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From: Simp on 15 Jul 2010 10:10
On 14 Lip, 14:31, someone2 <glenn.spig...(a)btinternet.com> wrote:
> I was wondering if anyone could help me understand issues of
> simultaneity under relativity.
>
> I was thinking of a circular track in space, which a vechile moving at
> 100mph takes 2 years to make a circuit, and there being a mirror in
> the middle of the track, and an observer 1.5 light years from the
> track,  travelling towards it at near light speed, while firing a beam
> of light at the mirror.
>
> Would I be correct in thinking that after 1 year the observer would be
> 0.5 light years from the track, and be receiving back a reflection of
> the fired light beam from the mirror, and the light from when the
> vehicle was quarter of the way around the track (assumes the vehicle
> to have started its journey from the start of the track, as the
> observer started start its journey towards the track & mirror).
>
> If I am not correct, could you please state what the correct
> observations would be expected to be.

Mirror| v=c<-----|R - observer

distance: M-R = L; t_start = 0;

travel time: v*t_t = L; t_t = L/v;
for v = c: t_t = L/c;

Mirror| c<-----light

distance: L = ct => t = L/c = t_t;

Observer don't see any reflection.
From: dlzc on 15 Jul 2010 10:29
Dear someone2:

On Jul 15, 3:02 am, someone2 <glenn.spig...(a)btinternet.com> wrote:
> On 15 July, 05:08,dlzc<dl...(a)cox.net> wrote:
> > On Jul 14, 5:13 pm, someone2 <glenn.spig...(a)btinternet.com> wrote:
> > > On 14 July, 20:56,dlzc<dl...(a)cox.net> wrote:
> > > > On Jul 14, 11:43 am, someone2 <glenn.spig...(a)btinternet.com> wrote:> On 14 July, 19:10,dlzc<dl...(a)cox.net> wrote:
> > > > > > On Jul 14, 9:40 am, someone2 <glenn.spig...(a)btinternet.com> wrote:
>
> > > > wrote:
> > > > ...
>
> > > > > > > I was thinking that since although the light
> > > > > > > would be travelling away from the observer at
> > > > > > > the speed of light, and thus be 0.75 light years
> > > > > > > in front of the observer after 0.75 of a year
>
> > > > > > ... just not the observer's year.
>
> > > > > > > (and thus have reached the mirror by the time
> > > > > > > the observer is about 0.75 light years from the
> > > > > > > mirror),
>
> > > > > > No.
>
> > > > > > > it would only have travelled 0.25 light years
> > > > > > > back towards the observer, after the 1 year.
>
> > > > > > At the speed you have indicated, the moving
> > > > > > observer is just some short distance from the
> > > > > > mirror (a few million miles, say) when the
> > > > > > light hits.
>
> > > > > Ah this might seem to be a misunderstanding of
> > > > > mine then. I thought that unlike a car on a
> > > > > motorway where if car A was travelling at speed
> > > > > x-y, and car B was travelling at speed x, car B
> > > > > would only seem to pull away from car A at y mph,
> > > > > with light I thought, the idea was that even if
> > > > > the space ship is going at near the speed of
> > > > > light, the fired beam would still appear to travel
> > > > > away from it at the speed of light.
>
> > > > It does.  But *in the ship's frame*, the
> > > > rest distance of 1 light year can be
> > > > measured (by the ships' instruments) by many
> > > > different methods to be about 3.9 light hours.
>
> > > > > Likewise if the observer was travelling towards
> > > > > an incoming light beam at the near the speed of
> > > > > light, the light would only appear to travel
> > > > > towards the observer at the speed of light.
>
> > > > Correct.
>
> > > > > Are you saying that this is not the case, and
> > > > > that if a beam of light is fired from an object
> > > > > moving at velocity x in the same direction as
> > > > > the light beam, that the light beam will only
> > > > > appear to move away from that object at c-x?
>
> > > > No.  What I am saying is that "time dilation" and
> > > > "length contraction" are two faces of the same
> > > > coin.  Please look at the link I provided,
> > > > at minimum.  Or here, if you have the stomach for
> > > > it.
>
> > <snip link now broken by Google.Groups>
>
> > > > Everyone sets themselves up for "frame jumps",
> > > > since "common sense" says that measurements by
> > > > different observers "must be" the same for
> > > > all frames.  It is sometimes close enough where
> > > > our "common sense" is trained, but it is a
> > > > serious mistake at speeds close to c.
>
> > > > It is even detectable at the speed of aircraft
> > > > (special relativity), and/or altitudes of a few
> > > > hundred feet (general relativity), if you can
> > > > integrate the effects over time.
>
> > > The question is about the frame of reference of
> > > the observer that is approaching the tracke. I'm
> > > not sure where this frame is ever left, so I'm
> > > not clear on where you are suggesting frame
> > > jumping is taking place. I thought you were
> > > saying that the observer approaching the track
> > > would find themselves only a small distance away
> > > from the track as the beam of light they fired
> > > reached the mirror, since you said:
>
> > > "At the speed you have indicated, the moving
> > > observer is just some short distance from the
> > > mirror (a few million miles, say) when the
> > > light hits."
>
> > ... the "few million miles" in the frame of the
> > track and car, of course.
>
> > > Though if they were, then I'm not clear how
> > > the light appeared to them to be leaving them
> > > at the spead of light, since it is only a little
> > > bit ahead of them.
>
> > ... because their clock is running slow, and
> > their lengths contracted, they can measure c
> > to be the same friendly constant...
>
> > > I am assuming that they will be using the vehicle
> > > going around the track as their clock,
>
> > Why would they do that?  Even at constant speed,
> > the rate of vehicle motion will be extraordinarily
> > fast when approaching the track, and
> > extraordinarily slow when departing from the other
> > side.  Much easier just to carry your own clock with
> > you.  Like "heartbeats" if nothing else.
>
> > > taking into account how long it takes the light
> > > to reach them at their given position. Or is it
> > > that if they did this, that the light would not
> > > seem to be leaving them at the speed of light,
> > > but only if they used a clock on board their
> > > ship, such that measured by that onboard clock
> > > the journey appears to have taken less time,
>
> > Bingo.
>
> Ah ok. So if they used the track as the clock,
> and took into account the time they would have
> expected the light to have reached them from
> the track,

