A constant speed of light in all reference frames? Surely you can't be serious. [Physics]

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From: BURT on 12 Mar 2010 19:23

On Mar 8, 6:35 am, Ste <ste_ro...(a)hotmail.com> wrote:
> On 7 Mar, 02:51, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au>
> wrote:
>
>
>
>
>
> > "Ste" <ste_ro...(a)hotmail.com> wrote in message
>
> >news:651a713d-7ae4-4048-bafb-f1b3219ee4fc(a)v20g2000yqv.googlegroups.com....
>
> > > On 6 Mar, 12:47, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au>
> > > wrote:
> > >> > This should make perfect sense to you. If a clock is running 2%
> > >> > slower, then it is running 2% slower regardless of distance. But if,
> > >> > as a result of running 2% slower, it falls behind 6 minutes after
> > >> > running a certain amount of time, then it will fall behind 12 minutes
> > >> > after running for twice as long.
>
> > >> Agreed.
>
> > >> The question now is, if we agree that both clocks suffer time dilation
> > >> in this way, then when they return to the start point, how do they
> > >> each reconcile the fact that (after accounting for the effects of
> > >> acceleration) it ought to be the other clock which is slow, when in
> > >> fact one clock (the one that went furthest from the start point) will
> > >> be slower than the other? And a third clock, left at the start point,
> > >> will be running ahead of both?
>
> > >> _________________________
> > >> They know that the operations were not symmetric. Only one clock remained
> > >> in
> > >> the same inertial reference frame throughout. The other two clocks spent
> > >> different amounts of time in at least 3 different inertial reference
> > >> frames.
> > >> Everybody can see this is true, and so nobody expects that the clocks
> > >> will
> > >> remain synchronised.
>
> > > Yes, but the important question here is whether they agree *after* the
> > > effects of acceleration are taken into account. I mean, if we said
> > > that each travelling clock slows by 2% when moving away from the start
> > > point at a certain speed, then by rights both travelling clocks should
> > > slow equally. Yes?
>
> > As I understand your thought experiment, no.
>
> > In SR, time dilation is a function of relative speed and the time for which
> > they are moving at the speed. It is not a function of accleration.
>
> > A doesn't move. B moves at speed v for time t, and its clock will read x
> > behind A. C moves at speed v for time 2t, and its clock will read 2x behind
> > A.
>
> The question is this. We'll deal with only the outbound trip (in other
> words, the clocks are on the move, but time 't' has not yet elapsed,
> so there has been no further acceleration). I agree with your answer
> above, as it concerns A's frame.
>
> The question is, from the frame of B, what will the slowdown be on
> clock C, *after* having accounted for the increased distances between
> them (i.e. having accounted for the increased propagation delays). It
> seems to me that the natural answer is to say "4%".- Hide quoted text -
>
> - Show quoted text -

The clocks change when accelerating and decelerating in space. Time
decelerates and accelerates when there is a change in speed in space.

Mitch Raemsch

From: mpc755 on 12 Mar 2010 20:01

On Mar 12, 7:23 pm, BURT <macromi...(a)yahoo.com> wrote:
> On Mar 8, 6:35 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
>
>
> > On 7 Mar, 02:51, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au>
> > wrote:
>
> > > "Ste" <ste_ro...(a)hotmail.com> wrote in message
>
> > >news:651a713d-7ae4-4048-bafb-f1b3219ee4fc(a)v20g2000yqv.googlegroups.com....
>
> > > > On 6 Mar, 12:47, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au>
> > > > wrote:
> > > >> > This should make perfect sense to you. If a clock is running 2%
> > > >> > slower, then it is running 2% slower regardless of distance. But if,
> > > >> > as a result of running 2% slower, it falls behind 6 minutes after
> > > >> > running a certain amount of time, then it will fall behind 12 minutes
> > > >> > after running for twice as long.
>
> > > >> Agreed.
>
> > > >> The question now is, if we agree that both clocks suffer time dilation
> > > >> in this way, then when they return to the start point, how do they
> > > >> each reconcile the fact that (after accounting for the effects of
> > > >> acceleration) it ought to be the other clock which is slow, when in
> > > >> fact one clock (the one that went furthest from the start point) will
> > > >> be slower than the other? And a third clock, left at the start point,
> > > >> will be running ahead of both?
>
> > > >> _________________________
> > > >> They know that the operations were not symmetric. Only one clock remained
> > > >> in
> > > >> the same inertial reference frame throughout. The other two clocks spent
> > > >> different amounts of time in at least 3 different inertial reference
> > > >> frames.
> > > >> Everybody can see this is true, and so nobody expects that the clocks
> > > >> will
> > > >> remain synchronised.
>
> > > > Yes, but the important question here is whether they agree *after* the
> > > > effects of acceleration are taken into account. I mean, if we said
> > > > that each travelling clock slows by 2% when moving away from the start
> > > > point at a certain speed, then by rights both travelling clocks should
> > > > slow equally. Yes?
>
> > > As I understand your thought experiment, no.
>
> > > In SR, time dilation is a function of relative speed and the time for which
> > > they are moving at the speed. It is not a function of accleration.
>
> > > A doesn't move. B moves at speed v for time t, and its clock will read x
> > > behind A. C moves at speed v for time 2t, and its clock will read 2x behind
> > > A.
>
> > The question is this. We'll deal with only the outbound trip (in other
> > words, the clocks are on the move, but time 't' has not yet elapsed,
> > so there has been no further acceleration). I agree with your answer
> > above, as it concerns A's frame.
>
> > The question is, from the frame of B, what will the slowdown be on
> > clock C, *after* having accounted for the increased distances between
> > them (i.e. having accounted for the increased propagation delays). It
> > seems to me that the natural answer is to say "4%".- Hide quoted text -
>
> > - Show quoted text -
>
> The clocks change when accelerating and decelerating in space. Time
> decelerates and accelerates when there is a change in speed in space.
>
> Mitch Raemsch

