John Berry wrote:
>
>
> On 3/4/07, *Stephen A. Lawrence* <[EMAIL PROTECTED]
> <mailto:[EMAIL PROTECTED]>> wrote:
>
> I will let you have the last shot; I won't be replying on this
> topic in this mailing list after this message.
>
>
> Fine with me, but you'd better read what I wrote as it took too long
> to type to be ignored.
Sigh ... You're being so polite after I've been so condescending, how
can I ignore it...
And besides the "paradoxes" are such fun... so here I am typing a
reply when I should be going to bed, and when I said I wasn't going
to.
>
> John Berry wrote:
> > On 3/3/07, *Stephen A. Lawrence* < [EMAIL PROTECTED]
> <mailto:[EMAIL PROTECTED]>
> > <mailto:[EMAIL PROTECTED] <mailto:[EMAIL PROTECTED]>>> wrote:
> >
> >
> >
> > John Berry wrote:
> >
> > > It is the only possible model as SR is illogical
> >
> > Well, that sure shoots down SR.
> >
> >
> > SR has many logical inconsistencies
> <snip>
> To learn relativity is to understand it, and if you did that,
> then you would see that it's not internally inconsistent.
> However, that takes a lot more effort than just calling it
> "illogical".
>
>
> I spent years learning it, believed it, looked for holes and failed
> to find any. Then when I saw them I was shocked at how obvious they
> were.
>
> Here is a simple one, it's called the twin paradox for a reason. If
> you have 2 twins and one stays on earth and the other one travels to
> a distant star and returns then SR states that the traveling twin
> still has youthful good looks while the other has long been pushing
> up daises after dying of old age.
Set C to 1, it simplifies things. Set the metric to diag(1,-1,-1,-1),
'cause it makes time-based intervals easier, but from now on we'll
drop two of the space dimensions.
If the distance is x and the velocity is v then arrival time in the
stay-at-home's frame is x/v. Starting coordinates are (0,0), arrival
coordinates are (x/v,x) and proper distance is sqrt((x/v)^2 - x^2),
or:
x*sqrt(1/v^2 - 1)
That's the age at arrival of the traveler.
Age at arrival of the stay-at-home twin, in the stay at home twin's
frame, is x/v, and the ratio of the traveler's age to the
stay-at-home's age is v*sqrt(1/v^2 - 1), or:
sqrt(1 - v^2)
As long as 0<v<1 (remember, we set c=1) that will be less than 1 and
greater than 0.
If the traveler comes home again the distances and times all double
but the ratio of ages doesn't change.
OK so far? (Note that we didn't need gamma for anything here -- I
just used the metric to find the proper distances.)
> Now what if these twins had some form of instantaneous communication
> between them, then we could easily measure the different rates of
> time each twin experiences, we could even find the stationary
> reference frame.
Instantaneous communication instantly leads to contradictions; see for
instance
http://www.physicsinsights.org/ccentipede.html
If you introduce FTL communication you also must introduce a preferred
reference frame. Instantaneous commmunication can be allowed in only
that frame. (I think this is equivalent to embedding the manifold in
a higher space and allowing certain shortcuts through the higher space
but I haven't worked out the details.) Unfortunately in GR there is
no globally flat frame, you can't generally even sensibly ask what's
happening far away "right now" let alone answer it, and reconciling
instantaneous communication with the theory would be stornrey hard, if
not impossible.
> Of course SR says that you can't have instantaneous communication
> and relies on the doppler effect which will effect any light based
> communication attempts, but doppler time shift is not time dilation
> but a separate effect, you could very well calculate the doppler
> effect and reconstruct the real rate of passage of time the other
> twin is experiencing.
I'm not sure I see what you're getting at here.
Of course you can measure the rate at which time appears to be passing
for a distant target, within suitable limits. Look at a clock with a
telescope and see how fast it's apparently ticking. Figure out how
fast he's going. Apply appropriate compensation factors. (Of course
that shows you what was going on some time in the past, due to travel
time delay in the light getting to you.)
In a gedanken experiment, you're also allowed to imagine space is
filled with observers all co-moving with your astronaut, who take
copious notes on what they see; they can put all their notes together
later on to produce measurements "after the fact".
> SR then says 'no no no, it's not the velocity difference it is the
> acceleration one twin faces that makes the difference'.
That is what breaks the symmetry. Acceleration pushes you off a
geodesic path, and in relativity geodesics are the paths of longest
(NOT shortest!) proper length. Non-geodesic motion produces a path
between two events of shorter proper length than a geodesic connecting
the events.
> And yes the twin who stays home may easily go under more
> acceleration by being on s spinning body orbiting a star, driving
> everywhere.