"the time they would have expected"
.... is that rest time, or knowing the the tools of relativity time?

> it wouldn't seem that light was travelling at c.

Only with a frame jump implied with using measurements from diffferent
frames.

> It would simply be that the laws of physics
> would propergate slower at high velocity (and
> perhaps under high gravity, I don't know),

Know that gravity need have nothing to do with this. It creates
orders of magnitude more math, for a correction of fractions of a
percent.

> such that the tick of a clock on board the
> ship might synchronise with a lap of the
> vehicle on the track while the ship is
> stationary, but when approacing the track at
> a higher speed it might synchronise with 5
> laps of the vehicle on the track for example.

Yes, given the correct closing speed and/or track length / speed.

> Since the propergation rate of the laws is
> referred to as t (time),

Laws don't propagate. Laws aren't really laws. Theories describe
sets of observations, including propagation.

> they talk about time going slower or faster
> in particular frames of reference. Have I
> understood this correctly?

Not quite. I think you have the concept, but the language is loose.
Everything in a frame appears to be self-consistent, not "time going
slower or faster". The differences in length and "time rate" occur
when viewing *another* frame with enough difference in velocity, that
relativity's effects are larger than the error-bars of measurement.

> > > and therefore the slight distance the light
> > > beam was ahead of them, given the time shown
> > > on the onboard clock,
>
> > ... and the contracted distance ...
>
> > > would indicate light travelling at c.
>
> > Correct.
>
> > > Purely from the frame of reference of the
> > > observer that fires the beam of light towards
> > > the track, what will they be expected to see
> > > at the time the light is reflected back to
> > > them (how far will the vechile have appeared
> > > to have gone around the track)?
>
> > You said the vehicle traveled 100mph, presumably
> > in its own frame, and it travels really close to
> > exactly one year, so:
> > 100 * 24 * 365.25 (just averaging the year) =
> > 876,600 miles, in just under 2 hours.  The
> > illusion of traveling FTL.  Outbound (the next
> > light year), it slows to an apparent 0.1 mph so
> > the next 3.9 hours takes it a light year away,
> > and the car travels less than a mile.
>
> "The illusion of traveling FTL."
>
> For it to be an illusion, wouldn't that mean the
> acceptance that they hadn't really travelled that
> distance in two hours, and that it was simply that
> the laws of physics propergated slower while they
> were at speed?

No "propagation of laws".

SHIP...
3.9 hours "really" passes onboard the ship. Only they travel 3.9
light hours to pass the mirror. And they measure c as c.

TRACK...
1 year passes on the track. They (might) see the distant ship leave
the "1 light year point" about 3.9 hours before the ship blows by
them. The light is 1 year in transit, and the ship just that little
bit longer. And c is measured as c in the track's frame.