Due to the change in the pressure associated with the aether.

From: BURT on 12 Mar 2010 20:17

On Mar 12, 5:01 pm, mpc755 <mpc...(a)gmail.com> wrote:
> On Mar 12, 7:23 pm, BURT <macromi...(a)yahoo.com> wrote:
>
>
>
>
>
> > On Mar 8, 6:35 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > On 7 Mar, 02:51, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au>
> > > wrote:
>
> > > > "Ste" <ste_ro...(a)hotmail.com> wrote in message
>
> > > >news:651a713d-7ae4-4048-bafb-f1b3219ee4fc(a)v20g2000yqv.googlegroups.com...
>
> > > > > On 6 Mar, 12:47, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au>
> > > > > wrote:
> > > > >> > This should make perfect sense to you. If a clock is running 2%
> > > > >> > slower, then it is running 2% slower regardless of distance. But if,
> > > > >> > as a result of running 2% slower, it falls behind 6 minutes after
> > > > >> > running a certain amount of time, then it will fall behind 12 minutes
> > > > >> > after running for twice as long.
>
> > > > >> Agreed.
>
> > > > >> The question now is, if we agree that both clocks suffer time dilation
> > > > >> in this way, then when they return to the start point, how do they
> > > > >> each reconcile the fact that (after accounting for the effects of
> > > > >> acceleration) it ought to be the other clock which is slow, when in
> > > > >> fact one clock (the one that went furthest from the start point) will
> > > > >> be slower than the other? And a third clock, left at the start point,
> > > > >> will be running ahead of both?
>
> > > > >> _________________________
> > > > >> They know that the operations were not symmetric. Only one clock remained
> > > > >> in
> > > > >> the same inertial reference frame throughout. The other two clocks spent
> > > > >> different amounts of time in at least 3 different inertial reference
> > > > >> frames.
> > > > >> Everybody can see this is true, and so nobody expects that the clocks
> > > > >> will
> > > > >> remain synchronised.
>
> > > > > Yes, but the important question here is whether they agree *after* the
> > > > > effects of acceleration are taken into account. I mean, if we said
> > > > > that each travelling clock slows by 2% when moving away from the start
> > > > > point at a certain speed, then by rights both travelling clocks should
> > > > > slow equally. Yes?
>
> > > > As I understand your thought experiment, no.
>
> > > > In SR, time dilation is a function of relative speed and the time for which
> > > > they are moving at the speed. It is not a function of accleration.
>
> > > > A doesn't move. B moves at speed v for time t, and its clock will read x
> > > > behind A. C moves at speed v for time 2t, and its clock will read 2x behind
> > > > A.
>
> > > The question is this. We'll deal with only the outbound trip (in other
> > > words, the clocks are on the move, but time 't' has not yet elapsed,
> > > so there has been no further acceleration). I agree with your answer
> > > above, as it concerns A's frame.
>
> > > The question is, from the frame of B, what will the slowdown be on
> > > clock C, *after* having accounted for the increased distances between
> > > them (i.e. having accounted for the increased propagation delays). It
> > > seems to me that the natural answer is to say "4%".- Hide quoted text -
>
> > > - Show quoted text -
>
> > The clocks change when accelerating and decelerating in space. Time
> > decelerates and accelerates when there is a change in speed in space.
>
> > Mitch Raemsch
>
> Due to the change in the pressure associated with the aether.- Hide quoted text -
>
> - Show quoted text -

In my model Aether flows over flowing energy and field without any
pressure. And time flows nowhere else that is empty but only over
particle and field.

