From: G on
On Dec 19, 8:53 am, "Jeckyl" <no...(a)nowhere.com> wrote:
> "G" <gehan_ameresek...(a)hotmail.com> wrote in message
>
> news:39f5bc69-8a82-438b-a2da-d744ed9a9217(a)d21g2000prf.googlegroups.com...
>
>
>
> > On Dec 19, 4:16 am, "Jeckyl" <no...(a)nowhere.com> wrote:
> >> "G" <gehan_ameresek...(a)hotmail.com> wrote in message
>
> >>news:b27f63a7-4e95-4425-805b-3c71f932e8b8(a)d4g2000prg.googlegroups.com...
>
> >> > Assume the case where two spaceships leave the earth at exactly the
> >> > same time, accelerate at 1g for 10 hours, and pass by two "space
> >> > buoys" SpaceBouy 1 and 2
>

[SNIP SNIP]

>
> > Is there an instant that is common to all frames?
>
> No

Surprising. If you take the timeline of all three there must be a
point that coincides on all three timelines?
>
> > If not, is there a
> > time value that is equivalent in magnitude in each frame?
>
> No
>

OK let me put it this way:given the relatrive velocity of the
spaceships, it is possible to calculate the time dilation and then
calculate the "display time"aboard the spaceships, as seen from earth
and from the other spaceship.

That is, if spaceship 1 transmits a picture of the clock display
taken at the time of the ship1's passing of the space buoy to earth
and spaceship2, will this picture show the same time or different
times?

> > In which
> > case you will still have the inequality.
>
> Yes .. there is inequality.
>
> > Why doesnt someone calculate the values?
>
> Which ones? Can't you do it yourself using the Lorentz transforms .. its
> not that difficult (if you do not use the more complicated acceleration
> case)
>
> > You cannot have two different
> > times in space ship1 and sapceship2

That is, earth sees the same time on the clockface of 1 and 2
However spaceship 1 sees a different value of spaceships 2 clock.

>
> No reason why not
>
> > Also, EA
>
> EA?

I meant AE
>
[SNIP]
From: Jeckyl on
"Sue..." <suzysewnshow(a)yahoo.com.au> wrote in message
news:72e5a0a7-bffb-407b-8b93-f5afa8feca6e(a)q3g2000hsg.googlegroups.com...
> You are presenting one after another argument
> on behalf of Newton's ether and his corpuscular
> light bullets. If you and Jeckyl are persistant
> enough, you may actually convince me that
> Einstein was wrong.

You are already convinced of that .. we're trying to convince you he was
right .. or rather, that you simply do not understand enough to make any
judgement.


From: Jeckyl on
"G" <gehan_ameresekere(a)hotmail.com> wrote in message
news:a13eaf0d-83cf-4573-8d15-dd457c545af0(a)s19g2000prg.googlegroups.com...
> On Dec 19, 8:53 am, "Jeckyl" <no...(a)nowhere.com> wrote:
>> "G" <gehan_ameresek...(a)hotmail.com> wrote in message
>>
>> news:39f5bc69-8a82-438b-a2da-d744ed9a9217(a)d21g2000prf.googlegroups.com...
>>
>>
>>
>> > On Dec 19, 4:16 am, "Jeckyl" <no...(a)nowhere.com> wrote:
>> >> "G" <gehan_ameresek...(a)hotmail.com> wrote in message
>>
>> >>news:b27f63a7-4e95-4425-805b-3c71f932e8b8(a)d4g2000prg.googlegroups.com...
>>
>> >> > Assume the case where two spaceships leave the earth at exactly the
>> >> > same time, accelerate at 1g for 10 hours, and pass by two "space
>> >> > buoys" SpaceBouy 1 and 2
>>
>
> [SNIP SNIP]
>
>>
>> > Is there an instant that is common to all frames?
>>
>> No
>
> Surprising. If you take the timeline of all three there must be a
> point that coincides on all three timelines?

It depends on what you mean by 'common'. Ceratinly two event that happen at
the same time at different locations in one frame, will happens at differnet
times in another frame.

>> > If not, is there a
>> > time value that is equivalent in magnitude in each frame?
>> No
> OK let me put it this way:given the relatrive velocity of the
> spaceships, it is possible to calculate the time dilation and then
> calculate the "display time"aboard the spaceships, as seen from earth
> and from the other spaceship.

Yes .. that's what physics is for

> That is, if spaceship 1 transmits a picture of the clock display
> taken at the time of the ship1's passing of the space buoy to earth
> and spaceship2, will this picture show the same time or different
> times?

Same or different to what?

>> > In which
>> > case you will still have the inequality.
>> Yes .. there is inequality.
>> > Why doesnt someone calculate the values?
>> Which ones? Can't you do it yourself using the Lorentz transforms .. its
>> not that difficult (if you do not use the more complicated acceleration
>> case)
>> > You cannot have two different
>> > times in space ship1 and sapceship2
> That is, earth sees the same time on the clockface of 1 and 2

Yes .. and both will be slower than the earth clock

> However spaceship 1 sees a different value of spaceships 2 clock.

Yes .. the ship1 will find that it is faster than the earth clock, and that
ship2 is slower than the earth clock.

There is no problem with that. It may seem strange when you try to think of
it in terms of how our world seem at our close-to-zero (relative to light)
speeds that we experience in our day to day life.



