Super Polymath Posted June 26, 2018 Report Share Posted June 26, 2018 Here's how it works, they say. If B was "naive" then he might think that A and C have a relative speed of .9c, just like they naively think. But he aint naive. He knows better. So he hauls out his trusty calculator, gets out his velocity addition formula, and punches some buttons. He determines that A and C can't have a relative speed of 9c, because his calculator says so. Upon hearing this, A and C both slow down to the extent necessary for their true speed to match the calculator's findings. Link to comment Share on other sites More sharing options...
A-wal Posted June 26, 2018 Report Share Posted June 26, 2018 As a broader issue here, the motion must be treated as absolute, not relative. If the GPS actually took the proposition that A and B are "both correct" when they make mutually exclusive claims about who is moving and whose clock is slower (as SR requires) seriously, the GPS would be dead on arrival. You could not draw any meaningful conclusions at all in that case.This is a lie. Any arbitrary frame can be used because GPS world still work if the Earth motion relative to the centre of the galaxy for example were different. So, ya have two objects--A and B. One is travelling at .99c. In that frame you can theoretically accelerate a particle to .99c. So how fast would that particle actually be going? To take it a step further, if that particle had a lab, then it could accelerate another particle, call it particle B, to another .99c from there. So how fast would particle B actually be going? Suppose particle B also has a lab...ad infinitum.Just use the velocity addition formula to add .9 to .9 and keep adding .9, obviously. I never got a coherent explanation for this confusing situation. Does anyone have one?Yes you have. You're just either incapable or unwilling to understand it, We have two objects, A and C. Let's look at this from A's perspective. He sees C moving away from him at .9c. And, of course, C will also see A receding from him at .9c, because that's their relative velocity, right? What will B see? He will see A receding from him at .45c, right? He has to, because that's how A sees him. That's their relative velocity.Like I told you yesterday, yes! He will also see C receding from him at the rate of .45c, but in the opposite direction, right?NO!!! Link to comment Share on other sites More sharing options...
Moronium Posted June 26, 2018 Report Share Posted June 26, 2018 (edited) NO!!! Like I said, I never did get any coherent explanation. Oh, wait, I see now....putting it in bold, underlining it, and adding three exclamation points makes it an explanation. My bad. Edited June 26, 2018 by Moronium Link to comment Share on other sites More sharing options...
A-wal Posted June 26, 2018 Report Share Posted June 26, 2018 No it makes it another ****ing strawman. It's nobody else's job to make you capable of grasping it, it might not even be possible. If you want to carry on thinking that your lack of understanding is the model's and not yours go ahead, it's hilarious. Link to comment Share on other sites More sharing options...
Farsight Posted June 26, 2018 Report Share Posted June 26, 2018 (edited) As I posted earlier, from Wikipedia: Galaxy GN-z11 is currently the oldest and most distant known galaxy in the observable universe. It is also the most red-shifted object we have observed. GN-z11 has a spectroscopic redshift of z = 11.09, which corresponds to a proper distance of approximately 32 billion light-years (9.8 billion parsecs)... So, these authors have taken it upon themselves to abandon the mathematics of Relativistic Doppler Redshift in favor of some other formula? I would like to know what formula they are using and how they can detect the redshift, or any light at all for that matter, from “galaxies that have, and always have had recession velocities greater than the speed of light”.I'm not telling you porkies, OceanBreeze. See https://arxiv.org/abs/astro-ph/0011070 . Note this: Here we show that galaxies with recession velocities faster than the speed of light are observable and that in all viable cosmological models, galaxies above a redshift of three are receding superluminally. Do your own research. You'll soon find that for z=1.46 the astronomical object has a recession velocity of c, and for z=6.5 the recession velocity is 2c. I wouldn't be surprised if it turned out that neutrinos travelled faster than photons. Edited June 26, 2018 by Farsight Link to comment Share on other sites More sharing options...
