Why are the 'Fixed Stars' so FIXED?

"Henri Wilson" HW@.... wrote in message
...
On Mon, 26 Feb 2007 20:24:05 -0000, "George Dishman"

wrote:
"Henri Wilson" HW@.... wrote in message
. ..
On Mon, 26 Feb 2007 00:15:09 -0000, "George Dishman"
....
so neither you
nor I know what it is that I'm supposed to be matching.

You are supposed to be matching the velocity curve. We
also know the phase which we can use next and may improve
the determination.

It isn't as easy as you think. See my latest post.
My green curve is a measure of acceleration not velocity.

I think this is your error.

Perhaps, I had assumed your green curve was derived in
such a way that it included the effects of both velocity
and acceleration. If not, then there could be an error
in that too while I have been thinking it was accurate.

I suppose you want to assume that the brightness curve of the pulsar can
be inferred from the bunching of the pulses...since their energy should
not
change even if their widths DO.

We can make that assumption but it isn't testable as
it will be smaller than can be measured so I treat
it as an output from the program.

If you would like to produce such a curve I will try to match it...the
I'll worry about getting the red curve right.

No, fix the maths for the red curve and match that to
the velocity curve which we do have, I'm not worried
about the brightness at all and your code for that is
already OK.

The red curve is OK. You are confusing the velocity curve with the
acceleration
curve.

No, I think you are confusing my comments about velocity.
Bear in mind there is a big difference between the actual
velocity shown by your blue curve and what you call the
"willusory" value shown by the red curve. That depends on
the true values of both radial speed and radial
acceleration.

Cool. The last several weeks of discussion will have been
totally pointless if you don't and I have better things to
do with my time (as the wife keeps telling me).

(mine is away at present. I think I need a new one.)

Nearly 30 years now and I'm very happy with mine

Some people are lucky.....

I've had about ten...they're all mad as far as I can see....different
brains
entirely...

Mine just read the above over my shoulder and left with
a grin.

You could have written your own program by now. ..and found the answer.

I had a look and I could, the only hard bit is doing
the velocity and acceleration equations for elliptical
orbits and the tedious stuff of the GUI and converting
units. However, I doubt you would trust something I
wrote anyway.

That's why I use Vbasic. It's very user friendly and quite fast.....or it
WAS
till the latest version... which is no better than Java.

EJS really cuts the work down.

Skype is becoming very popular. I don't know why the various computer
phone
companies don't try to standardise so they can be mutually compatible.

SIP is an international standard that they all should use. Skype
keeps their method proprietary to establish dominance - bad for
us but good for business. I needed a router anyway when I went
to ADSL but it's annoying I bought it a year or two before Skype
became so universal.

George

On 27 Feb 2007 02:27:42 -0800, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Mon, 26 Feb 2007 20:24:05 -0000, "George Dishman"

It isn't as easy as you think. See my latest post.
My green curve is a measure of acceleration not velocity.

I think this is your error.

Perhaps, I had assumed your green curve was derived in
such a way that it included the effects of both velocity
and acceleration. If not, then there could be an error
in that too while I have been thinking it was accurate.

It's complicated. But I'm sure we can work it out.


The red curve is OK. You are confusing the velocity curve with the
acceleration
curve.

No, I think you are confusing my comments about velocity.
Bear in mind there is a big difference between the actual
velocity shown by your blue curve and what you call the
"willusory" value shown by the red curve. That depends on
the true values of both radial speed and radial
acceleration.

Well I think we are starting to realise the big difference between the
classical view and the Bath. I think my red curve is correct. It is not related
to normal doppler.
Cool. The last several weeks of discussion will have been
totally pointless if you don't and I have better things to
do with my time (as the wife keeps telling me).

(mine is away at present. I think I need a new one.)

Nearly 30 years now and I'm very happy with mine

Some people are lucky.....

I've had about ten...they're all mad as far as I can see....different
brains
entirely...

Mine just read the above over my shoulder and left with
a grin.

beware if the ones that grin.....



That's why I use Vbasic. It's very user friendly and quite fast.....or it
WAS
till the latest version... which is no better than Java.

EJS really cuts the work down.

What's EJS? I would like to find a better program now that microsift as made
Vbasic completely unrealistic.

Skype is becoming very popular. I don't know why the various computer
phone
companies don't try to standardise so they can be mutually compatible.

