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The Least Complicated Proof
For the light path in frame K: y=CtFor the identical light path in frame K’: y’=Ct’Per SR, y’=y. Then: Ct=y=y’=Ct’ any two of these are equal.
Hi David.Albert Einstein (1879–1955). Relativity: The Special and General Theory. 1920.XI. The Lorentz TransformationParagraph 4:A. Einstein clearly states, y’=y and z’=z
For the light path in frame K: y=Ct
For the identical light path in frame K’: y’=Ct’
David and JeffreyH,The interest you both have is greatly appreciated.Nobel laureates Einstein and Feynman were professors at Caltech. I mention that because Caltech in cooperation with the Corporation for Public Broadcasting (CPB) produced the truly amazing physics series “The Mechanical Universe and Beyond”.Here is the link for the segment that explains the Lorentz transformation as developed by Lorentz and Einstein.http://www.learner.org/resources/series42.html?pop=yes&pid=611I guarantee you it will hold your interest and give you a deeper insight on the subject. Please let me know what you think.Butch
And the reason you made this mistake is that you neglected the momentum of the dilated frame otherwise time would be accelerated in the moving frame rather than slowed down.
Firstly, I already know about Lorentz transformations.
Secondly, do you believe I will find some error in relativity by studying this?
jeffreyH,You wrote:QuoteAnd the reason you made this mistake is that you neglected the momentum of the dilated frame otherwise time would be accelerated in the moving frame rather than slowed down.Here Einstein explains ‘frames’:http://www.bartleby.com/173/11.htmlAlbert Einstein (1879–1955). Relativity: The Special and General Theory. 1920.XI. The Lorentz TransformationParagraph 3:“A co-ordinate system K then corresponds to the embankment and a co-ordinate system K’ to the train.” – Frames K and K’Since a co-ordinate system has no mass, THERE IS NO MOMENTUM. Yet, you state:QuoteFirstly, I already know about Lorentz transformations.You wrote:QuoteSecondly, do you believe I will find some error in relativity by studying this?It was clearly stated “Lorentz transformation as developed by Lorentz and Einstein”So, quite obviously, the answer is NO.Again, thank you for your interest.Butch
Please be more specific. Exactly what is wrong with which step.Thank you,Butch
17. In 2 seconds, light propagates 600,000km.18. Relative to length in K the calculated length of the light path in K’ is 600,000km.
That calculation is wrong because the observer in each frame knows that the other is moving (because the received clock pulses are out of sync) and therefore applies the necessary relativistic correction to the signals he receives, and calculates 300,000 km . Every subsequent step is therefore incorrect since it begins with an incorrect assumption. In short, you are assuming relativity is incorrect in order to prove that it is incorrect. All that is necessary to prove that it is correct is to state that each observer knows that the other has an identical clock.
How is light then constant seeing as B has been monitoring it for less local time than A?
Here's one I thought about reading this, its the confusion about how the speed of light is constant for all observers, and time dilation.1. Observer A is stood at point A and shines a packet of light to a far away point C2. Observer B is on a rocket ship traveling at speed to point B somewhere between point A and C3. The packet of light will be travelling at c to both observers4. But if observer B is having time dilation, travels to point B and back to point A, time will have passed slower for him than observer A but if light hasn't got to C by the time he gets back to point A it will arrive at C simultaneously for both observers 5. How is light then constant seeing as B has been monitoring it for less local time than A?
B stops at A..
How does each of them know that the light has reached C?
No, it was all with reference to A.Actually wiht more investigation I found that B was not an inertial frame of reference as it undergoes acceleration.. So I rephrased my question to.If B is travelling at constant velocity towards AWhen A calculated B is going to arrive at that point (without stopping) in 1 year he emits a light signal towards a reciever 1 light year away at point C.At the exact point B reaches A the reciever detects the light the time of which takes 1 year form A's point of reference, but obviously less time from B's point of reference.
If I considered the ladder paradox, would this mean that the distance between A and C appears smaller to B? So that B can calculate the speed of light to be constant.So would that mean that the faster you travel, the closer things appear to be to each other?
Is this why its hard to find a parking space?