On Sat, 30 Jul 2005 00:39:43 +0000 (UTC), bz
wrote:

H@..(Henri Wilson) wrote in
:


usual nonsense snipped.




The clocks don't change within their own FoR [aside from drifts that are
much smaller than the observed realitivistic effects seen by comparing
clocks in different FoRs]

But you must be aware by now that the TO sees a change in the clock. He
is always in the frame of the clock.

He sees the rest of the universe slow down as his clock speeds up.


But can you not get it into your head that the universe doesn't really slow
down just because HE MOVES into orbit?
His clock rate has obviously physically quickened. It now gives out more ticks
per orbit that it did on the ground.

What could be more obvious?


He is
smart enough to realize that from the FoR of the rest of the universe, his
time has speeded up because he moved out of the earths gravity field. [not
all of it, just most of it].

No Bob. He is smart enough to realize that his clock is not perfect. It is
reading fast when in free fall and orbiting the earth.


red...............white
because the c-v photons continue to be red shifted, they move deeper
into the infra red.

No the effect stops after a certain distance. After that, any dopler
shift remains constant.

Why would the c-v photons stop losing energy? They should keep losing
energy until they reach 0 Hz in frequency.

I didn't say they would. I said the unification process asymptotes towards an
equilibrium.
Redshift through 'atom dragging' continues forever.


You will have to wait till I incorporate this into my program. It is a
little tricky.

I have been very patient with you, Henri.

the principle involved is simple. It shouldn't need further explaining.


Bob, here is a question I asked several years ago:

Two photons are emitted by two differently moving sources. One source
is at rest wrt Earth. The two photons end up traveling across space
side by side, as SR insists.
When they reach an observer here, one exhibits a doppler shift, the
other doesn't.
How come?
Why is one shifted and not the other?

One lost or gained energy wrt the observer due to the relative motion
between the source and the observer. The energy of motion of the source
[wrt the earth] adds/subtracts from the energy/photon which causes a
shift called 'doppler shift'.

Bob, the photon might have left its source a billion years ago.

might have.

Does it carry information about the relative velocity of its source
along with it?

It must have because we observe it has a wavelength/frequency/energy and a
velocity. We observe the velocity is c. We can measure the
wavelength/frequency/energy.

Would it not be easier if it just moved at c+v wrt us?
That would convey all the information we need.



----------------------------P1__________-_________E
----------------------------P2

Two photons are approaching Earth together. They know not from where or
whence they came.

They know what energy was imparted unto them.

How do they know that?
What makes one traveling photon's energy different from another's?

If you reply, "its frequency", please define that frequency.

When they arrive, one doppler shifts, the other doesn't.

How come?

One source had a non zero relative velocity wrt earth, the other's source
had a zero relative velocity wrt earth.

Good.


Two photons from the same source. One arrives at earth when earth is moving
toward the source, the other arrives when earth is moving away from the
source. Both are moving at c. They show different doppler shifts. How come?

Good.


We don't know 'how come' we just know that all photons, regardless of
doppler shift, seem to be moving at c regardless of the motion of source or
destination.

No we don't Bob. We DO know that doppler shift is a direct consequence of
light's relative velocity to its target.


Buy a police Lidar. It doesn't care about anything but the
RELATIVE velocities of the gun and the target.

....Phase shift 'beat rate'...small changes in light speed are irrelevant.


The statement "a stick occupies a length of space" doesn't require any
FoRs.

Yes it does, just like light speed must be measure wrt something. The
length is different when measured in the sticks FoR and when measured
from another FoR at motion wrt the first.

Read what I said Bob.

The statement "a stick occupies a length of space" doesn't require a
FoR.

No more or less than "There is a beam of light moving at c" requires a FoR.

You are trying to give it a value..... Hence not an equivalent statement.




Clocks can be synced within an individual FoR, then moved. If they are
perfect clocks, they will remain in synch forever.

When you move them from one FoR to another, you are Lorentz transforming
them.

Only if an absolute frame exists.


Like rod length, clock rate is not physically affected by velocity
changes. The proof is trivial, as I have shown many times. (if a rod is
given a short shove, does its length physically increase or decrease)

When it moves from one FoR to another, you are Lorentz transforming it.

That is an illusion unless an absolute frame exists.
The proof is trivial...see above.

HW.
www.users.bigpond.com/hewn/index.htm

Sometimes I feel like a complete failure.
The most useful thing I have ever done is prove Einstein wrong.

H@..(Henri Wilson) wrote in
:

On Sat, 30 Jul 2005 00:39:43 +0000 (UTC), bz
wrote:

H@..(Henri Wilson) wrote in
m:


usual nonsense snipped.




