Gravity slows light, why not the clocks inside 1971 plane experiment?
gu...@hotmail.com
I believe Gravity slows light and other moving objects, *****and by a
*PERCENTAGE* factor of the object's velocity?
I'm ***NOT*** talking about "gravity time dilation" which is based only
on the distance(height) of the object from the center of the Earth....
INSTEAD I'm talking about the SAME HEIGHT (thus the same gravity time
dilation(t = to (1+gr/c^2)) but with two SAME objects and each
traveling at a different velocity.
So a car moving at 100m/s will slow down by perhaps 1m/s where as a car
moving at 1000m/s will slow down by 10m/s (perhaps same percentage but
different amount 1m/s vs 10m/s)...and logically perhaps likewise for a
ticking cesium clock?
So for the above reason wouldn't the cesium clock inside the 1971 plane
experiment slow down more than the clock on Earth since the plane's
clock is relatively moving faster (in both directions) than the clock
on Earth?
(More of a Kinetic time dilation due to gravity's effect on different
velocities versus gravitational time dilation due to only height from
center of the Earth)
1971 Plane experiment at:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html
Dirk Van de moortel
<gu...@hotmail.com> wrote in message news:1110158868.2...@o13g2000cwo.googlegroups.com...
This should give a reply to your previous post
"Why isn't Earth's surface clock slower then one at Earth's Centre?"
as well.
(1) A few days ago I explained about mutual time dilation
of two inertial clocks, where (so-to speak) A finds B's
clock to slow down, and B finds A's clock to slow
down.
See message on the thread "Is the gamma reversed?"
http://groups-beta.google.com/group/sci.physics.relativity/msg/176223f0819d4d46
This is only valid when both clocks are inertial.
(2) This is not the case in this experiment where A is on
the ground of the spinning Earth and B is inside an airplane.
Here we have two clocks A and B that are not inertial:
they follow a circular path arounf the center of the Earth.
The only clock that is approximately inertial here, is an
imaginary clock C at the center of the earth, but *not*
turning around its axis together with the Earth.
It is of course not *really* inertial because it revolves
around the center of mass of Earth and Moon, and they
both revolve around the center of mass of the Earth,
Moon and Sun, etc...
But we can treat it as inertial to a very good approximation.
Furthermore, at this stage we also ignore the so-called
gravitational time dilation... That is a correction that can
be made later. By the way, this way of treating the problem
is valid only in a weak gravitational field. It would not
work when dealing with a person and a plane comparing
their clocks on a neutron star.
So this part is about so-called kinetic time dilation only,
in absence of a gravitational field. Just imagine that the
earth isn't even there!
In this case A and C will *not* find that "the other one
is running slow". The inertial clock at the center C will
indeed find A's clock to run slow, but this time the A
clock will not find the center clock to run slow as well.
On the contrary: A will find the C clock to run *fast*.
(Always be careful with the language).
So they both agree that the A clock runs slower than
the C-clock.
Likewise, the B-clock will run slower than the C clock.
For a very nice derivation of this, look at
http://www.physicsinsights.org/revolving_clock.html
(3) Now, if you finally want to compare the non-inertial
A-clock with the non-inertial B-clock, you can in principle
make the calculation, but it is *very* for the simple reason
that the frames are not inertial, and the relative speeds in
the gamma factor are not constant.
The easiest way to find the rate, is by comparing both
rates to the inertial clock C that sees both clocks A and B
to move with a constant speed. This calculation turns
out to be very *easy*. It is in fact trivial. The rate between
A and B is found by simply dividing the gamma rates of
A w.r.t. C by the gamma rate of B w.r.t. C.
Exactly the same reasoning goes for the GPS.
With all this in mind, have another look at the article:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html
hth.
Dirk Vdm
Sue...
Dirk Van de moortel
"Sue..." <suzyse...@yahoo.com.au> wrote in message news:1110194044....@o13g2000cwo.googlegroups.com...
And this one
http://groups-beta.google.com/group/sci.physics.relativity/msg/66a14ac909e2f7ac
Dirk Vdm
Sue...
At the very least the H&K experiment should have
included a light-clock outside the aircraft as a
control.
Even classical physic predicts that will slow with motion
through the dielectric of air.
Sigh... ignorance is bliss.
Sue...
Sue...
It's a good thing you math is better than your stalking.
Did you remember to send Dennis a valentine ?
Sue...
Dirk Van de moortel
"Sue..." <suzyse...@yahoo.com.au> wrote in message news:1110194741.1...@z14g2000cwz.googlegroups.com...
And don't forget these ones either:
http://www.dipmat.unipg.it/~bartocci/listafis.htm
http://www.dipmat.unipg.it/~bartocci/poll.html
http://mywebpages.comcast.net/Deneb/Steps.htm
Dirk Vdm
Sue...
You and FrediFizx need to combine your efforts.
Surely there is some commonality between your
work with message headers and Fredi's work
with magnetic monopoles.
You guys are probably just a sign or an exponent
from some great discovery... like the solution to
the equations for an inside-out wormhole.
How exciting! I'll alert the King of Sweeden
to set two extra places at the table.
Sue...
Dirk Van de moortel
"Sue..." <suzyse...@yahoo.com.au> wrote in message news:1110197573....@g14g2000cwa.googlegroups.com...
So, taken together, don't forget:
http://groups-beta.google.com/group/sci.physics.relativity/msg/66a14ac909e2f7ac
And don't forget:
Sue...
No way!
I only click on .edu URL's per your advise.
Sue...
whop...@csd.uwm.edu
> I believe Gravity slows light and other moving objects, *****and by a
> *PERCENTAGE* factor of the object's velocity?
There's no need for guess work. The amount of time that's elapsed for
a route can be determined from the metric, itself (assuming a
Schwarzschild exterior):
t = integral dT
where
dT^2 = dt^2 (1 - 2GM/(r c^2)) - dr^2/c^2 (1 - 2GM/(r c^2))^{-1}
- r^2 (d(theta)^2 - (sin(theta) d(phi))^2)
in terms of sphiercal coordinates centered on the earth's center with
the angular coordinates fixed with respect to the stars (i.e., theta =
0 at north pole, 90 degrees at equator, 180 degrees at south pole; but
phi does NOT equal longitude, since the Earth is rotating; instead it's
the longtitude, fixed with respect to the stars).
r is the distance from the Earth's center, which is basically fixed,
since the flight's not going out into space, so 2GM/(r c^2) = 2gr/c^2,
where g is the acceleration of gravity near the Earth's surface. t is
the time elapsed with respect to a clock rigidly connected to the fixed
frame (i.e., not a clock moving with the Earth's rotation); and may.
International standards, I believe, are defined relative to a
stationary frame, instead of with the rotating earth. So, this is
probably the same as Greenwich, plus or minus a time-0 setting.
The integral is a path integral taken with respect to the flight path:
r = r(t) ~~ constant;
theta = theta(t), phi = phi(t)
as t ranges over the time of the flight.
You will get DIFFERENT answers for flights going East, versus flights
going West, since the Earth is rotating. It, therefore, also depends
on your latitude -- as is already obvious by the formula above. The
ones going East faster than you are on the group, the ones going West
are STILL going East, but at a slower speed -- until they reach 1000
MPH (if on the equator), at which point, they're stationary.
gu...@hotmail.com
whop...@csd.uwm.edu wrote:
> gu...@hotmail.com wrote:
> > I believe Gravity slows light and other moving objects, *****and by
a
> > *PERCENTAGE* factor of the object's velocity?
