"David" wrote in message
...

There are two reference frames that have a relative velocity V=0.866c
along the x-axis. I use that value so that its simple to talk about
the clock rates. Einstein really didn't detail how to synchronize
clocks so I am using a simple method. At time t0 = 0, a light pulse is
emitted from a point on the x-axis. In their respective inertial
frames, the distance from each clock to this point is measured. When
the light pulse arrives at a clock, the clock is set to the time equal
to this distance divided by c. Now all clocks in their respective
frames run at the same rate and are set to the same time.

Here's the problem. Let one frame be called the rest frame and the
other frame be called the moving frame. With the relative velocity
between the two frames at 0.866c, each frame measures that clocks in
the other frame run at half the rate of clocks in their own frame. So
let's say observers in the moving frame want to place a clock in the
rest frame that runs at the same rate as all of their clocks in the
moving frame. Since "time" in the rest frame runs half as fast as in
their own frame, they modify the oscillator frequency of the clock by
doubling it. I'll call this clock Cd (d for double the rate of ideal
clocks). So now the moving observers have a clock in the rest frame
that runs at the same rate as all of their clocks. They place it in
the rest frame at time t=0 at the point where the pulse of light used
to synchronize all clocks originated.

The rest frame observers say that this clock is not setup correctly.
They say clocks in the moving frame run at half the rate of the rest
frame clocks, so they say that instead of doubling the oscillator
frequency, the clock in the rest frame that runs at the same rate as
all the moving frame clocks should have its oscillator refrequency cut
in half instaed of doubled. So the rest frame observers modify a
clock so that it runs at half the rate of an ideal clock. I'll call
this clock Ch (h for half the rate of ideal clocks). This clock is
also placed in the rest frame at time t=0 at the point where the pulse
of light used to synchronize all clocks originated.

Now let two clocks in the moving frame be separated by 866
light-seconds. I'll call these clock A and Clock B. And let clock A
be at the point of origin of the light pulse used to synchronize all
clocks. Let the moving frame be moving in the negative x-direction
relative to the rest frame. Clocks Cd and Ch will take 1000 seconds
(as measured in the moving frame) to reach clock B. Now we can apply
Einstein's equations, and find that the rest frame observers say that
the separation between these two moving frame clocks is 433
light-seconds and the second moving frame clock will be at the same
point in space as clocks Cd and Ch in only 500 seconds. So we find
that when the clocks meet, clock B reads 1000 seconds, clock Cd reads
1000 seconds (it is running at twice the rate as other rest frame
clocks), but clock Ch only reads 250 seconds (it is running at half
the rate as other rest frame clocks). So clock Ch is not synchronized
with all of the moving frame clocks as thought by the rest frame
observers.

I don't see how to resolve this problem. According to the rest frame
observers, clock Ch is running at the same rate as every moving frame
clock. Clock Ch got set to zero at the same time as Clock A in the
moving frame and Clock Cd were set to zero. According to the rest
frame observers, Clock Ch is running at the same rate as Clock A and
Clock B, so therefore this must have been some problem with the
initialization of Clock B. But I could not determine how the rest
frame observers say the initialization of Clock B should have been
done.


David, there are _two*_ relativistic effects going on he

(1) 'moving' clocks tick slower than 'rest' clocks according to the rest
frame. This means that if the rest observer watches any _one_ of the moving
clocks as it passes by a line of rest-frame-synchronized clocks, the moving
clock will lag more and more behind each successive rest-clock that it
passes.

(2) moving clocks that are strung along the x axis of the moving frame and
synchronized according to the moving frame are _not synchronized_ according
to the rest frame. This means that two different moving clocks do not read
the same time simultaneously according to the rest frame.

As others have pointed out, you are overlooking (2). When (2) is taken into
account, everything is consistent.

Tom

"Tom S." wrote in message
m...

"David" wrote in message
...

I don't see how to resolve this problem. According to the rest frame
observers, clock Ch is running at the same rate as every moving frame
clock. Clock Ch got set to zero at the same time as Clock A in the
moving frame and Clock Cd were set to zero. According to the rest
frame observers, Clock Ch is running at the same rate as Clock A and
Clock B, so therefore this must have been some problem with the
initialization of Clock B. But I could not determine how the rest
frame observers say the initialization of Clock B should have been
done.


David, there are _two*_ relativistic effects going on he

(1) 'moving' clocks tick slower than 'rest' clocks according to the rest
frame. This means that if the rest observer watches any _one_ of the
moving clocks as it passes by a line of rest-frame-synchronized clocks,
the moving clock will lag more and more behind each successive rest-clock
that it passes.