The clocks don't change within their own FoR [aside from drifts that
are much smaller than the observed realitivistic effects seen by
comparing clocks in different FoRs]

But you must be aware by now that the TO sees a change in the clock.
He is always in the frame of the clock.

He sees the rest of the universe slow down as his clock speeds up.


But can you not get it into your head that the universe doesn't really
slow down just because HE MOVES into orbit?

I know that. The universe doesn't really slow down and the clock doesn't
'physically change'

His clock rate has obviously physically quickened. It now gives out more
ticks per orbit that it did on the ground.

What could be more obvious?

The problem is with the definition of 'physical change'.

When you take the clock back to the launch pad and test it, you see that it
has not changed.

He is
smart enough to realize that from the FoR of the rest of the universe,
his time has speeded up because he moved out of the earths gravity
field. [not all of it, just most of it].

No Bob. He is smart enough to realize that his clock is not perfect. It
is reading fast when in free fall and orbiting the earth.

But I thought we were working with perfect clocks in this thought
experiment. A perfect clock CAN NOT physically change.



red...............white
because the c-v photons continue to be red shifted, they move deeper
into the infra red.

No the effect stops after a certain distance. After that, any dopler
shift remains constant.

Why would the c-v photons stop losing energy? They should keep losing
energy until they reach 0 Hz in frequency.

I didn't say they would. I said the unification process asymptotes
towards an equilibrium.

There is no mechanism that is statistically valid that will allow you to
ADD velocity to c'=c-v photons as they travel through the low temperature
gasses in interstellar space.

Redshift through 'atom dragging' continues forever.


You will have to wait till I incorporate this into my program. It is a
little tricky.

I have been very patient with you, Henri.

the principle involved is simple. It shouldn't need further explaining.

Unless you invoke magic, there is no explanation for 'extinction' of c'=c-v
photon. There is nothing that allows them to gain velocity in interstellar
space. There are mechanisms that will cause them to continue to lose
energy/velocity.

This is a fatal flaw for c'=c+/-v.

Bob, here is a question I asked several years ago:

Two photons are emitted by two differently moving sources. One
source is at rest wrt Earth. The two photons end up traveling across
space side by side, as SR insists.
When they reach an observer here, one exhibits a doppler shift, the
other doesn't.
How come?
Why is one shifted and not the other?

One lost or gained energy wrt the observer due to the relative motion
between the source and the observer. The energy of motion of the
source [wrt the earth] adds/subtracts from the energy/photon which
causes a shift called 'doppler shift'.

Bob, the photon might have left its source a billion years ago.

might have.

Does it carry information about the relative velocity of its source
along with it?

It must have because we observe it has a wavelength/frequency/energy and
a velocity. We observe the velocity is c. We can measure the
wavelength/frequency/energy.

Would it not be easier if it just moved at c+v wrt us?
That would convey all the information we need.

It would be easier if the earth were flat and there were only 4 elements.

----------------------------P1__________-_________E
----------------------------P2

Two photons are approaching Earth together. They know not from where
or whence they came.

They know what energy was imparted unto them.

How do they know that?

I have no idea.

What makes one traveling photon's energy different from another's?
If you reply, "its frequency", please define that frequency.

f = E/h

and here is wavelength too.

lamda = c h / E
[aside: is there a significance to c h = 1.986447*10^-25*joule*m ?]

When they arrive, one doppler shifts, the other doesn't.

How come?

One source had a non zero relative velocity wrt earth, the other's
source had a zero relative velocity wrt earth.

Good.


Two photons from the same source. One arrives at earth when earth is
moving toward the source, the other arrives when earth is moving away
from the source. Both are moving at c. They show different doppler
shifts. How come?

Good.

We don't know 'how come' we just know that all photons, regardless of
doppler shift, seem to be moving at c regardless of the motion of source
or destination.

No we don't Bob. We DO know that doppler shift is a direct consequence
of light's relative velocity to its target.

Of the emitter's relative velocity wrt the target.
We haven't observed any photons traveling at velocities other than c in a
vacuum.

Buy a police Lidar. It doesn't care about anything but the
RELATIVE velocities of the gun and the target.

...Phase shift 'beat rate'...small changes in light speed are
irrelevant.

Changes in light speed have never been observed in vacuum.

If you claim the doppler shift is due to the photon's velocity then doppler
radar and ALL doppler effects are due to the same effect. There is no
'classical doppler shift' plus 'relativistic doppler shift'. There is only
one doppler shift.

But the police lidar shows that it doesn't matter whether the source of the
target is moving. The important factor is the RELATIVE velocity.

