Jump to content

Leaderboard

Popular Content

Showing content with the highest reputation on 08/26/22 in all areas

  1. There are many minor conditions that are almost inescapable. Almost everyone over 65 starts to have some lactose intolerance, even if they didn't when younger. Almost everyone has some drop in natural melatonin production, which means needing to take more care to get enough sleep. Some loss of muscle mass is almost inevitable, which affects strength and resistance to cold. The immune system declines (with considerable variation between individuals) and recovery from illness is slower. Joints wear out, no matter how robust your overall health, and will need some extra attention. Other changes are common in western societies but less common in low tech pastoral societies where people do physical labor throughout life and eat less processed food and breathe less nasty city air, e.g. high blood pressure, acid reflux disease, diabetes, constipation, atherosclerosis, etc. So...walk everywhere, use your muscles, do chores minimizing machinery assist, eat lots of plants, pour on the olive oil and fish oil, do core strength exercises, and try to avoid biking behind a bus.
    1 point
  2. I understand that your are attempting "to further clarify your position", but your position has no evident relation to the statement I made regarding 20th century secularism; therefore I cannot regard it as a response.
    1 point
  3. It's a solid state device. There's no liquid or gas to leak.
    1 point
  4. ...or you can just look at the invariants of the Riemann tensor in that region, in particular the Kretschmann scalar. Since it exists and is regular and well defined on the horizon, spacetime must necessarily be smooth and continuous there.
    1 point
  5. This is a picture of a laser diode with the protective housing removed. The picture scale is less than 1 cm across. You can see tiny wires going to the laser and to its substrate - this puts the voltage across it and allows current to flow. These can break. There are also wires leading from the power supply to the laser assembly. These could possibly break, too
    1 point
  6. Here's the Kruskal-Szekeres change of charts: It's singular at r=RSchwarzschild=2GM because the Jacobian is zero there. I don't know where you got the 00 problem from. 00 is no problem if you can calculate the limit. The problem is the Jacobian is zero. This is not allowed for good reasons I'm not going to delve into. But:What is a chart? It's an assignment of coordinates: \[ \phi:\mathcal{U}\subseteq\mathcal{M}\rightarrow\mathbb{R}^{4} \] This could be the Schwarzschild chart. And in comes Kruskal: \[ \psi:\mathcal{U}\subseteq\mathcal{M}\rightarrow\mathbb{R}^{4} \] What's a change of charts? In this case from Schwarzschild to Kruskal: \[ \psi\circ\phi^{-1}:\mathbb{R}^{4}\rightarrow\mathbb{R}^{4} \] The initial chart is singular at the 2-surface r=2GM. Now Kruskal introduces a singular change of charts that restores smoothness on that region. It stands to reason that you must do something drastic on r=2GM if you want to restore smoothness. This is a particular feature of analytic continuation in this case. Now "stasis". You do not look at the metric in a particular coordinate set and infer anything. That's a sure way to make mistakes. What you do is what Markus told you: In this case, if one wants to prove that the metric is static, what one does is define a Lie derivative, and from there introduce time-like Killing fields. Then prove that your metric is invariant under those infinitesimal transformations.
    1 point
  7. OK I will give the answer as one member has asked for it ! Any triangle can be split into four equal and identical parts each being similar triangle ! Any quadrilateral polygon can be divided into two triangles !! [even the odd figures] Cutting these 2 triangles into 4 equal parts each enables you to join them into 4 pairs of identical shapes !!! In the case of polygons with no crossing lines two of these shapes will occur naturally connected together each being similar to the parent polygon and the other two are obtained by moving and pairing the one-fourth triangles !!!! That's it 🙂
    1 point
  8. joigus has beaten me to it with his excellent answer (+1). As I said earlier, the manifold and a particular coordinate chart chosen on it are not the same things at all - to put it succinctly, having a ‘hole’ in an embedding diagram does not necessarily imply that there is a corresponding ‘hole’ in the manifold, in a topological sense. These are different things. You can in fact have patches (or entire manifolds) without coordinates defined on them. For Schwarzschild spacetime, you need only transform the metric to a different, more complete coordinate basis to see this. But if you want to be absolutely sure and precise, it is always best to use tools that are coordinate-independent. Yes, indeed. Arriving at a precise value is actually not easy, also because external conditions play a role during the collapse. But I think the salient point is that there is such a limit, for any given level of degeneracy. It’s hypothetical to some degree, yes. But just as in the case of quantum gravity, there are good reasons to believe that the GUT domain is quite real, even if we don’t know for sure which of the numerous GUT candidate models will apply. That being the case, quarks and gluons are by-products of a broken GUT symmetry, so once energy levels are high enough, the strong interaction will cease to exist in its ordinary form. In more general terms, I very much agree that singularities are not real-world objects, but artefacts of our models being pushed beyond their domains of applicability.