SIP is an international standard that they all should use. Skype
keeps their method proprietary to establish dominance - bad for
us but good for business. I needed a router anyway when I went
to ADSL but it's annoying I bought it a year or two before Skype
became so universal.

Unfortunately, communication is one of those areas where competition is not
always a good thing. Governments should step in to at least set some common
standards.
Anyway Skype is still free...more or less


George


"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."
--Jonathan Swift.

On Feb 26, 6:53 pm, HW@....(Henri Wilson) wrote:
On 26 Feb 2007 13:12:25 -0800, "PD" wrote:


A capsule definition is that a frame of reference is a tool, a set of
coordinate axes that permeates a physical system, and in which the
user of the reference frame can specify the coordinates of spacetime
events or spatial positions of points on objects in the system
(whether moving or stationary relative to the origin of the reference
frame) as a function of time, as well as the spatial orientation of
extended objects in the system as a function of time.

That is really funny Draper. Big words impress small minds.

What big words? Did "orientation" throw you off?

Spacetime is not a physical entity.

Depends on what you mean by a physical entity. Is momentum a physical
entity?

Yes. A frame dependent one...but you don't know what that means Draper.

Sure I do, and on this small point we're in agreement.
So tell me, what makes momentum a physical entity and spacetime not?


A reference frame is no more or no less a physical entity than
spacetime.

Nobody claimed it was.

Then I'm not sure what your point was. I described what a frame of
reference in part by using the term "spacetime events" and you
commented that spacetime is not a physical entity. But since a frame
of reference is no less a physical entity than spacetime, I'm not sure
why that's important.


In the maths construction known as space/time, there is no movement.

That's why I included the word "or" between "spacetime events" and
"spatial positions". Reading comprehension problems, Henri? Is this
why you don't like to read books about relativity? Or books at all,
for that matter?

Draper, spacetime is completely static. There are no 'events'. Everything that
ever happened is laid out permanently in 4D. Static!!!!!
Even the future is static.... just not plotted yet.

A spacetime event, Henri, is something at a location (x, y, z, t) that
is distinguishable from neighboring locations. The snapping of fingers
is something that happens at a particular place and a particular time
and so marks a location in spacetime, called an "event".


....unless of course you want to agree with me that at least two orthogonal
time dimensions exist....

I asked you, where did you get the idea that a frame is everything at
rest with respect to a defined point?

Because it is Draper...including Wilsonian nort-holes.
Anything at rest wrt the defined point also defines the frame.

That doesn't tell me where you got the idea, Henri. You can also tell
me you got the idea that mammals all give live birth to their young
"because they do", but it wouldn't answer the question where you got
that ridiculous notion.

Draper, I have told you before, I'm not going to waste my time teaching YOU
elementary physics.
GO BACK TO SCHOOL... like geesey has.

Then tell me in which school, and under what professor, using which
textbook, you got the idea that a frame is everything at rest with
respect to a defined point.


Are you under the misguided impression that my definition precludes the
possibility that objects can move wrt a reference frame.

I didn't relate any impression, Henri. I asked you a question, which
you have yet to answer. You asked me a question about my definition,
and I answered that.

....with drivel...

Now once again: Where did you get the idea that a frame is everything
that is at rest with respect to a defined point?

You really can't answer a simple question, Henri? This isn't about me
going to school, it's about where YOU got this idea.


I answered your question. Now you answer mine. Are you an idiot or a
liar, Henri?

Learn some physics Draper..... you simply don't have any idea at present..

Happy to. Tell me what reference I should read about reference frames
that shows your definition, Henri.

Any real physicist knows what a reference frame is.

Sure they do. I just don't think it's the same as what you think it
is. The reason is that I can go to any reference written by "any real
physicist" and find a definition that doesn't have anything to do with
what you say it is. Unless of course, you've got an "any real
physicist" in mind that agrees with your definition. That is, unless
YOU are the only "any real physicist" you know....


IT'S A BLOODY REFERENCE FOR VELOCITIES.
EVERYTHING THAT IS MUTUALLY AT REST IS IN THE SAME FRAME.

As well as everything that is not mutually at rest -- all in the same
frame. Do you disagree with that?

PD

On Tue, 27 Feb 2007 09:22:01 -0000, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Mon, 26 Feb 2007 19:44:54 -0000, "George Dishman"

George, you are completely missing the point again. You are ignoring the
crucial factor of the delay in emission times. You are assuming that the
pulses
are emitted simultaneously from all points around the orbit.