The statement "a stick occupies a length of space" doesn't require
any FoRs.

Yes it does, just like light speed must be measure wrt something. The
length is different when measured in the sticks FoR and when measured
from another FoR at motion wrt the first.

Read what I said Bob.

The statement "a stick occupies a length of space" doesn't require a
FoR.

No more or less than "There is a beam of light moving at c" requires a
FoR.

You are trying to give it a value..... Hence not an equivalent
statement.

Measure a stick or light speed without a value? Both require values.

Clocks can be synced within an individual FoR, then moved. If they are
perfect clocks, they will remain in synch forever.

When you move them from one FoR to another, you are Lorentz transforming
them.

Only if an absolute frame exists.

No. If you have two FoRs and move something from one FoR to the other FoR
you are performing a physical lorentz transform.

Like rod length, clock rate is not physically affected by velocity
changes. The proof is trivial, as I have shown many times. (if a rod
is given a short shove, does its length physically increase or
decrease)

When it moves from one FoR to another, you are Lorentz transforming it.

That is an illusion unless an absolute frame exists.
The proof is trivial...see above.

Moving from one FoR to another transforms the object.

In fact, if you have a perfect rod, you can not give it a shove.
You can only create a perfect duplicate in another FoR and set that FoR in
motion. When you do that, you will find that the LT gives you the length of
the second rod.




--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

On Sun, 31 Jul 2005 04:42:07 +0000 (UTC), bz
wrote:

H@..(Henri Wilson) wrote in
:



He sees the rest of the universe slow down as his clock speeds up.


But can you not get it into your head that the universe doesn't really
slow down just because HE MOVES into orbit?

I know that. The universe doesn't really slow down and the clock doesn't
'physically change'

the clock DOES physically change.


His clock rate has obviously physically quickened. It now gives out more
ticks per orbit that it did on the ground.

What could be more obvious?

The problem is with the definition of 'physical change'.

When you take the clock back to the launch pad and test it, you see that it
has not changed.

Its rate should return to the original one. BUT Its reading will have changed.
That will be proof that its rate HAD phyically changed during the time it was
in orbit.
Thus we have an experiment that could clear up this controversy very easily.
But does any relativist want to DO this simple experiment.

OF COURSE NOT.

Why not Bob?


He is
smart enough to realize that from the FoR of the rest of the universe,
his time has speeded up because he moved out of the earths gravity
field. [not all of it, just most of it].

No Bob. He is smart enough to realize that his clock is not perfect. It
is reading fast when in free fall and orbiting the earth.

But I thought we were working with perfect clocks in this thought
experiment. A perfect clock CAN NOT physically change.

Whatever made you think that?
We are talking about GPS atomic clocks.



Why would the c-v photons stop losing energy? They should keep losing
energy until they reach 0 Hz in frequency.

I didn't say they would. I said the unification process asymptotes
towards an equilibrium.

There is no mechanism that is statistically valid that will allow you to
ADD velocity to c'=c-v photons as they travel through the low temperature
gasses in interstellar space.

I will look for an unknown mechanism then.


Redshift through 'atom dragging' continues forever.


You will have to wait till I incorporate this into my program. It is a
little tricky.

I have been very patient with you, Henri.

the principle involved is simple. It shouldn't need further explaining.

Unless you invoke magic, there is no explanation for 'extinction' of c'=c-v
photon. There is nothing that allows them to gain velocity in interstellar
space. There are mechanisms that will cause them to continue to lose
energy/velocity.

they are probably sufficient.
The 'speed gain' is not important for my theory...but I don't reject the idea.

If a relatively dense volume of space gas is moving away from a star, it is
possible that the gas constitutes a medium in which Maxwell's equations hold,
in which case, the volume as a whole might possess some kind of 'natural light
speed'.
I'm not saying it does.....but I am not rejecting the idea outright.


This is a fatal flaw for c'=c+/-v.

Bob, here is a question I asked several years ago:

Two photons are emitted by two differently moving sources. One
source is at rest wrt Earth. The two photons end up traveling across
space side by side, as SR insists.
When they reach an observer here, one exhibits a doppler shift, the
other doesn't.
How come?
Why is one shifted and not the other?

One lost or gained energy wrt the observer due to the relative motion
between the source and the observer. The energy of motion of the
source [wrt the earth] adds/subtracts from the energy/photon which
causes a shift called 'doppler shift'.

Bob, the photon might have left its source a billion years ago.

might have.

Does it carry information about the relative velocity of its source
along with it?

It must have because we observe it has a wavelength/frequency/energy and
a velocity. We observe the velocity is c. We can measure the
wavelength/frequency/energy.