Moronium Posted June 26, 2018 Report Share Posted June 26, 2018 Just use the velocity addition formula to add .9 to .9 and keep adding .9, obviously. Well, let's see here then...If I add .9 to .9, I get 1.8c. Then 2.7c, then 3.6c, and so on. OK, that makes sense. But I don't really think that's what the velocity addition formula would say, for some damn reason. Link to comment Share on other sites More sharing options...
Moronium Posted June 26, 2018 Report Share Posted June 26, 2018 (edited) I would still like to see your response to this, Moronium. Did you see post 185? I responded there, as a general matter. Truth be told, I didn't even pay much attention to your particular hypothetical, if that's what you want a specific answer to. I still haven't, but let's just say my answer would be "I don't know." Then what? Would that mean absolute motion is impossible? That's how I responded to it. P. S.: I just now noticed this post, or else I would have responded sooner. Edited June 26, 2018 by Moronium Link to comment Share on other sites More sharing options...
Farsight Posted June 26, 2018 Report Share Posted June 26, 2018 (edited) I never got a coherent explanation for this confusing situation. Does anyone have one? Yes, it's simple. We have two objects, A and C. Let's look at this from A's perspective. He sees C moving away from him at .9c. And, of course, C will also see A receding from him at .9c, because that's their relative velocity, right? Yes. Now then, let's say that A notices object B, which is between him and C. He sees B as going in the same direction as C, except that B is receding from him at the rate of only .45c. No problem there. What will B see? He will see A receding from him at .45c, right? He has to, because that's how A sees him. That's their relative velocity. Yes. He will also see C receding from him at the rate of .45c, but in the opposite direction, right? Yes, they have a relative velocity of .45c too. Going back to A's perspective, because he noticed B, A no longer sees C receding from him at .9c. A now sees C receding from him at the lesser speed of (approx) .75c, right? No. You're missing the trick. You should have arranged the scenario such that A and C are moving apart at 0.45c apiece from a common centre, and B is moving at 0.9c away from the common centre in the same direction as A. Then C doesn't measure B to be moving at 1.35c relative to C, he measures him to be moving at 0.96c relative to C. This measurement is wrong. Their relative motion is 1.35c. See Rod Nave's hyperphysics. What A-wal said earlier about a different scenario where the relative motion was 2c was correct. With respect to C, A's doppler readings have now changed, right? No, you got that wrong. But, if B suddenly explodes, and ceases to exist, A will once again see C receding from him at .9c, right? Wrong. A always saw C receding at .9c. Something in the middle doesn't make a blind bit of difference. Edited June 26, 2018 by Farsight Link to comment Share on other sites More sharing options...
Moronium Posted June 26, 2018 Report Share Posted June 26, 2018 (edited) As I posted earlier, from Wikipedia: I would like to know what formula they are using and how they can detect the redshift, or any light at all for that matter, from “galaxies that have, and always have had recession velocities greater than the speed of light” It makes no sense to me. I don't know how they did, but maybe the phrase "speed of light" is too reflexive and ambiguous. Perhaps they should have said "at a speed greater than 186,000 mps." As already noted, using the standard "co-moving" point of view, time and distance (and therefore speed) are measured differently. In that theoretical framework 186,000 mps is NOT the absolute, universal "speed of light." That particular speed is just what we measure it to be, in our frame, with distorted instruments. All other frames may measure it to be that too, but their measurements are wrong, that's all, even more wrong than ours, if they're going faster. Edited June 26, 2018 by Moronium Link to comment Share on other sites More sharing options...
Farsight Posted June 26, 2018 Report Share Posted June 26, 2018 Well, let's see here then...If I add .9 to .9, I get 1.8c. Then 2.7c, then 3.6c, and so on. OK, that makes sense. But I don't really think that's what the velocity addition formula would say, for some damn reason.It would say 0.994475138121547c. See http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html#c3'>hyperphysics and key in 0.9 for VA and 0.9 for VP. Link to comment Share on other sites More sharing options...