> Else why would they not be considering themselves
> to ahave observed FTL travel.

Because it involves a frame jump. Measurements made in different
frames when combined "prove" all sorts of thoughtless things. Just
keep track of who measures what.

By the way, you think your ship is "really fast"? Check this out...
http://www.fourmilab.ch/documents/OhMyGodParticle/

David A. Smith
From: Jonathan Doolin on 15 Jul 2010 13:57
On Jul 15, 9:29 am, dlzc <dl...(a)cox.net> wrote:
> Dear someone2:
>
> On Jul 15, 3:02 am, someone2 <glenn.spig...(a)btinternet.com> wrote:
>
>
>
>
>
> > On 15 July, 05:08,dlzc<dl...(a)cox.net> wrote:
> > > On Jul 14, 5:13 pm, someone2 <glenn.spig...(a)btinternet.com> wrote:
> > > > On 14 July, 20:56,dlzc<dl...(a)cox.net> wrote:
> > > > > On Jul 14, 11:43 am, someone2 <glenn.spig...(a)btinternet.com> wrote:> On 14 July, 19:10,dlzc<dl...(a)cox.net> wrote:
> > > > > > > On Jul 14, 9:40 am, someone2 <glenn.spig...(a)btinternet.com> wrote:
>
> > > > > wrote:
> > > > > ...
>
> > > > > > > > I was thinking that since although the light
> > > > > > > > would be travelling away from the observer at
> > > > > > > > the speed of light, and thus be 0.75 light years
> > > > > > > > in front of the observer after 0.75 of a year
>
> > > > > > > ... just not the observer's year.
>
> > > > > > > > (and thus have reached the mirror by the time
> > > > > > > > the observer is about 0.75 light years from the
> > > > > > > > mirror),
>
> > > > > > > No.
>
> > > > > > > > it would only have travelled 0.25 light years
> > > > > > > > back towards the observer, after the 1 year.
>
> > > > > > > At the speed you have indicated, the moving
> > > > > > > observer is just some short distance from the
> > > > > > > mirror (a few million miles, say) when the
> > > > > > > light hits.
>
> > > > > > Ah this might seem to be a misunderstanding of
> > > > > > mine then. I thought that unlike a car on a
> > > > > > motorway where if car A was travelling at speed
> > > > > > x-y, and car B was travelling at speed x, car B
> > > > > > would only seem to pull away from car A at y mph,
> > > > > > with light I thought, the idea was that even if
> > > > > > the space ship is going at near the speed of
> > > > > > light, the fired beam would still appear to travel
> > > > > > away from it at the speed of light.
>
> > > > > It does.  But *in the ship's frame*, the
> > > > > rest distance of 1 light year can be
> > > > > measured (by the ships' instruments) by many
> > > > > different methods to be about 3.9 light hours.
>
> > > > > > Likewise if the observer was travelling towards
> > > > > > an incoming light beam at the near the speed of
> > > > > > light, the light would only appear to travel
> > > > > > towards the observer at the speed of light.
>
> > > > > Correct.
>
> > > > > > Are you saying that this is not the case, and
> > > > > > that if a beam of light is fired from an object
> > > > > > moving at velocity x in the same direction as
> > > > > > the light beam, that the light beam will only
> > > > > > appear to move away from that object at c-x?
>
> > > > > No.  What I am saying is that "time dilation" and
> > > > > "length contraction" are two faces of the same
> > > > > coin.  Please look at the link I provided,
> > > > > at minimum.  Or here, if you have the stomach for
> > > > > it.
>
> > > <snip link now broken by Google.Groups>
>
> > > > > Everyone sets themselves up for "frame jumps",
> > > > > since "common sense" says that measurements by
> > > > > different observers "must be" the same for
> > > > > all frames.  It is sometimes close enough where
> > > > > our "common sense" is trained, but it is a
> > > > > serious mistake at speeds close to c.
>
> > > > > It is even detectable at the speed of aircraft
> > > > > (special relativity), and/or altitudes of a few
> > > > > hundred feet (general relativity), if you can
> > > > > integrate the effects over time.
>
> > > > The question is about the frame of reference of
> > > > the observer that is approaching the tracke. I'm
> > > > not sure where this frame is ever left, so I'm
> > > > not clear on where you are suggesting frame
> > > > jumping is taking place. I thought you were
> > > > saying that the observer approaching the track
> > > > would find themselves only a small distance away
> > > > from the track as the beam of light they fired
> > > > reached the mirror, since you said:
>
> > > > "At the speed you have indicated, the moving
> > > > observer is just some short distance from the
