Discussion:
Testing relativity of simultaneity using GPS satellites
(too old to reply)
PengKuan Em
2019-10-26 15:38:55 UTC
Permalink
In Special Relativity relativity of simultaneity is the fact that 2 simulta=
neous events occurring in a stationary frame does not appear simultaneous i=
n a moving frame. For example, in Einstein's train thought experiment 2 sim=
ultaneous flashes of light on the platform do not appear simultaneous for t=
he observer in the train. But relativity of simultaneity has never been tes=
ted with real simultaneous events.

For testing relativity of simultaneity we need 2 synchronized clocks moving=
at high speed and we will read them in a stationary frame. Fortunately, we=
have at hand many GPS satellites which carry precision clocks and broadcas=
t their time, with which we can check relativity of simultaneity.=20

Suppose that 2 satellites are separated by the distance L in the same orbit=
. Their clocks are synchronized with one clock on Earth, that is, the event=
"time of the satellite 1 is t0" and the event "time of the satellite 2 is =
t0" occur simultaneously on Earth at the time te. In the frame of these 2 s=
atellites, due to relativity of simultaneity, these same events occur at ti=
me t1 on the satellite 1 and t2 on the satellite 2 and the difference of ti=
me is dt=3D t2- t1.

Suppose that we have n satellites equally spaced in the same orbit which is=
circular. The nth satellite is the last satellite and the (n+1)th satellit=
e is the first satellite, which complete the circle of the orbit.=20

Due to relativity of simultaneity, the difference of time is always dt from=
one satellite to the next and the difference of time between the ith satel=
lite and the first satellite equals (i-1)*dt. Then, the difference of time =
between the (n+1)th satellite and the first satellite equals n*dt. So, The =
time of the (n+1)th satellite is t1+n*dt.

We notice that the (n+1)th satellite is the first satellite but its time, t=
1+n*dt, is different from t1 the time of the first satellite. How can the =
time of a satellite is not the time of itself?=20

I explain this phenomenon in the article below.
PDF: https://pengkuanonphysics.blogspot.com/2019/10/testing-relativity-of-s=
imultaneity.html=20
Word: https://www.academia.edu/40736335/Testing_relativity_of_simultaneity_=
using_GPS_satellites =20
Jonathan Thornburg [remove -animal to reply]
2019-10-30 08:26:37 UTC
Permalink
Post by PengKuan Em
Suppose that we have n satellites equally spaced in the same orbit
which is circular. The nth satellite is the last satellite and the
(n+1)th satellite is the first satellite, which complete the circle of
the orbit.
Due to relativity of simultaneity, the difference of time is
always dt fromone satellite to the next and the difference of time
between the ith satellite and the first satellite equals (i-1)*dt.
Then, the difference of time between the (n+1)th satellite and the
first satellite equals n*dt. So, The time of the (n+1)th satellite
is t1+n*dt.
We notice that the (n+1)th satellite is the first satellite but its
time, t> 1+n*dt, is different from t1 the time of the first satellite.
How can the time of a satellite is not the time of itself?
I think you've rediscovered the Sagnac effect:
https://en.wikipedia.org/wiki/Sagnac_effect
https://link.springer.com/article/10.12942/lrr-2003-1#Sec2

The latter of those links includes a direct experimental test of this:

# In 1984 GPS satellites 3, 4, 6, and 8 were used in simultaneous
# common view between three pairs of earth timing centers, to accomplish
# closure in performing an around-the-world Sagnac experiment. The
# centers were the National Bureau of Standards (NBS) in Boulder, CO,
# Physikalisch-Technische Bundesanstalt (PTB) in Braunschweig, West
# Germany, and Tokyo Astronomical Observatory (TAO). The size of the
# Sagnac correction varied from 240 to 350 ns. Enough data were
# collected to perform 90 independent circumnavigations. The actual
# mean value of the residual obtained after adding the three pairs
# of time differences was 5 ns, which was less than 2 percent of the
# magnitude of the calculated total Sagnac effect [4].