(2) moving clocks that are strung along the x axis of the moving frame
and synchronized according to the moving frame are _not synchronized_
according to the rest frame. This means that two different moving clocks
do not read the same time simultaneously according to the rest frame.

As others have pointed out, you are overlooking (2). When (2) is taken
into account, everything is consistent.


David. In many previous posts you have demonstrated that you have the
ability to correctly use the Lorentz transformation equations. Why don't
you use these equations to answer your question for yourself? I would guess
that virtually every serious student of SR has done something like the
following.

Cut out two long strips of paper. Along one strip (the 'rest strip') draw
circles representing the faces of stopwatches spaced at equal intervals of
distance, say 0.866 light-seconds. Use a pencil to draw all the second
hands pointing to zero on these watches. This strip represents the watches
in the rest frame at time t = 0 in the rest frame.

The other strip will represent the 'moving' frame as viewed by the rest
frame at time t = 0 in the rest frame. According to your scenario, the
moving frame is moving in the negative x direction relative to the rest
frame with a speed 0.866c. For convenience, let the clocks in this frame be
spaced at 2 times 0.866c according to the moving frame. Due to length
contraction, you will need to draw these watches spaced at only 0.866
light-seconds since we are considering how things look from the point of
view of the rest frame. So, the clocks will be drawn with the same spacing
on both strips. Assume the middle watch on each strip marks the origin of
each frame. Use the LT to determine the reading of each watch in the moving
frame at time t = 0 according to the rest frame.

Place the strips along side each other so the origins are adjacent. This
represents the state of affairs at time t = 0 according to the rest frame.
You will see how the rest frame claims that the moving clocks are not
synchronized. Now let time in the rest frame advance by one second. You
will need to advance all of the second-hands of the moving clocks by 0.5
seconds (time dilation). These moving clocks will of course still be
unsynchronized with one another according to the rest frame. Finally, you
need to slide the moving strip in the negative x direction relative to the
rest strip by an interval of 0.866 light-seconds (i.e., equal to the spacing
of the clocks). Now you have the state of affairs at time t = 1 sec
according to the rest frame.

Repeat this as many times as you wish to advance forward in time according
to the rest frame. You can also add two more clocks at the origin of the
rest frame, one ticking at half the rate of a rest clock and one ticking at
twice the rate of a rest clock. These represent your Ch and Cd clocks.
You will discover that the Cd clock always reads the same time as whatever
moving clock is coincident with Cd even though, according to the rest frame,
each moving clock is ticking at the same rate as Ch.

Tom

On Fri, 6 Apr 2007 13:21:41 -0500, "Tom S." wrote:


"Tom S." wrote in message
om...

"David" wrote in message
...

I don't see how to resolve this problem. According to the rest frame
observers, clock Ch is running at the same rate as every moving frame
clock. Clock Ch got set to zero at the same time as Clock A in the
moving frame and Clock Cd were set to zero. According to the rest
frame observers, Clock Ch is running at the same rate as Clock A and
Clock B, so therefore this must have been some problem with the
initialization of Clock B. But I could not determine how the rest
frame observers say the initialization of Clock B should have been
done.


David, there are _two*_ relativistic effects going on he

(1) 'moving' clocks tick slower than 'rest' clocks according to the rest
frame. This means that if the rest observer watches any _one_ of the
moving clocks as it passes by a line of rest-frame-synchronized clocks,
the moving clock will lag more and more behind each successive rest-clock
that it passes.

(2) moving clocks that are strung along the x axis of the moving frame
and synchronized according to the moving frame are _not synchronized_
according to the rest frame. This means that two different moving clocks
do not read the same time simultaneously according to the rest frame.

As others have pointed out, you are overlooking (2). When (2) is taken
into account, everything is consistent.


David. In many previous posts you have demonstrated that you have the
ability to correctly use the Lorentz transformation equations. Why don't
you use these equations to answer your question for yourself? I would guess
that virtually every serious student of SR has done something like the
following.

Cut out two long strips of paper. Along one strip (the 'rest strip') draw
circles representing the faces of stopwatches spaced at equal intervals of
distance, say 0.866 light-seconds. Use a pencil to draw all the second
hands pointing to zero on these watches. This strip represents the watches
in the rest frame at time t = 0 in the rest frame.