No, I am assuming your program already includes that.

It is the initial delay that largely determines the time taken for one
pulse to
catch the next. The fact that one pulse catches the preceding one does not
indicate an 'infinite velocity' at all, as you seem to believe.

Three things determine the time between arrivals, the
time between emissions of 2.95ms, the reduced radial
distance due to radial speed which gives the classical
Doppler and the delta-v between the emissions which
produces a catch-up effect dependent on observer
distance and the extinction distance. I am assuming
your program already includes at least the first three
and you alter the observer distance to simulate the
combination of actual observer distance and extinction
distance if you haven't included extinction explicitly.

Yes you expressed that well.

To deal with the odd situation of simultaneous pulses
that occurs at and beyond the critical distance, you
can assume the Doppler equation is:

I wont be going beyond the critical distance.

Sensible, but the point is that using this equation
copes with it anyway. Just some advice, it's your
choice, but doing that would avoid crashes due to
division by zero and allow a simple test to prove
your code is correct.

No, I have already gone to a lot of trouble to avoid divisions by zero.

f'/f = c/(c-v)

or using the time between arrivals and turning it round
to find v:

v = c(1 - t'/t)

Distance does not appear in this equation.

Correct, the distance influences t' by affecting how
much catch-up effect occurs.

where t is the time between pulse emissions and t' is
the time between arrivals. That gives v=c for t'=0.

If I guess correctly, your brightness curve is

b = t/t'

so

v = c(1 - 1/b)

gives you a direct conversion from your existing green
curve (base ratio, not the log version) but the former
formulation avoids the infinity in brightness at t'=0.

"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."

So what do you call a genius who needs one of the dunces
to do his maths for him ;-)

...but you keep oversimplifying and solving the wrong problem.

I am making an assumption that your current code is correct
and includes all the effects in the way it calculates the
pulse arrival times.

No, there are many complications.
I'm quite confident my program is correct.

This is how I will have to approach it.

My program assumes that identical pulses are emitted at regular intervals
from
some 30000 points around the orbit. (the pulsar does much the same) The
true
radial velocities towards the observer are known and are stored in an
array.
The emission 'lag' between consecutive pulses is period/30000. ...that
provides
the value of t.

Yes, so far so good.

The program calculates the maximum and minimum travel times over the
observer
distance of the pulses emitted in one cycle . (I actually use three period
cycles to avoid end effects but we will only talk about 1 period here.)
The max and min are divided into 167 equal time divisions, into which each
pulse is appropriately placed as it arrives..

"as it arrives" is the key phrase there. I am assuming you
take into account both the location of the source in its
orbit and the radial speed in order to calculate the time
taken. If you are ignoring the location and just dividing
the distance between the barycentres of the binary system
and the solar system by the c+v speed then you will have
another error that I wasn't previously aware of.

Both Androcles and I soon realised that it is not necessary to include the
Rsin(theta) term because it is always very small compared with relative
movement of pulses over many lightyears.
Ignore it George.
....although, it might be significant for your pulsar.

..so some divisions might end up
with say 27 others with 116, depending on the bunching. I then plot these
167
numbers, equally spaced, across the screen. This gives the bunching
pattern of
the pulses over one period, in 'space' at the observer distance. A spatial
scan
of that pattern give brightness vs time according to the basic equation,
dn/dt
= dn/dx.dx/dt. (where dx/dt = ~c)

Assuming no nett radial velocity of the barycentre, the period of the
brightness curve is the same as that of the source star...so 30000/167 or
180
pulses are emitted per time division at the source end. The time interval
spanned by an interval is period/167 (seconds) at both source and
receiver.

[ Because the divisions are sufficiently small, I can assume an average
source
velocity for each....which I can calculate and graph (blue curve). I can
also
track and average the source velocities of all the pulses that arrive in
each
time division. Note: assume no pulse actually overtakes another or even
goes
close.]

If 'n' pulses arrive in a particular time division at distance D, then
n/180 is
an indication of the bunching there.

Right, or say b = n/180 and you plot b as the green curve.

Yes.....well, I actually plot n with a scaling factor.

If you calculate v = c(1 - 1/b) then you get the "willusory"
radial speed so v = c(1 - 180/n). Just plot v as the red
curve and the job is done.

