Jump to content

Iggy

Senior Members
  • Content Count

    1607
  • Joined

  • Last visited

Everything posted by Iggy

  1. Perhaps. Let me think about it... If this is something that has never been explored then I think it would make an interesting problem to solve. For example, if fish under an amber or red light don't school together by color then we'd know it is a visual cue. Otherwise it may be a behavioral cue and fish may not actually know what color they are. I wonder if it is a question that has been answered.
  2. I have heard that genetic studies and other research has disproved the idea that physical and behavioral qualities are a function of breeding populations. Here are some applicable quotes: Keita, S O Y et al. 2004. "Conceptualizing human variation". Nature 36 (11s):"Modern human biological variation is not structured into phylogenetic subspecies ('races'), nor are the taxa of the standard anthropological 'racial' classifications breeding populations. The 'racial taxa' do not meet the phylogenetic criteria. 'Race' denotes socially constructed units as a function of the incorrect usage of the t
  3. A couple years ago 6 goldfish were born from the same batch of eggs and from the same female in my outdoor pond. 4 are black, 2 are orange, and one is yellow. They are all the same size. As they swim, the four black ones always stick together (usually touching or no more than a couple inches apart)—same for the 2 orange. The yellow one is usually by itself. My question is, how does a goldfish know what color it is? I couldn't find a quick answer online.
  4. I do. It's quite approximate and weak-field-limit of me, but that's just the way I talk. Who can't talk in terms over c squared? Quite dielectric.
  5. So... you aren't just mistaking my meaning, you are literally mistaking my words? How fun! If we were involved in something dialectic then it would suggest that there were a difference of opinion, but I assure you that is not the case. It's just that you said two wildly different things, and that is *dielectric*. Perhaps you could look the word up, and consider your position, and get back to me, because either "Schwarzschild coordinates are the standard", or "the coordinate speed of light isn't physical". You'll have to pick one or the other, and I guarantee you the dictionary says t
  6. This sentence is meaningless. Dielectric means that you can't get from one to the other. My hope for you faded fast. We don't understand each other. I'll welcome your solution to Shapiro delay as proffered in post 3, or I'll welcome your insistence that nothing non-local is physical. I'll welcome either thing you've said, but they are fiercely dielectric. You can't get from one to the other, you know? The forces of the universe will keep those assertions apart despite your best efforts.
  7. I apologize profusely. Is the variable speed of light the unphysical entity, or the "standard answer", both of which you'd said. Could you please just say one is right and one is wrong so I can understand what you mean? You were right before and wrong now, or right now and wrong before? I can't tell until you say it explicitly. Is it that the "variability is applied to an unphysical entity" or is it that "Schwarzschild coordinates are standard"? I'm happy with either. They are physically dielectric. Can't you choose one thing of which you've said?
  8. I'm sorry... the "coordinate speed of light" is unphysical, or "Schoolchild coordinates are the standard". Which were you saying? It looks above like you just negated something you said earlier. Were you ok with that as long as you're interjecting with me now? It's a bit difficult to tell. Can you just quickly say "the coordinate speed of light is unphysical, and therefore the earlier thing I said about it being the standard answer is just my best attempt at a pail stain on the carpet". Can't you just say that? Or some humerus variation of it? It would end our confrontation immediate
  9. Correction: the coordinate speed of light is variable in Schwarzschild coordinates. That isn't a correction. That is something logically equivalent to what I said. I'm sure Euclid made some point about things logically equivalent to a mutual party are logically equivalent to themselves. The speed of light in a certain set of coordinates doesn't need corrected as the certain coordinate speed of light in coordinates. That is just you skipping the issue as always, I hope. I hope for you as always. Please derive Shapiro delay in a way that reflects nothing of Schwarzschil
  10. Tell me about it. The speed of light is variable in Schwarzschild coordinates. How wrong can one person be? Which of the two above quotes would you like to retract? I'll be happy with either, but you still haven't answered.
  11. Yes, I'm perfectly capable of doing that. If, then the radial coordinate velocity of light is, [math]v_c = \pm \left(1- \frac{r_s}{r} \right)[/math] and in terms of acceleration of gravity, g, and two radial points, A and B... [math]v_{BA} = \left(1- \frac{gr}{2c^2} \right)[/math] (where A is further than B and [math]v_{BA}[/math] is the average velocity of light between the two as measured by A) If true then we get Shapiro delay for a round trip A->B->A... [math]\Delta \tau = \Delta t_{flat} \left(1 + \frac{gr}{2c^2} \right)[/math]
  12. Distance is constant in the Rindler metric a priori. A rindler observer at x=1 is always at x=1, and the same with an observer at x=2. The distance between them is constant one by definition (it's assumed by the metric). Constant distance is proven in the derivation of the metric, not with the metric. It's like asking someone to use a meter stick to prove that the meter stick is one meter. It's a confused question.
  13. I appreciate your feedback, unfortunately we are no closer to a common understanding. Both your metric and mine are Rindler, and they are physically equivalent. I will explain in a moment. I should first say that the t variable on the right hand side of your equation is not proper time, but rather coordinate time. This is common to all metrics (the t and r on the right hand side of the Schwarzschild metric, for example, are coordinate time and radial distance). In fact, you rightly solve the metric above to get the following: If [math]d \zeta[/math] and [math]d \tau[/math]