The citation [4] refers to

Allan, D.W., Weiss, M., and Ashby, N.,
"Around-the-World Relativistic Sagnac Experiment",
Science, 228, 69--70, (1985).
https://science.sciencemag.org/content/228/4695/69

ciao,
--
-- "Jonathan Thornburg [remove -animal to reply]" <***@astro.indiana-zebra.edu>
Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
currently on the west coast of Canada
"There was of course no way of knowing whether you were being watched
at any given moment. How often, or on what system, the Thought Police
plugged in on any individual wire was guesswork. It was even conceivable
that they watched everybody all the time." -- George Orwell, "1984"
PengKuan Em
2019-10-31 20:55:42 UTC
Permalink
Post by Jonathan Thornburg [remove -animal to reply]
https://en.wikipedia.org/wiki/Sagnac_effect
https://link.springer.com/article/10.12942/lrr-2003-1#Sec2
# In 1984 GPS satellites 3, 4, 6, and 8 were used in simultaneous
# common view between three pairs of earth timing centers, to accomplish
# closure in performing an around-the-world Sagnac experiment. The
# centers were the National Bureau of Standards (NBS) in Boulder, CO,
# Physikalisch-Technische Bundesanstalt (PTB) in Braunschweig, West
# Germany, and Tokyo Astronomical Observatory (TAO). The size of the
# Sagnac correction varied from 240 to 350 ns. Enough data were
# collected to perform 90 independent circumnavigations. The actual then
# satellite 4 retransmits the wave to satellite 6,
# mean value of the residual obtained after adding the three pairs
# of time differences was 5 ns, which was less than 2 percent of the
# magnitude of the calculated total Sagnac effect [4].
The Sagnac effect according to
https://en.wikipedia.org/wiki/Sagnac_effect is the shift of phase
between 2 opposite lights. If GPS satellites 3, 4, 6, and 8 were really
used to test Sagnac effect, one should send 2 EM waves in opposite
direction:
1. EM wave A is sent from satellite 3 to satellite 4, then
satellite 4 retransmits the wave to satellite 6, then satellite 6
retransmits the wave to satellite 8, then satellite 8 retransmits the
wave to satellite 3.
2. At the same time, EM wave B is sent from satellite 3 to
satellite 8, then satellite 8 retransmits the wave to satellite 6, then
satellite 6 retransmits the wave to satellite 4, then satellite 4
retransmits the wave to satellite
3.
At satellite 3 , we compare the phases of EM wave A and B. This is a
test of Sagnac effect.

But the 1984 GPS satellites experiment is actually a test of relativity
of simultaneity around the world, because " In 1984 GPS satellites 3, 4,
6, and 8 were used in simultaneous common viewâ=80|". They compared the
time of satellite 3 with the time retransmitted around the world by GPS
satellites 3, 4, 6, and 8 and find 5 ns of difference.

Although the experiment is not a test of Sagnac effect, the result is
actually great, because it proves my prediction: The relativity of
simultaneity in GPS satellites is zero, and thus, relativity of
simultaneity in the moving frame of the satellites is incorrect. I
explain my prediction below.

What does mean clock 1 is synchronized with clock 2? I think it means
that, when clock 1 indicates t=0, in the same frame clock 2 indicates
t=0 too. In Earth Centered Inertial frame, let the clock of satellite 1
be synchronized with the clock on Earth and the clock of satellite 2 be
synchronized with the clock on Earth too.
1. In Earth Centered Inertial frame, the clock on Earth indicates
te=0, the clock of satellite 1 indicates t1=0.
2. In Earth Centered Inertial frame, the clock on Earth indicates
te=0, the clock of satellite 2 indicates t2=0.
3. In Earth Centered Inertial frame, the clock on Earth indicates
te=0, the clock of satellite 1 indicates t1=0, the clock of satellite 2
indicates t2=0. Then, the 3 clocks are synchronized with one another.

Because Earth Centered Inertial frame does not rotate, the satellites 1
and 2 are in relative motion with respect to Earth Centered Inertial
frame. Suppose their velocity is v and their instantaneous positions are
x1 and x2 at te=0. Let us agree that t1=0 and t2=0 are the times in the
Earth Centered Inertial frame seen by an observer at the instantaneous
positions x1.

In the Earth Centered Inertial frame t1=0 and t2=0 for te=0. What is the
time of the satellite 2 in the frame moving with the satellites 1?
According to Lorentz: t'=gamma(t-x*v/c^2)
Let x1=0. Then t'1=gamma(0-0*v/c^2) =0 Because x2=/=0, we have
t'2=gamma*(0+x2*v/c^2). So, t'2 is not zero. This means that an observer
moving with the satellites 1 must see that the clock of satellite 2
indicates t'2= gamma*x2*v/c^2, so they are not synchronized in the
moving frame.

