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Natural Science Forum / Physics / Relativity / December 2004another paradox ???
Versus the system K the relatively moving system K' is moving at .5 C from the origin of K' (at moment of the two origins meet) a ball is send at the speed of .5 C, as measured in K', in the same direction of the moviment of K'. The ball travels and hits a wall at rest in K'. for the composition of the speed, the forwards speed of the ball as seen from K is (.5+.5)/(1+(.5*.5))= .8 C The ball is perfectly rigid, also the wall (of a much greater mass), the reflection is perfectly elastic, so in K' we expect that the ball going forwards at .5 C is reflected back at -.5 C. A local speed of -.5 C in K' as seen from K, gives a composite speed of (-.5+.5)/(1+(-.5*.5))= 0 C So as seen from K a perfectly rigid ball going at .8 C hitting a perfectly rigid wall going at .5 C would never come back, why ??? Where is the strangeness ????? I expect a deluge of profanities, please don't, just let me know where is the bug. best regards beda pietanza > Versus the system K the relatively moving system K' is moving at .5 C > [quoted text clipped - 16 lines] > So as seen from K a perfectly rigid ball going at .8 C hitting a perfectly > rigid wall going at .5 C would never come back, why ??? Because, as seen from K, the collision brings the ball to a stop. > Where is the strangeness ????? There is nothing strange about it at all; you can do the same thing at non-relativistic speeds where simple addition can be used for composition of velocities. Paul Cardinale > > Versus the system K the relatively moving system K' is moving at .5 C > > [quoted text clipped - 33 lines] > > Paul Cardinale You mean that a ball of insignificant mass traveling at .8 C hitting a wall of very large mass traveling in the same direction at .5 C would result in a velocity of 0 C ?? Is like saying that a little ball of glass that travels at 8 mt./s hitting a heavy wall going at 5 mt/s in the same direction, after the impact would stop ????? Are you sure ??????? It looks like to me that the little ball would be bounced backwards at the speed of 3 mt /s. So would do a ball traveling at .8 C after hitting the wall at .5 C, it should be bounced back at the speed of .3 C !!!!! would you please verify once more ??, thanks best regards beda pietanza > > > Versus the system K the relatively moving system K' is moving at .5 C > > > [quoted text clipped - 46 lines] > It looks like to me that the little ball would be bounced backwards at the > speed of 3 mt /s. That would be 2 m/s forwards. See below. > So would do a ball traveling at .8 C after hitting > the wall at .5 C, it should be bounced back at the speed of .3 C !!!!! [quoted text clipped - 4 lines] > > beda pietanza Small ball mass m and velocity v before collision and v' after collision. Big ball mass M and velocity V and V' after collision. Positive velocities indicate movement to the right. Negative velocities indicate movement to the left. At the end I'll turn the big ball into a wall. Conservation of momentum gives m v + M V = m v' + M V' Conservation of energy gives 1/2 m v^2 + 1/2 M V^2 = 1/2 m v'^2 + 1/2 M V'^2 Putting k = M/m this becomes v + k V = v' + k V' v^2 + k V^2 = v'^2 + k V'^2 Solving for v' and V' gives v' = (2 k V + v - k v ) / (k+1) V' = (k V + 2 v -V ) / (k+1) An infinitely massive wall means infinite k, so taking the limits: limit( k --> infinity, v' ) = 2 V - v limit( k --> infinity, V' ) = V Little ball has initial velocity v = 8 m/s Big ball (the wall!) has initial velocity V = 5 m/s After the collision, the wall keeps going at V'(limit) = 5 m/s, and the little ball has v'(limit) = 2 V - v = 2 m/s, which is positive, so still going forward. Dirk Vdm > > > > Versus the system K the relatively moving system K' is moving at .5 C > > > > [quoted text clipped - 86 lines] > > Dirk Vdm I thank you very much. Looks like that in order to stop the ball should have a speed of 10 mt/s, taking the wall as reference the ball approaches it at 5 mt/s and of course is bounced back at -.5 mt/s versus to the wall, that makes the final speed of the ball 0 mt/s. In my original post the ball (in the