> > > > mirror (a few million miles, say) when the
> > > > light hits."
>
> > > ... the "few million miles" in the frame of the
> > > track and car, of course.
>
> > > > Though if they were, then I'm not clear how
> > > > the light appeared to them to be leaving them
> > > > at the spead of light, since it is only a little
> > > > bit ahead of them.
>
> > > ... because their clock is running slow, and
> > > their lengths contracted, they can measure c
> > > to be the same friendly constant...
>
> > > > I am assuming that they will be using the vehicle
> > > > going around the track as their clock,
>
> > > Why would they do that?  Even at constant speed,
> > > the rate of vehicle motion will be extraordinarily
> > > fast when approaching the track, and
> > > extraordinarily slow when departing from the other
> > > side.  Much easier just to carry your own clock with
> > > you.  Like "heartbeats" if nothing else.
>
> > > > taking into account how long it takes the light
> > > > to reach them at their given position. Or is it
> > > > that if they did this, that the light would not
> > > > seem to be leaving them at the speed of light,
> > > > but only if they used a clock on board their
> > > > ship, such that measured by that onboard clock
> > > > the journey appears to have taken less time,
>
> > > Bingo.
>
> > Ah ok. So if they used the track as the clock,
> > and took into account the time they would have
> > expected the light to have reached them from
> > the track,
>
> "the time they would have expected"
> ... is that rest time, or knowing the the tools of relativity time?
>
> > it wouldn't seem that light was travelling at c.
>
> Only with a frame jump implied with using measurements from diffferent
> frames.
>
> > It would simply be that the laws of physics
> > would propergate slower at high velocity (and
> > perhaps under high gravity, I don't know),
>
> Know that gravity need have nothing to do with this.  It creates
> orders of magnitude more math, for a correction of fractions of a
> percent.
>
> > such that the tick of a clock on board the
> > ship might synchronise with a lap of the
> > vehicle on the track while the ship is
> > stationary, but when approacing the track at
> > a higher speed it might synchronise with 5
> > laps of the vehicle on the track for example.
>
> Yes, given the correct closing speed and/or track length / speed.
>
> > Since the propergation rate of the laws is
> > referred to as t (time),
>
> Laws don't propagate.  Laws aren't really laws.  Theories describe
> sets of observations, including propagation.
>
> > they talk about time going slower or faster
> > in particular frames of reference. Have I
> > understood this correctly?
>
> Not quite.  I think you have the concept, but the language is loose.
> Everything in a frame appears to be self-consistent, not "time going
> slower or faster".  The differences in length and "time rate" occur
> when viewing *another* frame with enough difference in velocity, that
> relativity's effects are larger than the error-bars of measurement.
>
>
>
>
>
> > > > and therefore the slight distance the light
> > > > beam was ahead of them, given the time shown
> > > > on the onboard clock,
>
> > > ... and the contracted distance ...
>
> > > > would indicate light travelling at c.
>
> > > Correct.
>
> > > > Purely from the frame of reference of the
> > > > observer that fires the beam of light towards
> > > > the track, what will they be expected to see
> > > > at the time the light is reflected back to
> > > > them (how far will the vechile have appeared
> > > > to have gone around the track)?
>
> > > You said the vehicle traveled 100mph, presumably
> > > in its own frame, and it travels really close to
> > > exactly one year, so:
> > > 100 * 24 * 365.25 (just averaging the year) =
> > >  876,600 miles, in just under 2 hours.  The
> > > illusion of traveling FTL.  Outbound (the next
> > > light year), it slows to an apparent 0.1 mph so
> > > the next 3.9 hours takes it a light year away,
> > > and the car travels less than a mile.
>
> > "The illusion of traveling FTL."
>
> > For it to be an illusion, wouldn't that mean the
> > acceptance that they hadn't really travelled that
> > distance in two hours, and that it was simply that
> > the laws of physics propergated slower while they
> > were at speed?
>
> No "propagation of laws".
>
> SHIP...
> 3.9 hours "really" passes onboard the ship.  Only they travel 3.9
> light hours to pass the mirror.  And they measure c as c.
>

