) Arxiv submission by Hyat Huang and Jinbo Yang, https://arxiv.org/abs/1909.04603, generalizes the Ellis Wormhole to one with charge. AFAI can see, it is a proper treatment and there are many references.


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In the Scharzschild solution, we can write the geodesic equation in the form of the equations of motion $$ r^2 \dot\phi = h = \mathrm {const}$$ $$ {\dot r}^2 = {2\mu\over r} - \bigg(1-{2m\over r}\bigg){h^2\over r^2} $$ Einstein showed that solving these equations perturbs the Newtonian equations and results in orbital precession. For comparison, the first ...


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There are two steps to be carried out. First the anti-symmetrization $[cd;e]$ means you consider every permutation of the indices, and multiply by the sign of the permutation. In this case we have, $$R^a_{bcd;e} -R^a_{bce;d} + R^a_{bec;d}-R^a_{ebc;d} + R^a_{ebd;c}-R^a_{bed;c} = 0$$ where I have ignored the $\frac{1}{n!}$ factor ($n=\# \, \mathrm{indices}$) ...


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1) Given vectors $v$ in a vector space $V$ a dual vector is simply a function $f$ such that $f(v)$ is a scalar. It is easy to see that the set of these functions forms a vector space itself, $V^*$, called the space dual to $V$. Addition etc are defined as usual $(f+g)(v)=f(v)+g(v)$ etc. 2) Now, if $V^*$ is a vector space, one can find a basis such that any $...


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To add one more point of view, think about how the vector is defined in differential geometry (GR). The vector at point P can be defined as an equivalence class of curves going through point P given by the relations $$x^i(\gamma_1(0))=x^i(\gamma_2(0))$$ $$\left.\frac{d}{dt}x^i(\gamma_1(t))\right|_{t=0}=\left.\frac{d}{dt}x^i(\gamma_2(t))\right|_{t=0}$$ that ...


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You have a basis ${\bf e}_i$ in some vector space. The contravariant components of a vector ${\bf v}$ are given by ${\bf v}=v^i{\bf e_i}$, as Charles Francis says. The covariant components of a vector ${\bf v}$ are given by $v_i=\mathbf v\cdot\mathbf e_i$ I think that's a more basic way of thinking about them than going in to their transformation ...


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We don't talk of covariant and contravariant bases. Start with the basis $\{\mathbf e_i\}$. Then a general vector can be written $$\mathbf v = v^i \mathbf e_i$$ Now if you double the length of a basis vector, you must halve the length of the componant. The components are said to be contravariant, because they change opposite to the basis. In index notation ...


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I know you basically said this, but here is the answer. If you have a metric which is a sum of a background metric and a small perturbation, $$ g_{\mu \nu} = \overline{g}_{\mu \nu} + h_{\mu \nu} $$ then the inverse metric is $$ g^{\mu \nu} = \overline{g}^{\mu \nu} - h^{\mu \nu} $$ which can be confirmed by checking $g_{\mu \nu} g^{\nu \rho} = \delta_\mu^\...


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I assume you define the translational generators $P_\mu$ as proportional to the covariant derivatives $\nabla_\mu$. Then the commutativity of translational generators is indeed equivalent to the nullness of Riemann's curvature tensor. To see this recall the Ricci identity which relates the commutator of covariant derivatives to Riemann's curvature tensor: $...


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First some review, to establish context. Second order tensors come in three types: $(0, 2)$, $(1, 1)$, and $(2, 0)$, depending on the number of upper and lower indices. In general, an $(n, m)$ tensor is a multilinear real-valued function of $n$ one-forms and $m$ vectors. However, by currying the function, an $(n, m)$ tensor can also be equivalently viewed as ...


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I recommend you use Mathematica. The package GRQUICK.m https://library.wolfram.com/infocenter/MathSource/8329/ is a good one to compute the Ricci tensor and scalar. If you develop your own code and learn how to use the package you will be able to compute geometric objects very fast.


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For physical intuition, it can be helpful to think of vectors as describing velocities through a space (geometrically represented as an arrow), and one-forms as describing the rate at which a quantity varies across the space (for a two-dimensional space and a single-valued quantity, this can be geometrically represented as a small plane angled with respect ...


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Matter/energy =m/e. Space time. As described by Stephen Hawking, in a forward time frame stated in the theory of everything. Before the Big Bang, the fog of energy, as it was, we realise that space does not need m/e to be space. However m/e needs space to be space so m/e can be in it’s current forms of m/e. With space, you are always in it, but can never ...


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Introducing the dual space of linear maps allows you to work with co- and contra-variant indices even without a metric being defined. As Charles Francis answered earlier, in that case column and row vectors are a nice way to think about things. On the other hand, as it seems you may have noticed, in a metric space with inner product there is really no need ...