Would it not be easier if it just moved at c+v wrt us?
That would convey all the information we need.

It would be easier if the earth were flat and there were only 4 elements.

But the Earth IS flat....look out your window......


----------------------------P1__________-_________E
----------------------------P2

Two photons are approaching Earth together. They know not from where
or whence they came.

They know what energy was imparted unto them.

How do they know that?

I have no idea.

What makes one traveling photon's energy different from another's?
If you reply, "its frequency", please define that frequency.

f = E/h

and here is wavelength too.

lamda = c h / E
[aside: is there a significance to c h = 1.986447*10^-25*joule*m ?]

well then, please define E.


When they arrive, one doppler shifts, the other doesn't.

How come?

One source had a non zero relative velocity wrt earth, the other's
source had a zero relative velocity wrt earth.

Good.


Two photons from the same source. One arrives at earth when earth is
moving toward the source, the other arrives when earth is moving away
from the source. Both are moving at c. They show different doppler
shifts. How come?

Good.

We don't know 'how come' we just know that all photons, regardless of
doppler shift, seem to be moving at c regardless of the motion of source
or destination.

No we don't Bob. We DO know that doppler shift is a direct consequence
of light's relative velocity to its target.

Of the emitter's relative velocity wrt the target.
We haven't observed any photons traveling at velocities other than c in a
vacuum.

Because we haven't looked, have we Bob?


Buy a police Lidar. It doesn't care about anything but the
RELATIVE velocities of the gun and the target.

...Phase shift 'beat rate'...small changes in light speed are
irrelevant.

Changes in light speed have never been observed in vacuum.

Because we haven't looked, have we Bob?


If you claim the doppler shift is due to the photon's velocity then doppler
radar and ALL doppler effects are due to the same effect. There is no
'classical doppler shift' plus 'relativistic doppler shift'. There is only
one doppler shift.

But the police lidar shows that it doesn't matter whether the source of the
target is moving. The important factor is the RELATIVE velocity.

Police radar doesn't use doppler shift. Even though it is called 'doppler
radar...it isn't actually based on the signal's doppler shift.

It is based on the way the phasing between the outgoing and return signals
changes with time.
The small change in reflected light speed (c+2v) makes no significant
difference to the outcome.


No more or less than "There is a beam of light moving at c" requires a
FoR.

You are trying to give it a value..... Hence not an equivalent
statement.

Measure a stick or light speed without a value? Both require values.

A stick occupies a length of space whether or not you give it a value.


Clocks can be synced within an individual FoR, then moved. If they are
perfect clocks, they will remain in synch forever.

When you move them from one FoR to another, you are Lorentz transforming
them.

Only if an absolute frame exists.

No. If you have two FoRs and move something from one FoR to the other FoR
you are performing a physical lorentz transform.

LTs generally don't exist, physical or illusory.


Like rod length, clock rate is not physically affected by velocity
changes. The proof is trivial, as I have shown many times. (if a rod
is given a short shove, does its length physically increase or
decrease)

When it moves from one FoR to another, you are Lorentz transforming it.

That is an illusion unless an absolute frame exists.
The proof is trivial...see above.

Moving from one FoR to another transforms the object.

In fact, if you have a perfect rod, you can not give it a shove.
You can only create a perfect duplicate in another FoR and set that FoR in
motion. When you do that, you will find that the LT gives you the length of
the second rod.

Now you are being silly again.


HW.
www.users.bigpond.com/hewn/index.htm

Sometimes I feel like a complete failure.
The most useful thing I have ever done is prove Einstein wrong.

H@..(Henri Wilson) wrote in
:

On Sun, 31 Jul 2005 04:42:07 +0000 (UTC), bz
wrote:

H@..(Henri Wilson) wrote in
m:



He sees the rest of the universe slow down as his clock speeds up.


But can you not get it into your head that the universe doesn't really
slow down just because HE MOVES into orbit?

I know that. The universe doesn't really slow down and the clock doesn't
'physically change'

the clock DOES physically change.

How does it change? Does the size change? Does the shape change?
Can you see anything that changes WITHOUT LOOKING OUTSIDE the clocks FoR?

It is meaningless to say that it physically changes when no one in the
clocks FoR can tell that it changed W/o looking outside the FoR.

His clock rate has obviously physically quickened. It now gives out
more ticks per orbit that it did on the ground.

What could be more obvious?

The problem is with the definition of 'physical change'.

When you take the clock back to the launch pad and test it, you see that
it has not changed.

Its rate should return to the original one. BUT Its reading will have
changed. That will be proof that its rate HAD phyically changed during
the time it was in orbit.