Mitch Raemsch

From: mpc755 on 12 Mar 2010 20:25

On Mar 12, 8:17 pm, BURT <macromi...(a)yahoo.com> wrote:
> On Mar 12, 5:01 pm, mpc755 <mpc...(a)gmail.com> wrote:
>
>
>
> > On Mar 12, 7:23 pm, BURT <macromi...(a)yahoo.com> wrote:
>
> > > On Mar 8, 6:35 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > On 7 Mar, 02:51, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au>
> > > > wrote:
>
> > > > > "Ste" <ste_ro...(a)hotmail.com> wrote in message
>
> > > > >news:651a713d-7ae4-4048-bafb-f1b3219ee4fc(a)v20g2000yqv.googlegroups..com...
>
> > > > > > On 6 Mar, 12:47, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au>
> > > > > > wrote:
> > > > > >> > This should make perfect sense to you. If a clock is running 2%
> > > > > >> > slower, then it is running 2% slower regardless of distance. But if,
> > > > > >> > as a result of running 2% slower, it falls behind 6 minutes after
> > > > > >> > running a certain amount of time, then it will fall behind 12 minutes
> > > > > >> > after running for twice as long.
>
> > > > > >> Agreed.
>
> > > > > >> The question now is, if we agree that both clocks suffer time dilation
> > > > > >> in this way, then when they return to the start point, how do they
> > > > > >> each reconcile the fact that (after accounting for the effects of
> > > > > >> acceleration) it ought to be the other clock which is slow, when in
> > > > > >> fact one clock (the one that went furthest from the start point) will
> > > > > >> be slower than the other? And a third clock, left at the start point,
> > > > > >> will be running ahead of both?
>
> > > > > >> _________________________
> > > > > >> They know that the operations were not symmetric. Only one clock remained
> > > > > >> in
> > > > > >> the same inertial reference frame throughout. The other two clocks spent
> > > > > >> different amounts of time in at least 3 different inertial reference
> > > > > >> frames.
> > > > > >> Everybody can see this is true, and so nobody expects that the clocks
> > > > > >> will
> > > > > >> remain synchronised.
>
> > > > > > Yes, but the important question here is whether they agree *after* the
> > > > > > effects of acceleration are taken into account. I mean, if we said
> > > > > > that each travelling clock slows by 2% when moving away from the start
> > > > > > point at a certain speed, then by rights both travelling clocks should
> > > > > > slow equally. Yes?
>
> > > > > As I understand your thought experiment, no.
>
> > > > > In SR, time dilation is a function of relative speed and the time for which
> > > > > they are moving at the speed. It is not a function of accleration..
>
> > > > > A doesn't move. B moves at speed v for time t, and its clock will read x
> > > > > behind A. C moves at speed v for time 2t, and its clock will read 2x behind
> > > > > A.
>
> > > > The question is this. We'll deal with only the outbound trip (in other
> > > > words, the clocks are on the move, but time 't' has not yet elapsed,
> > > > so there has been no further acceleration). I agree with your answer
> > > > above, as it concerns A's frame.
>
> > > > The question is, from the frame of B, what will the slowdown be on
> > > > clock C, *after* having accounted for the increased distances between
> > > > them (i.e. having accounted for the increased propagation delays). It
> > > > seems to me that the natural answer is to say "4%".- Hide quoted text -
>
> > > > - Show quoted text -
>
> > > The clocks change when accelerating and decelerating in space. Time
> > > decelerates and accelerates when there is a change in speed in space.
>
> > > Mitch Raemsch
>
> > Due to the change in the pressure associated with the aether.- Hide quoted text -
>
> > - Show quoted text -
>
> In my model Aether flows over flowing energy and field without any
> pressure. And time flows nowhere else that is empty but only over
> particle and field.
>
> Mitch Raemsch

In AD, the pressure associated with the aether displaced by massive
objects is gravity. The faster a clock moves with respect to the
aether the greater the pressure associated with the aether on the
clock the slower the clock ticks.