Then you have two twins both undergoing acceleration, and you need to
know all the details to determine who ages more. Neither followed a
geodesic so there's no simple yes/no test, like "the one on the
geodesic will be older".
You also can't introduce orbits around bodies without stepping outside
the realm of applicability of SR -- it doesn't handle gravity. You
can use rocket engines to produce big accelerations, though, if you
need them in the gedanken.
> Also the thing about thought experiments is they aren't limited to
> what is comfortable or practical only what is technically possible
> in order to exaggerate something to make a point. So the traveling
> twin may accelerate based on either twins clocks to the final
> velocity (let's say .99c) in a mere fraction of a second.
That's correct, of course. Intuitively, it shouldn't surprise you
that any effect due to acceleration might very well tend to go as
accleration*duration; in order to halve the duration of the
acceleration you must double its intensity.
> Furthermore there are a number of ways to have instantaneous
> communication. (or near instantaneous communication that has no
> Doppler time shift)
Not in SR. If you disagree you need to be very precise about what you
mean.
> One way is to have two parallel almost infinatly long trains in
> space, they start of stationary with the twins in opposite
> carriages, but first a few details. Each cabin has a clock (which
> as designed to be easily read by those of the other train even at
> relativistic speeds), it is generally accepted that there are many
> ways by which a number of observes at a distance in the same
> reference frame may synchronize their watches Also each cabin how
> it's own propulsion unit and again it is able to reach near
> lightspeeds in under a second by any observers watch.
>
> Now one of the twins accelerates, each can keep an eye on the rate
> of time the other train is observing, if they each sees the other as
> experiencing time slower then themselves then when the trains are
> stopped each will have different expectations, they simply can't
> match.
You have big, big problems in this scenario, which you have not fully
worked out. Work out all the details and the timekeeping problems go
away, but the details are a mess.
First of all a long object cannot accelerate simultaneously along its
length, because as soon as it starts to accelerate its parts
(stretched out along its length) no longer share a single inertial
frame. Furthermore, the whole thing must shrink due to Fitzgerald
contraction, which means that either different parts accelerate at
different rates or the thing breaks up into pieces (this is related to
Bell's paradox, which has two accelerating space ships, one in front
of the other, connected by a long string -- if they both acclerate at
the same rate, eventually the string must break). The longer it is, the
less it can accelerate without breaking up; if it accelerates at 1g,
for instance, and it's more than 1 light year long, the back end will
break off no matter how hard it tries to keep up: there's what's
called a Rindler horizon about 1 ly behind someone accelerating at 1g.
In general, if a very long train tries to start "simultaneously" along
its length it will break up as all the cars shrink -- but that's just
what the people on the platform see. The people on the train see
something entirely different. In that FOR the train doesn't shrink,
of course. Instead, they see that as soon as it starts to move the
clocks go out of sync, and as far as anyone on the train can
determine, the locomotive started up before the caboose and hence
broke the train apart.
If you just want to find the durations each party experiences, just
find the proper distances as I did, above, for the twins. Proper
distance is a Lorentz invariant, so you'll get the same answer no
matter what frame you work the problem in. If you want to determine
exactly what everybody actually sees at each moment you have a lot
more work to do.
> There is another way however, you can have one twin stay of earth
> and the other twin orbit the earth at near light speed. You see the
> twin paradox always assume the twins are moving away from each other
No it doesn't. In fact the original "paradox" had one fly away and
then come back, so half the time they were moving toward each other.
> but in the case of orbit where they can constantly communicate or
> for that matter merely flying by where they get a moment to observe
> the rate of time the other experience and communicate without
> Doppler distortion.
The distant orbiting astronaut situation is very cool. See, for
instance:
http://www.physicsinsights.org/revolving_astronaut.html
Dig that clock going backwards.
Note that there's a Rindler horizon in there somewhere which keeps
anyone from ever actually _seeing_ a clock go backwards.
Counter-orbiting twins are entertaining too, and tricky to work out:
http://www.physicsinsights.org/revolving_twins.html
The simple question of what you "see" when you accelerate is pretty
nifty, too. This came up recently on this list; I mentioned then that
I had done up a rather elementary page on it, here:
http://www.physicsinsights.org/porthole_view_1.html
Sorry for the rerun.
> There are yet more problems.
I haven't seen any real problems in what you've described so far.
Surprising results, yes. Counterintuitive, yes. Contradictory? No.
> Let's say we now have 3 parallel close trains with open beds, we'd
> better put them on earth so no one suffocates.
>
> We will have 2 flash bulbs of each train, each a set distance apart
> and an observees on each train positioned in the middle of the 2
> flash bulbs, if the bulbs go off at the same time the short sharp
> photon pulse reach the observes and he sees a single bright flash.