The reading will have changed, agreed. The 'physical' is meaningless.

Thus we have an experiment that could clear up this controversy very
easily. But does any relativist want to DO this simple experiment.

OF COURSE NOT.

No objection to doing the experiment. In fact it has already been done. The
times changes as predicted.

Does the change 'prove' that something physically changed? No.


Why not Bob?

No one objects.

He is
smart enough to realize that from the FoR of the rest of the universe,
his time has speeded up because he moved out of the earths gravity
field. [not all of it, just most of it].

No Bob. He is smart enough to realize that his clock is not perfect.
It is reading fast when in free fall and orbiting the earth.

But I thought we were working with perfect clocks in this thought
experiment. A perfect clock CAN NOT physically change.

Whatever made you think that?
We are talking about GPS atomic clocks.

Then we know what they do, already.

Why would the c-v photons stop losing energy? They should keep losing
energy until they reach 0 Hz in frequency.

I didn't say they would. I said the unification process asymptotes
towards an equilibrium.

There is no mechanism that is statistically valid that will allow you to
ADD velocity to c'=c-v photons as they travel through the low
temperature gasses in interstellar space.

I will look for an unknown mechanism then.

Good luck.

Redshift through 'atom dragging' continues forever.


You will have to wait till I incorporate this into my program. It is
a little tricky.

I have been very patient with you, Henri.

the principle involved is simple. It shouldn't need further
explaining.

Unless you invoke magic, there is no explanation for 'extinction' of
c'=c-v photon. There is nothing that allows them to gain velocity in
interstellar space. There are mechanisms that will cause them to
continue to lose energy/velocity.

they are probably sufficient.
The 'speed gain' is not important for my theory...but I don't reject the
idea.

If a relatively dense volume of space gas is moving away from a star, it
is possible that the gas constitutes a medium in which Maxwell's
equations hold, in which case, the volume as a whole might possess some
kind of 'natural light speed'.
I'm not saying it does.....but I am not rejecting the idea outright.

If the photons pass through a cloud of gas that is moving at speeds faster
than light, then the c'=c-v photons could gain velocity and get up to c.
The problem is that such a cloud will also speed up the c photons and the
c'=c+v photons.

This is a fatal flaw for c'=c+/-v.
.....
Would it not be easier if it just moved at c+v wrt us?
That would convey all the information we need.

It would be easier if the earth were flat and there were only 4
elements.

But the Earth IS flat....look out your window......

.....

What makes one traveling photon's energy different from another's?
If you reply, "its frequency", please define that frequency.

f = E/h

and here is wavelength too.

lamda = c h / E
[aside: is there a significance to c h = 1.986447*10^-25*joule*m ?]

well then, please define E.

E=mcc+KE

When they arrive, one doppler shifts, the other doesn't.

How come?

One source had a non zero relative velocity wrt earth, the other's
source had a zero relative velocity wrt earth.

Good.


Two photons from the same source. One arrives at earth when earth is
moving toward the source, the other arrives when earth is moving away
from the source. Both are moving at c. They show different doppler
shifts. How come?

Good.

We don't know 'how come' we just know that all photons, regardless of
doppler shift, seem to be moving at c regardless of the motion of
source or destination.

No we don't Bob. We DO know that doppler shift is a direct consequence
of light's relative velocity to its target.

Of the emitter's relative velocity wrt the target.
We haven't observed any photons traveling at velocities other than c in
a vacuum.

Because we haven't looked, have we Bob?

We are looking all the time, Henri. The HST is studing stars all the time.
Many things would show that the photons were traveling at a velocity other
than c.

Buy a police Lidar. It doesn't care about anything but the
RELATIVE velocities of the gun and the target.

...Phase shift 'beat rate'...small changes in light speed are
irrelevant.

Changes in light speed have never been observed in vacuum.

Because we haven't looked, have we Bob?

We are looking all the time, Henri.

If you claim the doppler shift is due to the photon's velocity then
doppler radar and ALL doppler effects are due to the same effect. There
is no 'classical doppler shift' plus 'relativistic doppler shift'. There
is only one doppler shift.

But the police lidar shows that it doesn't matter whether the source of
the target is moving. The important factor is the RELATIVE velocity.

Police radar doesn't use doppler shift. Even though it is called
'doppler radar...it isn't actually based on the signal's doppler shift.

Henri, I hate to tell you, but the beat between the outgoing and return
signals is due to the difference between those frequencies. The difference
is due to the doppler shift.

It is called doppler radar / lidar because the output signal is the direct
effect of the doppler shift.

