art@ACC.ARPA (11/10/85)
> If someone travels at near light speed, isn't elapsed time for she/he > less than the elapsed time for someone traveling at a much slower speed. The difference in the rate at which time passes for two different inertial frames of reference is determined by the Lorentz Transformation: --------- / v**2 v = Velocity of one frame with respect to \ / 1 - ---- the other. \/ c**2 c = Speed of light. As one approaches the speed of light, the rate that time passes in one frame of reference (like a spaceship) as "observed" from the other (say on Earth) approaches zero. Some thought on this leads to the classic "Twin Paradox". I'll let someone else point that solution out. "Art Berggreen"<Art@ACC.ARPA> ------
slerner@sesame.UUCP (Simcha-Yitzchak Lerner) (11/14/85)
> > > The difference in the rate at which time passes for two different inertial > frames of reference is determined by the Lorentz Transformation: > > --------- > / v**2 v = Velocity of one frame with respect to > \ / 1 - ---- the other. > \/ c**2 c = Speed of light. > > As one approaches the speed of light, the rate that time passes in one > frame of reference (like a spaceship) as "observed" from the other > (say on Earth) approaches zero. What happens if two ships leave with opposite vectors, and they both approach the speed of light relative to their initial frame. The v above, relative to each other, would approach 2*c, giving a non-real answer. Where am I goofing? (Or is it time to invest in a FTL ship? :-) ) -- Opinions expressed are public domain, and do not belong to Lotus Development Corp. ---------------------------------------------------------------- Simcha-Yitzchak Lerner {genrad|ihnp4|ima}!wjh12!talcott!sesame!slerner {cbosgd|harvard}!talcott!sesame!slerner talcott!sesame!slerner@harvard.ARPA
matt@utastro.UUCP (Matt Wood) (11/16/85)
> > > > > > The difference in the rate at which time passes for two different inertial > > frames of reference is determined by the Lorentz Transformation: > > > > --------- > > / v**2 v = Velocity of one frame with respect to > > \ / 1 - ---- the other. > > \/ c**2 c = Speed of light. > > > > As one approaches the speed of light, the rate that time passes in one > > frame of reference (like a spaceship) as "observed" from the other > > (say on Earth) approaches zero. > > What happens if two ships leave with opposite vectors, and they both approach > the speed of light relative to their initial frame. The v above, relative to > each other, would approach 2*c, giving a non-real answer. Where am I goofing? > (Or is it time to invest in a FTL ship? :-) ) > > > -- > Opinions expressed are public domain, and do not belong to Lotus > Development Corp. > ---------------------------------------------------------------- > > Simcha-Yitzchak Lerner > > {genrad|ihnp4|ima}!wjh12!talcott!sesame!slerner > {cbosgd|harvard}!talcott!sesame!slerner > talcott!sesame!slerner@harvard.ARPA Velocity transformation isn't as simple. Assume there's two reference frames, K1 and K2, that K1 is at rest, and that K2 is moving at velocity v along the x-axis. Your question then is: if I have a space ship in K1 moving in the -x direction at nearly the speed of light, there's a ship in K2 at rest, and K2 is moving in the +x direction (as seen from K1) at nearly the speed of light (i.e., v = (nearly) c), then what is the velocity of ship 1 as measured by ship 2. Let u1 = the velocity of ship 1 as measured in K1. We want to know the expression of u2 = the the velocity of ship 1 as measured in K2. The equation is: u2 = (u1 + v) / (1 + (v*u1) / c**2 ) For v = u1 = 0.990000 c, we get u2 = 0.999949 c You can see that in the limit, u2 = c. If you want, see the book "Special Relativity" by Resnick. It's written for an undergrad jr.-level course for science majors, and is small, unintimidating, and clearly written. "If this is SU UMa, then my great, great, great grandchild must be dead." -- Matt A. Wood Astronomy Dept, University of Texas, Austin TX 78712 {allegra,ihnp4}!{ut-sally,noao}!utastro!matt (UUCP) matt@astro.UTEXAS.EDU. (Internet)
carroll@uiucdcsb.CS.UIUC.EDU (11/18/85)
It is possible to see two things receeding from each other at 2c from YOUR point of view. The people on the ships, however, see something different. To transform their speed in YOUR reference frame, you have to use the transform (a+b)/(1+(ab/c*c)) where a,b are the speeds observed in YOUR reference frame. Note that if a,b = c (speed of light), then each ship sees the other receeding at (c+c)/(1+(c*c/c*c)) = c. So, no one ever sees something moving away from him at a speed greater than c, i.e. the speed of an object in an inertial frame is always <= c. This is in the same vein as the question "What happens if some ship is going at .9c and fires something out the front at .9c? Doesn't it go at 1.8c?". No, it doesn't. The ship sees it go at .9c, but the "stationary" observer sees it move at something less than c. (For an exact answer, switch to the ship reference frame, and compute the above as if the "stationary" observer and the object are moving at .9c in opposite directions).
henry@utzoo.UUCP (Henry Spencer) (11/19/85)
> What happens if two ships leave with opposite vectors, and they both approach > the speed of light relative to their initial frame. The v above, relative to > each other, would approach 2*c, giving a non-real answer. Where am I goofing? Velocity addition follows different rules at relativistic speeds. The velocity of one ship as observed from the other never exceeds (or reaches) the speed of light. So the FTL velocity isn't "real". Note that imaginary numbers aren't necessarily the kiss of death for a theory; the elaborate body of theory surrounding the still-hypothetical tachyon has them everywhere, but they turn out to be unobservable. > (Or is it time to invest in a FTL ship? :-) ) Only if the dealer demos it for you first! -- Henry Spencer @ U of Toronto Zoology {allegra,ihnp4,linus,decvax}!utzoo!henry
mouse@mcgill-vision.UUCP (der Mouse) (11/22/85)
[[ This is getting off of space issues, so I am cross-posting to net.physics and directing followups to net.physics only. For net.physics people, this started in net.space. ]] OK, since nobody else responded (or is news just slow to here?) >> The difference in the rate at which time passes for two different inertial >> frames of reference is determined by the Lorentz Transformation: >> --------- >> / v**2 v = Velocity of one frame with respect to >> \ / 1 - ---- the other. >> \/ c**2 c = Speed of light. >> > What happens if two ships leave with opposite vectors, and they both > approach the speed of light relative to their initial frame. The v > above, relative to each other, would approach 2*c, giving a non-real > answer. Where am I goofing? (Or is it time to invest in a FTL ship? > :-) "v = Velocity of one frame with respect to the other". That is, as observed by an observer attached to the other (frame). The velocity of frame X as observed by (an observer in) frame Y is (necessarily) less than c, so the result doesn't go imaginary. The 2c velocity is obtained by measuring from a third intertial frame; if you want to get results for two frames based on quantities observed in a third, you have to use more complicated equations. -- der Mouse USA: {ihnp4,decvax,akgua,etc}!utcsri!mcgill-vision!mouse philabs!micomvax!musocs!mcgill-vision!mouse Europe: mcvax!seismo!cmcl2!philabs!micomvax!musocs!mcgill-vision!mouse Hacker: One responsible for destroying / Wizard: One responsible for recovering it afterward