  14. I forgot to answer this part, sorry. It seems you're finding the coordinate speed of light where the t axis is made by the worldlines of the accelerating ships, and the x axis is made by the lines of simultaneity in the inertial frame. The problem I tried to touch on in a rush yesterday is that time dilation isn't accounted for in that case. As an analogy, if someone sat near the event horizon of a black hole with rockets firing at enormous acceleration in order to hold their position then they would see the universe above them unfold at an extremely rapid rate. Light would move fr
  15. I think the thing is that accelerated coordinates usually transform away from the horizontal Cartesian-like time coordinate. My understanding is that they should look more like... from this site. Or... From wiki. I think everything you're saying about the spatial coordinate is good, but the time coordinate would spread out as it fans right to reflect how clocks tick faster further from the origin. A light ray at 45 degrees around the middle of these diagrams would cross about one unit of t for every unit of x, but if you imagine traveling far from the origin where
  16. Apologies for the delay in responding. I've been away from the site awhile. Hectic Holidays. Agreed. Besides making the gamma factor an easy rational number, there's nothing significant about 0.6 I didn't try (and fail) to calculate the v=.6 distance relative to the lab frame, nor did I actually calculate the v=.6 distance relative to the lab frame. I very clearly showed (with JVNY's diagram), calculated (with equations), and explained (with words) that the *proper distance* remained constant, not the spatial distance in the lab frame. Let me find the
  17. Yeah, that's what I had No, it apparently stayed constant in the v=.6c frame, not the frame of the launch pad. JVNY calculated it earlier, and Md verifies the math. Also, Born rigidity has been around for about 100 years, I don't think you're about to disprove it. So... No, not a function of tau. It is well-established science that the proper distance between accelerating observers stays constant if their acceleration is inversely proportional to their distance. I, nevertheless, appreciate your insistence otherwise. It's like Christopher Hitchens said in Canada I think
  18. Md65536, do you wanna help us out? A spatial distance of 0.625 in our frame is what in a frame of relative velocity v=.6c?
  19. Fair enough. Let's answer your Shapiro delay question in terms of Schwarzschild coordinates: Q & A: How does Shapiro delay work The rule about the speed of light being constant only applies locally to patches of spacetime small enough to be effectively flat, i.e. ones which can be described by special relativity. On a bigger scale, with gravity involved, phrases like "the same distance" become ambiguous. Let's think of light from some distant star. There's an extra delay in how long it takes to reach us when the light happens to pass near the sun on its way. How come? We can
  20. I will consider carefully what you say. The ships constitute a Born-rigid system, so the proper distance between them is not a function of [math]\tau[/math]. But, that isn't what I'm asking. I'm saying that the following is an inertial frame: in which the spatial distance of the blue line is 0.5, and the spatial distance of the green line is 0.625. What is the distance of the green line in a v=0.6 inertial frame? You say there is no number, but I assure you the question I just asked has a numerical answer. I've spoken of speeds in part
  21. I'm not trying to talk down. I gave my reasoning and cited three sources to answer your Shapiro delay question in post 73. Before before you "pointed out quite a few errors" in that post I asked a very straightforward question. Given that the blue line is 0.5... What is the distance of the green line in a v=0.6c frame? You have told me repeatedly that my use of the length contraction formula won't work, so I'm very curious what solution you find. Can you give me a number. This is literally the simplest question you could get in special relativity, and I'm just looking for
  22. The blue line is 0.5 in the rest frame Since you disagree with my calculation of the proper distance between the first and the third worldlines, could you please calculate the distance of the green line in a v=0.6 frame? Thank you. I should also mention -- I appreciate your persistence and dedication in neg-reppring all of my posts. Occasionally, I think you might forget one, but you are very meticulous and I can appreciate that quality. Any case... the length of the green line?
  23. A site that says Einstein meant "velocity" rather than "speed" when he said "geschwindigkeit" is, itself, playing word games. I just corrected the site. Whenever Einstein says "velocity" in his early publications, one should never assume that he must be referring to a direction-dependent quantity. Plus I gave two other quotes saying the same thing that use the english word "speed". I'm sorry this appears to you as a word game. I didn't explain Shapiro delay -- I said that it verifies a variable speed of light. It is shown to do that because the speed of light approaches zero as
  24. Velocity and speed are equivalent in German because the same word in German means both. The site JVNY quoted said "Since Einstein talks of velocity (a vector quantity: speed with direction) rather than speed alone...", but like I said, that site doesn't realize that the quote was originally written in German in 1916, and translated to English in 1920. Einstein's quote says "the law of the constancy of the velocity of light in vacuo... cannot claim any unlimited validity", and the word "velocity" there is a translator's choice. It could have just as easily and correctly been translated as "
  25. I don't see anything supporting your claim, why don't you explain, with math, in your own words , your claim. Please do so. The part for which you asked -- the part about maintaining a constant proper distance -- can be quickly proved. Given JVNY's derivation (and illustration) of his thought experiment: The proper distance at t=0 is given along the x axis as: [math]\sigma = 1 - 0.5 = 0.5[/math] The proper distance at the dotted line between the accelerated hyperbolic worldlines is given from the numbers in this frame by length contraction: [math]\sigma = L_{0
×
×
  • 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.