This means that in the frame moving with the satellite 1, t'1=0 and t'2=
gamma*x2*v/c^2. So, delta t' = t'2-t'1=gamma*x2*v/c^2. Also, between the
3rd and the 2d satellites, delta t' = t'3-t'2=gamma*x2*v/c^2 and etc.

Let us have n satellites. The clock of the (n+1)th satellite should
indicate in the frame moving with the satellite 1 t'(n+1)=
t'(n+1)-t'n+t'n-t'(n-1)+ â=80| + t'2-t'1=n*gamma*x2*v/c^2. So, t'(n+1)=
n* gamma*x2*v/c^2 in the frame moving with the satellite 1

But the orbit is a circle, the (n+1)th satellite is the first satellite.
Then, the clock of the first satellite should indicate
n*gamma*x2*v/c^2=7204 ns. But we started with t'1=0. Hence
contradiction.

In fact, I predict that the clock of the (n+1)th satellite should
indicate 0 in the frame moving with the satellite 1. Because relativity
of simultaneity gives time of the (n+1)th satellite t'(n+1)= 7204 ns,
relativity of simultaneity could be incorrect, that is Lorentz
transformations could be incorrect.

PK
r***@gmail.com
2019-10-31 20:56:12 UTC
Permalink
Post by PengKuan Em
In Special Relativity relativity of simultaneity is the fact that 2
simultaneous events occurring in a stationary frame does not appear
simultaneous in a moving frame. For example, in Einstein's train thought
experiment 2 simultaneous flashes of light on the platform do not appear
simultaneous for the observer in the train. But relativity of
simultaneity has never been tested with real simultaneous events.
For testing relativity of simultaneity we need 2 synchronized clocks moving
at high speed and we will read them in a stationary frame. Fortunately, we
have at hand many GPS satellites which carry precision clocks and broadcast
their time, with which we can check relativity of simultaneity.=20
Suppose that 2 satellites are separated by the distance L in the same
orbit. Their clocks are synchronized with one clock on Earth, that is,
the event "time of the satellite 1 is t0" and the event "time of the
satellite 2 is t0" occur simultaneously on Earth at the time te. In the
frame of these 2 satellites, due to relativity of simultaneity, these
same events occur at time t1 on the satellite 1 and t2 on the satellite
2 and the difference of time is dt=3D t2- t1.
Suppose that we have n satellites equally spaced in the same orbit which
is circular. The nth satellite is the last satellite and the (n+1)th
satellite is the first satellite, which complete the circle of the
orbit.

Due to relativity of simultaneity, the difference of time is always dt
from one satellite to the next and the difference of time between the
ith satellite and the first satellite equals (i-1)*dt. Then, the
difference of time between the (n+1)th satellite and the first satellite
equals n*dt. So, The time of the (n+1)th satellite is t1+n*dt.

We notice that the (n+1)th satellite is the first satellite but its
time, t1+n*dt, is different from t1 the time of the first satellite.
How can the time of a satellite is not the time of itself?=20

I explain this phenomenon in the article below.
PDF: https://pengkuanonphysics.blogspot.com/2019/10/testing-relativity-of-simultaneity.html=20
Word: https://www.academia.edu/40736335/Testing_relativity_of_simultaneity_using_GPS_satellites =20

Special Relativity only applies to inertial frames in flat space-time.
Satelites in orbit are not in flat space-time, even though they are in
locally inertial frames. To properly analyze this case you need GR,
which I assume would handle it correctly.

