What is or is not a paradox?
On Dec 31 2012, 1:48 am, Sylvia Else wrote:
On 31/12/2012 5:04 PM, Koobee Wublee wrote:
From the Lorentz transformations, you can write down the following
equation per Minkowski spacetime. Points #1, #2, and #3 are
observers. They are observing the same target.
** c^2 dt1^2 ds1^2 = c^2 dt2^2 ds2^2 = c^2 dt3^2 ds3^2
** dt1 = Time flow at Point #1
** dt2 = Time flow at Point #2
** dt3 = Time flow at Point #3
** ds1 = Observed target displacement segment by #1
** ds2 = Observed target displacement segment by #2
** ds3 = Observed target displacement segment by #3
The above spacetime equation can also be written as follows.
** dt1^2 (1 B1^2) = dt2^2 (1 B2^2) = dt3^2 (1 B3^2)
** B^2 = (ds/dt)^2 / c^2
When #1 is observing #2, the following equation can be deduced from
the equation above.
** dt1^2 (1 B1^2) = dt2^2 . . . (1)
** B2^2 = 0, #2 is observing itself
Similarly, when #2 is observing #1, the following equation can be
** dt1^2 = dt2^2 (1 B2^2) . . . (2)
** B1^2 = 0, #1 is observing itself
According to relativity, the following must be true.
** B1^2 = B2^2
Thus, equations (1) and (2) become the following equations
** dt1^2 (1 B^2) = dt2^2 . . . (3)
** dt1^2 = dt2^2 (1 B^2) . . . (4)
** B^2 = B1^2 = B2^2
(2) doesn't become (4) just be writing B for B2.
Are you complaining about the typo? It is corrected above. shrug
The only time the equations (3) and (4) can co-exist is when B^2 = 0.
The symmetry is everything about the twins paradox. shrug
In the classical twins paradox, there is no symmetry. The travelling
twin has to change velocities in order to be able to get back to the
stay at home twin.
[snip more nonsense]
This is the second time, you are asked to show the math that shows
this acceleration breaking the symmetry nonsense. There is no way you
can, and that is because you are totally wrong just like Born.
So, you believe in the nonsense of Born? He was the first one to
propose acceleration thing breaking the symmetry. Can you show any
mathematics that support your/Borns claim? No self-styled physicists
have now believed in such nonsense. shrug
The symmetry can be broken without acceleration though to bring an
actual person back then involves cloning. It's simpler to forget the
twin, and just take a clock whose time is copied onto another clock
going in the opposite direction halfway through the travel.
But the symmetry is still broken, and once that happens, you have no
You have no idea what you are talking about, and there is no need to
discuss any further. shrug