Timeline for The Four-Clock Special Relativity Conundrum
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20 events
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| Jan 9, 2013 at 5:15 | comment | added | Retarded Potential | In the extreme case, if X had "frame velocity" c, Earth would see X leave Jupiter at 1:00, and X would arrive also at 1:00, "apparent velocity": infinite. | |
| Jan 9, 2013 at 5:14 | comment | added | Retarded Potential | There were not two moving objects in my supernova example, so I guess neither of us understands what the other is talking about. Sort of special relativistic who's-on-first routine. Okay: say Mars and Jupiter are in conjunction, 15 light-minutes and 1 light-hour away, resp. (made up numbers). Midnight, X leaves Jupiter straight towards the Earth with "frame velocity" $\frac{3}{5}c$. 1:00, Earth sees X leave Jupiter. 1:15, X passes Mars. 1:30, Earth sees X pass Mars. 1:40, X arrives on Earth. Earth sees 40 mins between X leaving and arriving, "apparent velocity": $\frac{3}{2}c$. | |
| Jan 9, 2013 at 0:07 | comment | added | Terry Bollinger | @SpacelikeCadet, specifically on that supernova example: I'm with Art Brown on that one: You don't see even apparent superluminal velocities that way. The issue of apparent superluminal velocities has indeed come up for near-light axial jets, but I recall even without looking it up again that it was a more complicated scenario. What you will observe for the situation you just described is two objects both moving away at velocities very close to c. One is just "a bit closer" to c in its apparent velocity. | |
| Jan 6, 2013 at 23:16 | comment | added | Retarded Potential | @Art: As I say above, what we actually see is not our simultaneous space, but our past light cone. Does that help at all? If someone were directly between us and the supernova, then we would see them seeing the supernova at the same time as we would see the supernova itself. (And thanks for the handle compliments.) | |
| Jan 6, 2013 at 21:56 | comment | added | Art Brown | @SpacelikeCadet: Indeed I have, I suppose; I have no idea what you mean. Are you saying that we see a supernova in the Andromeda galaxy simultaneously with someone in that galaxy? (by the way, I do like your handle.) | |
| Jan 6, 2013 at 21:21 | comment | added | Retarded Potential | @Art: Then you've missed the whole point of APPARENT. If something is coming directly at you at lightspeed, nevermind the Minkowski coordinates, what do you actually see? You see it arrive immediately, its APPARENT speed is infinite. | |
| Jan 6, 2013 at 20:57 | comment | added | Art Brown | "This is still less than the apparent speed of oncoming light, which is infinite." This statement is incorrect. The whole point of relativity is that lightspeed $=c$ in any frame. | |
| Jan 6, 2013 at 20:55 | comment | added | Retarded Potential | I've added the neutral observer O. As a general note about Dingle, one of his criticisms (at least as summarised here) was "all phenomena generally associated with Relativity – relative contraction of rods and distances when approaching light-speed, relative slowing down of clocks etc. – are not matters of observation, but are wholly concerned with coordinate systems..." and so I am talking wholly in terms of direct observables, and things add up just fine. | |
| Jan 6, 2013 at 20:49 | history | edited | Retarded Potential | CC BY-SA 3.0 |
add neutral observer
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| Jan 6, 2013 at 19:39 | comment | added | Terry Bollinger | Spacelike Cadet, thanks. Both A and B will see the others as having a time slope, even if the query process is done using unambiguous direct-contact data transfer methods. That scenario is in effect the very first thought problem in Einstein's original SR paper, and it's still delightful fun to read. It's particularly insightful how he derived both time and length contraction by first defining simultaneity and how it cannot be invariant across frames. (And I agree with others: That's a really cool name you picked, especially for SR topics.) | |
| Jan 6, 2013 at 7:24 | comment | added | Retarded Potential | Turn the question around: you are not moving. Event A is that your watch shows a certain time. Event B is that your watch shows a later time. There are two observers heading towards you with equal speed from different (not even necessarily opposite) directions. What time and distance will they see between A and B? | |
| Jan 6, 2013 at 7:18 | comment | added | Retarded Potential | Notice they disagree about the direction. Yes, they all think one event happens at the front, and the other later at the back of their own train. But the trains have opposite orientation! | |
| Jan 6, 2013 at 4:09 | comment | added | Terry Bollinger | Ouch. Keeping the discussion on apparent velocities was my original intent. That's why I had radars at the front of each train, to give velocities less than c. You are asserting $T_{\Delta{A}}=T_{\Delta{B}}$, yes? Even though $T1$ and $T2$ define two localized events in spacetime? Where my brain keeps hiccuping on that answer is any two such events should be separated by an invariant interval. So how in the heck do two quite different frames observe that interval, yet still arrive at the same pure-time separation figures within their frames? Am I the only one deeply troubled by that? | |
| Jan 5, 2013 at 17:43 | comment | added | Retarded Potential | The "frame" velocity (i.e. in Minkowski co-ordinates) can be calculated from the formula I gave, $w=\frac{cv}{c-v}$. So $v=-\frac{3}{2}c$ and $v=+\frac{3}{8}c$ become $w=\pm\frac{3}{5}c$. But as I say, none of the observers directly measure these frame velocities, they measure the "apparent" velocities. | |
| Jan 5, 2013 at 17:35 | comment | added | Retarded Potential | Correct. And as far as I can see the math is fine. Remember that I am talking exclusively about what each observer actually sees, I am never using simultaneous spaces. If it helps, insert "apparent" before all occurances of speeds, rates, and distances. An oncoming object certainly can have an apparent speed of $\frac{3}{2}c$. This is still less than the apparent speed of oncoming light, which is infinite. | |
| Jan 5, 2013 at 1:41 | comment | added | Terry Bollinger | Please verify your math notations before I comment any further. Early in your analysis you invoke a velocity of "$\frac{3}{2}c$", which is decidedly uncommon in this universe. | |
| Jan 5, 2013 at 1:15 | comment | added | Terry Bollinger | Nice math, but let's get back to the actual questions for a moment. If I read through your answer rightly, your answers are: (1) The experimental setup is valid; (2) $T_{\Delta{A}}$ = $T_{\Delta{B}}$, always; and (3) Not applicable. Is that correct? Also, am I correct that you are completely fine with asserting that the delay between $T1$ and $T2$ is independent of the frame from which it is observed? | |
| S Jan 4, 2013 at 23:12 | history | answered | Retarded Potential | CC BY-SA 3.0 | |
| S Jan 4, 2013 at 23:12 | history | made wiki | Post Made Community Wiki by Retarded Potential |