Of course, from the ship's perspective, the ship is stationary, and it
takes the mirror 3.9 hours to pass them.

> TRACK...
> 1 year passes on the track.  They (might) see the distant ship leave
> the "1 light year point" about 3.9 hours before the ship blows by
> them.  The light is 1 year in transit, and the ship just that little
> bit longer.  And c is measured as c in the track's frame.
>
> > Else why would they not be considering themselves
> > to ahave observed FTL travel.
>
> Because it involves a frame jump.  Measurements made in different
> frames when combined "prove" all sorts of thoughtless things.  Just
> keep track of who measures what.
>

I would recommend the following procedure for keeping track of
things. Measure all the event coordinates from the time that the ship
passes the earth. So for instance, the time that the ship passes the
earth is t=0, x=0. That way the two frames can measure everything
from that shared event. Figure out when and where the other events
happen in the earth's frame. For instance the beaming of the ray from
the oncoming ship, the arrival of the ray at the mirror, the return of
the ray to the oncoming ship.

Then do the lorentz transformation around the shared event. Despite
the frame jump, as long as the two frames share an origin event
t=t'=0, x=x'=0, you shouldn't have a problem with "proving"
thoughtless things.


> By the way, you think your ship is "really fast"?  Check this out...http://www.fourmilab.ch/documents/OhMyGodParticle/
>

Scary.