>
> There's no need for guess work. The amount of time that's elapsed
for
> a route can be determined from the metric, itself (assuming a
> Schwarzschild exterior):
> t = integral dT
> where
> dT^2 = dt^2 (1 - 2GM/(r c^2)) - dr^2/c^2 (1 - 2GM/(r
c^2))^{-1}
> - r^2 (d(theta)^2 - (sin(theta) d(phi))^2)
The equation is to calculate how time slows down both kineticaly and
gravitationaly.
But what I wrote was about the VELOCITY of light slowing down due to
gravity (not time slowing down due to gravity).
Therefore the answer you gave I think is irrelevant to the question I
posed which is if gravity slows down the VELOCITY of light:
1. Couldn't it then also slow down the VELOCITY of a clock?
2. And obviously it does not slow down light by the same amount as the
clock therefore couldn't a faster moving clock slow down more than a
faster moving clock (clock's x axis VELOCITY = +/- 10m/s but inside a
plane moving at 1000m/s the clock's VELOCITY = 1010 & 990 m/s)?
gu...@hotmail.com
Ok C will find A's clock to run slow (which makes sense to me)
but the kinetic Time dilation calculations section of the 1971 Plane
experiment at:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html
SAYS THE OPPOSITE OF WHAT YOU WROTE (It says: A will find C's clock to
run slow)??
Quote: "Taking a "proper time" at the earth's center as if the master
clock were there, the time measured by a clock on the surface would be
****LARGER****"
and "Ts = To gamma" (Where To = clock at Earth's center and gamma's v =
angular velocity of Earth's surface)
Since gamma >= 1 therefore they say Ts > To thus To (C) will be slower
than Ts (A)?
Dirk Van de moortel
<gu...@hotmail.com> wrote in message news:1110321583.7...@g14g2000cwa.googlegroups.com...
Oh, yes. I obviously made a mistake, although I explicitly
added the warning "(Always be careful with the language)".
I should have said:
In this case A and C will *not* find that "the other one
is running slow". The clock at the surface A will indeed
find C's clock to run slow, but this time the C-clock
will not find the surface clock to run slow as well.
On the contrary: C will find the A clock to run *fast*.
(Always be careful with the language!!!).
So they both agree that the A clock runs faster than
the C-clock.
Likewise, the B-clock will run faster than the C clock.
The essential part is the lack of symmetry between the
frames. One is inertial, the other is not.
Tom Roberts
> So for the above reason wouldn't the cesium clock inside the 1971 plane
> experiment slow down more than the clock on Earth since the plane's
> clock is relatively moving faster (in both directions) than the clock
> on Earth?
In GR, one simply integrates the metric over the path of the clock to
determine its elapsed proper time. Do this for the clocks in the
airplanes, and for the clocks at USNO to which they were compared.
In the weak-field approximation suitable here on earth, when computed in
the ECI frame, that splits into two terms: a function of speed relative
to the ECI with a minus sign, and a function of altitude with a plus
sign. So at a given altitude, higher speed leads to less elapsed proper
time; for a given speed, higher altitude leads to more elapsed proper time.
For the Haefle and Keating experiment, both terms are important.
Tom Roberts tjro...@lucent.com
Tom Roberts
> But what I wrote was about the VELOCITY of light slowing down due to
> gravity (not time slowing down due to gravity).
If you put a speed-of-light measuring apparatus in an airplane, and
measured the speed of light for various altitudes and airspeeds, the
measurements would always obtain the value c (to within current
accuarcies of a few parts per billion),. This is independent of altitude
and airspeed.
Tom Roberts tjro...@lucent.com
gu...@hotmail.com
Ok but I do not understand the logic behind it since C is at rest and
inertial and A is spinning about it. So why is spinnin A's clock faster
where as in a cyclotron the spinning particle's clock(tau) runs slower?
gu...@hotmail.com
"independent of altitude" are you saying that the force of gravity does
not slow down light because I believe I read somewhere that it did??
> and airspeed.
>
gu...@hotmail.com
Tom Roberts wrote:
> gu...@hotmail.com wrote:
> > So for the above reason wouldn't the cesium clock inside the 1971
plane
> > experiment slow down more than the clock on Earth since the plane's
> > clock is relatively moving faster (in both directions) than the
clock
> > on Earth?
>
> In GR, one simply integrates the metric over the path of the clock to
> determine its elapsed proper time.
Yes but my question is not about GR in terms of TIME but rather
**VELOCITY**.
So my question is not how velocity affects time but rather isn't it
true that the *VELOCITY* of light slows down due to gravity and why?
Dirk Van de moortel
<gu...@hotmail.com> wrote in message news:1110487564.9...@z14g2000cwz.googlegroups.com...
Well, never mind. I was kind of waiting for you to protest
because I'm not at all happy with the terminology to begin
with. If someone tells me that this still is all wrong, I will
agree. I really don't even try to get this kind of language
right. Talking about clocks running fast or slow, is only
really safe when the clocks have been reunited or, if not
reunited, are at rest w.r.t. each other in the same inertial
frame. So I gladly distantiate myself from the two things
I wrote above.
The right way to express what happens is this: suppose
the circularly moving clock has 1 second (tau) between
two of its ticks. If the central observer would then measure
the time (T) on his own non-revolving clock between the
two revolving ticks, he will find more than a second: T > tau.
T = tau / sqrt(1-v^2)
Also, when the central clock has 1 second (T) between
two of its ticks, then if the revolving observer would measure
the time (tau) on his own clock between the central ticks, he
will find less than a second between them: tau < T
tau = T sqrt(1-v^2)
This says something about what will be measured, and it
is safe. Have a long and close look at Taylor and Wheeler's
http://www.eftaylor.com/pub/projecta.pdf
and at Sal's
http://www.physicsinsights.org/revolving_clock.html
but be careful with the usage of going slower and faster
in the introduction.
Try not to think in terms of slower or faster. Whenever
you hear or read it, shut your mind for it. As you can
see, it is utterly confusing.
Dirk Vdm
Tom Roberts
> Tom Roberts wrote:
>>If you put a speed-of-light measuring apparatus in an airplane, and
>>measured the speed of light for various altitudes and airspeeds, the
>>measurements would always obtain the value c (to within current
>>accuarcies of a few parts per billion),. This is independent of
>
> "independent of altitude" are you saying that the force of gravity does
> not slow down light because I believe I read somewhere that it did??
I am saying precisely what I said above. Specifically: for a measurement of the
local speed of light, as described above, the "force of gravity" does not affect
the measured speed, nor does the altitude or airspeed of the airplane. "reading
somewhere" that the "force of gravity" does affect the "speed of light" is so
loose and poorly stated that either your source was confused or your memory is
omitting important details.
There most definitely is an effect of gravitational potential (not force) on the
frequency and wavelength (not speed) of light rays. This gravitational redshift
is well known and extremely well established.
If you want to use locally-non-inertial coordinates you can turn
changes in frequency and/or wavelength into changes in speed. The
local speed of light is always c in locally-inertial coordinates,
but for other coordinate systems anything goes....
