24 Matching Annotations
  1. Jul 2019
    1. here exists a se-quence of initial data that satisfy all the hypothesis of item (i) and suchthat in the limit the equality in (3) is achieved. In this limit, the radius,the charge and the total mass of this sequence tend to zero.
  2. Jun 2019
    1. A standard computation using the Gauss equation shows that∂f∂t(0,t′) =ddt|Σ0|g(t)(t′) =−∫Σ0(R−Ric(ν,ν))dμ=−4πχ(Σ0)−∫Σ0(Ric(ν,ν) +|A|2)dμ,where all geometric quantities are computed with respect tog(t′).
  3. arxiv.org arxiv.org
    1. A computation in coordinates shows that the Ricci tensor ofhis given byRich(X,X) =−(1V∆gV)h(X,X),Rich(X,Z) = 0,Rich(Y,Z) =Ricg(Y,Z)−1V(HessgV)(Y,Z)
    2. The structure of the metrichnear the singular set clearly implies thatgeodesics realizing the distance between a point inNand a component of∂Mmeets∂Morthogonally. The proof of this fact is essentially the sameas the proof of the Gauss’ Lemma.
  4. Mar 2019
    1. Evolução da carga

      $$ \begin{aligned} Q(t) & \equiv Q_{\nabla\phi}(\Sigma) := {1 \over 4\pi} \int_{\Sigma} \langle \nabla\phi, \nu \rangle d\sigma_g \\ %% & = {1 \over 4\pi} \int_{\Sigma} d\phi \cdot \nu d\sigma = {1 \over 4\pi} \int_{\Sigma} \frac{\partial\phi}{\partial\nu} d\sigma \end{aligned} $$

      $$ \begin{aligned} \Longrightarrow \frac{dQ}{dt} = {1 \over 4\pi} \int_{\Sigma} \left[ d(\partial_t \phi) \cdot \nu + d\phi \cdot \partial_t \nu + d\phi \cdot \nu \frac{tr_{\Sigma} \partial_t g}{2} \right] d\sigma \end{aligned} $$

      Tomando \( \alpha = 2 \), obtemos: $$ \begin{aligned}

      • {tr_{\Sigma} \partialt g \over 2} & = R - Rc(\nu, \nu) - \alpha \left( |\nabla \phi|^2 - (\partial{\nu} \phi)^2 \right) \ & = R - Rc(\nu, \nu) - 2 \left( |\nabla \phi|^2 - (\partial_{\nu} \phi)^2 \right) \end{aligned} $$
    2. the maximum principle above, yieldsSmin(t)≥Smin(0)1−2tmSmin(0)(5.3)for allt≥0 as long as the flow exists
    3. Theorem 4.4Let(g(t),φ(t))solve(RH)αwithα(t)≡α >0. ThenSandSdefined as above satisfy thefollowing evolution equations∂∂tS=△S+ 2|Sij|2+ 2α|τgφ|2,∂∂tSij=△LSij+ 2ατgφ∇i∇jφ.(4.14)Proof.This follows directly by combining the evolution equations from Proposition4.2withthose from Proposition4.3.Remark.Note that in contrast to the evolution of Rc,R,∇φ⊗∇φand|∇φ|2the evolutionequations in Theorem4.4for the combinations Rc−α∇φ⊗∇φandR−α|∇φ|2donotdepend on the intrinsic curvature ofN.

      Note que,

      $$\alpha = 2 \Longrightarrow S = R - 2 |\nabla \phi|^2,$$

      que é justamente a função que precisamos estimar (veja prova do corolário 5.2), no caso particular de um campo gradiente.

      Haja visto que no na aproximação eletrostática do eletromagnetismo clássico, o campo elétrico (em domínios simplesmente conexos) é gerado por um potencial escalar, isso sugere que, pelo menos nessa aproximação particular, podemos utilizar esse fluxo \((RH)_{\alpha}\), tomando o pontical elétrico como dado inicial para o fluxo do calor para mapas harmônicos.

      Essa ideia é inspirada nas ideias das seções 2 e 3, desse artigo do Benhard List, onde ele observa que soluções estáticas desse fluxo, com \(\alpha\) escolhido adequadamente, coincide com as soluções estáticas para a equação de Einstein no vácuo.