> David A. Smith- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -

From: xxein on 15 Jul 2010 18:42
On Jul 15, 5:43 am, someone2 <glenn.spig...(a)btinternet.com> wrote:
> On 15 July, 04:08, xxein <xxxx...(a)gmail.com> wrote:
>
>
>
>
>
> > On Jul 14, 8:13 pm, someone2 <glenn.spig...(a)btinternet.com> wrote:
>
> > > On 14 July, 20:56, dlzc <dl...(a)cox.net> wrote:
>
> > > > Dear someone2:
>
> > > > On Jul 14, 11:43 am, someone2 <glenn.spig...(a)btinternet.com> wrote:> On 14 July, 19:10, dlzc <dl...(a)cox.net> wrote:
> > > > > > On Jul 14, 9:40 am, someone2 <glenn.spig...(a)btinternet.com> wrote:
>
> > > > wrote:
> > > > ...
>
> > > > > > > I was thinking that since although the light
> > > > > > > would be travelling away from the observer at
> > > > > > > the speed of light, and thus be 0.75 light years
> > > > > > > in front of the observer after 0.75 of a year
>
> > > > > > ... just not the observer's year.
>
> > > > > > > (and thus have reached the mirror by the time
> > > > > > > the observer is about 0.75 light years from the
> > > > > > > mirror),
>
> > > > > > No.
>
> > > > > > > it would only have travelled 0.25 light years
> > > > > > > back towards the observer, after the 1 year.
>
> > > > > > At the speed you have indicated, the moving
> > > > > > observer is just some short distance from the
> > > > > > mirror (a few million miles, say) when the
> > > > > > light hits.
>
> > > > > Ah this might seem to be a misunderstanding of
> > > > > mine then. I thought that unlike a car on a
> > > > > motorway where if car A was travelling at speed
> > > > > x-y, and car B was travelling at speed x, car B
> > > > > would only seem to pull away from car A at y mph,
> > > > > with light I thought, the idea was that even if
> > > > > the space ship is going at near the speed of
> > > > > light, the fired beam would still appear to travel
> > > > > away from it at the speed of light.
>
> > > > It does.  But *in the ship's frame*, the rest distance of 1 light year
> > > > can be measured (by the ships' instruments) by many different methods
> > > > to be about 3.9 light hours.
>
> > > > > Likewise if the observer was travelling towards
> > > > > an incoming light beam at the near the speed of
> > > > > light, the light would only appear to travel
> > > > > towards the observer at the speed of light.
>
> > > > Correct.
>
> > > > > Are you saying that this is not the case, and
> > > > > that if a beam of light is fired from an object
> > > > > moving at velocity x in the same direction as
> > > > > the light beam, that the light beam will only
> > > > > appear to move away from that object at c-x?
>
> > > > No.  What I am saying is that "time dilation" and "length contraction"
> > > > are two faces of the same coin.  Please look at the link I provided,
> > > > at minimum.  Or here, if you have the stomach for it.
>
> > > >http://www.motionmountain.net/mmdownload.php?f=motionmountain-volume2...
>
> > > > Everyone sets themselves up for "frame jumps", since "common sense"
> > > > says that measurements by different observers "must be" the same for
> > > > all frames.  It is sometimes close enough where our "common sense" is
> > > > trained, but it is a serious mistake at speeds close to c.
>
> > > > It is even detectable at the speed of aircraft (special relativity),
> > > > and/or altitudes of a few hundred feet (general relativity), if you
> > > > can integrate the effects over time.
>
> > > The question is about the frame of reference of the observer that is
> > > approaching the tracke. I'm not sure where this frame is ever left, so
> > > I'm not clear on where you are suggesting frame jumping is taking
> > > place. I thought you were saying that the observer approaching the
> > > track would find themselves only a small distance away from the track
> > > as the beam of light they fired reached the mirror, since you said:
>
> > > "At the speed you have indicated, the moving observer is just some
> > > short distance from the mirror (a few million miles, say) when the
> > > light hits."
>
> > > Though if they were, then I'm not clear how the light appeared to them
> > > to be leaving them at the spead of light, since it is only a little
> > > bit ahead of them. I am assuming that they will be using the vehicle
> > > going around the track as their clock, taking into account how long it
> > > takes the light to reach them at their given position. Or is it that
> > > if they did this, that the light would not seem to be leaving them at
> > > the speed of light, but only if they used a clock on board their ship,
> > > such that measured by that onboard clock the journey appears to have
> > > taken less time, and therefore the slight distance the light beam was
> > > ahead of them, given the time shown on the onboard clock, would
> > > indicate light travelling at c.
>
> > > Purely from the frame of reference of the observer that fires the beam
> > > of light towards the track, what will they be expected to see at the
> > > time the light is reflected back to them (how far will the vechile
> > > have appeared to have gone around the track)?
>
> > > (I took a look at the link you provided btw to the FAQ, and it is a
> > > useful link, I'm not clear what I should have looked at for the answer
> > > to the question though, perhaps if you just gave me the answer and
> > > explained how you arrived at it, it would help).- Hide quoted text -
>
> > > - Show quoted text -
>
> > xxein:  To understand this better, don't allow gravity to have an
> > affect.  It will confuse your issue of particular understanding.
> > Worry about gravity later.  I can help you there.
>
> > Lorentz was very capable to handle inertial frames.  Einstein tried to
> > make a shortcut math to make it simpler to compute.  But Einsein left
> > out the physic of it all.  Lorentz did not say that the speed of light
> > was a constant c in all frames.  There are moving frames.  What does
> > that mean for observers?  It means that if observers (in the simplest
> > terms) are moving at the same velocity along the same line, that they
> > will observe each other as if their shared inertial frame was moving
> > or not.  That is all there is to the observational aspect of it.
>
> > Now.  There is a physic of how this happens.  It is contained in the
> > Lorentz formulas.  Einstein thought there was a math shortcut that
> > superceded the physic.  Silly man (or his wife).  But everyone liked
> > it.  Just as the Einstein followers you read here.
>
> > c is a physical constant --- not a math constant just because it is
> > observed that way.  A few posts tried to tell you that you can observe
> > light going away from you.  No such thing.  You can only observe light
> > coming directly toward you.  How else would light come to your
> > observation?  How else could you travel at a speed that was
> > differentiated from the speed of light?  These guys aren't real
> > physicists.  They are math beleivers - just like Einstein.  Just for
> > fun, ask them to describe the evolution of a universe from energy
> > released to what is now.  They have no coherent explanation.  It might
> > be an unfair question but ask them how the energy got released and/or
> > where it came from.
>
> > Nevertheless, the term of 'inertial frame' is ill-defined except for
> > the purpose of a theory.  Just because someone believes a partiular
> > theory, it does not mean that the definition is physically a correct
> > one.
>
> > But let's get back to what you want to understand.  Observers A and B
> > and/or what will A observe if moving wrt B (and vv).  See?  We already
> > have moving frames.  They are in a physical respect to c already.  How
> > can any moving frame claim that c is c to him?  Only by what each
> > observes.  It becomes paradoxical to a physic, but not to a physics
> > made from observation.  So?  Do we have knowledge of the physic or
> > only a belief in a math theory to make a physics?  The dilemna is that
> > the math theory tells us what we will physically measure but not
> > physically why it happens that way.
>
> > Lorentz puts out the physics and Einstein puts out a short math.
> > Please learn the difference.
>
> > I'm too lazy right now to look up my old notes to show how this works,
> > but please learn that a math expression, written to give the same math
> > result but with different form, can give the same value.  Which form
> > is correct to express the physical meaning?  One describes this and
> > the other describes that.  That the values are the same does not mean
> > they describe the physical process as the same in it's proper logical
> > and/or physical context.  Perhaps I'll get unlazy some day and give
> > you a concrete example.
>
> > In the meantime, study Lorentz and wonder why gravity exists (even
> > though he didn't know or addres it).- Hide quoted text -
>
> Thanks for the reply. It might be useful if you could say from the
> frame of reference of the observer that fires the beam of light
> towards the track, what they would be expected to see at the time the
> light is reflected back to them (how far will the vechile have
> appeared to have gone around the track)- Hide quoted text -
>
> - Show quoted text -

xxein: I won't deal with anyone that only wants a specific answer to
a conditional question. That's not how you learn to understand the
physic.