Tom ROberts tjro...@lucent.com
gu...@hotmail.com
Tom Roberts wrote:
> gu...@hotmail.com wrote:
> > Tom Roberts wrote:
> >>If you put a speed-of-light measuring apparatus in an airplane, and
> >>measured the speed of light for various altitudes and airspeeds,
the
> >>measurements would always obtain the value c (to within current
> >>accuarcies of a few parts per billion),. This is independent of
> >>altitude and airspeed.
> >
> > "independent of altitude" are you saying that the force of gravity
does
> > not slow down light because I believe I read somewhere that it
did??
>
> I am saying precisely what I said above. Specifically: for a
measurement of the
> local speed of light, as described above, the "force of gravity" does
not affect
of gravitational potential (not force) on the > frequency and
wavelength (not speed) of light rays. This gravitational redshift > is
well known and extremely well established.
(Web links at the bottom)
That's what I believe but it's backed by Harvard science department and
others recent research... which they say light slows down in any medium
including gravity...(which is why they now always specify light
traveling in a space vaccum).
They also say the gravitational force of black holes can also SLOW down
light and not only force light to re-direct towards it's center.
QUOTE:"Physicists may soon create artificial black holes in the
laboratory, analogous to the ones expected to lurk in distant
space."...."Tiny tornadoes within wispy clouds of gas, they would snag
photons through their remarkable ability to SLOW light pulses."
I personally believe their research maybe bogus??? For light slowing
down in a medium perhaps these mediums are not only
refractive/transparent but also behave as opaque mediums (meaning some
atoms of these refractive mediums perhaps?absorb/release (= perhaps
timegap???) photons/energy thus making it seem that light coming out of
the refractive medium is slow down?
Harvard claims that they can slow down light ...I did not read their
Full Research papers but the first thing I think they should always
combine with such claims is to prove that light passing through their
gas is not simply being absorbed for a specific time period and then
emitted(released) (which is indeed what they seem to be saying) thus
never really slowed down: meaning it always traveling at c unless it's
temporarily absorbed into a gazeous atom's atomic spin. So it's never
really slowed down but only stored temporarily and then travels at c.
Actually their papers are more inclined on the capability of future
light storage for computing tasks more than on the velocity of light
slowing down???
QUOTE:"That may change soon. Last year, Lene V. Hau, now of Harvard
University, and her colleagues used a stationary, laser-manipulated
atom cloud to limit light to an astoundingly sluggish 17 meters per
second-roughly bicycle speed"
Also Harvard adds a third element to their model experiment which is
another laser who's task is to uncondense the gas thus making the gas
transparent (vs photon absorbing opaque atoms) ....and since a laser is
also light (photons) perhaps it also affects their initial photon
emission test...but obviously it seems??? they're simply absorbing and
re-emitting the photons which is not the same as slowing down the
photons. Example a ball travels at 10m/s, is stopped by atoms for 3
seconds and then released at 10m/s does not mean the ball (photon) has
constistantly slowed down to 7m/s through it's trajectory in space.
http://www.sciencenews.org/articles/20000205/fob4.asp
http://www.phschool.com/science/science_news/articles/light_stands_still.html
-------------------------------------------------
Present belief is that gravity slows down clocks of SAME ALTITUDE by
the same amount REGARDLESS of the clock's(plane's) velocity where as
below perhaps the clocks higher velocity does also get affected more
due to gravity?
The below is not due to gravity's altitude effect on time but on
VELOCITY...therefore the TWO CLOCKS below are at the SAME altitude but
at different velocities only.
So if gravity can as they say above slow down the "VELOCITY" of light
(not just redshift) then also perhaps a faster moving clock (than
Earth's clock) will have it's velocity also slowed down by gravity and
by a larger amount than Earth's clock(Earth's sways at say +/- 10ms
where as clock in plane travels(sways/ticks) at say 1010 and 990 m/s
thus perhaps? affected by the same percentage as Earth's clock but by a
larger amount due to it's higher velocity in the plane).
Example: (Values exagerated to demonstrated difference) Earth because
of gravity effect on velocity (not time) Earth clock's VELOCITY slows
to +/-10m/s instead of +/-12m/s (exagerated) but on plane it slows to
20% of 1010 = 808 and 792 (instead of 1010 and 990m/s)....
thus perhaps (808-792)/2 = +/-8m/s instead of +/-10m/s... or sometype
of calculation like that??
The logic above is that logically light must be slowed down by a
percentage factor of it's velocity and not a constant rate to all
moving objects...exagerated example: velocity c is slowed by 1/10c due
to gravity but impossible for Earth clock (or slower propagating wave)
which ticks at +/- 10m/s to also be slowed down by 1/10c and thus
gravity must affect velocity by a percentage factor? (clock sways at
+10m/s - 1/10c due to gravity = impossible thus gravity should effect %
of clock's velocity?)
Tom Roberts
> Tom Roberts wrote:
>>I am saying precisely what I said above. Specifically: for a
>> measurement of the
>>local speed of light, as described above, the "force of gravity" does
>> not affect
>>the measured speed, (snip)......
>
> others recent research... which they say light slows down in any medium
> including gravity...(which is why they now always specify light
> traveling in a space vaccum).
You have to keep track of what is being discussed. I described a LOCAL
measurement inside an airplane, which is manifestly much smaller than
the earth (which is the scale of the gravitational field for this).
BTW most physicists do not consider gravity to be a "medium". Because
the gravitational field of the earth does not affect measurements of the
speed of light in laboratories on earth. Or airplanes (as above).
In GR the speed of light is c for any LOCAL measurement. Here "local"
means "on a scale very small compared to any variations in the
gravitational fields". For instance, the length of an airplane or
laboratory compared to the size of the earth.
> They also say the gravitational force of black holes can also SLOW down
> light and not only force light to re-direct towards it's center.
>
> QUOTE:"Physicists may soon create artificial black holes in the
> laboratory, analogous to the ones expected to lurk in distant
> space."...."Tiny tornadoes within wispy clouds of gas, they would snag
> photons through their remarkable ability to SLOW light pulses."
They are presumably discussing a "laboratory" measurement in which the
black hole is quite small, and the speed of light is measured over a
NON-LOCAL distance in the laboratory (for a path close to the miniature
black hole but much larger than its size, which is the relevant scale of
the gravitational field for this). This would be directly analogous to
the Shapiro time delay in which light signals earth->mars->earth are
measurably slowed when the path grazes the sun.
That is assuming they are really talking about a laboratory sized black
hole, which I _strongly_ doubt -- I believe the researchers are actually
discussing ACOUSTIC black holes, not real ones (which AFAIK nobody has a
clue how to construct in a laboratory, much less contain and control).
Such ACOUSTIC black holes would not have a measurable gravitational
effect on light pulses (but their index of refraction would affect
light, which is what I believe that quote is refering to). They would
also have a profound effect on SOUND pulses (including a horizon, which
is the whole point).
Bottom line: sound bites and popular science magazines are not enough to
understand physics.
> [...]