An atomic clock 'ticks' based upon the aether pressure it exists in.
The speed of a GPS satellite with respect to the aether causes it to
displace more aether and for that aether to exert more pressure on the
clock in the GPS satellite than the aether pressure associated with a
clock at rest with respect to the Earth. This causes the GPS satellite
clock to "result in a delay of about 7 ìs/day". The aether pressure
associated with the aether displaced by the Earth exerts less pressure
on the GPS satellite than a similar clock at rest on the Earth
"causing the GPS clocks to appear faster by about 45 ìs/day". The
aether pressure associated with the speed at which the GPS satellite
moves in the aether and the aether pressure associated with the aether
displaced by the Earth causes "clocks on the GPS satellites tick
approximately 38 ìs/day faster than clocks on the ground."
(quoted text from http://en.wikipedia.org/wiki/Effects_of_relativity_on_GPS).

From: Bruce Richmond on 12 Mar 2010 21:02

On Mar 11, 7:42 am, mpc755 <mpc...(a)gmail.com> wrote:
> On Mar 10, 11:30 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote:
>
>
>
>
>
> > On Mar 10, 9:18 pm, mpc755 <mpc...(a)gmail.com> wrote:
>
> > > On Mar 10, 8:45 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote:
>
> > > > On Mar 10, 8:10 pm, mpc755 <mpc...(a)gmail.com> wrote:
>
> > > > > On Mar 10, 7:45 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote:
>
> > > > > > On Mar 10, 9:36 am, mpc755 <mpc...(a)gmail.com> wrote:
>
> > > > > > > On Mar 10, 8:05 am, "Inertial" <relativ...(a)rest.com> wrote:
>
> > > > > > > > "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote in message
>
> > > > > > > >news:4b970c19$0$8039$afc38c87(a)news.optusnet.com.au...
>
> > > > > > > > > I know I still have a long way to go but my goal here is to truely
> > > > > > > > > understand SR, not to just parrot explainations. LET helped me see
> > > > > > > > > that the math of SR is correct, but I also realize it has become a
> > > > > > > > > hiderence in understanding SR.
>
> > > > > > > > > ________________________________
> > > > > > > > > Good. There is one key insight which makes the jump from LET to SR a
> > > > > > > > > little easier (in my opinion).
>
> > > > > > > > > For all the talk of relative motion against the ether in LET, the
> > > > > > > > > equations work out exactly the same whatever you choose as the rest frame
> > > > > > > > > of the ether. So the actual rest frame of the ether cannot be detected
> > > > > > > > > within LET.
>
> > > > > > > > That's right. That's what Dono doesn't get.
>
> > > > > > > > > Its only a small hop, skip and jump from saying that "it cannot be
> > > > > > > > > detected" to "it doesn't exist".
>
> > > > > > > > Or at least 'it doesn't matter'.
>
> > > > > > > > Once you go beyond just the aether frame, and relating frames directly to
> > > > > > > > it, LET becomes more of a hinderance than a help
>
> > > > > > > > LET tells you (for instance) that even though objects at rest in frame A may
> > > > > > > > be more length compressed and time slowed than those in frame B (where A
> > > > > > > > moves faster in the aether frame than B) .. yet A will see objects at rest
> > > > > > > > in B as being more contracted and time dilated than its own.. Which really
> > > > > > > > confuses those who use the simple 'motion in the aether shrinks and slows
> > > > > > > > things' idea of LET as a way to 'understand' into a spin. You end up with a
> > > > > > > > strange combination of real compression and apparent contraction, real
> > > > > > > > slowing and apparent time dilaton. Its not really helpful :):)
>
> > > > > > > It is helpful in that it gets 'us' closer to understanding what occurs
> > > > > > > to objects as they move with respect to the aether.
>
> > > > > > > The issue with LET is everything is relative.
>
> > > > > > > For example, "the state of the [ether] is at every place determined by
> > > > > > > connections with the matter and the state of the ether in neighbouring
> > > > > > > places" - Albert Einstein.
>
> > > > > > You like Einstein quotes about the ether so try this one:
>
> > > > > >http://www-groups.dcs.st-and.ac.uk/~history/Extras/Einstein_ether.html
>
> > > > > > "We may assume the existence of an ether; only we must give up
> > > > > > ascribing a definite state of motion to it" - Albert Einstein.
>
> > > > > "If the existence of such floats for tracking the motion of the
> > > > > particles of a fluid were a fundamental impossibility in physics - if,
> > > > > in fact nothing else whatever were observable than the shape of the
> > > > > space occupied by the water as it varies in time, we should have no
> > > > > ground for the assumption that water consists of movable particles.
> > > > > But all the same we could characterise it as a medium."
>
> > > > > "[extended physical objects to which the idea of motion cannot be
> > > > > applied] may not be thought of as consisting of particles which allow
> > > > > themselves to be separately tracked through time."