I could, of course, show you how to figure out an answer but it seems
like you do not want to learn about the physic.
From: Androcles on 15 Jul 2010 18:50

"xxein" <xxxxein(a)gmail.com> wrote in message
news:b682dfc0-3e86-44ba-ab64-9e1af3bb31ec(a)y11g2000yqm.googlegroups.com...
On Jul 15, 5:43 am, someone2 <glenn.spig...(a)btinternet.com> wrote:
> On 15 July, 04:08, xxein <xxxx...(a)gmail.com> wrote:
>
>
>
>
>
> > On Jul 14, 8:13 pm, someone2 <glenn.spig...(a)btinternet.com> wrote:
>
> > > On 14 July, 20:56, dlzc <dl...(a)cox.net> wrote:
>
> > > > Dear someone2:
>
> > > > On Jul 14, 11:43 am, someone2 <glenn.spig...(a)btinternet.com> wrote:>
> > > > On 14 July, 19:10, dlzc <dl...(a)cox.net> wrote:
> > > > > > On Jul 14, 9:40 am, someone2 <glenn.spig...(a)btinternet.com>
> > > > > > wrote:
>
> > > > wrote:
> > > > ...
>
> > > > > > > I was thinking that since although the light
> > > > > > > would be travelling away from the observer at
> > > > > > > the speed of light, and thus be 0.75 light years
> > > > > > > in front of the observer after 0.75 of a year
>
> > > > > > ... just not the observer's year.
>
> > > > > > > (and thus have reached the mirror by the time
> > > > > > > the observer is about 0.75 light years from the
> > > > > > > mirror),
>
> > > > > > No.
>
> > > > > > > it would only have travelled 0.25 light years
> > > > > > > back towards the observer, after the 1 year.
>
> > > > > > At the speed you have indicated, the moving
> > > > > > observer is just some short distance from the
> > > > > > mirror (a few million miles, say) when the
> > > > > > light hits.
>
> > > > > Ah this might seem to be a misunderstanding of
> > > > > mine then. I thought that unlike a car on a
> > > > > motorway where if car A was travelling at speed
> > > > > x-y, and car B was travelling at speed x, car B
> > > > > would only seem to pull away from car A at y mph,
> > > > > with light I thought, the idea was that even if
> > > > > the space ship is going at near the speed of
> > > > > light, the fired beam would still appear to travel
> > > > > away from it at the speed of light.
>
> > > > It does. But *in the ship's frame*, the rest distance of 1 light
> > > > year
> > > > can be measured (by the ships' instruments) by many different
> > > > methods
> > > > to be about 3.9 light hours.
>
> > > > > Likewise if the observer was travelling towards
> > > > > an incoming light beam at the near the speed of
> > > > > light, the light would only appear to travel
> > > > > towards the observer at the speed of light.
>
> > > > Correct.
>
> > > > > Are you saying that this is not the case, and
> > > > > that if a beam of light is fired from an object
> > > > > moving at velocity x in the same direction as
> > > > > the light beam, that the light beam will only
> > > > > appear to move away from that object at c-x?
>
> > > > No. What I am saying is that "time dilation" and "length
> > > > contraction"
> > > > are two faces of the same coin. Please look at the link I provided,
> > > > at minimum. Or here, if you have the stomach for it.
>
> > > >http://www.motionmountain.net/mmdownload.php?f=motionmountain-volume2...
>
> > > > Everyone sets themselves up for "frame jumps", since "common sense"
> > > > says that measurements by different observers "must be" the same for
> > > > all frames. It is sometimes close enough where our "common sense" is
> > > > trained, but it is a serious mistake at speeds close to c.
>
> > > > It is even detectable at the speed of aircraft (special relativity),
> > > > and/or altitudes of a few hundred feet (general relativity), if you
> > > > can integrate the effects over time.
>
> > > The question is about the frame of reference of the observer that is
> > > approaching the tracke. I'm not sure where this frame is ever left, so
> > > I'm not clear on where you are suggesting frame jumping is taking
> > > place. I thought you were saying that the observer approaching the
> > > track would find themselves only a small distance away from the track
> > > as the beam of light they fired reached the mirror, since you said:
>
> > > "At the speed you have indicated, the moving observer is just some
> > > short distance from the mirror (a few million miles, say) when the
> > > light hits."
>
> > > Though if they were, then I'm not clear how the light appeared to them
> > > to be leaving them at the spead of light, since it is only a little
> > > bit ahead of them. I am assuming that they will be using the vehicle
> > > going around the track as their clock, taking into account how long it
> > > takes the light to reach them at their given position. Or is it that
> > > if they did this, that the light would not seem to be leaving them at
> > > the speed of light, but only if they used a clock on board their ship,
> > > such that measured by that onboard clock the journey appears to have
> > > taken less time, and therefore the slight distance the light beam was