Tom Roberts tjro...@lucent.com
gu...@hotmail.com
I just noticed they interchanged/reversed T_earthcenter with
T_earthsurface for the very same equation:
QUOTE:"Now this relationship is just the reverse of the actual
experiment, since we have assumed that the clock is at the center of
the earth, whereas the actual clocks are in the frames which are moving
with respect to the center."
thus TA-TS= To (2RWv+v^2)/2c^2 was changed to TA-TS= -TS
(2RWv+v^2)/2c^2
thus TA = TS - TS(2RWv+v^2)/2c^2
where W is there Earth's angular surface velocity which I believe they
are permitting themselves to simply add small velocities together
(earth's velocity + plane's velocity instead of (u+v)/(1+uv/c^2))
But it still doesn't explain why in the 1971 experiment,the WEST
plane's clock ticks faster than Earth clock?
The reasoning that Earth's clock ticking faster than Earth's center
clock is probably irrelevant since they never meet to compare. Actually
it's very confusing about what they're using as inertial? they seem to
be interchanging between both?:
Not me but they gave the equation: T_surface = T_earthscenter gamma
thus from the equation the clock on the surface must tick faster than a
clock on the center.....BUT THAT'S OK since as you said neither clock
accelerates to meet the other and can be used as only a reference to
help determine the clock's of the planes which are the ones who
accelerate/deccelerate from the Earth's surface clock?
....ACTUALLY I think I know what they did....Because the Earth is
spinning even though it's angular velocity is constant...it's linear
velocity on the other hand is always ALSO accelerating/deccelerating
thus all three are w.r.t. Earth's center: Earth's surface and both
planes.
Therefore the ABOVE perhaps explains why all the velocities are w.r.t.
Earth's center BUT since they compare their clock readings from Earth's
surface ONLY instead of Earth's center therefore the Equation only (not
the velocities inside the equation) has been modified to w.r.t Earth's
surface instead of Earth's center?????
Which is perhaps why the plane going West has a clock ticking faster
than Earth's surface clock?
(So if Earth's surface was moving at a constant LINEAR velocity instead
of a constant ANGULAR velocity THEN ONLY they would not need Earth's
center as an inertial frame of reference perhaps since even if the
planes velocities where given w.r.t. Earth's center it would still be
impossible for any departing plane to have it's clock tick faster than
the platform from which it departed(accelerated))....you would simply
use w = (u+v)/(1+uv/c^2) to modify the velocities so that their values
become relative to the Earth's surface instead of Earth's center???)
That's the only logical reasoning I can think even after reading the
links you gave me?, which I also posted comments about what I believe I
read and understood below here, next to each given link.
>So I gladly distantiate myself from the two things
> I wrote above.
> The right way to express what happens is this: suppose
> the circularly moving clock has 1 second (tau) between
> two of its ticks. If the central observer would then measure
> the time (T) on his own non-revolving clock between the
> two revolving ticks, he will find more than a second: T > tau.
> T = tau / sqrt(1-v^2)
> Also, when the central clock has 1 second (T) between
> two of its ticks, then if the revolving observer would measure
> the time (tau) on his own clock between the central ticks, he
> will find less than a second between them: tau < T
> tau = T sqrt(1-v^2)
> This says something about what will be measured, and it
> is safe. Have a long and close look at Taylor and Wheeler's
> http://www.eftaylor.com/pub/projecta.pdf
I believe they say from what I read that the GPS's KINETIC clock is
always slower than Earth's. And that Schwarzschild metric simply
combines the gravitational and KINETIC time dilation together (and
possibly the curvature of motion since they are moving in circular
motion (v/r) and not linearly)?
They say altitude wise (gravitational time dilation) the GPS is 50,000
nanosec ticking FASTER at QUERY4 and that Kinetic time dilation the GPS
is ticking SLOWER by 11,000 than EARTH's surface clock (thus the total
time dilation difference together makes 39,000 nanosec
(50,000-11,000=39,000) at QUERY9?
QUOTE:"Therefore we expect the effect of motion (KINETIC TIME DILATION)
to reduce the amount by which the satellite clock runs fast
(GRAVITATIONAL/ALTITUDE TIME DILATION) compared to the Earth clock. In
brief, when velocity effects are taken into account, we
expect the satellite clock to run faster than the Earth clock by *LESS*
than the
estimated 50 000 nanoseconds per day."
-----------------------------------------------------------
But the above link explanation is not the same for some reason as the
plane's clocks? since in the 1971 experiment, the plane's clock going
WEST is actually ***KINETICLY*** TICKING FASTER (not slower like the
GPS) than Earth's surface clock??
> and at Sal's
> http://www.physicsinsights.org/revolving_clock.html
> but be careful with the usage of going slower and faster
> in the introduction.
> Try not to think in terms of slower or faster. Whenever
> you hear or read it, shut your mind for it. As you can
> see, it is utterly confusing.
>
> Dirk Vdm
>
> >
> >
> >
> > > Likewise, the B-clock will run faster than the C clock.
> > >
> > > The essential part is the lack of symmetry between the
> > > frames. One is inertial, the other is not.
> > >
> > >
> > > >
> > > >
> > > > > (Always be careful with the language).
> > > > > So they both agree that the A clock runs slower than
> > > > > the C-clock.
> > > > > Likewise, the B-clock will run slower than the C clock.
> > > > >
> > > > > For a very nice derivation of this, look at
> > > > > http://www.physicsinsights.org/revolving_clock.html
>From what I believe I read it says the same as the GPS link, quote:
1. Altitude/Gravitational Time dilation, clocks tick always *FASTER*
with height:
"The effects of the gravitational field on an orbiting clock are
straightforward and easy to understand, at least at this level: the
deeper in the well you go the slower you go."
2. Motion/Kinetic Time dilation, revolving clocks tick always *SLOWER*:
"But the issue considered on this page is just the effect of the
clock's motion, which causes the clock to go more slowly than a
stationary clock in the middle of the circle."
"The revolving clock is moving relative to the stationary clock, so it
goes more slowly."
---------------------------------------------------------------------------------------------
The link also speaks of the Momentarily Comoving Reference Frame (MCRF)
because the trajectories (velocity and distance) are curved and only
momentarily linear...but I couldn't figure out if any of that would
give logic as to why the WEST plane's clock is ticking faster than the
Surface clock?
---------------------------------------------------------------------------
Dirk Van de moortel
<gu...@hotmail.com> wrote in message news:1110757759.3...@f14g2000cwb.googlegroups.com...
>
> Dirk Van de moortel wrote:
> > <gu...@hotmail.com> wrote in message
> news:1110487564.9...@z14g2000cwz.googlegroups.com...
[snip]
Yes, since u and v are so small.
Note that the ground's velocity u is taken positive as seen in the
Earth centered inertial frame. This way the velocity v of the East
trip is also positve, while the velocity of the West trip is taken
negative.
>
> But it still doesn't explain why in the 1971 experiment,the WEST
> plane's clock ticks faster than Earth clock?
>
> The reasoning that Earth's clock ticking faster than Earth's center
> clock is probably irrelevant since they never meet to compare. Actually
> it's very confusing about what they're using as inertial? they seem to
> be interchanging between both?:
[snip]
The article does not do a good job at explaining, so I will cut it
short right here. In stead, let's have a look at cleaner calculation
that gives the same results.
(Using units where c=1 and G=1)
Normally you use the Schwarzschild time rate
Schw(va,ra,vb,rb) = sqrt(1-2 M/ra-va^2) / sqrt(1-2 M/rs-vb^2)
to make the calculations (although even this is not exact
- see section 5 in reference projectA).
