Known physics defeated by simple puzzle?
in [Relativity]
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From: Inertial on 4 Mar 2010 07:53 "JT" <jonas.thornvall(a)hotmail.com> wrote in message news:2e2fd274-f957-4d5e-b092-9ec061c6ba34(a)c16g2000yqd.googlegroups.com... [snip for brevity[ > No you did not even answer in the simplest scenario with one emitter > and the two receivers, there is two possible answers. > > Either the timings (B-A) for both experiment X and Y is the ***same*** > or ***not***. > And you did not even answer that. Here is a copy /paste from my earlier answer (that you claim I never made) === > there is two spatially > separated object A,B at relative rest Yes .. that's pretty much what you said above > and one emitter C in the common > vector This is new .. you just said it had 'free velocity; whatever that meant. What is this new 'common vector' terminology you have invented? Do you mean it is colinear with A and B and moving in the direction of the vector between them? you really need to learn how to phrase things so they actually make sense. > of objects that fire lightpulses against the two objects at two > different moments, and with two velocities the distance between > objects do not matter at this time. C->-------A-------B > > Will (B-A) moment/emission X be the same as (B-A) moment emisson Y. In that scenario, as long as C remains on the same side of A B, yes .. if the second pulse happens when it gets past A, then no. > Still confused, No .. you are just very sloppy at describing scenarios and make up your own terminology and expect others to work out what it means. Perhaps if you actually study physics (and learn some) you'd avoid the embarrassment of having to re-explain things that you got wrong. It would also help if you actually DID explain things the first time, instead of adding in new conditions each time. > maybe you need help from the kindergarten group Sam > likes to take help from when confronted with examples that SR can not > handle? There is nothing about it the SR does not handle. Light travels at c from C to A and then on to B. So the difference in time is determined by a) the difference in clock settings (as you said they were not synchronized, but assume are ticking at the same rate .. so it would be constant) and b) that light travels at c for the fixed distance between A and B (so that is constant). That makes the difference constant. The velocity of C makes no difference. === AS you can see .. I answered it
From: Inertial on 4 Mar 2010 08:05 "JT" <jonas.thornvall(a)hotmail.com> wrote in message news:2e2fd274-f957-4d5e-b092-9ec061c6ba34(a)c16g2000yqd.googlegroups.com... [snip for brevity[ > No you did not even answer in the simplest scenario with one emitter > and the two receivers, there is two possible answers. > > Either the timings (B-A) for both experiment X and Y is the ***same*** > or ***not***. > And you did not even answer that. Here is a copy /paste from my earlier answer (that you claim I never made) === > there is two spatially > separated object A,B at relative rest Yes .. that's pretty much what you said above > and one emitter C in the common > vector This is new .. you just said it had 'free velocity; whatever that meant. What is this new 'common vector' terminology you have invented? Do you mean it is colinear with A and B and moving in the direction of the vector between them? you really need to learn how to phrase things so they actually make sense. > of objects that fire lightpulses against the two objects at two > different moments, and with two velocities the distance between > objects do not matter at this time. C->-------A-------B > > Will (B-A) moment/emission X be the same as (B-A) moment emisson Y. In that scenario, as long as C remains on the same side of A B, yes .. if the second pulse happens when it gets past A, then no. > Still confused, No .. you are just very sloppy at describing scenarios and make up your own terminology and expect others to work out what it means. Perhaps if you actually study physics (and learn some) you'd avoid the embarrassment of having to re-explain things that you got wrong. It would also help if you actually DID explain things the first time, instead of adding in new conditions each time. > maybe you need help from the kindergarten group Sam > likes to take help from when confronted with examples that SR can not > handle? There is nothing about it the SR does not handle. Light travels at c from C to A and then on to B. So the difference in time is determined by a) the difference in clock settings (as you said they were not synchronized, but assume are ticking at the same rate .. so it would be constant) and b) that light travels at c for the fixed distance between A and B (so that is constant). That makes the difference constant. The velocity of C makes no difference. === As you can see .. I answered it
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From: Sam Wormley on 5 Mar 2010 15:50 On 3/5/10 2:45 PM, JT wrote: > Nah it was not the puzzle that defeated SR it was inertial, the honest > one about SR, he actually proved the theory incorrect by boldly > stipulating the new term separation velocity and guess!!!! > Wrong -- Nobody has show SR to be incorrect. There has NEVER been an observation that has contradicted a prediction of SR.
From: JT on 5 Mar 2010 17:08
On 5 mar, 21:50, Sam Wormley <sworml...(a)gmail.com> wrote: > On 3/5/10 2:45 PM, JT wrote: > > > Nah it was not the puzzle that defeated SR it was inertial, the honest > > one about SR, he actually proved the theory incorrect by boldly > > stipulating the new term separation velocity and guess!!!! > > Wrong -- Nobody has show SR to be incorrect. There has NEVER been > an observation that has contradicted a prediction of SR. http://www.youtube.com/user/JmanNo42?feature=mhw4#p/a/f/1/evcIPAXPhNY The candy man rules... Oooooh it feels just like it should..... JT |