But I don't have to.
The program has already recorded the true radial velocities of all the pulses
that arrive in a certain time division. It takes the average and plots that.

I concede that when the number of pulses that arrive in a division becomes
quite high, for instance for magnitude changes above about 2, the 'average
speed' will not be a very accurate indicator of what is happening. So at
present my red curve is not to be taken seriously for large star magnitude
variations. It is quite accurate for mag changes below about 1, which covers
most cases we want to investigate.

If I want greater accuracy, I can increase the number of sample points around
the orbit, shorten the arrival time divisions and lengthen the graph on the
screen. My method is still 100% sound. It MUST give the same answer as your
equation.

We also know the last pulse left the
source nt seconds after the first one.
If we know the source velocities of the pulses at each end of a division,
we
can calculate how much compression there has been.
Conversely, knowing the compression, we can calculate the DIFFERENCE in
source
speeds of the end two pulses

Over distance D (time D/c), the last pulse has moved from a distance ~=
c.n.t
away from the first to ~c.180.t away. (c+v is close enough to c, here).
The
speed difference is ct(n-180).D/c = (n-180)/D. My green curve is a scaled
plot
of this.

So my green curve is a graph of acceleration vs time. I have to integrate
that
wrt time to get a velocity curve.

Would you like to comment before I go on?

The last part seems to be adding an unnecessary layer of
complexity.

My approach if I was to do it your way would be slightly
simpler. You are tracking N=30000 pulses emitted equal
times T apart to take account of the Keplerian orbit. For
each point n you find the distance to the observer and
the speed at emission. From those and the extinction
distance you calculate the time of arrival t(n). Then the
compression ratio is:

[You can assume the distance is the same for all points.
Travel times across the orbit are generally negligible.]

r = ( t(n) - t(n-1) ) / T

That is related to what I do....but you need lots of sample points for
elliptical orbits.

or better if you want to remove a small phase error:

r = ( t(n+1) - t(n-1) ) / (2T)

yes....better..

The brightness is

b(n) = 1 / r(n)

and you can trap the infinity at r(n)=0

The "willusory" velocity is

v(n) = c(1 - r(n))

which is always finite.

Then just plot b(n) as the green curve and v(n) as the
red curve versus t(n).


I'm not convinced the method works....I don't think it takes into account the
emission time lag around the orbit.
All you are really doing is calculating the compression that adjacent pulses
emitted at a certain point on the orbit will experience at distance D. This
doesn't tell you the overall picture...but it might work for small magnitude
changes.

The reason I used the current method was that I wanted to be able to
investigate what happens when the critical distance is exceeded. I don't need
to now because it obviously never IS, due to extinction.



George



"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."
--Jonathan Swift.

On Feb 26, 5:51 pm, HW@....(Henri Wilson) wrote:
On Mon, 26 Feb 2007 19:44:54 -0000, "George Dishman"
wrote:


For those that are interested, here is a basic description of the
mathematical foundation of Henri's simulation of an orbiting binary.

One of the interesting remarks about halfway down is the assumption
that no light pulse ever catches up with a preceding pulse, no matter
how far away the observer is. This somehow gets accomplished by
something resembling extinction, but apparently all he knows is that
the extinction has to happen on a distance much smaller than the
observer distance, to preserve that assumption.

I'm sure Henri is willing to entertain questions about his algorithm.

PD

This is how I will have to approach it.

My program assumes that identical pulses are emitted at regular intervals from
some 30000 points around the orbit. (the pulsar does much the same) The true
radial velocities towards the observer are known and are stored in an array.
The emission 'lag' between consecutive pulses is period/30000. ...that provides
the value of t.

The program calculates the maximum and minimum travel times over the observer
distance of the pulses emitted in one cycle . (I actually use three period
cycles to avoid end effects but we will only talk about 1 period here.)
The max and min are divided into 167 equal time divisions, into which each
pulse is appropriately placed as it arrives....so some divisions might end up
with say 27 others with 116, depending on the bunching. I then plot these 167
numbers, equally spaced, across the screen. This gives the bunching pattern of
the pulses over one period, in 'space' at the observer distance. A spatial scan
of that pattern give brightness vs time according to the basic equation, dn/dt
= dn/dx.dx/dt. (where dx/dt = ~c)

Assuming no nett radial velocity of the barycentre, the period of the
brightness curve is the same as that of the source star...so 30000/167 or 180
pulses are emitted per time division at the source end. The time interval
spanned by an interval is period/167 (seconds) at both source and receiver.