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To keep it simple, think of vectors (contravariant vectors) as column matrices and think of one-forms (covariant vectors) as row matrices (the dual space), and the inner product as a multiplication between row matrices and column matrices.


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To calculate the proper acceleration, you need to calculate all the components of the four vector, especially the radial one: (be careful to make a good covariant derivative) $$a^r = u^t(\partial_t u^r + \Gamma^r_{t\alpha}u^\alpha)$$ The proper radial velocity of this observer is 0. The time component can be calculated by making use of the fact that $u_\mu ...


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If the coordinate transformation removes h terms to first order then the metric is just left as $$ ds^2 = -c^2dt^2+dx^2+dy^2+dz^2 \tag{1} $$ which is clearly just the Minkowski metric in Cartesian coordinates. i.e. Its the choice of coordinates used initially that caused it to look 'non-flat' and so with a simple coordinate transformation, you recover flat ...


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The main idea answering this question was from this and here I'd like to answer in detail: We only have this: $\nabla_\lambda \sqrt{-g} g^{\mu\nu}=0$ ------(1 I will need some formulas which I will deduce firstly. $\delta^\mu_\nu=g^{\mu\alpha} g_{\alpha \nu} \implies \nabla_\lambda \delta^\mu_\nu=g_{\alpha \nu}\nabla_\lambda g^{\mu\alpha} +g^{\mu\alpha} ...


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The metric tensor, being a tensor, depends only on the space and not on the coordinate system. You can only get its components when you choose a coordinate system to describe the space. For example, both $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \text{and} \begin{bmatrix} 1 & 0 & 0 \\ 0 & r^2 &...


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I would say that you want a mathematical proof. You can see in Reign's answer a particular case, but here I will post a more general one. As you may know, a Lorentz transformation can be written as a 4-dimensional matrix, $\Lambda$, so that $x' = \Lambda x$. This matrix must fulfill this relation, \begin{equation} \Lambda^T \eta \Lambda = \eta \end{...


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You can provide a model of GR that does not require specifying a geometry - hence it is not a "geometric theory". See Henri Poincare convent


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Let $\{ \gamma_0, \gamma_1, \gamma_2, \gamma_3 \}$ be an orthonormal basis for Minkowski spacetime, with inner product $$\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu}$$ and timelike basis vector $\gamma_0$. Every lightlike vector can be expressed as a Lorentz transformation of $\gamma_0 + \gamma_3$, to which only two spacelike vectors are orthogonal: $\...


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He then goes on to state that there are three mutually orthogonal space-like vectors. Sure, the proof is by example: $$\epsilon_1^\mu = (0, 1, 0, 0), \quad \epsilon_2^\mu = (0, 0, 1, 0), \quad \epsilon_3^\mu = (0, 0, 0, 1)$$ where the time component is on the left. At one point he makes the statement that for given light-like vector, there are only two ...


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He can't see a Euclidean metric, except by restricting to space. I think the writer did not mean what he appears to say, as quoted in the comment. I think he only intended to give two examples of diagonal metrics, not to give two examples of metrics which an observer can actually see.


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I recommend you to first have a good handle of Spinors in GR. I personally took this route in getting a grip on Spinors: A Child's Guide to Spinors : http://www.weylmann.com/spinor.pdf An Introduction to Spinors : https://arxiv.org/abs/1312.3824 Penrose, Roger (1960). "A spinor approach to general relativity" The third one is going to be a smooth if you ...


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In differential geometry $\partial_i$ is usually tangent vector (or vector field) to the coordinate curve of $i$-th coordinate. The curve is given by the parametric equations: $$x^j(t)=x_0^j+\delta^j_i t$$ so the vector $\partial_1$ has components $\delta^j_1.$ The completion is by using the fact that $$ g(\dot{x},\partial_1)=g_{ij}\dot{x}^i\delta_1^j=g_{...


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Just use the definition. All off-diagonal elements of the metric vanish, so $$\eta^{\mu\nu} \eta_{\mu\nu} = \eta^{00} \eta_{00} + \eta^{11} \eta_{11} + \eta^{22} \eta_{22} + \eta^{33} \eta_{33} = 1 + 1 + 1 + 1 = 4.$$ Also, the question you asked before this one is about the "permutation invariance of the maximal helicity violating gravitational amplitude in ...


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Simple explanation: Its the only type of metric that conserves the speed of light. Mathematical explanation: Let us assume an inertial observer, $O$, measures the speed of light as $dx/dt = c$. Let us assume another inertial observer, $O'$, measuring the speed of light in his coordinates, $dx'/dt' = c$. Naturally at this point one can ask, What is the ...


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You need to be clearer about the distinction between coordinates and measured quantities. Coordinates are just a map, and like a map of the world, they contain scaling distortions described by the metric. Here is an illustration of a comoving coordinate system for an FLRW universe with positive curvature. The galaxies on geodesics maintain constant ...