frame K)would travel at .8 C and the wall at .5 C (at rest in the frame K'(K' moving at .5 C in the same direction)). After the ball hits the wall, SR tells us that the ball stops in the frame K while travels back at -.5 C in the frame K'. In K' the ball travel forth at .5 C hits the wall and comes back at -.5 C. In K the ball travels forwards at .8 C hits the wall at .5 C and stops. And applying your approach using the wall as reference the ball is travelling at .5 C approaching the wall that makes obvious that the ball would be bounced back at - .5 C versus the wall that brinks the ball to a stop versus the frame K, perfect, all fits. I still want to make out what the Esynchro has to do with this result. Anyway,.8 C versus .5 C in SR stands as 10mt/s versus 5 mt/s in low speed classical mechanics. Very grateful to you.. beda pietanza > > "beda pietanza" <beda-pietanza@libero.it> wrote in message > news:IkkAd.6600$H%6.292827@twister1.libero.it... [quoted text clipped - 117 lines] > -.5 mt/s versus to the wall, that makes the final speed of the ball 0 > mt/s. Yes, in the non-relativistic case, to make v' = 2 V - v zero and thus stop the ball, v must be the double of V. > In my original post the ball (in the frame K)would travel at .8 C and > the wall at .5 C (at rest in the frame K'(K' moving at .5 C in the same [quoted text clipped - 17 lines] > Anyway,.8 C versus .5 C in SR stands as 10mt/s versus 5 mt/s in low > speed classical mechanics. Yes, and like 9.9999999999999972 m/s versus 5 m/s in SR itself. Ask yourself this... is there any reason why one would prefer 10 m/s over 9.9999999999999972 m/s ? How would you measure the difference? > Very grateful to you.. > > beda pietanza No problem. I just gave this as a little showcase of the power of the mathematical model we use to do physics. Paul's explanation using the Galilean transformation leads you to the same result directly, but of course it only works in the case of an infinitely massive wall, and you don't get the intermediate results of the heavy ball. Dirk Vdm >>>Versus the system K the relatively moving system K' is moving at .5 C >>> [quoted text clipped - 51 lines] > of very large mass traveling in the same direction at .5 C would result in > a velocity of 0 C ?? Yes. > Is like saying that a little ball of glass that travels at 8 mt./s hitting a > heavy wall going at 5 mt/s in the same direction, after the impact would > stop ????? No, because 0.8 c is very different from 8 m/s. A ball going at 8 m/s would drop dead if it hits a wall going at 4 m/s. > Are you sure ??????? > > It looks like to me that the little ball would be bounced backwards at the > speed of 3 mt /s. Wrong. See Dirk's response. A somewhat simpler approach is to transform the speed to the wall-frame, and then transform it back, using the Galilean transform since the speeds are so low. In the wall frame, the ball is approaching the wall with the speed 3 m/s, and is bouncing back at the speed -3 m/s. Transformed back, the speed is thus - 3 m/s + 5 m/s = 2 m/s > So would do a ball traveling at .8 C after hitting > the wall at .5 C, it should be bounced back at the speed of .3 C !!!!! According to Galilean relativity, it would go at 0.2 c in the same direction. According to SR, it will drop dead. Paul beda pietanza: >Versus the system K the relatively moving system K' is moving at .5 C >I expect a deluge of profanities, please don't, just let me know where is >the bug. Why don't you go struggle with it for a few days or weeks? You haven't gotten anything from being told the answer other than another opportunity to create a new dilemna based on the same misconceptions. Learning involves frustration and figuring out things rather than being handed the answer. > beda pietanza: > >Versus the system K the relatively moving system K' is moving at .5 C [quoted text clipped - 6 lines] > to create a new dilemna based on the same misconceptions. Learning involves > frustration and figuring out things rather than being handed the answer. I accept you objections, I just try to make the best of this ng. Thanks to many here I'm learning a lot. regards beda pietanza |
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