tomclarke wrote:Two comments Johan:
(1) You have not replied to my proof above that the two twins clock's must read different times. I was expecting you to say which link in the reasoning you disagree with. I can only suppose you now accept this.
Thankyou.
Which proof? I did not see a valid proof. Maybe what you think is a proof I did not see as a proof. I then apologise: Would you mind to prove to me why two clocks which keep the exact same same time rate all the time during the whole journety of the twins can measure two different time durations for the same journey? I just cannot see it.
(2) The debate about clock synchronisation was because teahive asked you how you could calculate "time rate". He said this required comparison of two clocks. I pointed out (I'm sure he would too) that needs two synchonisations.
No it does not, since two synchronisations only reset the time origin twice in succession. What you have to do is to prove that after you have synchronised the two clocks they are actually keeping the same time at any instance in time on any one of the clocks. And that I have done over and over and over on this thread. As I pointed out the x-coordinate of the clock at the origin of Kp, within K is x=v*t; while the xp coordinate of the clock at the origin of K, within kp is v*tp. And at Any instance in time on either of the two clocks clocks there can only be single distance between them: So that v*t MUST be equal to v*tp and this means that t=tp as it must be according to Einstein's first postulate.
What you see when you transform events from Kp to K (for example) is not what actually occurs within Kp but a relativistic "distortion" of what occurs within Kp. Thus, when your transformation gives you a slow-down in time rate, this slow-down IS NOT actually occurrring on the clock within Kp. Just as the the path of an object within Kp is distorted when transformed into K. This does not mean that this distorted path within K is the path that is followed within Kp. To thus conclude that one twin is remaining younger sine "his time is dilated" within the other twin's reference frame is pure poppycock. This situation is symmetric and remains symmetric under the Lorentz transformation: And this symmetry is also demanded by Einstein's postulates. Since the clocks of both twins keep the exact same time rate, both twins age at exactly the same rate and will thus be exactly the same age when they meet up again.
Now you are saying that time rate is always the same by definition, so there is no need to measure it physically!
Not by definition but by the postulates that Einstein used to derive the Lorebntz transformation. If the time rtae is not always the same Einstein's postulates must be wrong, and if the Lorentz transformation is determined by these postulates and you use this transformation to prove the postulates wrong, then i suspect that your logic must be flawed.
That makes your whole argument circular. You have agreement that time rate of a local clock relative to local frame physical processes is constant. That is the definition of a clock, and that is what is implied by Einstein's postulate.
This is not so when there is no gravity.
You can't extend Einstein's postulate to a comparison of time in different frames without defining how you compare time rates in different frames.
Are the laws of physics the same within any inertial refrence frame? yes or no? What determines the clock rate of an atomic clock? The laws of physics? yes or no. This demands that for Einstein's Special Theory of Relativity to be valid, clock rates must be exactly the same on ALL clocks when there is no gravity.
But when I ask you, you tell me no comparison is needed because the time rates by definition are the same.
I did not just do that I also showed by by a direct impeccable proof that v*t=v*tp and therefore afer synchronisation when the clock within K shows a time t, the clock within Kp must at that same instant in time within K show a time tp=t within Kp. Is the algebra too complicated for you?
This argument of yours gets 0 respect from me,
Thus when v*t=v*tp one cannot conclude that t=tp? Shall we call this tomclarke algebra?