In the case of Hafele-Keating (and GPS) where M, va and
vb are very small, one can use a combination of two effectively
independent approximations.
One approximation is the so-called Kinematic time rate
Kine(va,vb) = sqrt(1-va^2) / sqrt(1-vb^2)
and it depends on relative motion only. It is a result of
special relativity - valid in absence of gravity.
The va and vb are seen in an Earth centered non-rotating
inertial frame.
The other approximation is the so called Gravitational
potential time rate
Grav(ra,rb) = sqrt(1-2 M/ra) / sqrt(1-2 M/rb)
and it depends on location only. It would be valid for
non-moving objects only. It is an idealized result from
general relativity.
For small M/ra, M/rb, va, vb, the Taylor expansion will
tell you that
Schw(va,ra,vb,rb) =~ Kine(va,vb) + Grav(ra,rb)
The daily kinematic time dilation is then calculated as
DayKine = Oneday - Oneday Kine(va,vb)
= Oneday ( 1 - Kine(va,vb) )
where
Oneday = 24*60*60 seconds as seen on the ground.
You can also verify that we can approximate:
Kine(va,vb) =~ 1 - 1/2 va^2 + 1/2 vb^2
= 1 - 1/2 (ra W)^2 + 1/2 (rb W)^2
= 1 - 1/2 (ra W)^2 + 1/2 (ra W + v)^2
= 1 + ( 2 ra W v + v^2 ) / 2
where v = relative velocity as seen from the ground.
So the daily dilation will be
DayKine = Oneday ( 1 - Kine(va,vb) )
=~ - Oneday ( 2 ra W v + v^2 ) / 2
This corresponds to the article's result
TA - TS = - TS ( 2 R W v + v^2 ) / (2 c^2 )
where ra = R and TS = Oneday (and c=1).
For the daily dilation of the East and West trips
you need vEast and vWest
DayKineEast = Oneday ( 1 - Kine( vGround, vEast ) )
DayKIneWest = Oneday ( 1 - Kine( vGround, vWest ) )
where
Oneday = 24*60*60 seconds
vGround = rGround W (and divide by c!)
and for the East bound trip:
vEast = vGround + 2 pi rEast / (fEast*60*60)
rEast = rGround + 8900 (average height)
fEast = 41.2*60*60 (time of flight)
and for the West bound trip:
vWest = vGround - 2 pi rEast / (fWest*60*60)
rWest = rGround + 9400 (average height)
fWest = 48.6*60*60 (time of flight)
Note the plus sign and the minus sign in the expressions
for vEast and vWest. And also note the v in the article's
expression. For the Eastbound trip v is positive. It is
negative for the Westbound trip.
Try it out and you'll find (unless I made a mistake somewhere)
DayKineEast = -156 nanoseconds
DayKineWest = 77 nanoseconds
with both calculations.
This nicely fits the -184 and 96 nanoseconds of the
article where a lot more factors have been taken into
account.
Likewise, you can calculate the daily gravitational dilation
DayGrav = Oneday ( 1 - Grav(ra,rb) )
You will find
DayGravEast = 143 nanoseconds
DayGravWest = 178 nanoseconds
Taken together:
DayKineGravEast = -14 nanoseconds
DayKineGravWest = 255 nanoseconds
which is nicely in the ballpark of the actual results.
[snip]
>
> >So I gladly distantiate myself from the two things
> > I wrote above.
> > The right way to express what happens is this: suppose
> > the circularly moving clock has 1 second (tau) between
> > two of its ticks. If the central observer would then measure
> > the time (T) on his own non-revolving clock between the
> > two revolving ticks, he will find more than a second: T > tau.
> > T = tau / sqrt(1-v^2)
> > Also, when the central clock has 1 second (T) between
> > two of its ticks, then if the revolving observer would measure
> > the time (tau) on his own clock between the central ticks, he
> > will find less than a second between them: tau < T
> > tau = T sqrt(1-v^2)
> > This says something about what will be measured, and it
> > is safe. Have a long and close look at Taylor and Wheeler's
> > http://www.eftaylor.com/pub/projecta.pdf
>
> I believe they say from what I read that the GPS's KINETIC clock is
> always slower than Earth's. And that Schwarzschild metric simply
> combines the gravitational and KINETIC time dilation together (and
> possibly the curvature of motion since they are moving in circular
> motion (v/r) and not linearly)?
See above.
I will snip here.
Dirk Vdm
Dirk Van de moortel
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message news:p4fZd.1585$1H7...@news.cpqcorp.net...
[snip]
> Schw(va,ra,vb,rb) = sqrt(1-2 M/ra-va^2) / sqrt(1-2 M/rs-vb^2)
[snip]
> Kine(va,vb) = sqrt(1-va^2) / sqrt(1-vb^2)
[snip]
> Grav(ra,rb) = sqrt(1-2 M/ra) / sqrt(1-2 M/rb)
[snip]
>
> For small M/ra, M/rb, va, vb, the Taylor expansion will
> tell you that
> Schw(va,ra,vb,rb) =~ Kine(va,vb) + Grav(ra,rb)
Ah, sicne we are dealing with rates, this should of course
be multiplication in stead of addition.
Schw(va,ra,vb,rb) =~ Kine(va,vb) Grav(ra,rb)
Addition is again approximately valid with:
(1 - Schw(va,ra,vb,rb) ) =~
(1 - Kine(va,vb) ) + (1 - Grav(ra,rb) )
Proof:
Since K and G are very close to 1, we get
K = 1+k (with very small k)
G = 1+g (with very small g)
so we have
1 - S = 1 - K G
= 1 - (1+k) (1+g)
=~ - k - g (ignoring the ultra small product k g)
= 1-(1+k) + 1-(1+g)
= (1-K) + (1-G)
Dirk Vdm
gu...@hotmail.com
So Either DayKine or TA-TS above
But it doesn't explain the "logic" of "why" it isn't instead Ta =
Ts/gamma?
Daykine equation is very different from Ta = Ts/gamma....instead it's
the difference between Ta and Ts or Oneday and Kine (va,vb)?
And is it because of (1)Gravity or instead because of the planes and
Earth's surface (2)CIRCULAR Trajectories (I believe it because of (2))?
Meaning if instead of 2 planes, if it was 2 Rockets (still affected by
GRAVITY) that instead of flying in circular orbits around Earth instead
both flew away straight and in opposite direction from planet Earth?
Would Daykine still be used or Ta=Ts/gamma or neither of those two:
Instead it would perhaps be Ta w.r.t Earth's center (not Ts but Tc) and
therefore not Daykine (nor Ta-Ts) nor Ta=Ts/gamma but another equation
would be used?
(Meaning would the equation have the rocket's Ta w.r.t. Earth's center
and substracted from Ts since they still depart from Earth's surface
which is spinning in circular orbits?)
Dirk Van de moortel
<gu...@hotmail.com> wrote in message news:1110982659.2...@l41g2000cwc.googlegroups.com...
It *could* be done with Ta = Ts/gamma, but that would be
extremely difficult since (1) this gamma is not constant while
A and S are moving, and (2) because they are not inertially
moving to begin with. A correction factor (dependent on the
proper acceleration) would be needed in addition to gamma.
At each point the rate would be gamma times this factor, and
the intergral over a period of time would be very difficult,
to calculate.