[ Because the divisions are sufficiently small, I can assume an average source
velocity for each....which I can calculate and graph (blue curve). I can also
track and average the source velocities of all the pulses that arrive in each
time division. Note: assume no pulse actually overtakes another or even goes
close.]

If 'n' pulses arrive in a particular time division at distance D, then n/180 is
an indication of the bunching there. We also know the last pulse left the
source nt seconds after the first one.
If we know the source velocities of the pulses at each end of a division, we
can calculate how much compression there has been.
Conversely, knowing the compression, we can calculate the DIFFERENCE in source
speeds of the end two pulses

Over distance D (time D/c), the last pulse has moved from a distance ~= c.n.t
away from the first to ~c.180.t away. (c+v is close enough to c, here). The
speed difference is ct(n-180).D/c = (n-180)/D. My green curve is a scaled plot
of this.

So my green curve is a graph of acceleration vs time. I have to integrate that
wrt time to get a velocity curve.

Would you like to comment before I go on?

On 27 Feb 2007 07:11:36 -0800, "PD" wrote:

On Feb 26, 6:53 pm, HW@....(Henri Wilson) wrote:
On 26 Feb 2007 13:12:25 -0800, "PD" wrote:



Depends on what you mean by a physical entity. Is momentum a physical
entity?

Yes. A frame dependent one...but you don't know what that means Draper.

Sure I do, and on this small point we're in agreement.
So tell me, what makes momentum a physical entity and spacetime not?

Momentum has physical rammifications. Spacetime is just a method of viewing a
physical situation.

A reference frame is no more or no less a physical entity than
spacetime.

Nobody claimed it was.

Then I'm not sure what your point was. I described what a frame of
reference in part by using the term "spacetime events" and you
commented that spacetime is not a physical entity. But since a frame
of reference is no less a physical entity than spacetime, I'm not sure
why that's important.

I don't wish to discuss your imaginary 'spacetime'. We were defining 'frame'.

You blundewred...now you want to change the subject....a typical SRian ploy.

In the maths construction known as space/time, there is no movement.

That's why I included the word "or" between "spacetime events" and
"spatial positions". Reading comprehension problems, Henri? Is this
why you don't like to read books about relativity? Or books at all,
for that matter?

Draper, spacetime is completely static. There are no 'events'. Everything that
ever happened is laid out permanently in 4D. Static!!!!!
Even the future is static.... just not plotted yet.

A spacetime event, Henri, is something at a location (x, y, z, t) that
is distinguishable from neighboring locations. The snapping of fingers
is something that happens at a particular place and a particular time
and so marks a location in spacetime, called an "event".

Oh crap Draper.
You blokes like to rave about this 'spacetime' when you haven't a clue what you
are even talking about.

Try plotting the snapping of fingers in a 4D representation Draper.
Better still, try something simple like a ball falling to ground. You can plot
that in 2D.
There is NO movement in the 2D representation. It is hardly an 'event'.

....unless of course you want to agree with me that at least two orthogonal
time dimensions exist....

I asked you, where did you get the idea that a frame is everything at
rest with respect to a defined point?

Because it is Draper...including Wilsonian nort-holes.
Anything at rest wrt the defined point also defines the frame.

That doesn't tell me where you got the idea, Henri. You can also tell
me you got the idea that mammals all give live birth to their young
"because they do", but it wouldn't answer the question where you got
that ridiculous notion.

Draper, I have told you before, I'm not going to waste my time teaching YOU
elementary physics.
GO BACK TO SCHOOL... like geesey has.

Then tell me in which school, and under what professor, using which
textbook, you got the idea that a frame is everything at rest with
respect to a defined point.

Everything at rest wrt a particular point or frame is in that frame and can
define that frame.
Other objects can move wrt that frame but are not in it.

Are you under the misguided impression that my definition precludes the
possibility that objects can move wrt a reference frame.

I didn't relate any impression, Henri. I asked you a question, which
you have yet to answer. You asked me a question about my definition,
and I answered that.

....with drivel...

Now once again: Where did you get the idea that a frame is everything
that is at rest with respect to a defined point?

You really can't answer a simple question, Henri? This isn't about me
going to school, it's about where YOU got this idea.

stop raving Draper. You are a waste of time...