Rich L.
Jonathan Thornburg [remove -animal to reply]
2019-11-06 08:08:26 UTC
Permalink
Post by PengKuan Em
Suppose that we have n satellites equally spaced in the same orbit
which is circular. The nth satellite is the last satellite and the
(n+1)th satellite is the first satellite, which complete the circle of
the orbit.
Due to relativity of simultaneity, the difference of time is
always dt fromone satellite to the next and the difference of time
between the ith satellite and the first satellite equals (i-1)*dt.
Then, the difference of time between the (n+1)th satellite and the
first satellite equals n*dt. So, The time of the (n+1)th satellite
is t1+n*dt.
We notice that the (n+1)th satellite is the first satellite but its
time, t> 1+n*dt, is different from t1 the time of the first satellite.
How can the time of a satellite is not the time of itself?
I think you've rediscovered the Sagnac effect:
https://en.wikipedia.org/wiki/Sagnac_effect
https://link.springer.com/article/10.12942/lrr-2003-1#Sec2

The latter of those links includes a direct experimental test of this:

# In 1984 GPS satellites 3, 4, 6, and 8 were used in simultaneous
# common view between three pairs of earth timing centers, to accomplish
# closure in performing an around-the-world Sagnac experiment. The
# centers were the National Bureau of Standards (NBS) in Boulder, CO,
# Physikalisch-Technische Bundesanstalt (PTB) in Braunschweig, West
# Germany, and Tokyo Astronomical Observatory (TAO). The size of the
# Sagnac correction varied from 240 to 350 ns. Enough data were
# collected to perform 90 independent circumnavigations. The actual
# mean value of the residual obtained after adding the three pairs
# of time differences was 5 ns, which was less than 2 percent of the
# magnitude of the calculated total Sagnac effect [4].

The citation [4] refers to

Allan, D.W., Weiss, M., and Ashby, N.,
"Around-the-World Relativistic Sagnac Experiment",
Science, 228, 69--70, (1985).
https://science.sciencemag.org/content/228/4695/69

ciao,
--
-- "Jonathan Thornburg [remove -animal to reply]" <***@astro.indiana-zebra.edu>
Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
currently on the west coast of Canada
"There was of course no way of knowing whether you were being watched
at any given moment. How often, or on what system, the Thought Police
plugged in on any individual wire was guesswork. It was even conceivable
that they watched everybody all the time." -- George Orwell, "1984"
PengKuan Em
2019-11-06 08:08:26 UTC
Permalink
Post by r***@gmail.com
Special Relativity only applies to inertial frames in flat space-time.
Satelites in orbit are not in flat space-time, even though they are in
locally inertial frames. To properly analyze this case you need GR,
which I assume would handle it correctly.
Rich L.
The local frame at any point in general relativity is a locally inertial
frame. In such frame, relativity of simultaneity applies, so the time of
point 1 and 2 are different. Suppose they are t'1 and t'2 and
dt'=t'2-t'1 in frame 1. then, in frame 2 next to frame 1, we have
dt'=t'3-t'2. So, we have
t'2=dt'+t'1 in frame 1. Suppose t'1=0. Then,t'2=dt'. Then, we have successively,
t'3=dt'+t'2=2*dt' in frame 2
t'4=dt'+t'3=3*dt' in frame 3
...
t'(n+1)=dt'+t'n=n*dt' in frame n

So, the time of the satellite n+1 is n*dt' even in GR. But satellite n+1
is satellite 1 and its time should be 0. Hence, even in GR, relativity
of simultaneity leads to contradiction.

PK
Tom Roberts
2019-11-06 21:03:40 UTC
Permalink
Post by PengKuan Em
The local frame at any point in general relativity is a locally inertial
frame. In such frame, relativity of simultaneity applies, so the time of
point 1 and 2 are different. Suppose they are t'1 and t'2 and
dt'=t'2-t'1 in frame 1. then, in frame 2 next to frame 1, we have
dt'=t'3-t'2. So, we have
t'2=dt'+t'1 in frame 1. Suppose t'1=0. Then,t'2=dt'. Then, we have successively,
t'3=dt'+t'2=2*dt' in frame 2
t'4=dt'+t'3=3*dt' in frame 3
...
t'(n+1)=dt'+t'n=n*dt' in frame n
So, the time of the satellite n+1 is n*dt' even in GR. But satellite n+1
is satellite 1 and its time should be 0. Hence, even in GR, relativity
of simultaneity leads to contradiction.
As I said before, each of those differences is in a DIFFERENT LOCALLY
INERTIAL FRAME. You cannot combine them like that.

You fooled yourself by erroneously using a single prime everywhere, when
in fact the differences are:
dt' = t'2-t'1
dt'' = t''3-t''2
dt''' = t'''4-t'''3
... etc.
(the number of primes indicates which clock's instantaneously
co-moving inertial frame is used)
Attempting to add dt' to dt'' makes no sense as it involves frame
jumping. Ditto for each of the other pairs.