>
> > > > > "The special theory of relativity forbids us to assume the ether to
> > > > > consist of particles observable through time, but the hypothesis of
> > > > > ether in itself is not in conflict with the special theory of
> > > > > relativity. Only we must be on our guard against ascribing a state of
> > > > > motion to the ether."
>
> > > > > "But this ether may not be thought of as endowed with the quality
> > > > > characteristic of ponderable media, as consisting of parts which may
> > > > > be tracked through time. The idea of motion may not be applied to it."
>
> > > > > Once you are willing to understand how Einstein defined motion, as
> > > > > particles which can be separately tracked through time, maybe you can
> > > > > advance from your statement.
>
> > > > > p.s. You still haven't answered how it is the train is length
> > > > > contacted because it is moving relative to the aether and the
> > > > > embankment is more at rest with respect to the embankment but at the
> > > > > same time LET has everything being relative. The answer is both the
> > > > > Observer at M and the Observer at M' will determine the train to be
> > > > > length contracted and for the clocks on the train to be ticking slower
> > > > > than the clocks on the embankment.
>
> > > > > > > This means the aether is more at rest with
> > > > > > > respect to the embankment than it is with respect to the train. The
> > > > > > > train is moving relative to the aether so it will be length contracted
> > > > > > > while the embankment will not. The ruler the Observer on the
> > > > > > > embankment uses to measure the length of the train is not length
> > > > > > > contracted. The ruler the Observer on the train uses to measure the
> > > > > > > length of the embankment is length contracted. The Observer on the
> > > > > > > embankment and the Observer on the train conclude the embankment is
> > > > > > > longer than the train.
>
> > > > > > > The same holds true for the clocks on the train and on the embankment.
> > > > > > > Since the train is moving relative to the aether while the embankment
> > > > > > > is more at rest with respect to the aether there will be a greater
> > > > > > > pressure associated with the aether on the clock on the train causing
> > > > > > > it to tick slower. If the Observers on the embankment and on the train
> > > > > > > where able to 'see' each others clocks as the M and M' pass each other
> > > > > > > both the Observer on the train and the Observer on the embankment
> > > > > > > would conclude the clock on the train ticks slower than the clock on
> > > > > > > the embankment.
>
> > > > You know for a while you were making progress. (I'm sure some here
> > > > are thinking the same about me ;) You managed to get away from each
> > > > frame having its own ether to having them share a single ether (for EM
> > > > waves anyway). Now if you could just get away from trying to attach
> > > > one of the frames to the ether...
>
> > > > Did you ever get anywhere with that diagram I made to explain RoS to
> > > > you. Einstein presented the train experiment from the point of view
> > > > of the tracks, but he never said that the tracks were at rest WRT the
> > > > ether.
>
> > > > - Hide quoted text -
>
> > > > > > > - Show quoted text -
>
> > > > > > According to both SR and LET there is no experiment that can reveal
> > > > > > which frame is at rest WRT the ether, so there is no way to know which
> > > > > > frame is more at rest WRT the ether.
>
> > > > > The clock which ticks the fastest is most at rest with respect to the
> > > > > aether.
>
> > > > But you have no way of knowing which clock is ticking faster. To
> > > > measure the tick rate of a moving clock requires more than one clock
> > > > at rest. And then you end up making assumptions to sychronize them.
> > > > Those assumptions affect your measurements.
>
> > > The two clocks are synchronized at some point in time. Then the clock
> > > at M and the clock at M' travel past one another. The Observer on the
> > > train and the Observer on the embankment have enough time to determine
> > > which clock is ticking faster. The clock which is ticking faster when
> > > M and M pass each other is the clock most at rest with respect to the
> > > aether.- Hide quoted text -
>
> > > - Show quoted text -
>
> > You need at least one more clock to measure a tic rate. Given clock
> > B, you compare the time on clock M to that on clock M' when they
> > pass. You cannot compare them a second time because M' is moving. So
> > you compare M' to B when they pass. With that comparison you can
> > decide whether the clock at M' has gained or lost time, but that
> > calculation assumes the clocks at M and B read the same. And
> > assumptions were required when those clocks were synchronized.
>
> Why can't you measure the clocks at M and M' a second time?- Hide quoted text -
>
> - Show quoted text -

M'----->
M

M'----->
M

If M' just went by M when do you think they are going to be facing
each other a second time?