I answered your question. Now you answer mine. Are you an idiot or a
liar, Henri?

Learn some physics Draper..... you simply don't have any idea at present..

Happy to. Tell me what reference I should read about reference frames
that shows your definition, Henri.

Any real physicist knows what a reference frame is.

Sure they do. I just don't think it's the same as what you think it
is. The reason is that I can go to any reference written by "any real
physicist" and find a definition that doesn't have anything to do with
what you say it is. Unless of course, you've got an "any real
physicist" in mind that agrees with your definition. That is, unless
YOU are the only "any real physicist" you know....

A reference frame is simply something that can be used to compare velocities.

IT'S A BLOODY REFERENCE FOR VELOCITIES.
EVERYTHING THAT IS MUTUALLY AT REST IS IN THE SAME FRAME.

As well as everything that is not mutually at rest -- all in the same
frame. Do you disagree with that?

You are a moron Draper. Are you thinking of an ant walking across a framed
picture?

Objects that are moving wrt a frame ARE NOT in that frame. The ant is NOT in
the frame of the picture.


PD


"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."
--Jonathan Swift.

On Feb 27, 5:23 pm, HW@....(Henri Wilson) wrote:
On 27 Feb 2007 07:11:36 -0800, "PD" wrote:

On Feb 26, 6:53 pm, HW@....(Henri Wilson) wrote:
On 26 Feb 2007 13:12:25 -0800, "PD" wrote:

Depends on what you mean by a physical entity. Is momentum a physical
entity?

Yes. A frame dependent one...but you don't know what that means Draper.

Sure I do, and on this small point we're in agreement.
So tell me, what makes momentum a physical entity and spacetime not?

Momentum has physical rammifications. Spacetime is just a method of viewing a
physical situation.

Which "physical rammifications" do you have in mind for momentum.
Do you think that 3D space has "physical rammifications"? Does
Euclidean geometry count as a "physical rammification"?


A reference frame is no more or no less a physical entity than
spacetime.

Nobody claimed it was.

Then I'm not sure what your point was. I described what a frame of
reference in part by using the term "spacetime events" and you
commented that spacetime is not a physical entity. But since a frame
of reference is no less a physical entity than spacetime, I'm not sure
why that's important.

I don't wish to discuss your imaginary 'spacetime'. We were defining 'frame'.

Ah, OK, then you didn't have a point. That's not unusual.


You blundewred...now you want to change the subject....a typical SRian ploy.

"Blundewred"? What "blundewr" was that, Henri?


In the maths construction known as space/time, there is no movement.

That's why I included the word "or" between "spacetime events" and
"spatial positions". Reading comprehension problems, Henri? Is this
why you don't like to read books about relativity? Or books at all,
for that matter?

Draper, spacetime is completely static. There are no 'events'. Everything that
ever happened is laid out permanently in 4D. Static!!!!!
Even the future is static.... just not plotted yet.

A spacetime event, Henri, is something at a location (x, y, z, t) that
is distinguishable from neighboring locations. The snapping of fingers
is something that happens at a particular place and a particular time
and so marks a location in spacetime, called an "event".

Oh crap Draper.
You blokes like to rave about this 'spacetime' when you haven't a clue what you
are even talking about.

Try plotting the snapping of fingers in a 4D representation Draper.

Sure. It's a little blot on a 4D space, kind of like the period at the
end of this sentence.

Better still, try something simple like a ball falling to ground. You can plot
that in 2D.

That's not a spacetime event, Henri. That's a sequence of spacetime
events. Now, the ball *landing* on the ground, that's a spacetime
event.

There is NO movement in the 2D representation. It is hardly an 'event'.


....unless of course you want to agree with me that at least two orthogonal
time dimensions exist....

I asked you, where did you get the idea that a frame is everything at
rest with respect to a defined point?

Because it is Draper...including Wilsonian nort-holes.
Anything at rest wrt the defined point also defines the frame.

That doesn't tell me where you got the idea, Henri. You can also tell
me you got the idea that mammals all give live birth to their young
"because they do", but it wouldn't answer the question where you got
that ridiculous notion.

Draper, I have told you before, I'm not going to waste my time teaching YOU
elementary physics.
GO BACK TO SCHOOL... like geesey has.

Then tell me in which school, and under what professor, using which
textbook, you got the idea that a frame is everything at rest with
respect to a defined point.