This is one common starting point for deriving the Lorentz
transforms.
> Now let the middle train not move, let the bulbs flash at 12:00 and
> at 12:01 (it's slow light ok ;)
You mean the flash is at 12:00, the arrival is at 12:01, I think.
Right?
> the observer on that train see a bright flash from each bulb
> simultaneously. However at 12:00 as the bulbs go off the train to
> the right was moving down the track, at that exact moment the
> observer on the right train passes the observer on the middle
> stationary train.
>
> The observer on the right train expects to see the bulbs flashes
> simultaneously because he was in the middle when they went off
NO HE WAS NOT!
Sorry for shouting.
You have neglected relativity of simultaneity. In the moving
observer's FOR, the bulbs did _NOT_ flash at the same time, and he was
_NOT_ in the middle "when they flashed".
The term "when they flashed" is FRAME RELATIVE.
> (and if there were a bunch of censors along the right train they
> would demand to see the pulses from each bulb advance from detector
> detector and hence must meet in the middle).
So what do they see? Don't just wave your hands, work it out.
Use the Lorentz transforms, that's what they're there for. Don't try
to "intuit" this one, it won't work.
> Furthermore in case you are unsure
I'm not. Been there, done that, this is a standard example, almost
identical to Einstein's two lightning flashes beside the train
tracks. He does that in "Relativity: The special and general
theories" IIRC.
> you could have (different
> colour?) bulbs on the moving right train that go off simultaneously
> and right next to the bulbs on the middle train, obviously the
> observer on the right train would insist on seeing the bulbs on it's
> own train similtaniously.
OK, let's work the darn thing out (just one bulb color, sorry). I'll
pick some distances, and I'm just going to do it in one moving train
with one set of stationary observers.
C=1. We'll have the bulbs go off at +/- 1 on the X axis, at time 0 in
the stationary frame.
Coordinate (0,0) matches in the two frames, and that's when the bulbs
go off in the stationary frame.
So far so good?
Train is moving in the +x direction at velocity "v". What time is it,
according to riders on the train (who all have synchronized watches),
when the flashes go off? For this I'll need gamma, g=1/sqrt(1-v^2).
If the train is the prime frame, the track is unprimed, then the
transforms from the track to the train are
t' = g*(t - x*v)
x' = g*(x - v*t)
In our particular case the events are at x=+/-1, time t=0, so we have
t' = +/- g*v
x' = +/- g*x
The flashes are equidistant from the origin, **BUT** they didn't take
place at time 0, **AND** they didn't take place at the same time. The
time lag between the flashes, according to the clocks on the train, is
2*g*v.
Again, simultaneity is frame-relative in SR.
> Now here is the really tricky part, if that didn't convince you SR
> is flawed
It convinced me that you don't understand relativity of simultaneity,
and you don't understand relativistic clock skew.
> as it require the same pulses to pass each other multiple
> times, in different places depending on the reference frame.
No it doesn't. You only think that because you didn't realize that
the flashes are only simultaneous in ONE of the reference frames. In
the frame where they're not simultaneous the moving observer who was
in the middle when the flashes went off (according to the stationary
observers) doesn't see them in the middle, _and_ doesn't see them
simultaneously, _and_ isn't surprised by this. The flashes actually
meet each other at the same event in all frames, but that event is at
different times and different x coordinates in all frames.
The rest of this is all more of the same. You did not work anything
out, you just waved your hands, and you neglected clock skew. Work it
out using the Lorentz transforms. Then if you get a contradiction, it
might be interesting -- but you won't, if you do it right, because the
theory is consistent.
>
> Now in the above example where the right train sees the impulses
> from the stationary frames bulbs simultaneousness, as does the
> observer on the stationary frame (but not a second observer on the
> stationary frame/train where the observer on the right train does
> experience the pulses simultaneously), indeed we could have the left
> train we have ignore until now also moving.
>
> Now what happens when just as the observer on the left train detects
> the simultaneous pulses the train stops (or the observer jumps) and
> finds them right next to a second observer on the stationary train,
> this observer would have recieved the pulse from the closer bulb but
> not yet the more distant one.
>
> So now we have to observers in the same place and frame, one who has
> seen 2 pulses one who has seen just one, so when the always
> stationary observer sees the second more distant bulbs pulse what of
> the previously moving observer does he see this pulse a second time?
>
> BTW I am well aware of length contraction also however it doesn't
> effect any of the above experiments, it might require there to be
> different sets of bulbs or trains to be replaced by individual
> vehicles.
>
> Ok, so you got me to reply.
That wasn't actually the intent :-)