It is based on the way the phasing between the outgoing and return
signals changes with time.
The small change in reflected light speed (c+2v) makes no significant
difference to the outcome.


No more or less than "There is a beam of light moving at c" requires a
FoR.

You are trying to give it a value..... Hence not an equivalent
statement.

Measure a stick or light speed without a value? Both require values.

A stick occupies a length of space whether or not you give it a value.

Yep. But once you decide to give it a value, the value you are going to
measure is going to depend of your FoR.

Clocks can be synced within an individual FoR, then moved. If they
are perfect clocks, they will remain in synch forever.

When you move them from one FoR to another, you are Lorentz
transforming them.

Only if an absolute frame exists.

No. If you have two FoRs and move something from one FoR to the other
FoR you are performing a physical lorentz transform.

LTs generally don't exist, physical or illusory.

An equation exists. Denial of the existance of the transform is contrary to
reality.

Like rod length, clock rate is not physically affected by velocity
changes. The proof is trivial, as I have shown many times. (if a rod
is given a short shove, does its length physically increase or
decrease)

When it moves from one FoR to another, you are Lorentz transforming
it.

That is an illusion unless an absolute frame exists.
The proof is trivial...see above.

Moving from one FoR to another transforms the object.

In fact, if you have a perfect rod, you can not give it a shove.
You can only create a perfect duplicate in another FoR and set that FoR
in motion. When you do that, you will find that the LT gives you the
length of the second rod.

Now you are being silly again.

No sillier than Henri.




--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

bz wrote:
H@..(Henri Wilson) wrote in
:
Police radar doesn't use doppler shift. Even though it is called
'doppler radar...it isn't actually based on the signal's doppler shift.

Henri, I hate to tell you, but the beat between the outgoing and return
signals is due to the difference between those frequencies. The difference
is due to the doppler shift.

It is called doppler radar / lidar because the output signal is the direct
effect of the doppler shift.

Right. "Rate of change of relative phase" and "doppler shift"
are the same thing. Henri is very confused. He thinks that because
the frequency shift is measured in terms of phase change vs
time, that somehow it's not a frequency shift.

It's a frequency shift. To be precise, an incoming signal
is compared to the outgoing signal exp(j*2*pi*f*t), which
has phase 2*pi*f*t vs. time. The incoming signal is found
to have a phase difference which is linear in time:
2*pi*f*t + a*t = 2*pi*(f + a/(2*pi)) t

So measurement of the phase slope a is equivalent to
measuring the frequency shift a/(2*pi), which just
differs by a constant. Measuring the phase slope IS
how you measure doppler shift. Henri is flat out wrong.

My employer builds Doppler radars, and I've simulated them.
I know more than a little about this subject.

- Randy

"bz" wrote in message 98.139...
H@..(Henri Wilson) wrote in
:
snip

We don't know 'how come' we just know that all photons, regardless of
doppler shift, seem to be moving at c regardless of the motion of source
or destination.

No we don't Bob. We DO know that doppler shift is a direct consequence
of light's relative velocity to its target.

I don't belive you know that.
*Projectiles* have *targets*.
Light works more like this:
http://www.physics.yorku.ca/undergra...ch/Feynm4.html

Sue...

snip

"Randy Poe" wrote in news:1122862445.222877.40820
@g47g2000cwa.googlegroups.com:


bz wrote:
H@..(Henri Wilson) wrote in
:
Police radar doesn't use doppler shift. Even though it is called
'doppler radar...it isn't actually based on the signal's doppler
shift.

Henri, I hate to tell you, but the beat between the outgoing and return
signals is due to the difference between those frequencies. The
difference
is due to the doppler shift.

It is called doppler radar / lidar because the output signal is the
direct
effect of the doppler shift.

Right. "Rate of change of relative phase" and "doppler shift"
are the same thing. Henri is very confused. He thinks that because
the frequency shift is measured in terms of phase change vs
time, that somehow it's not a frequency shift.

It's a frequency shift. To be precise, an incoming signal
is compared to the outgoing signal exp(j*2*pi*f*t), which
has phase 2*pi*f*t vs. time. The incoming signal is found
to have a phase difference which is linear in time:
2*pi*f*t + a*t = 2*pi*(f + a/(2*pi)) t

So measurement of the phase slope a is equivalent to
measuring the frequency shift a/(2*pi), which just
differs by a constant. Measuring the phase slope IS
how you measure doppler shift. Henri is flat out wrong.

My employer builds Doppler radars, and I've simulated them.
I know more than a little about this subject.

Yep. And I used to fix radars for a living.

BTW, The formula can get a bit more complex if one is chirping the outgoing
signal, which one might do to prevent jamming.