The other strip will represent the 'moving' frame as viewed by the rest
frame at time t = 0 in the rest frame. According to your scenario, the
moving frame is moving in the negative x direction relative to the rest
frame with a speed 0.866c. For convenience, let the clocks in this frame be
spaced at 2 times 0.866c according to the moving frame. Due to length
contraction, you will need to draw these watches spaced at only 0.866
light-seconds since we are considering how things look from the point of
view of the rest frame. So, the clocks will be drawn with the same spacing
on both strips. Assume the middle watch on each strip marks the origin of
each frame. Use the LT to determine the reading of each watch in the moving
frame at time t = 0 according to the rest frame.

Place the strips along side each other so the origins are adjacent. This
represents the state of affairs at time t = 0 according to the rest frame.
You will see how the rest frame claims that the moving clocks are not
synchronized. Now let time in the rest frame advance by one second. You
will need to advance all of the second-hands of the moving clocks by 0.5
seconds (time dilation). These moving clocks will of course still be
unsynchronized with one another according to the rest frame. Finally, you
need to slide the moving strip in the negative x direction relative to the
rest strip by an interval of 0.866 light-seconds (i.e., equal to the spacing
of the clocks). Now you have the state of affairs at time t = 1 sec
according to the rest frame.

Repeat this as many times as you wish to advance forward in time according
to the rest frame. You can also add two more clocks at the origin of the
rest frame, one ticking at half the rate of a rest clock and one ticking at
twice the rate of a rest clock. These represent your Ch and Cd clocks.
You will discover that the Cd clock always reads the same time as whatever
moving clock is coincident with Cd even though, according to the rest frame,
each moving clock is ticking at the same rate as Ch.

Tom
Thank you for the posting. And I do follow that the rest frame
doesn't agree that the clocks strung out in the moving frame are
synchronized. But that's not what I don't follow. Here's what I'm
trying to understand. The rest frame observer measures that all
clocks in the moving frame all run at the same rate, albeit they might
not all be set to the time as measured by the rest frame observer.

Now the moving frame observer puts a clock into the rest frame with
its rate set to twice the normal clock rate. That clock is
initialized at the starting point to t=0, and it matches the reading
of every moving frame clock that passes it.

Now the rest frame observer puts a clock into the rest frame at the
same point but has the clock rate set to half the rate of a normal
clock (clock Ch). He says this clock is running at the same rate as
all the moving frame clocks, and it is also synchronized (running at
the same rate and initialized to the same time) as the one moving
clock that was co-located with it at time t0.

Every moving frame clock that passes this clock (clock Ch) does not
agree with the time shown on clock Ch. The rest frame observer says
all the moving frame clocks are running at the same rate as clock Ch,
and therefore each moving frame clock that Ch passes was not set to
the correct initial time. But when I tried to figure out what method
should be used, according to the rest frame observer, to set the
initial time of all the moving frame clocks so that the reading of
clock Ch agrees with each clock it passes I was unable to come up with
an answer.

So if the rest frame observer at time t=0 initializes clock Ch to zero
and there iare moving frame clocks with one at some distance L1 and
the other at a distance L2 from this point in space as measured in the
rest frame, what initial time should each of those clocks be set to so
that when either of those clocks pass clock Ch they agree with the
reading on clock Ch. The rest frame observer measures all these
clocks to be running at the same rate.

By the way, I posted a thread "Nuts and bolts". Perhaps you could add
your insight there.

Thanks again for your posting.
David

On Apr 7, 7:24 am, David wrote:
On Fri, 6 Apr 2007 13:21:41 -0500, "Tom S." wrote:

"Tom S." wrote in message
om...

"David" wrote in message
. ..

I don't see how to resolve this problem. According to the rest frame
observers, clock Ch is running at the same rate as every moving frame
clock. Clock Ch got set to zero at the same time as Clock A in the
moving frame and Clock Cd were set to zero. According to the rest
frame observers, Clock Ch is running at the same rate as Clock A and
Clock B, so therefore this must have been some problem with the
initialization of Clock B. But I could not determine how the rest
frame observers say the initialization of Clock B should have been
done.

David, there are _two*_ relativistic effects going on he

(1) 'moving' clocks tick slower than 'rest' clocks according to the rest
frame. This means that if the rest observer watches any _one_ of the
moving clocks as it passes by a line of rest-frame-synchronized clocks,
the moving clock will lag more and more behind each successive rest-clock
that it passes.

(2) moving clocks that are strung along the x axis of the moving frame
and synchronized according to the moving frame are _not synchronized_
according to the rest frame. This means that two different moving clocks
do not read the same time simultaneously according to the rest frame.

As others have pointed out, you are overlooking (2). When (2) is taken
into account, everything is consistent.