    1 point
  9. if everything is information, the randomness of radioactive decay could be explained by the mechanism of description of individual atoms. If every atom is a specific algorithm set, consisting of strings of numbers at it's most fundamental level, perhaps the algorithm contains largely similar number sets, with small variations, like junk DNA in living cells. One or two variations in a cell's DNA might have no significant bearing on the identity or functioning of a particular cell. Similarly, small variations in protons and neutrons mathematical makeup might cause the randomness of certain radioactive decays...yet have them remain as functioning particles in normal relationships with other fundamental particles. Their sub-components of quarks are likely simpler arrangements of numbers and have few variations within them, but added together in the 3 quark components of protons or neutrons, the cumulative effects add complexity to the situation for each individual atom. If this is true, then a direct readout of the complete algorithmic structure of individual atoms could allow a prediction of the life span of a particular atom, and make the radioactive decay more predictable...ps to peter, if we don't "go there" into mysticism, we won't de-mystify it...
    1 point
  10. Not only. Proton emission also, double proton emission. It's easy to make mistake that will remain unnoticed if we will be doing such shortcuts. (in above example one electron and one proton will be free)
    1 point
  11. The electron mass is only an issue for beta plus decays, where the total number of electron-mass particles changes. Otherwise they cancel in the atomic mass measurements. The change in electron/atomic binding energy of the system is a few eV, which can be ignored.
    1 point
  12. Do you know how to calculate Decay Energy of unstable radioactive isotope? First, find parent isotope mass. They're in f.e. wikipedia articles such as http://en.wikipedia.org/wiki/Isotopes_of_hydrogen (last portion of link is name of element). For instance Carbon-14 has 14.0032419894 u Then, you need to multiply it by 1u = 931.494 MeV to receive total energy of nucleus + electrons. Then you need to get rid of mass-energy of electrons. Multiplying 0.510999 MeV by quantity of protons/electrons. Subtract total energy of electrons from energy of isotope. Repeat it for daughter isotope. In Carbon-14 case it's Nitrogen-14. Subtract N-14 nucleus energy from C-14 nucleus energy. Subtract electron energy (it's emitted during decay together with antineutrino) Result will be Decay Energy of isotope (kinetic energy of particles + energy of neutrino in our case). Decay Energy must be positive value. If you will calculate Decay Energy from stable isotope such as Deuterium, Helium-3, Helium-4 for all possible decay modes you will see that D.E. would be negative value. Thus these particles are stable. For Carbon-14 it's 0.156 MeV. As you can see this value does match article: http://en.wikipedia.org/wiki/Carbon-14 Other examples: Fusion of proton-proton, decay of carbon-11 and decay of Mangan-54: I wrote computer application which has built-in database of isotopes and calculates possible decay modes and released energy.
    1 point
  13. I can't decide if this is hilarious, or just plain sad ... Shaquille O'Neal Says His Flat-Earth Comments Are 'Just a Theory' While Questioning If Earth Spins (msn.com) We pay these people millions of dollars, and kids look up to them. The internet/TV/Hollywoood has made these people 'influencers', and they are contributing to the 'dumbing down' of society. Of course, I would not say any of this to Shaq's face; he would crush me.
    0 points
  14. Thanks for that video about people getting drunk together in Russia. Totally relevant to this thread about how the US should respond to Russia’s illegal invasion, theft, and murder of Ukraine and Ukrainians.
    0 points
  15. On the issue of drunkenness in Russia. Drunks on the streets of Moscow sometimes meet
    -1 points
  16. Well, if you insist on snipping the sentence "to further clarify my position.", then I can only conclude, your misunderstandings are deliberate, in order to slip the punch/question perhaps...
    -1 points
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.