--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

bz wrote:
"Randy Poe" wrote in news:1122862445.222877.40820
@g47g2000cwa.googlegroups.com:
My employer builds Doppler radars, and I've simulated them.
I know more than a little about this subject.

Yep. And I used to fix radars for a living.

BTW, The formula can get a bit more complex if one is chirping the outgoing
signal, which one might do to prevent jamming.

In the systems I've seen, the (chirped) radars use a series
of pulse returns to construct a range-doppler space. The
doppler dimension is calculated by a Fourier transform
across the pulses. It took me a little bit of time
when I first started this work, to understand why that
constituted a doppler estimation.

- Randy

"Randy Poe" wrote in
oups.com:


bz wrote:
"Randy Poe" wrote in
news:1122862445.222877.40820 @g47g2000cwa.googlegroups.com:
My employer builds Doppler radars, and I've simulated them.
I know more than a little about this subject.

Yep. And I used to fix radars for a living.

BTW, The formula can get a bit more complex if one is chirping the
outgoing signal, which one might do to prevent jamming.

In the systems I've seen, the (chirped) radars use a series
of pulse returns to construct a range-doppler space. The
doppler dimension is calculated by a Fourier transform
across the pulses. It took me a little bit of time
when I first started this work, to understand why that
constituted a doppler estimation.

Cool! I wondered exactly how it was done.

I knew that the beat would be constant in frequency [provided the target
was constant in relative velocity] [if the chirp were at a constant chirp
rate]. Things must get even more complex with evasive targets and modulated
chirp rates.



--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

On Mon, 1 Aug 2005 00:23:33 +0000 (UTC), bz wrote:

H@..(Henri Wilson) wrote in
:

On Sun, 31 Jul 2005 04:42:07 +0000 (UTC), bz
wrote:

H@..(Henri Wilson) wrote in
:



He sees the rest of the universe slow down as his clock speeds up.


But can you not get it into your head that the universe doesn't really
slow down just because HE MOVES into orbit?

I know that. The universe doesn't really slow down and the clock doesn't
'physically change'

the clock DOES physically change.

How does it change? Does the size change? Does the shape change?
Can you see anything that changes WITHOUT LOOKING OUTSIDE the clocks FoR?

The clock changes as seen in the original frame by he original observer. It
also changes by the same amount as seen by an obserevr who remains with th
clock constantly during the expeiment.


It is meaningless to say that it physically changes when no one in the
clocks FoR can tell that it changed W/o looking outside the FoR.

What a ridiculous statement.
How the hell can the TO possibly tell if the clock rate has changed if he
doesn't compare it with a known fixed standard. Any internal standard would
probably change by the same amount and would be useless.

You cannot conclude that something doesn't change because everything else you
can compare it with changes in the same way.

You really should realize that physics is not your forte, Bob.

His clock rate has obviously physically quickened. It now gives out
more ticks per orbit that it did on the ground.

What could be more obvious?

The problem is with the definition of 'physical change'.

When you take the clock back to the launch pad and test it, you see that
it has not changed.

Its rate should return to the original one. BUT Its reading will have
changed. That will be proof that its rate HAD phyically changed during
the time it was in orbit.

The reading will have changed, agreed. The 'physical' is meaningless.

Thus we have an experiment that could clear up this controversy very
easily. But does any relativist want to DO this simple experiment.

OF COURSE NOT.

No objection to doing the experiment. In fact it has already been done. The
times changes as predicted.

don't kid yourself.


Does the change 'prove' that something physically changed? No.


Why not Bob?

No one objects.

He is
smart enough to realize that from the FoR of the rest of the universe,
his time has speeded up because he moved out of the earths gravity
field. [not all of it, just most of it].

No Bob. He is smart enough to realize that his clock is not perfect.
It is reading fast when in free fall and orbiting the earth.

But I thought we were working with perfect clocks in this thought
experiment. A perfect clock CAN NOT physically change.

Whatever made you think that?
We are talking about GPS atomic clocks.

Then we know what they do, already.

They sped up slightly when in free fall and cutting the earth's fields.


Why would the c-v photons stop losing energy? They should keep losing
energy until they reach 0 Hz in frequency.

I didn't say they would. I said the unification process asymptotes
towards an equilibrium.

There is no mechanism that is statistically valid that will allow you to
ADD velocity to c'=c-v photons as they travel through the low
temperature gasses in interstellar space.

I will look for an unknown mechanism then.

Good luck.

Redshift through 'atom dragging' continues forever.


You will have to wait till I incorporate this into my program. It is
a little tricky.