Moronium Posted June 26, 2018 Report Share Posted June 26, 2018 (edited) With respect to C, A's doppler readings have now changed, right? No, you got that wrong. But, if B suddenly explodes, and ceases to exist, A will once again see C receding from him at .9c, right? Wrong. A always saw C receding at .9c. Something in the middle doesn't make a blind bit of difference. Thanks. Maybe it wasn't apparent, but those were facetious claims--but one's that seemed to be implied by A-wal's claim. I'm still trying to figure out why you say I "missed the trick," though. Edited June 26, 2018 by Moronium Link to comment Share on other sites More sharing options...
Moronium Posted June 26, 2018 Report Share Posted June 26, 2018 (edited) No. You're missing the trick. You should have arranged.... Well, maybe I should have, but I didn't. What is the answer to my question as posed? How will B calculate the relative speed of A and C? According to A-wal it would be approximately .75c, not .9c. What we end up with, in my example, is an object in the middle who sees two objects (A and C) moving away from him at (what he sees to be) .45c each, in opposite directions. Of course B considers himself to be at rest. So B serves as what you're calling the "common centre" (I take it that you're British). As far as B is concerned A and C just keep getting farther and farther away from him while he remains motionless, always staying smack dab halfway between the two. Edited June 27, 2018 by Moronium Link to comment Share on other sites More sharing options...
Farsight Posted June 26, 2018 Report Share Posted June 26, 2018 Well, maybe I should have, but I didn't. What is the answer to my question as posed? How will B calculate the relative speed of A and C? According to A-wal it would be approximately .75c, not .9c.I gave you the answer. You posed the wrong scenario. Look at hyperphysics again: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html#c3 If I launched a ship with you in it at 0.45c, and you then launched a forward projectile at what you thought was 0.45c relative to you, the projectile would be moving away from me at 0.748c. Link to comment Share on other sites More sharing options...
Moronium Posted June 26, 2018 Report Share Posted June 26, 2018 I gave you the answer. You posed the wrong scenario. Look at hyperphysics again: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html#c3 If I launched a ship with you in it at 0.45c, and you then launched a forward projectile at what you thought was 0.45c relative to you, the projectile would be moving away from me at 0.748c. OK, I'll take your word for that, but I'm not asking about a projectile launched forward. I don't want to talk about that. I want to talk about the scenario I posed. I have added some more to the post you just quoted, btw. Link to comment Share on other sites More sharing options...
Moronium Posted June 26, 2018 Report Share Posted June 26, 2018 (edited) He will also see C receding from him at the rate of .45c, but in the opposite direction, right? Yes, they have a relative velocity of .45c too. In the meantime I will note that A-wal answered with a resounding "No!!!" to this question. As always, he refused to say why he answered "no," but... Edited June 26, 2018 by Moronium Link to comment Share on other sites More sharing options...
Moronium Posted June 26, 2018 Report Share Posted June 26, 2018 I must admit, Farsight, that I'm a little frustrated with your answers so far. It's kinda like I asked you "what color is your dog," with you responding: "You asked me the wrong question. You should have asked me "what color is the sky." The answer is blue. So, I say, well, that wasn't my question, it was "what color is your dog?" Then you say: "I already answered that." Link to comment Share on other sites More sharing options...
A-wal Posted June 26, 2018 Report Share Posted June 26, 2018 Well, maybe I should have, but I didn't. What is the answer to my question as posed? How will B calculate the relative speed of A and C? According to A-wal it would be approximately .75c, not .9c.No, it would be .9c in B's frame. In the meantime I will note that A-wal answered with a resounding "No!!!" to this question. As always, he refused to say why he answered "no," but...Because you'd already given the velocity between A and C as .9c and the velocity of one of them relative to B as .45c so the velocity between the other one and B would be more than .45c. OK, I'll take your word for that, but I'm not asking about a projectile launched forward. I don't want to talk about that. I want to talk about the scenario I posed. I have added some more to the post you just quoted, btw.It's exactly the same! You've got three objects, one central and the other two moving at .45c relative to it in the central objects frame. So the two outside objects are moving at .75c relative to each other in their frames. So the velocity between B and A would be .3c in C's frame and the velocity between B and C would be .3c in A's frame. Link to comment Share on other sites More sharing options...
Recommended Posts