There is no contradiction, this is just the well-known Sagnac effect --
it directly implies that you cannot consistently synchronize clocks
around the circumference of a rotating system. Of course the GPS never
does that. Nor does anyone who understands SR.

Tom Roberts
r***@gmail.com
2019-11-06 21:13:18 UTC
Permalink
Post by PengKuan Em
Post by r***@gmail.com
Special Relativity only applies to inertial frames in flat space-time.
Satelites in orbit are not in flat space-time, even though they are in
locally inertial frames. To properly analyze this case you need GR,
which I assume would handle it correctly.
Rich L.
The local frame at any point in general relativity is a locally inertial
frame. In such frame, relativity of simultaneity applies, so the time of
point 1 and 2 are different. Suppose they are t'1 and t'2 and
dt'=t'2-t'1 in frame 1. then, in frame 2 next to frame 1, we have
dt'=t'3-t'2. So, we have
t'2=dt'+t'1 in frame 1. Suppose t'1=0. Then,t'2=dt'. Then, we have successively,
t'3=dt'+t'2=2*dt' in frame 2
t'4=dt'+t'3=3*dt' in frame 3
...
t'(n+1)=dt'+t'n=n*dt' in frame n
So, the time of the satellite n+1 is n*dt' even in GR. But satellite n+1
is satellite 1 and its time should be 0. Hence, even in GR, relativity
of simultaneity leads to contradiction.
PK
The satellites are not in an inertial frame as required by SR, at any
instant they can be considered in an inertial frame at some velocity,
but a moment later they are in a different frame. In SR boosts to
different frames is not a linear process. You can't just add the time
differences as you are suggesting. Look up Thomas Precession.

Consider viewing the satellites from the north pole, with all the
satellites in an equatorial orbit. If all the satellites are at the
same altitude the (GR) gravitational red shift will be equal for all of
them. They are all moving at the same speed in this orbit, so the SR
time dilation will also be the same. You could synchronize the clocks
with a single pulse from the north pole. The orbiting clocks are now
synchronized and running at the same speed. Doesn't that counter your
argument?

Rich L.
Tom Roberts
2019-11-08 10:04:06 UTC
Permalink
Post by r***@gmail.com
Consider viewing the satellites from the north pole, with all the
satellites in an equatorial orbit. If all the satellites are at the
same altitude the (GR) gravitational red shift will be equal for all of
them. They are all moving at the same speed in this orbit, so the SR
time dilation will also be the same. You could synchronize the clocks
with a single pulse from the north pole. The orbiting clocks are now
synchronized and running at the same speed.
Whenever you say "synchronized" you must also say in which frame it
applies (not doing that is what confused PK). This synchronization is in
the frame of the north pole. The clocks are "running at the same speed"
when measured in the frame of the north pole.

That is close to what the GPS does. GPS clocks are synchronized in the
ECI frame, not the instantaneously co-moving inertial frame of any
satellite.
Post by r***@gmail.com
[to PK] Doesn't that counter your argument?
Since synchronization is frame dependent, no. But the frame-jumping he
performs does invalidate his argument.

Tom Roberts
PengKuan Em
2019-11-09 06:47:38 UTC
Permalink
Post by r***@gmail.com
The satellites are not in an inertial frame as required by SR, at any
instant they can be considered in an inertial frame at some velocity,
but a moment later they are in a different frame. In SR boosts to
different frames is not a linear process. You can't just add the time
differences as you are suggesting. Look up Thomas Precession.
Consider viewing the satellites from the north pole, with all the
satellites in an equatorial orbit. If all the satellites are at the
same altitude the (GR) gravitational red shift will be equal for all of
them. They are all moving at the same speed in this orbit, so the SR
time dilation will also be the same. You could synchronize the clocks
with a single pulse from the north pole. The orbiting clocks are now
synchronized and running at the same speed. Doesn't that counter your
argument?
Rich L.
Synchronization of moving satellites is not like synchronizing fixed
clocks. The link is my reply to you with drawing:
https://pengkuanonphysics.blogspot.com/2019/11/synchronizing-moving-gps-clocks.html PDF

1. Light pulse synchronization
Can we synchronize clocks of a moving frame? Let us see Figure 1 where
we have stationary frame F1 and moving frame F'2. The 2 clocks in the
frame F1 are synchronized with the master clock through a light pulse.

