Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.
|
|
|
|
Re: Simplified Twin Paradox Resolution.
Posted:
Jan 7, 2013 1:43 PM
|
|
On Jan 7, 10:02 am, Absolutely Vertical wrote:
> On 1/5/2013 10:24 AM, Vilas Tamhane wrote:
> > It is perfectly symmetrical. Note that SR does not seek to find who
> > actually fired the rocket. Between the two spaceships A and B, A can
> > accelerate or B can accelerate or both can accelerate. SR deals with
> > uniform motion after acceleration.
>
> that's not so.
That was what PD said years ago. Stupid PD, an ex-professor of
physics at a university in Texas. <shrug>
> if both accelerate, there is no time difference.
After both have done their acceleration, they continue to move away
from each other. What is their relative speed? Does the Lorentz
transform not say time dilation? At this moment, who is actually
moving, and who is not? If time dilation is building up, how does it
evaporate? <shrug>
> the fact that one accelerates and the other doesn't is the reason there
> is a difference.
Actually not according to the Lorentz transform. You cannot make up
your own laws of physics. You are no god. <shrug>
> sr accounts for the difference in that case. if you
> thought that sr just ignores acceleration then you thought wrong and the
> twin example was designed to elicit that mistake.
In this case, both accelerate with a coasting period to allow for
mutual time dilation building up. Shouldn?t the magic effect of
acceleration cancel out? If not, why not? Just what part of this
simple scenario do you not understand, PD? <shrug>
Let?s recap the mathemagic trick Einstein dingleberries like to pitch
when one accelerates and the other does not.
** dt1 = dt2 / sqrt(1 ? B^2)
And
** dt2 = dt1 sqrt(1 ? B^2)
Where
** B c = Relative speed between 1 and 2
When both accelerate, well they will probably say the following.
** dt1 = dt2
And
** dt2 = dt1
It is indeed interesting what type of mathemagic trick they are going
to pull out when both 1 and 2 are coasting away or towards each
other. <shrug>
For reference, the Lorentz transform always says the following
regardless who is accelerating or no:
** dt1 = dt2 sqrt(1 ? B^2)
And
** dt2 = dt1 sqrt(1 ? B^2)
The only time when there is no contradiction is when (B^2 = 0).
<shrug>
|
|
|
|
