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elfmotat

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Everything posted by elfmotat

  1. Okay tar, I'll make this as simple as I possibly can: Bob is on the ground in front of some train tracks, and Alice is standing in the middle of a train that's moving with constant velocity. On each end of the train there is a light bulb. At the moment when Alice is directly in front of Bob (i.e. when the distance between Alice and Bob is the smallest), each light bulb emits a flash of light. Bob is the same distance from each light bulb when the flashes go off, and each flash spreads out at the constant velocity c, so we conclude that Bob sees the two flashes at the same time. Bob sees Alice moving toward one flash, and away from the other. That means one flash will reach her before the other. So Alice must see the two flashes occur at different times.
  2. No it isn't. You should try rereading it. It takes a shortcut by using the fact that the non-Newtonian additional deflection is due to spacial curvature, and the factor contributing to the extra gravity is the same as the Doppler-shift factor. Also, there's no such thing as a "proof" of a theory. This is science, not math.
  3. I don't even know what you're trying to say. Where did we mention anything about time dilation? What about the cherry-picked quote you just gave makes you think it has anything whatsoever to do with interpretations of time dilatation?
  4. I can think of a few ways to do it. One way would be with a metric theory with the field equation: [math]\nabla^2 h_{00} = \frac{8 \pi G}{c^2} \rho [/math] where [math]g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}[/math]. The field equation has no time derivatives in it, so the speed of gravity remains infinite just like Newtonian gravity. (But this is irrelevant for the OP's problem since the gravitational field of the Sun is assumed to be static.) For a spherical body like the sun, the metric in spherical coordinates would be: [math]ds^2=-c^2 \left (1-\frac{2GM}{c^2 r} \right ) dt^2 +dr^2 +r^2 d \Omega^2[/math] Plugging that into geodesic equation, the equations of motion are equivalent to those for Newtonian gravity. But now you can use the fact that ds2 = 0 for photons to calculate their trajectories, so it's Newtonian gravity + constant speed of light.
  5. You should read Enthalpy's post again. It's not the average rotational speed that's important, it's the speed as a function of radius, v(r ), which is important. The fact that v(r ) ≈ const. remains true regardless of whether or not you Doppler-shift the frequency of rotation.
  6. What? That's just logic. If what we're discussing is not well-defined then there is nothing objective or falsifiable to have a discussion about. Which means we're not doing science. This is a science board after all. And while we're talking about definitions, you should look up the definition of "censorship."
  7. Please explain what your picture means. Because as far as I can tell it looks like some random shapes plus some random words arranged in a nonsensical way.
  8. How about this: http://www.iisc.ernet.in/currsci/apr102005/1155.pdf The first part shows the Newtonian calculation.
  9. Lazarus, you're under the impression that when the calculations using Newtonian gravity were done, they neglected to include the fact that light always travels at c. I don't know where you got this silly idea from, but it's completely untrue.
  10. I don't think you understand. That's exactly how the calculations were done - using Newtonian gravity and the fact that light always travels at c, you get a prediction that is 1/2 of what it should be. This is because Newtonian gravity does not give accurate predictions when you're considering objects moving at or near c, or when the gravitational field is very strong. You need GR in those cases. GR gives the correct result: twice the Newtonian prediction.
  11. I'm not really sure how someone's opinion can count as censorship. You're more than welcome to believe in whatever you want.
  12. I'm not sure what you mean by this. Rotating a graph doesn't physically change anything.
  13. http://www.scienceforums.net/topic/85716-blacks-holes-yesno/
  14. This looks like the classic "God of the gaps" argument. People always say "God must be the cause of X, because we don't know how it works." The problem is, eventually we do figure out how it works. And when we do, God shrinks. Also, great name:
  15. I stand corrected. My point was that we haven't been looking for DE for very long, only about ~15 years, and that the direct effects of DE on our solar system would be extremely tiny. The "reason" is because of the Einstein field equations, not because of Hubble's law. If you take the Einstein field equations with nonzero cosmological constant in the limit that everything moves slowly compared to light and gravitational fields are weak, then you get the modified Poisson equation as the field equation. You might find this post to be helpful: http://www.scienceforums.net/topic/72570-how-can-galaxies-exist-with-the-expansion-of-space/?p=728733
  16. The error was so massive because, referring again to that force law I posted, when two massive bodies are near each other the GM/r2 term completely dominates cosmological constant term. Hubble's Law is actually the special case where Λc2r/3 >> GM/r2 ≈ 0. To see this you can let the GM/r2 term in the force law go to zero, and solve the corresponding differential equation: [math]\frac{d^2 r}{dt^2} = \frac{\Lambda c^2}{3} r = H^2 r[/math] where I've defined H2 = Λc2/3. The solution to this equation is: [math]r=k e^{Ht}[/math] for some constant k. The velocity is therefore: [math]v= Hk e^{Ht}=Hr[/math].
  17. Hubble's Law only works on bodies that are not gravitationally bound to us. So it's not surprising that it spits out strange values when you attempt to apply it where it doesn't belong.
  18. I'm not sure what you mean by this. What exactly did you calculate? Let's assume the Earth and the Moon are the only two bodies in the entire universe. With a linear approximation of GR with non-zero cosmological constant, we can derive the following force law: [math]\frac{\mathbf{F}}{m}=-\frac{GM}{|\mathbf{r}|^3}\mathbf{r}+\frac{c^2\Lambda}{3}\mathbf{r}[/math] (For more information about where this equation came from, see my post here: http://www.scienceforums.net/topic/72550-antigravity-is-the-source-of-dark-energy-accelerating-expansion/?p=728909 .) The value of the cosmological constant is of the order 10-52 m-2. Plug that into the second term, along with average Earth-Moon distance, and you get a correction to the gravitational acceleration between them of the order of 10-27 m/s2, which is about four orders of magnitude smaller than the secular pole motion acceleration. Additionally, the recession of the Moon from the Earth was calculated based on centuries worth of data, and it cannot be measured over short time scales.
  19. It's too small to be measurable at small scales. We "know" it must be happening because the cosmological constant is nonzero.
  20. You're not wrong in this post, but you were wrong in your previous post. You said that the space between the Sun and Earth doesn't expand. But it does. Just because the distance between them isn't changing doesn't mean that metric expansion isn't going on. It's going on everywhere.
  21. Sort of. It's kind of like putting a coin on a sheet of rubber, and stretching the sheet. The coin isn't pulled apart by the stretching because the forces holding the coin together are far too strong. But if you put two coins on the sheet, they aren't bound together by any forces, so they will move apart from each other. Sorry, I don't understand your question.
  22. To be fair, off the top of my head Faraday's relationship with Maxwell was pretty much exactly what sunshaker was talking about. Faraday had a bunch of ideas about electric and magnetic fields and how they behaved, but he couldn't express his ideas mathematically. Maxwell had to do it for him. But, Faraday was working with physics he could do at home. Current theoretical physics requires particle accelerators and powerful satellite telescopes to gather data, so it's unlikely that someone like Faraday would pop up nowadays.
  23. Your understanding is wrong.
  24. Unruh's response: http://www.iflscience.com/physics/physicist-claims-have-proven-mathematically-black-holes-do-not-exist
  25. No it isn't. The metric is generally not static. The FLRW metric, for example, isn't time independent. gtt could be time-dependent, sure. That's no more recursive than the fact that grr in, for example, the Schwarzschild metric is dependent on r. In other words, if the distance between two points in space can depend on where you are in space, then why couldn't the time between two events depend on where you are in time?
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