Re: Hard SR questions?
- From: jt64@xxxxxxxx
- Date: 13 Aug 2006 07:09:58 -0700
The Ghost In The Machine skrev:
In sci.physics.relativity, jt64@xxxxxxxx
<jt64@xxxxxxxx>
wrote
on 7 Aug 2006 01:56:49 -0700
<1154941009.075265.131900@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
The Ghost In The Machine skrev:
On Sun, 06 Aug 2006 06:18:01 -0700, jt64 wrote:
Ship (A) travel 0.99c approaching earth.
At a distance of 300 000 km as seen from planet (B) inertial frame the
ship start a framed TV transmission.
Considering that the planet probably won't know the ship is even there
prior to the transmission
Well it will there is a pod there at rest that register when the ship
pass by, using doppler shift.
That pod is going to have its own issues, even assuming that it can
remain motionless with respect to the center of the Earth (as opposed to
orbiting it).
Well it is really not earth this is a very small planet inhibited by
ants. But the have very nice transmiting and receiving technologies for
signals.
there are some issues with the above statement.
However, one can of course ask the question as to when the ship should
start broadcasting in order to get 60 frames in before it passes by or
impacts the planet.
The ship inertial frame do not send 60 frames/hz
Depends on who's doing the measurement. The ship's
equipment is indeed sending at 60 Hz. (BTW, that's more
correctly represented "60 frames/second".)
No it is not ->i told you<-, the observers of planet know that during
the 300 000km journey towards earth the ship leave 60 frames.(*there
are thousands of tv receivers along the way*)
That do not equal a framerate of 60 hz at the ship there is some
dilation *and doppler* to consider.
The planets inertial view do not receive 60 frames/hz
Correct, because of the time compression.
Mainly because relativities<->"doppler effect" is one of the most ugly
mathematical artefacts i ever seen.
You see if the ship traveled really close to c. The doppler effect
slowly would reach towards infinity. But unfortunatly that would also
mean that no frames would be sent out due to the time dilation, it is
the most ugly artefact of Lorentz transformation ever ;)...
Between 300 000 to 0 km the ship send out 60 frames/hz as measured from
the fram of earth.
Let us say it is digital 1 bit sent 300 000 km away, last bit sent
0,00...1 km from planet.
Frames are fine; it wouldn't make any difference.
Oh it would you, just do not understand the example, you see frames as
sent out from ship(frame) and frames as measured at planet(frame)
during a second. *Really differs* from frames that leave the ship
during 300 000 km travel towards planet(frame) and time planet(frame)
as registered by the thousands of receivers placed along the way.
Hint:There is some doppler and dilation to consider.
The answer is obvious: (0,-1)_S. (In other words, 1
second before flyby or impact -- assuming the ship knows when it will
flyby or impact, ship time.)
Because we're using SR, and therefore the Lorentz, this transforms into
(-7.0179239, -7.088812)_E
Since Earth is incapable of seeing things 7 light-seconds from its
position, we need to adjust this a little, by adding abs(x_E) to t_E
(basically, the light needs to move a little); therefore, we get the event
(0,-0.0708881)_E
as the start of the transmission. Those 60 frames will compress into a
packet of 0.0708881 second, for a frequency of 846.4 frames per second, as
observed by Earth.
Note that this is not limited to frames; if the ship is trying to
broadcast at a frequency of 100 MHz, the Earth will receive it at the
frequency 1.4107 GHz.
As the ship flies by the Earth the situation shifts, and everything slows
wa-y-y-y-y down, as observed from Earth. This can be computed by looking
at the event
(0,1)_S
which is after the ship has broadcast 20 frames. By the Lorentz this
transforms into
(7.0179239, 7.088812)_E
and now the second cluster of 60 frames will occupy 14.1067 seconds of
Earth time for an effective frequency of 4.2533 frames/second and a
broadcast frequency of only 7.0888 MHz.
The transmission is such that 60 frames will be sent from ship (A)
during the distance of 300 000 km relative earths inertial frame.
Not quite that simple, sport.
Well it is look above, they will be sent from ship not from within the
ships inertial frame.
And if the ship is not within its own inertial frame, where is it? :-)
Yes the *ship* is but the *signal* travel through relativities doppler
frame, and then there are some time dilation on ship to consider.
And i told you it is the most awful, ugly mathematical artefact i ever
seen.
Now my question.
1. How long time will it take to travel to planet using (ship point of
view).
This question is extremely confused, as you've not yet established
precisely where/when the ship is. Using my assumptions, of course, the
first wavefront will take 7.0179239 seconds from ship to Earth as one
might measure from Earth were it at all possible, but the time skew screws
things up so badly there's no way to know what happened *prior* to the
0.0708881 seconds before impact, as determined from Earth.
The ship pass *exactly* a earth synchronised pod 300 000 km away. The
exact moment ship pass by is measured by doppler shift.
It will take 1 second for the Earth to know where the ship is
from this beacon.
No it actually pass earth after 1.01 seconds the clock at pod and earth
are synchronised and clock of earth-pod actually show 1.01 this is real
values.
I would say it will take 1.01 seconds from earths point of view and
then take on the calculation from there. To travel there the first
frame is received *exactly* after one second and the last frame must be
received
after 1.0999... just before the ship pass.
Then you apply the Lorentz transformation to find out how it look from
the ship point of view.
You can do that, if you like. In order for Earth to receive 60 frames
at 60 Hz, one calculates the following
(0,-1)_E = (-7.0179239, -7.088812)_S
In order for a beam to reach this point it must start
7.0179239 seconds earlier from the ship's origin,
or -14.106736 seconds. Ergo, the ship sends using a
frame rate of about 4.25329 Hz, so that Earth can
properly receive it.
And as i told you 60 frames pass in 0,01 seconds, this would be 6000
frames in one second (due to the most ugly mathematical artefact
->relativity doppler<-).
second to travel to earth sending out the 60 frames, actually just takeFrom ship point of view time actually slow down by a factor of 7, their
0,142857....
as measured from within ship.
So from the ship point of view the framerate 60/0,142857 =420 fps
So the answer is 6000 fps/hz received at planet.
And 420 fps/hz sent out from ship.
This is a very easily solved problem logically however, the results
above is just artefacts of relativity. And have nothing in common with
reality.
[rest snipped]
--
#191, ewill3@xxxxxxxxxxxxx
Windows Vista. Because it's time to refresh your hardware. Trust us.
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