The 2 clocks of the frame F'2 are moving. At the time t0 in the frame
F1, the 2 clocks in the moving frame F'2 coincide with that of the frame
F1, but due to relativity of simultaneity the time of the moving clocks
are different, dt'=t'2-t'1=/=0. So, the light pulse does not synchronize
the clocks in the moving frame.

2. Fixed clocks in orbit
GPS satellites are moving, so they suffer from relativity of
simultaneity. Let us imagine a disk centered at the center of the Earth
whose rim is the GPS orbit and on the rim are equally spaced clocks, see
Figure 2. The disk does not rotate with the Earth and the frame of the
disk is the frame F1. These clocks can be synchronized because they are
fixed and they are synchronized with the master clock at the North Pole
of the Earth through a light pulse. They all show the time t0 when each
GPS satellite pass in front of each corresponding fixed clock. But the
clocks of the GPS satellites are not synchronized with the master clock
because they are moving.

3. Time in GPS satellites
The event the satellite 1 coincides with a fixed clock and the satellite
2 coincides with the next fixed clock are simultaneous in the frame F1.
But these 2 events are not simultaneous in the frame F'2, which is the
frame moving with the satellite 1 and containing the satellite 2, see
Figure 3.

In the frame F'2, the clock of the satellite 1 reads t'1 and the clock
of the satellite 2 reads t'2 because of non-simultaneity. So, we have
the gap of time dt'=t'2-t'1=/=0.

In the same way, the event the satellite 2 coincides with a fixed clock
and the satellite 3 coincides with the next fixed clock are
simultaneous in the frame F1, but not in the frame F'3, which is the
frame moving with the satellite 2 and containing the satellite 3, see
Figure 3. Also for the same reason, the gap of time in the frame F'3 is
dt'=t'3-t''2=/=0.

Note that here in the frame F'3, the time of the satellite 2 is denoted
by t''2 with double prime to distinguish it from the time of the
satellite 2 in the frame F'2.

4. Time of the satellite 2
We have 2 different notations of the time of the satellite 2: t'2 with
single prime in the frame F'2 and t''2 with double prime in the frame
F'3. The satellite 2 has only one time which is shown by its clock. For
the event the satellite 2 coincides with the fixed clock in Figure 3,
its value in the frame F'2 is t'2. Now we ask: does the reading of the
clock of the satellite 2 change to a different value t''2 when seen in
the frame F'3? Logically no, because t'2 is the time of the satellite 2
while t''2 is the time of the same satellite at the same event. So, we
must have t'2 = t''2. In this case, we have t'3 =
t''2+dt'=t'2+dt'=t'1+dt'+dt'=t'1+2dt'

5. Time of the satellite n
For the same reason, the double-primed t''3 equals the single-primed t'3
and each double-primed t''i equals each corresponding single-primed t'i.
Then, the n+1th satellite has the time: t'(n+1) =t'1+ndt'.

Because ndt' is not zero, t'(n+1) =/=t'1 while the satellite n+1 being
the satellite 1, hence contradiction.

6. If t'(n+1) =t'1?
We can assert that the satellite n+1 being the satellite 1 and having
the same time, that is, t'(n+1) =t'1. In this case, we first suppose
that the single-primed t'2 does not equal the double-primed t''2, but
instead their different cancels dt', that is, t''2=t'2-dt'. So,
t'3=t''2+dt'=t'2-dt'+dt'=t'1+dt' .

In this case, we will have for the n+1th satellites t'(n+1) =t'1+dt' .

However, we still do not have t'(n+1) =t'1. For achieving this equality,
we have to suppose the difference of time between single-primed and
double-primed time to make t''2 = t'2 -dt'(1+1/n). This way, we will
have at the n+1th satellite: t'(n+1) =t'1+dt'-dt'*n/n=t'1

But this assumption is absurd and cannot be true. So, due to relativity
of simultaneity, the satellite n+1 and the satellite 1 are the same
satellite but cannot have the same time.

The only way for the satellite n+1 and the satellite 1 to have the same
time is that relativity of simultaneity is zero, that is, dt'=0.

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