# Tag Info

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The main result of Nottale is well known as just a consistency postulate of quantum gravity: that if the electromagnetic renormalisation of electron mass is cut off at Planck Scale, the correction is of the same order of magnitude that the electron mass itself. This is remarked eg in Polchiski string theory book. Over this consistency postulate, Nottale ...

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It is an analogy, as a 4 dimensional equivalent would be hard to draw, and if it was 3 dimensional, you couldn't see what's inside!

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You might be confusing some issues. In special relativity, space and time do not stretch or compress. It really comes down to measurements with clocks and rulers made by people that are moving uniformly with respect to each other. One option that is consistent with observations for SR is that there is one family whose clocks and rulers are right and ...

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I have heard three theories for how space-time is shaped, flat, sphere-like, or saddle-like. Flat is the most likely, as all our measurements implies that space time has curvature close to 0. Inflation makes it so that a sphere like or saddle like spacetime evolves into a sphere like or saddle like spacetime that has a curvature very very close to zero. ...

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This is just a short answer, from Wikipedia, which maybe familiar to you already. It may be useful as an analogy of what you are trying to do, if I have understood your question correctly. Lattice Field Theory In physics, lattice field theory is the study of lattice models of quantum field theory, that is, of field theory on a spacetime that has ...

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Firstly, you could be discrete without being a lattice, for instance you could make a bunch of different shaped tetrahedra such that each region of space was inside a tetrahedra. If the tetrahedra were randomly sheared but all had a kind of regularity where they were all big enough to contain a sphere of radius A and be contained in a sphere of radius B, ...

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I also don't think I can give a definitive answer. But maybe no one can because maybe no one knows enough about the theory and about why people don't know about the theory. But I will do my best. As mentioned by Nikolajs maybe fee people think they can contribute to it or make use of it. This is partly because of the success of quantum mechanics and ...

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The fact of a wave doesn't needfully mean there's tension or elasticity in the sense of those phenomena in an acoustic medium. General Relativity describes how spacetime's geometry depends on the distribution of energy within it[1]. But GTR does not tell us anything about the microscopic "machinery" of spacetime that gives rise to this behavior: tension ...