I have been very patient with you, Henri.

the principle involved is simple. It shouldn't need further
explaining.

Unless you invoke magic, there is no explanation for 'extinction' of
c'=c-v photon. There is nothing that allows them to gain velocity in
interstellar space. There are mechanisms that will cause them to
continue to lose energy/velocity.

they are probably sufficient.
The 'speed gain' is not important for my theory...but I don't reject the
idea.

If a relatively dense volume of space gas is moving away from a star, it
is possible that the gas constitutes a medium in which Maxwell's
equations hold, in which case, the volume as a whole might possess some
kind of 'natural light speed'.
I'm not saying it does.....but I am not rejecting the idea outright.

If the photons pass through a cloud of gas that is moving at speeds faster
than light, then the c'=c-v photons could gain velocity and get up to c.
The problem is that such a cloud will also speed up the c photons and the
c'=c+v photons.

Your statement is meaningless.
The expression "a cloud of gas that is moving at speeds faster
than light" must have a reference object.


This is a fatal flaw for c'=c+/-v.
....
Would it not be easier if it just moved at c+v wrt us?
That would convey all the information we need.

It would be easier if the earth were flat and there were only 4
elements.

But the Earth IS flat....look out your window......

....

What makes one traveling photon's energy different from another's?
If you reply, "its frequency", please define that frequency.

f = E/h

and here is wavelength too.

lamda = c h / E
[aside: is there a significance to c h = 1.986447*10^-25*joule*m ?]

well then, please define E.

E=mcc+KE

Please define KE.


When they arrive, one doppler shifts, the other doesn't.

How come?

One source had a non zero relative velocity wrt earth, the other's
source had a zero relative velocity wrt earth.

Good.


Two photons from the same source. One arrives at earth when earth is
moving toward the source, the other arrives when earth is moving away
from the source. Both are moving at c. They show different doppler
shifts. How come?

Good.

We don't know 'how come' we just know that all photons, regardless of
doppler shift, seem to be moving at c regardless of the motion of
source or destination.

No we don't Bob. We DO know that doppler shift is a direct consequence
of light's relative velocity to its target.

Of the emitter's relative velocity wrt the target.
We haven't observed any photons traveling at velocities other than c in
a vacuum.

Because we haven't looked, have we Bob?

We are looking all the time, Henri. The HST is studing stars all the time.
Many things would show that the photons were traveling at a velocity other
than c.

They DO show up but they are interpreted wrongly.


But the police lidar shows that it doesn't matter whether the source of
the target is moving. The important factor is the RELATIVE velocity.

Police radar doesn't use doppler shift. Even though it is called
'doppler radar...it isn't actually based on the signal's doppler shift.

Henri, I hate to tell you, but the beat between the outgoing and return
signals is due to the difference between those frequencies. The difference
is due to the doppler shift.

It is called doppler radar / lidar because the output signal is the direct
effect of the doppler shift.

Bob, the phase difference between incoming and outgoing signals continuously
changes linearly with the object's speed.
The doppler shift of the signal due to that speed is negligible.
If it wasn't, then the device would be useless because its reading would depend
on absolute distance as well as car speed. Think about it.

It is based on the way the phasing between the outgoing and return
signals changes with time.
The small change in reflected light speed (c+2v) makes no significant
difference to the outcome.


No more or less than "There is a beam of light moving at c" requires a
FoR.

You are trying to give it a value..... Hence not an equivalent
statement.

Measure a stick or light speed without a value? Both require values.

A stick occupies a length of space whether or not you give it a value.

Yep. But once you decide to give it a value, the value you are going to
measure is going to depend of your FoR.

Only if light is used to measure it.
If instruments ar used, its length will remain constant.


Clocks can be synced within an individual FoR, then moved. If they
are perfect clocks, they will remain in synch forever.

When you move them from one FoR to another, you are Lorentz
transforming them.

Only if an absolute frame exists.

No. If you have two FoRs and move something from one FoR to the other
FoR you are performing a physical lorentz transform.

LTs generally don't exist, physical or illusory.

An equation exists. Denial of the existance of the transform is contrary to
reality.

Rubbish.
Charged particles obey a similar law, but for a different reason.



Moving from one FoR to another transforms the object.

In fact, if you have a perfect rod, you can not give it a shove.
You can only create a perfect duplicate in another FoR and set that FoR
in motion. When you do that, you will find that the LT gives you the
length of the second rod.

Now you are being silly again.

No sillier than Henri.


HW.
www.users.bigpond.com/hewn/index.htm

Sometimes I feel like a complete failure.
The most useful thing I have ever done is prove Einstein wrong.