David. In many previous posts you have demonstrated that you have the
ability to correctly use the Lorentz transformation equations. Why don't
you use these equations to answer your question for yourself? I would guess
that virtually every serious student of SR has done something like the
following.

Cut out two long strips of paper. Along one strip (the 'rest strip') draw
circles representing the faces of stopwatches spaced at equal intervals of
distance, say 0.866 light-seconds. Use a pencil to draw all the second
hands pointing to zero on these watches. This strip represents the watches
in the rest frame at time t = 0 in the rest frame.

The other strip will represent the 'moving' frame as viewed by the rest
frame at time t = 0 in the rest frame. According to your scenario, the
moving frame is moving in the negative x direction relative to the rest
frame with a speed 0.866c. For convenience, let the clocks in this frame be
spaced at 2 times 0.866c according to the moving frame. Due to length
contraction, you will need to draw these watches spaced at only 0.866
light-seconds since we are considering how things look from the point of
view of the rest frame. So, the clocks will be drawn with the same spacing
on both strips. Assume the middle watch on each strip marks the origin of
each frame. Use the LT to determine the reading of each watch in the moving
frame at time t = 0 according to the rest frame.

Place the strips along side each other so the origins are adjacent. This
represents the state of affairs at time t = 0 according to the rest frame.
You will see how the rest frame claims that the moving clocks are not
synchronized. Now let time in the rest frame advance by one second. You
will need to advance all of the second-hands of the moving clocks by 0.5
seconds (time dilation). These moving clocks will of course still be
unsynchronized with one another according to the rest frame. Finally, you
need to slide the moving strip in the negative x direction relative to the
rest strip by an interval of 0.866 light-seconds (i.e., equal to the spacing
of the clocks). Now you have the state of affairs at time t = 1 sec
according to the rest frame.

Repeat this as many times as you wish to advance forward in time according
to the rest frame. You can also add two more clocks at the origin of the
rest frame, one ticking at half the rate of a rest clock and one ticking at
twice the rate of a rest clock. These represent your Ch and Cd clocks.
You will discover that the Cd clock always reads the same time as whatever
moving clock is coincident with Cd even though, according to the rest frame,
each moving clock is ticking at the same rate as Ch.

Tom

Thank you for the posting. And I do follow that the rest frame
doesn't agree that the clocks strung out in the moving frame are
synchronized. But that's not what I don't follow. Here's what I'm
trying to understand. The rest frame observer measures that all
clocks in the moving frame all run at the same rate, albeit they might
not all be set to the time as measured by the rest frame observer.

Now the moving frame observer puts a clock into the rest frame with
its rate set to twice the normal clock rate. That clock is
initialized at the starting point to t=0, and it matches the reading
of every moving frame clock that passes it.

Now the rest frame observer puts a clock into the rest frame at the
same point but has the clock rate set to half the rate of a normal
clock (clock Ch). He says this clock is running at the same rate as
all the moving frame clocks, and it is also synchronized (running at
the same rate and initialized to the same time) as the one moving
clock that was co-located with it at time t0.

Every moving frame clock that passes this clock (clock Ch) does not
agree with the time shown on clock Ch. The rest frame observer says
all the moving frame clocks are running at the same rate as clock Ch,
and therefore each moving frame clock that Ch passes was not set to
the correct initial time. But when I tried to figure out what method
should be used, according to the rest frame observer, to set the
initial time of all the moving frame clocks so that the reading of
clock Ch agrees with each clock it passes I was unable to come up with
an answer.

So if the rest frame observer at time t=0 initializes clock Ch to zero
and there iare moving frame clocks with one at some distance L1 and
the other at a distance L2 from this point in space as measured in the
rest frame, what initial time should each of those clocks be set to so
that when either of those clocks pass clock Ch they agree with the
reading on clock Ch. The rest frame observer measures all these
clocks to be running at the same rate.

By the way, I posted a thread "Nuts and bolts". Perhaps you could add
your insight there.

Thanks again for your posting.
David- Hide quoted text -

- Show quoted text -

I assume you mean that the two moving frame clocks are at the rest-
frame distances L1 and L2 at the _same_ time t = 0 according to the
rest frame. If so, then the two clocks should both be set to zero at
this instant in order to have them pass Ch with the same reading as
Ch. Of course, the moving observer will say that these two clocks are
now not synchronized.

Equivalently, if a moving clock is located at coordinate xprime in the
moving frame, then initialize this clock to zero at the instant it has
an x-coordinate in the rest frame equal to xprime/gamma. Then the
clock will agree with Ch when it passes Ch.

Tom