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It's quite common to use time to parameterise an equation. For example suppose you have a particle moving in a circle (or radius $r$). One way of describing its motion would be to say the trajectory describing its motion is: $$x^2 + y^2 = r^2 \tag{1}$$ but an alternative description would be to use the pair of equations: \begin{align} x &= ... 2 Solutions to G_{\mu\nu}=0 are called vacuum solutions in GR, it follows mathematically that this happens if and only if the Ricci tensor vanishes, i.e. the solutions are exactly the Ricci flat Lorentzian manifolds. In most known explicit examples only some region is Ricci flat (e.g. around Schwarzschild or Kerr black holes), but some global vacuum ... 2 Even though calculus was invented in order to study a physics problem, it is a mathematical tool that is very general, that also describes discrete situations. But if I am not mistaken a fundamental principle of the quantum world is that things can only take on disjointed, discrete values You are confusing necessary and sufficient. It is sufficient to ... 2 One simplification of what gravitational waves are and how they're created, can be understood by visualizing a massive body and its gravitational field when stationary and what happens when the body accelerates. Imagine what happens if such a body were to start accelerate to the right for a while, and then stop. Just like the time-delay with light, it will ... 2 The math in that article is based in Cartesian space. Note specifically figure 4, where a portion of a Cartesian plane is pinched in at one side to show the supposed warping due to gravity. Using the shown transformation, she concludes that space is compressed near a black hole rather than stretched. The diagrams after that along with the process ... 2 Yes. You can make a model where you have coordinates t, x,y,z where for any x,y,z the universe looks the same. The metric ends up looking e.g. likeds^2=dt^2-(a(t))^2(dx^2+dy^2+dz^2)$$and you can move your x,y,z to have any value and everything looks the same (those things do loom different for different cues of t). You end up with the densities ... 2 You are approaching the question from the wrong end. The expansion of the universe is described by a particular solution to Einstein's equation called the FLRW metric. To derive this metric we have to make some assumptions, and the key assumptions are that the universe is isotropic and homogeneous i.e. that it is the same everywhere. So the universe being ... 2 In a certain narrow sense, I think the answer to your question is believed by most physicists to be an emphatic no. That is, if by curvature you mean the fourth rank, valence \left(\begin{array}{c}1\\3\end{array}\right) tensor \mathbf{Riemann} that appears (through its contracted rank 2 covariant version the Ricci tensor) in the Einstein field equations. ... 1 The axis of simultaneity, or in other words, the set of events which are simultaneous as measured in the rest frame of the ship, does indeed change suddenly when we turn back. This is because it depends on your reference frame. There isn't a single inertial frame that stays with the ship for the whole journey; you can either accept that the frame is ... 1 Perhaps it's more illuminating to look at the whole thing in a spacetime diagram. we have the earth frame with coordinates (t,x), and its trajectory through spacetime is the blue line. The trajectory of the spaceship is the red one. Straight worldlines are inertial frames of reference, curved or non-straight worldlines are non-inertial frames of ... 1 It just means the Schwarzschild coordinate system is faulty at the horizon because it assigns the same coordinates (t,r,\theta, \phi)=(\infty,2m,\theta,\phi) to multiple events that are actually distinct events. If you look at the Kruskal-Szekeres coordinate system you can pick an event on the horizon and then draw the past light cone and those are ... 1 The mathematics of general relativity is clear and unambiguous. The trouble comes when you try and describe what is going on it non-mathematical terms, because there is no precise way to do it. Kip Thorne is attempting to talk about black holes in non-mathematical terms, and he is adopting a different perspective from (probably) most of us. That doesn't mean ... 1 Imagine a 1d space and a little vector in it that can change its location but not its length. It can't continuously change in a way that turns it around. If that vector was the momentum this says in 1d you can't continuously change you direction without your momentum being zero. A similar thing happens in relativity. A massive object has a location in 4d ... 1 Due to gravitational time dilation, for an observer of the planet, the frequency of electromagnetic radiation would be slower. Visible light emitted from the planet would appear as infrared or micro-waves. The amplitude of the radiation would not change. Since frequency decreases while amplitude remains constant, the radiometer would receive less ... 1 Sébastien : I can't give a definitive answer, just an opinion. And I would say that to get anywhere, this theory has to be consistent with relativity, which is one of the best-tested theories we've got. See http://arxiv.org/abs/1403.7377. It has to persuade the "relativists" first, and only then can it gain credence amongst the wider physics community and be ... 1 You are not travelling faster than light in the sense that if you send some light to your destination it gets there before you do. It can be faster than light in the sense that if space is isotropically and homogeneously distributed with energy and such then there is an obvious global frame and distance in the global frame between two points can decrease ... 1 One should clarify that there exist different frameworks for answering this question. The classical and the quantum mechanical. The reply by Xeren addresses the classical framework, i.e where , as with the E and B fields of classical electromagnetic radiation there exists the G field , a tensor field, and in a similar way to electromagnetism builds the ... 1 Marek didn't really make a mistake, but you did. But Marek might have been unclear by jumping over some steps. So first let's clarify. The metric gives the differential squared interval ds^2=-dt^2+dx^2 from which you can get the differential proper time d\tau=\sqrt{dt^2-dx^2}. So for the blue straight line you have \tau=\int ... 1 Calculus works in two senses and fails in a third one: as a mathematical theory, it is well-founded, independently of Newton and Leibniz invention. For operations with space and velocity, it also works, because these quantities are not even discrete in usual quantum mechanics, and the only argument can come if you go "atomistic"; then it must be substituted ... 1 The s-orbital is spherically symmetric and thus its angular momentum should not generate any sort of gravitational waves. According to your theory, atoms should never be able to excite out of an s-orbital. That's experimentally wrong. If the s-orbital is not spherically symmetric (contrary to what experiments show), but involves a classical particle ... 1 Light cannot exist without spacetime. Every light ray has one place of emission and one place of absorption. You might agree that between both places there must be a spatial distance, otherwise there would be no light ray and no transport of momentum. Light in vacuum (v=c) is associating this spatial interval with a zero spacetime interval, because it is a ... 1 OP is looking for the vacuum solutions to the Einstein field equations. Including only the cosmological constant \Lambda, the EFE become the Lambdavacuum field equations,$$ R_{\mu\nu}=\left(\frac12R-\Lambda\right)g_{\mu\nu}\tag{1}  with $R_{\mu\nu}$ the Ricci curvature tensor and $R$ the Ricci scalar. Solutions to this depend on the sign/value of ...

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