  5. Aug 2018
  6. Apr 2018
    1. the scalar curvatureRofds2is given byR= (1−u−2)Rρ+u−2n−1∑i,jR0ijij+ 2n−1∑i=1Rnini= (1−u−2)Rρ+u−2R0−2u−1∆ρu+ 2u−3∂u∂ρH0whereR0is the scalar curvature ofNwith respect tods20andRρis the scalar curvatureof Σρwith the induced metric.
  7. Sep 2017
    1. Let Σ0be a compact strictly convex hypersurface inRn,Xbe the position vector ofa point on Σ0, and letNbe the unit outward normal of Σ0atX. Let Σrbe the convexhypersurface described byY=X+rN, withr≥0. The Euclidean space outside Σ0canbe represented by(Σ0×(0,∞),dr2+gr)wheregris the induced metric on Σr. Consider the following initial value problem(2.1)2H0∂u∂r= 2u2∆ru+ (u−u3)Rron Σ0×[0,∞)u(x,0) =u0(x)whereu0(x)>0 is a smooth function on Σ0,H0andRrare the mean curvature and scalarcurvature of Σrrespectively, and ∆ris the Laplacian operator on Σr.

      Note que de agora em diante o autor se detém a estudar esse caso particular, onde estão inteiramente determinadas as geometrias intrínseca e extrínseca das folhas do semi cilindro, obtido folheando-se pelas paralelas o exterior da hipersuperfície estritamente convexa dada a priori.

    2. Given a functionRonN, we want to find the equation forusuch that(1.2)ds2=u2dρ2+gρhas scalar curvatureR.

      O papel da aplicação \( u: N \longrightarrow \mathbb{R} \) é distorcer as fibras do semi cilindro \( N \), por dilatações e torções, deixando a geometria intrínseca das folhas invariante, de tal forma que o resultado seja um semi cilindro com a curvatura escalar prescrita \( \mathcal{R} \).

    3. Let Σ be a smooth compact manifold without boundary with dimensionn−1 and letN= [a,∞)×Σ equipped with a Riemannian metric of the form(1.1)ds20=dρ2+gρfor a point (ρ,x)∈N. Heregρis the induced metric on Σρwhich is the level surfaceρ=constant

      Isso significa que a construção a seguir é feita a partir de um semi cilindro em que a geometria das folhas é dada a priori.

      Esse artigo não trata da construção desse semi cilindro inicial.

  8. May 2017
    1. However, this re-scaling of the space and time coordinates is in fact the result of Einstein ignoring the principle of the constancy of the speed of light by applying the usual vectorial velocity addition and then subsequently trying to compensate for this error by making a further error and changing the given length and time definitions (see my page regarding the Speed of Light).
    1. if the corresponding distances are identical in both systems, then the delay times will also be identical and there won't be any difference in the clock readings afterwards
  9. Feb 2016
    1. An all-star international team of astrophysicists used an exquisitely sensitive, $1.1 billion set of twin instruments known as the Laser Interferometer Gravitational-wave Observatory, or LIGO, to detect a gravitational wave generated by the collision of two black holes 1.3 billion light-years from Earth.
  10. Dec 2015
    1. Lemma 2.3.(2.1) has a unique solutionufor allrwhich satisfies the estimates in Lemma2.2.
    2. Let Σ0be a smooth compact strictly convex hypersurface inRn. Letrbe the distance function from Σ0. Then the metric on the exteriorNof Σ0is given bydr2+gr, wheregris the induced metric on Σr, which is the hypersurface with distancerfrom Σ0. The functionuwith prescribed scalar curvatureR= 0 is given by2H0∂u∂r= 2u2∆ru+ (u−u3)RrwhereH0is the mean curvature of Σr,Rris the scalar curvature of ΣrandR0is the scalarcurvature of Σrwith the induced metric fromRnand ∆ris the Laplacian on Σr.
  11. Nov 2015
    1. If the space doctor’s ideas were wrong, your phone wouldn’t be able to tell where it was.

      If all of the space boats travel at the same velocity and the same distance from the Earth's surface, the time shift of the on-board clock due to relativity would be the same. The difference between the clocks would be almost the same. How would the trouble appear?

  12. Mar 2015
    1. θ dμ ≥ p 16 π | Σ |

      Qual a relação dessa desigualdade com a dita desigualdade de Penrose Riemanniana provada por Huisken-Ilmanen e Bray?

    2. GIBBONS-PENROSE INEQUALITY

      Qual a relação dessa desigualdade com a dita desigualdade de Penrose Riemanniana provada por Huisken-Ilmanen e Bray?

  13. Nov 2013
    1. "the correct perception"-which would mean "the adequate expression of an object in the subject"-is a contradictory impossibility. For between two absolutely different spheres, as between subject and object, there is no causality, no correctness, and no expression; there is, at most, an aesthetic relation:

      No absolute truth, all subjective and relative. Reality much more complex

  14. Sep 2013
    1. For the truth is, Socrates, that you, who pretend to be engaged in the pursuit of truth, are appealing now to the popular and vulgar notions of right, which are not natural, but only conventional

      Right and wrong are convention