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See Nuclear Shell Theory by Amos de Shalit and Igal Talmi. The book is constructed from two parts and one for tensors a total of 586 pages. My thought is he dived into some confusing details which should be read carefully. The following parts consider applications that makes a good deal with mathematics.


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You are doing it correctly. In reference to your comment, to find the minkowski metric in cylindrical coordinates, it is easiest to transform the line element so: $$ ds^2=g_{\mu\nu}dx^\alpha dx^\beta \tag{1}, $$ Where usually the minkowski metric is denoted by $\eta_{\mu\nu}$ Which is $$ ds^2 = -c^2dt^2+dx^2+dy^2+dz^2 \tag{2} $$ In Cartesian. If you ...


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If $\pi_{ijk}$ is the ordinary Levi-Civita symbol, then the Levi-Civita tensor has components $\epsilon_{ijk}=\sqrt{g}\pi_{ijk}$ (I am assuming the metric is positive definite, but if not then replace $g$ with $|g|$), where $g$ is the determinant of the metric tensor. Therefore, we have for $C_{i}=\epsilon_{ijk}A^j B^k$ $$ C_1=\sqrt g(A^2B^3-A^3B^2) \\ C_2=\...


1

Assume you have $\mathbf p = p^{\alpha'} \mathbf e_\alpha$. Then $$ \mathbf p\cdot\mathbf e_\beta = p^{\alpha'} \mathbf e_\alpha \cdot\mathbf e_\beta = p^{\alpha'}\eta_{\alpha\beta}$$ so $$ p^{\alpha'} = \eta^{\alpha\beta}\mathbf p\cdot\mathbf e_\beta$$ In particular $$ p^{0'} = \eta^{00} \mathbf p\cdot\mathbf e_0 = -\mathbf p\cdot\mathbf e_0$$ And yes, you ...


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If the metric gμν is dimensionless and gravitons are quantum excitations of the metric does that mean that gravitons themselves are dimensionless? It has dimensions , at least look here to a particular metric , the matrix elements have the dimensions of meter square. Is graviton energy included in the stress-energy tensor Tμν? In phsyics there are ...


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First question:If gravitons do exist , according to the Standard Model which describes all particles they must have certain properties -) Average rest mass >>>> mass of any particle ( because gravity is increadibly weak). -) Infinite decay lifetime ( they do not decay ). -) Charge = 0 -) Quantum Spin = 2 ( only attraction ) But the gravitons ( if they ...


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The derivation you refer to is done locally in a set of coordinates in which $g_{ik,l} = 0$, which can be also characterized as related to a set of Riemann normal coordinates by a constant linear transform. This means, that you can make operations such as this one $$g^{ij} (g_{kl,mn}) = (g^{ij}g_{kl,m})_{,n}$$ Another useful identity is $g^{ij}g_{jk,l} = -g^{...


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It says that the metric behaves as a constant with respect to covariant differentiation, meaning that vectors have constant magnitude under parallel displacement. In other words, if you displace a metre rule, it is still a metre rule, and if you displace a clock which measures one second per second (without disturbing the mechanism), it will still measure ...


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I think I have a fairly intuitive answer that is grounded in geometry and physics. If the covariant derivative acting on the metric tensor vanishes, it means that during the parallel transport of any two vectors $u$, $v$ (that is, vectors living on the tangent space at a point of the manifold) along any curve, the inner product between them is covariantly ...


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Schwarzschild coordinates work Most astrophysical calculations involve things that are outside the event horizon. In these cases Schwarzschild coordinates work just fine. Observed phenomena like gravitational lensing, orbital changes relative to Newtonian gravity, and gravitational redshift can all be understood using Schwarzschild coordinates. ...


1

To calculate its apparent position, would we have to calculate the path of a beam of light from the projectile to an observer? Wouldn't we also have to do that in Schwarzschild coordinates? This seems like a non-trivial calculation mathematically. This is exactly what you have to do to show this. Fortunately, this is not a very hard calculation. Outward ...


1

Schwarzschild (Droste) coordinates have a number of advantages over other coordinates systems for the Schwarzschild metric such as Eddington-Finkelstein (EF) or Gullstand-Painlevé (GP) coordinates. 1. Symmetry One is that in Schwarzschild coordinates the time translation and axial symmetries are explicitly manifest. In these coordinates the vector fields $\...


2

If cosmic matter would not move WRT such a system, how would it be possible to notice the expansion of the universe ? Observationally we see the light is redshifted so this is caused by the expansion of the universe. Other theories such as static universe or tired light theory has been ruled out. And you can show that in a static matter-filled universe (...


1

Whether a “reasonable” spacetime is geodesically connected depends on the definition of “reasonable”. If we consider globally hyperbolic spacetime as reasonable, then there is a theorem by Avez (1963) & Seifert (1967), that states that globally hyperbolic spacetimes are causally geodesically connected. Note, that the two points here must be causally ...


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