From: paparios on
On 20 dic, 21:16, G <gehan_ameresek...(a)hotmail.com> wrote:
> On Dec 19, 8:53 am, "Jeckyl" <no...(a)nowhere.com> wrote:
>
> > "G" <gehan_ameresek...(a)hotmail.com> wrote in message
>
> >news:39f5bc69-8a82-438b-a2da-d744ed9a9217(a)d21g2000prf.googlegroups.com...
>
> > > On Dec 19, 4:16 am, "Jeckyl" <no...(a)nowhere.com> wrote:
> > >> "G" <gehan_ameresek...(a)hotmail.com> wrote in message
>
> > >>news:b27f63a7-4e95-4425-805b-3c71f932e8b8(a)d4g2000prg.googlegroups.com...
>
> > >> > Assume the case where two spaceships leave the earth at exactly the
> > >> > same time, accelerate at 1g for 10 hours, and pass by two "space
> > >> > buoys" SpaceBouy 1 and 2
>
> [SNIP SNIP]
>
>
>
> > > Is there an instant that is common to all frames?
>
> > No
>
> Surprising. If you take the timeline of all three there must be a
> point that coincides on all three timelines?
>
>
>
> > > If not, is there a
> > > time value that is equivalent in magnitude in each frame?
>
> > No
>
> OK let me put it this way:given the relatrive velocity of the
> spaceships, it is possible to calculate the time dilation and then
> calculate the "display time"aboard the spaceships, as seen from earth
> and from the other spaceship.
>
> That is, if spaceship 1 transmits a picture of the clock display
> taken at the time of the ship1's passing of the space buoy to earth
> and spaceship2, will this picture show the same time or different
> times?
>

As I mentioned before, that picture will show the time as it was when
the information left the ship. Depending on the distance that
information must travel to reach the other ship, it can take years in
reaching its destination and so, of course, local time at destination
will be different.

Let us see this graphically. Let us take Earth as the inertial frame
used as reference (you can choose either of the ships as well as
reference, the results being the same), and locate Earth at
coordinates (x,y,z)=(0,0,0) measured in light years. Ship 1 travels
away at v=0.6c into the +x direction, and Ship 2 travels away at
v=0.6c into the -x direction. After 10 years, Ship 1 will be at
(6,0,0) light years and Ship 2 will be at (-6,0,0) lightyears.
However, both ships will experience time dilation, so both ships
clocks will read 8 years instead of 10. Let us set the trip so each
ship, and Earth, sends a light signal every new year. What does the
other ship receives?
After 1 year travelling, measured or calculated from Earth local
clock, both Ships will be 0.6 light years away from Earth. If someone
on Earth sends a light signal to both ships, that signal will have to
cover a lot of space before reaching the ships, since both are moving
away very fast. So, for instance, Ship 1 will receive that first year
light signal from Earth only when Ship 1 is 1.5 light years away and
its local clock reads 2 years. Similarly, the first year light signal
sent from Ship 2, according to its own local clock, will arrive to
Earth when Earth clock reads 2 years, and will only catch Ship 1 when
its local clock reads 4 years.
So you see that, from the point of view of everyone involved, clocks
are running slow. Earth sees light signals from both Ship 1 and Ship 2
arriving every 2 years (they all know they are sending the signals
once per year according to their local clocks). Ship 1 sees light
signals from Earth arriving every 2 years, and light signals from Ship
2 arriving every 4 years. Finally, Ship 2 sees light signals from
Earth arriving every 2 years, and light signals from Ship 1 arriving
every 4 years. Note, however, that both local clocks at the ships are
running at the same rate (which is slow in the gamma factor with
respect to the Earth clock, we are using as reference).

The Minkowski diagram in http://en.wikipedia.org/wiki/Twin_paradox
helps in understanding this.

Miguel Rios

From: Sue... on
On Dec 20, 8:27 pm, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote:
> On 20 dic, 21:16, G <gehan_ameresek...(a)hotmail.com> wrote:
>
>
>
>
>
> > On Dec 19, 8:53 am, "Jeckyl" <no...(a)nowhere.com> wrote:
>
> > > "G" <gehan_ameresek...(a)hotmail.com> wrote in message
>
> > >news:39f5bc69-8a82-438b-a2da-d744ed9a9217(a)d21g2000prf.googlegroups.com....
>
> > > > On Dec 19, 4:16 am, "Jeckyl" <no...(a)nowhere.com> wrote:
> > > >> "G" <gehan_ameresek...(a)hotmail.com> wrote in message
>
> > > >>news:b27f63a7-4e95-4425-805b-3c71f932e8b8(a)d4g2000prg.googlegroups.com...
>
> > > >> > Assume the case where two spaceships leave the earth at exactly the
> > > >> > same time, accelerate at 1g for 10 hours, and pass by two "space
> > > >> > buoys" SpaceBouy 1 and 2
>
> > [SNIP SNIP]
>
> > > > Is there an instant that is common to all frames?
>
> > > No
>
> > Surprising. If you take the timeline of all three there must  be a
> > point that coincides on all three timelines?
>
> > > > If not, is there a
> > > > time value that is equivalent in magnitude in each frame?
>
> > > No
>
> > OK let me put it this way:given the relatrive velocity of the
> > spaceships, it is possible to calculate the time dilation and then
> > calculate the "display time"aboard the spaceships, as seen from earth
> > and from the other spaceship.
>
> > That is, if spaceship 1 transmits a picture of the clock display
> > taken at the time of the ship1's passing of the space buoy to earth
> > and spaceship2, will this picture show the same time or different
> > times?
>
> As I mentioned before, that picture will show the time as it was when
> the information left the ship. Depending on the distance that
> information must travel to reach the other ship, it can take years in
> reaching its destination and so, of course, local time at destination
> will be different.
>
> Let us see this graphically. Let us take Earth as the inertial frame
> used as reference

That is Newton's way.
Einstein said we take the heavnly bodies as an
*inertial* reference.
http://nobelprize.org/nobel_prizes/physics/laureates/1921/einstein-lecture.html

Sue...
[...]