For the time rate of an accelerated clock as seen in an
accelerated frame, look at
http://www.geocities.com/slithytove5/AccelClocks.htm
where you find the rate in equation (18).
For an example of an integrated timerate of an accelerated
clock in an inertial frame, see
http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
In the case of circular motion around a common point, the
intergrals are extremely *easy* to calculate.
See at the end of
http://groups-beta.google.com/group/sci.physics.relativity/msg/73404447e1e7326d
where I explained something similar to someone else.
>
> Daykine equation is very different from Ta = Ts/gamma....instead it's
> the difference between Ta and Ts or Oneday and Kine (va,vb)?
Kine(va,vb) = sqrt(1-va^2) / sqrt(1-vb^2) is the relative rate
between A and B. It is useful because it is constant throughout
circular paths. It actually is (1/gamma_a) / (1/gamma_b).
From the equation
DayKine = Oneday ( 1 - Kine(va,vb) )
which is the integrated rate over one day, you see that DayKine
is the time difference between one day at the surface (given
by Oneday) and one day in the plane, given by the product
Oneday * Kine(va,vb).
>
> And is it because of (1)Gravity or instead because of the planes and
> Earth's surface (2)CIRCULAR Trajectories (I believe it because of (2))?
Yes, it works because they both are *approximately* circular.
>
> Meaning if instead of 2 planes, if it was 2 Rockets (still affected by
> GRAVITY) that instead of flying in circular orbits around Earth instead
> both flew away straight and in opposite direction from planet Earth?
>
> Would Daykine still be used or Ta=Ts/gamma or neither of those two:
> Instead it would perhaps be Ta w.r.t Earth's center (not Ts but Tc) and
> therefore not Daykine (nor Ta-Ts) nor Ta=Ts/gamma but another equation
> would be used?
>
> (Meaning would the equation have the rocket's Ta w.r.t. Earth's center
> and substracted from Ts since they still depart from Earth's surface
> which is spinning in circular orbits?)
The circular rate Kine(va,vb) would not work in the case
of the rockets flying away in opposite directions.
In stead of the rate
Kine(va,vb) = sqrt(1-va^2) / sqrt(1-vb^2)
you would (still in the absence of large masses) have to use
the "straight rate"
Stra(va,vb) = sqrt(1-vab^2)
where you see the composed velocity between A and B:
vab = ( va + vb ) / ( 1 + va vb ).
With a bit of algebra you can verify that this gives the
symmetrical result:
Stra(va,vb) = sqrt(1-va^2) * sqrt(1-vb^2) / (1 + va vb )
satisfying the reciprocity of time dilation:
Stra(va,vb) = Stra(vb,va) ,
whereas the circular rate is not reciprocal at all:
Kine(va,vb) # Kine(vb,va)
Dirk Vdm
Dirk Van de moortel
<gu...@hotmail.com> wrote in message news:1110982659.2...@l41g2000cwc.googlegroups.com...
It *could* be done with Ta = Ts/gamma, but that would be
extremely difficult since (1) this gamma is not constant while
A and S are moving, and (2) because they are not inertially
moving to begin with. A correction factor (dependent on the
proper acceleration) would be needed in addition to gamma.
At each point the rate would be gamma times this factor, and
the intergral over a period of time would be very difficult,
to calculate.
For the time rate of an accelerated clock as seen in an
accelerated frame, look at
http://www.geocities.com/slithytove5/AccelClocks.htm
where you find the rate in equation (18).
For an example of an integrated timerate of an accelerated
clock in an inertial frame, see
http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
In the case of circular motion around a common point, the
intergrals are extremely *easy* to calculate.
See at the end of
http://groups-beta.google.com/group/sci.physics.relativity/msg/73404447e1e7326d
where I explained something similar to someone else.
>
> Daykine equation is very different from Ta = Ts/gamma....instead it's
> the difference between Ta and Ts or Oneday and Kine (va,vb)?
Kine(va,vb) = sqrt(1-va^2) / sqrt(1-vb^2) is the relative rate
between A and B. It is useful because it is constant throughout
circular paths. It actually is (1/gamma_a) / (1/gamma_b).
From the equation
DayKine = Oneday ( 1 - Kine(va,vb) )
which is the integrated rate over one day, you see that DayKine
is the time difference between one day at the surface (given
by Oneday) and one day in the plane, given by the product
Oneday * Kine(va,vb).
>
> And is it because of (1)Gravity or instead because of the planes and
> Earth's surface (2)CIRCULAR Trajectories (I believe it because of (2))?
Yes, it works because they both are *approximately* circular.
>
> Meaning if instead of 2 planes, if it was 2 Rockets (still affected by
> GRAVITY) that instead of flying in circular orbits around Earth instead
> both flew away straight and in opposite direction from planet Earth?
>
> Would Daykine still be used or Ta=Ts/gamma or neither of those two:
> Instead it would perhaps be Ta w.r.t Earth's center (not Ts but Tc) and
> therefore not Daykine (nor Ta-Ts) nor Ta=Ts/gamma but another equation
> would be used?
>
> (Meaning would the equation have the rocket's Ta w.r.t. Earth's center
> and substracted from Ts since they still depart from Earth's surface
> which is spinning in circular orbits?)
The circular rate Kine(va,vb) would not work in the case
of the rockets flying away in opposite directions.
In stead of the rate
Kine(va,vb) = sqrt(1-va^2) / sqrt(1-vb^2)
gu...@hotmail.com
Ok it makes sense although I "think" in cyclotrons TAU = Ta = Tc/gamma
but of course that relation is not between any two particles
circulating at different velocities (such as plane and earth's surface)
but instead with the center clock (and outside)of the cyclotron.
> A correction factor (dependent on the
> proper acceleration) would be needed in addition to gamma.
> At each point the rate would be gamma times this factor, and
> the intergral over a period of time would be very difficult,
> to calculate.
> For the time rate of an accelerated clock as seen in an
> accelerated frame, look at
> http://www.geocities.com/slithytove5/AccelClocks.htm
> where you find the rate in equation (18).
> For an example of an integrated timerate of an accelerated
> clock in an inertial frame, see
> http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
>
A lot of arithmatics!
OK but I'm not familiar with Stra only with gamma equations
Would it mean you can't use Trocket = Ts/gamma since the Ts in circular
accelerating motion?
Dirk Van de moortel
<gu...@hotmail.com> wrote in message news:1111339879.3...@o13g2000cwo.googlegroups.com...
>
> Dirk Van de moortel wrote:
> > <gu...@hotmail.com> wrote in message
> news:1110982659.2...@l41g2000cwc.googlegroups.com...
> > >
[snip]
> > > But it doesn't explain the "logic" of "why" it isn't instead Ta =
> > > Ts/gamma?
> >
> > It *could* be done with Ta = Ts/gamma, but that would be
> > extremely difficult since (1) this gamma is not constant while
> > A and S are moving, and (2) because they are not inertially
> > moving to begin with.
>
> Ok it makes sense although I "think" in cyclotrons TAU = Ta = Tc/gamma
> but of course that relation is not between any two particles
> circulating at different velocities (such as plane and earth's surface)
> but instead with the center clock (and outside)of the cyclotron.
Indeed, it is between the center and a single uniformly
rotating particle - extremely simple.