> > > ahead of them, given the time shown on the onboard clock, would
> > > indicate light travelling at c.
>
> > > Purely from the frame of reference of the observer that fires the beam
> > > of light towards the track, what will they be expected to see at the
> > > time the light is reflected back to them (how far will the vechile
> > > have appeared to have gone around the track)?
>
> > > (I took a look at the link you provided btw to the FAQ, and it is a
> > > useful link, I'm not clear what I should have looked at for the answer
> > > to the question though, perhaps if you just gave me the answer and
> > > explained how you arrived at it, it would help).- Hide quoted text -
>
> > > - Show quoted text -
>
> > xxein: To understand this better, don't allow gravity to have an
> > affect. It will confuse your issue of particular understanding.
> > Worry about gravity later. I can help you there.
>
> > Lorentz was very capable to handle inertial frames. Einstein tried to
> > make a shortcut math to make it simpler to compute. But Einsein left
> > out the physic of it all. Lorentz did not say that the speed of light
> > was a constant c in all frames. There are moving frames. What does
> > that mean for observers? It means that if observers (in the simplest
> > terms) are moving at the same velocity along the same line, that they
> > will observe each other as if their shared inertial frame was moving
> > or not. That is all there is to the observational aspect of it.
>
> > Now. There is a physic of how this happens. It is contained in the
> > Lorentz formulas. Einstein thought there was a math shortcut that
> > superceded the physic. Silly man (or his wife). But everyone liked
> > it. Just as the Einstein followers you read here.
>
> > c is a physical constant --- not a math constant just because it is
> > observed that way. A few posts tried to tell you that you can observe
> > light going away from you. No such thing. You can only observe light
> > coming directly toward you. How else would light come to your
> > observation? How else could you travel at a speed that was
> > differentiated from the speed of light? These guys aren't real
> > physicists. They are math beleivers - just like Einstein. Just for
> > fun, ask them to describe the evolution of a universe from energy
> > released to what is now. They have no coherent explanation. It might
> > be an unfair question but ask them how the energy got released and/or
> > where it came from.
>
> > Nevertheless, the term of 'inertial frame' is ill-defined except for
> > the purpose of a theory. Just because someone believes a partiular
> > theory, it does not mean that the definition is physically a correct
> > one.
>
> > But let's get back to what you want to understand. Observers A and B
> > and/or what will A observe if moving wrt B (and vv). See? We already
> > have moving frames. They are in a physical respect to c already. How
> > can any moving frame claim that c is c to him? Only by what each
> > observes. It becomes paradoxical to a physic, but not to a physics
> > made from observation. So? Do we have knowledge of the physic or
> > only a belief in a math theory to make a physics? The dilemna is that
> > the math theory tells us what we will physically measure but not
> > physically why it happens that way.
>
> > Lorentz puts out the physics and Einstein puts out a short math.
> > Please learn the difference.
>
> > I'm too lazy right now to look up my old notes to show how this works,
> > but please learn that a math expression, written to give the same math
> > result but with different form, can give the same value. Which form
> > is correct to express the physical meaning? One describes this and
> > the other describes that. That the values are the same does not mean
> > they describe the physical process as the same in it's proper logical
> > and/or physical context. Perhaps I'll get unlazy some day and give
> > you a concrete example.
>
> > In the meantime, study Lorentz and wonder why gravity exists (even
> > though he didn't know or addres it).- Hide quoted text -
>
> Thanks for the reply. It might be useful if you could say from the
> frame of reference of the observer that fires the beam of light
> towards the track, what they would be expected to see at the time the
> light is reflected back to them (how far will the vechile have
> appeared to have gone around the track)- Hide quoted text -
>
> - Show quoted text -

xxein: I won't deal with anyone that only wants a specific answer to
a conditional question. That's not how you learn to understand the
physic.

I could, of course, show you how to figure out an answer but it seems
like you do not want to learn about the physic.
=================================================
Did you explain to him that the physic is an artefactual/superficially
imposed yin-yang of sorts?



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