With the GPS or Hafele-Keating it would be extremely
difficult.
>
> > A correction factor (dependent on the
> > proper acceleration) would be needed in addition to gamma.
> > At each point the rate would be gamma times this factor, and
> > the intergral over a period of time would be very difficult,
> > to calculate.
> > For the time rate of an accelerated clock as seen in an
> > accelerated frame, look at
> > http://www.geocities.com/slithytove5/AccelClocks.htm
> > where you find the rate in equation (18).
> > For an example of an integrated timerate of an accelerated
> > clock in an inertial frame, see
> > http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
> >
>
> A lot of arithmatics!
Yes, if we would do it without mathematics, it would take
thousands of pages to explain.
[snip]
> >
> > The circular rate Kine(va,vb) would not work in the case
> > of the rockets flying away in opposite directions.
> > In stead of the rate
> > Kine(va,vb) = sqrt(1-va^2) / sqrt(1-vb^2)
> > you would (still in the absence of large masses) have to use
> > the "straight rate"
> > Stra(va,vb) = sqrt(1-vab^2)
> > where you see the composed velocity between A and B:
> > vab = ( va + vb ) / ( 1 + va vb ).
> >
> > With a bit of algebra you can verify that this gives the
> > symmetrical result:
> > Stra(va,vb) = sqrt(1-va^2) * sqrt(1-vb^2) / (1 + va vb )
> OK but I'm not familiar with Stra only with gamma equations
It *is* a gamma relation for the velocity vab:
Stra(va,vb) = sqrt(1-vab^2)
= 1 / gamma(vab)
which is easily integrated to
Trocket = Thome / gamma(vab)
provided Thome is the 'home time' of an inertial frame.
>
> Would it mean you can't use Trocket = Ts/gamma since the Ts
> in circular accelerating motion?
Indeed, if Ts still represents the 'surface time', then you would
need the correction factor and a nasty integral.
But in case of the classical 'twin paradox' and someone
staying at home on the surface of our spinning Earth, there is
only a *very* small difference between the 'home time' of the
inertial frame and the 'surface time' when compared to the
time dilation as calculated with the gamma(vab) factor.
Dirk Vdm
Dirk Van de moortel
>
> <gu...@hotmail.com> wrote in message news:1111339879.3...@o13g2000cwo.googlegroups.com...
> >
> > Dirk Van de moortel wrote:
[snip]
>
> [snip]
>
> > >
> > > The circular rate Kine(va,vb) would not work in the case
> > > of the rockets flying away in opposite directions.
> > > In stead of the rate
> > > Kine(va,vb) = sqrt(1-va^2) / sqrt(1-vb^2)
> > > you would (still in the absence of large masses) have to use
> > > the "straight rate"
> > > Stra(va,vb) = sqrt(1-vab^2)
> > > where you see the composed velocity between A and B:
> > > vab = ( va + vb ) / ( 1 + va vb ).
> > >
> > > With a bit of algebra you can verify that this gives the
> > > symmetrical result:
> > > Stra(va,vb) = sqrt(1-va^2) * sqrt(1-vb^2) / (1 + va vb )
>
> > OK but I'm not familiar with Stra only with gamma equations
>
> It *is* a gamma relation for the velocity vab:
> Stra(va,vb) = sqrt(1-vab^2)
> = 1 / gamma(vab)
> which is easily integrated to
> Trocket = Thome / gamma(vab)
> provided Thome is the 'home time' of an inertial frame.
That should be of course:
It *is* a gamma relation for the velocity vab:
Stra(va,vb) = sqrt(1-vab^2)
= 1 / gamma(vab)
which is easily integrated to
TrocketA = TrocketB / gamma(vab)
> >
> > Would it mean you can't use Trocket = Ts/gamma since the Ts
> > in circular accelerating motion?
>
> Indeed, if Ts still represents the 'surface time', then you would
> need the correction factor and a nasty integral.
> But in case of the classical 'twin paradox' and someone
> staying at home on the surface of our spinning Earth, there is
> only a *very* small difference between the 'home time' of the
> inertial frame and the 'surface time' when compared to the
> time dilation as calculated with the gamma(vab) factor.
and this should be
... time dilation as calculated with the gamma(va) factor.
Dirk Vdm
Dirk Van de moortel
<gu...@hotmail.com> wrote in message news:1111434490....@g14g2000cwa.googlegroups.com...
>
> Dirk Van de moortel wrote:
> > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
> wrote in message
> > news:OFk%d.42072$BA6.3...@phobos.telenet-ops.be...
> > >
> > > <gu...@hotmail.com> wrote in message
> news:1111339879.3...@o13g2000cwo.googlegroups.com...
> > > >
> > > > Dirk Van de moortel wrote:
> >
> > [snip]
> >
> > >
> > > [snip]
> > >
<unsnip>
| "Meaning if instead of 2 planes, if it was 2 Rockets (still affected
| by GRAVITY) that instead of flying in circular orbits around Earth
| instead both flew away straight and in opposite direction from planet
| Earth?
|
| Would Daykine still be used or Ta=Ts/gamma or neither of those two:
| Instead it would perhaps be Ta w.r.t Earth's center (not Ts but Tc)
| and therefore not Daykine (nor Ta-Ts) nor Ta=Ts/gamma but another
| equation would be used?
|
| Meaning would the equation have the rocket's Ta w.r.t. Earth's
| center and substracted from Ts since they still depart from Earth's
| surface which is spinning in circular orbits?)"
</unsnip>
> > > > >
> > > > > The circular rate Kine(va,vb) would not work in the case
> > > > > of the rockets flying away in opposite directions.
> > > > > In stead of the rate
> > > > > Kine(va,vb) = sqrt(1-va^2) / sqrt(1-vb^2)
> > > > > you would (still in the absence of large masses) have to use
> > > > > the "straight rate"
> > > > > Stra(va,vb) = sqrt(1-vab^2)
> > > > > where you see the composed velocity between A and B:
> > > > > vab = ( va + vb ) / ( 1 + va vb ).
> > > > >
> > > > > With a bit of algebra you can verify that this gives the
> > > > > symmetrical result:
> > > > > Stra(va,vb) = sqrt(1-va^2) * sqrt(1-vb^2) / (1 + va vb )
> > >
> > > > OK but I'm not familiar with Stra only with gamma equations
> > >
> > > It *is* a gamma relation for the velocity vab:
> > > Stra(va,vb) = sqrt(1-vab^2)
> > > = 1 / gamma(vab)
> > > which is easily integrated to
> > > Trocket = Thome / gamma(vab)
>
> I believe va for Vrocket and vb for Vhome?
va for rocket A away from home in one direction
vb for rocket B away from home in opposite direction
>
> Therefore is the reason why we use vab because va and vb are both
> related to the center of the Earth instead of each other?
>
> vab = ( va + vb ) / ( 1 + va vb )
>
> what's the logic/reason for adding the velocities together in the above
> manner?
Standard special relativity velocity composition.
If
Z has velocity v w.r.t. Y
Y has velocity w w.r.t. X
then
Z has velocity (v+w)/(1+v w) w.r.t X
In this case
Z = rocket B
Y = home on Earth
X = rocket A
(apart from the signs - we are working with speeds here)
>
> as well I'm only familiar with va+vb = ( va + vb ) / ( 1 + va vb "/
> c^2" )
Don't ever write it like that. It is formally wrong.
Write this:
comp(va,vb) = ( va + vb ) / ( 1 + va vb / c^2 )
or, if you prefer infix notation, something like this:
va [+] vb = ( va + vb ) / ( 1 + va vb / c^2 )
I use velocites that have swallowed c.
Some people call this "using units where c = 1".
>
>
> (As well I believe that va vb are already related to each other instead
> of the center of the Earth since both planes flying say +/-500 km/s
> (East & West), their velocities are alreay w.r.t Earth's surface and
> not Earth's center?)
In this case you asked about rockets flying away in
opposite directions away from Earth.
See unsnipped part at top of this post.
Dirk Vdm
gu...@hotmail.com
I believe va for Vrocket and vb for Vhome?
Therefore is the reason why we use vab because va and vb are both
related to the center of the Earth instead of each other?
vab = ( va + vb ) / ( 1 + va vb )
what's the logic/reason for adding the velocities together in the above
manner?
as well I'm only familiar with va+vb = ( va + vb ) / ( 1 + va vb "/
c^2" )
(As well I believe that va vb are already related to each other instead
of the center of the Earth since both planes flying say +/-500 km/s
(East & West), their velocities are alreay w.r.t Earth's surface and
not Earth's center?)
gu...@hotmail.com
What is exatctly the relation between Stra(va,vb) and Trocket, is
Stra(va,vb) = Trocket w.r.t. Thome?
Which is strange since it uses velocities w.r.t. the other rocket
instead of home (Earth's surface) and as well Time of the rocket is
w.r.t. home (instead of the other rocket) = not one but two reference
frames for one single equation?
What if there was only one rocket instead of two (therefore no vb),
then how would the STRA and Trocket equation look like?
> >
> > I believe va for Vrocket and vb for Vhome?
>
> va for rocket A away from home in one direction
> vb for rocket B away from home in opposite direction
>
Ok
Ok Thanks
Dirk Van de moortel
<gu...@hotmail.com> wrote in message news:1111617195.9...@g14g2000cwa.googlegroups.com...
>
> Dirk Van de moortel wrote:
> > <gu...@hotmail.com> wrote in message
> news:1111434490....@g14g2000cwa.googlegroups.com...
[snip]
> > > > > > OK but I'm not familiar with Stra only with gamma equations
> > > > >
> > > > > It *is* a gamma relation for the velocity vab:
> > > > > Stra(va,vb) = sqrt(1-vab^2)
> > > > > = 1 / gamma(vab)
> > > > > which is easily integrated to
> > > > > Trocket = Thome / gamma(vab)
>
> What is exatctly the relation between Stra(va,vb) and Trocket, is
> Stra(va,vb) = Trocket w.r.t. Thome?
I corrected this on March 21:
http://groups-beta.google.com/group/sci.physics.relativity/msg/a0d189b91154159a
Dirk Vdm
gu...@hotmail.com
> <gu...@hotmail.com> wrote in message
news:1111617195.9...@g14g2000cwa.googlegroups.com...
> >
> > Dirk Van de moortel wrote:
> > > <gu...@hotmail.com> wrote in message
> > news:1111434490....@g14g2000cwa.googlegroups.com...
>
> [snip]
>
> > > > > > > OK but I'm not familiar with Stra only with gamma
equations
> > > > > >
> > > > > > It *is* a gamma relation for the velocity vab:
> > > > > > Stra(va,vb) = sqrt(1-vab^2)
> > > > > > = 1 / gamma(vab)
> > > > > > which is easily integrated to
> > > > > > Trocket = Thome / gamma(vab)
> >
> > What is exatctly the relation between Stra(va,vb) and Trocket, is
> > Stra(va,vb) = Trocket w.r.t. Thome?
>
> I corrected this on March 21:
>
http://groups-beta.google.com/group/sci.physics.relativity/msg/a0d189b91154159a
>
> Dirk Vdm
Ok I re-read it, you said forget Kine for the rockets and use Stra
instead.
And then you said Stra is the same as:
1. Trocket = Tc / gamma (vab) (Tc = Tcenter_earth for "Thome" and if
we use Ts = Tsurface_earth then the equation would be much more
complex?)
Tc = Thome = Tcenter of Earth and vab = va[+]vb so that their velocity
is w.r.t. Tc
...?since otherwise if the Earth was not spinning it would be Trocket =
Tc /gamma(va) ?
--------------------------------------------------------
For #2 it's a little confusing and this is what I can make out?
2. and then you said: TrocketA= TrocketB /gamma (vab) ....is not very
practical since we do not know the value for two variables in one
equation, TrocketB as well as TrocketA (as well since the goal is to
compare the clocks with Ts (Tsurface_Earth)?
-------------------------------------
So I'm not sure anymore if Stra (va,vb) means TrocketA = TrocketB
/gamma (vab) since the main goal is to compare either clocks (A or B)
with Ts (Tsurface_earth) ?
-------------------------------------------------------------------------------
Is Stra(va,vb) = TrocketA = TrocketB and both w.r.t Tc (or both w.r.t.
Ts?)
(confusing since in #2 above TrocketA is not = TrocketB) ?
Dirk Van de moortel
<gu...@hotmail.com> wrote in message news:1111703370....@g14g2000cwa.googlegroups.com...
That was the part that I corrected on March 21:
http://groups-beta.google.com/group/sci.physics.relativity/msg/a0d189b91154159a
You aren't paying attention.
It's all there and it's up to you now.
Dirk Vdm
gu...@hotmail.com
> > > >
> > > > The circular rate Kine(va,vb) would not work in the case
> > > > of the rockets flying away in opposite directions.
> > > > In stead of the rate
> > > > Kine(va,vb) = sqrt(1-va^2) / sqrt(1-vb^2)
> > > > you would (still in the absence of large masses) have to use
> > > > the "straight rate"
> > > > Stra(va,vb) = sqrt(1-vab^2)
> > > > where you see the composed velocity between A and B:
> > > > vab = ( va + vb ) / ( 1 + va vb ).
> > > >
> > > > With a bit of algebra you can verify that this gives the
> > > > symmetrical result:
> > > > Stra(va,vb) = sqrt(1-va^2) * sqrt(1-vb^2) / (1 + va vb )
> >
> > > OK but I'm not familiar with Stra only with gamma equations
> >
> > It *is* a gamma relation for the velocity vab:
> > Stra(va,vb) = sqrt(1-vab^2)
> > = 1 / gamma(vab)
> > which is easily integrated to
> > Trocket = Thome / gamma(vab)
Oh I thought Stra(va,vb) = Trocket but re-reading it says Stra(va,vb) =
1/gamma(vavb)
thus Trocket = Thome * Stra(va,vb) (#1)
> > provided Thome is the 'home time' of an inertial frame.
>
> That should be of course:
>
> It *is* a gamma relation for the velocity vab:
> Stra(va,vb) = sqrt(1-vab^2)
> = 1 / gamma(vab)
> which is easily integrated to
> TrocketA = TrocketB / gamma(vab)
>
Thus TrocketA = TrocketB * Stra(va,vb) ....correct?
(and this time TrocketA is w.r.t. TrocketB where as #1 above Trocket
(TrocketA or TrocketB is w.r.t. Thome......correct?)