Everything at rest wrt a particular point or frame is in that frame and can
define that frame.
Other objects can move wrt that frame but are not in it.

And that's ridiculous, Henri.

See? You can't answer a simple question. I asked you where you got
this notion, and all you can do is repeat it.
HW: "2+2=7.43"
PD: "Why Henri, wherever did you get that idea?"
HW: "Don't be an idiot, PD. 2+2=7.43"


Are you under the misguided impression that my definition precludes the
possibility that objects can move wrt a reference frame.

I didn't relate any impression, Henri. I asked you a question, which
you have yet to answer. You asked me a question about my definition,
and I answered that.

....with drivel...

Now once again: Where did you get the idea that a frame is everything
that is at rest with respect to a defined point?

You really can't answer a simple question, Henri? This isn't about me
going to school, it's about where YOU got this idea.

stop raving Draper. You are a waste of time...

Every time I ask you a question you don't want to answer (or you can't
remember what the answer is), you say it's raving.

Every time you say something stupid and I point it out to you, you say
that any "real physicist" understands what you are talking about and
agrees with it, and then when I ask you what "real physicist" that is,
you suddenly say I am raving. It's like a little pattern with you. I
can ring a little bell and you'll salivate.


I answered your question. Now you answer mine. Are you an idiot or a
liar, Henri?

Learn some physics Draper..... you simply don't have any idea at present..

Happy to. Tell me what reference I should read about reference frames
that shows your definition, Henri.

Any real physicist knows what a reference frame is.

Sure they do. I just don't think it's the same as what you think it
is. The reason is that I can go to any reference written by "any real
physicist" and find a definition that doesn't have anything to do with
what you say it is. Unless of course, you've got an "any real
physicist" in mind that agrees with your definition. That is, unless
YOU are the only "any real physicist" you know....

A reference frame is simply something that can be used to compare velocities.

Really? How can a reference frame be something that can be used to
compare velocities, if the only things that are in the frame are
things that are mutually at rest?

Or are you saying that the only way to compare the velocity of a stop
sign and a truck is to find the two reference frames in which each of
these things is at rest and then compare the two reference frames?
Sort of a "Have your people talk to my people" kind of thing?

Are you really this dense?

And what "real physicist" says that a reference frame is defined as
everything that is at rest with respect to a defined point? And which
"real physicist" says that a reference frame is simply something that
can be used to compare velocities? You still haven't answered either
of these questions.


IT'S A BLOODY REFERENCE FOR VELOCITIES.
EVERYTHING THAT IS MUTUALLY AT REST IS IN THE SAME FRAME.

As well as everything that is not mutually at rest -- all in the same
frame. Do you disagree with that?

You are a moron Draper. Are you thinking of an ant walking across a framed
picture?

Objects that are moving wrt a frame ARE NOT in that frame. The ant is NOT in
the frame of the picture.


Of course the ant is not in the frame of the picture. The frame of a
picture forms a boundary around a picture. What does this have to do
with a reference frame, Henri? Were you thinking that a reference
frame is a box around a collection of objects?

PD

On 27 Feb 2007 16:01:11 -0800, "PD" wrote:

On Feb 27, 5:23 pm, HW@....(Henri Wilson) wrote:
On 27 Feb 2007 07:11:36 -0800, "PD" wrote:


Yes. A frame dependent one...but you don't know what that means Draper.

Sure I do, and on this small point we're in agreement.
So tell me, what makes momentum a physical entity and spacetime not?

Momentum has physical rammifications. Spacetime is just a method of viewing a
physical situation.

Which "physical rammifications" do you have in mind for momentum.
Do you think that 3D space has "physical rammifications"? Does
Euclidean geometry count as a "physical rammification"?

No



Oh crap Draper.
You blokes like to rave about this 'spacetime' when you haven't a clue what you
are even talking about.

Try plotting the snapping of fingers in a 4D representation Draper.

Sure. It's a little blot on a 4D space, kind of like the period at the
end of this sentence.

Hahahojhohoh!

Better still, try something simple like a ball falling to ground. You can plot
that in 2D.

That's not a spacetime event, Henri. That's a sequence of spacetime
events. Now, the ball *landing* on the ground, that's a spacetime
event.

Hahahahwhwhahwahwahawhawhohohohohohoh!


Then tell me in which school, and under what professor, using which
textbook, you got the idea that a frame is everything at rest with
respect to a defined point.

Everything at rest wrt a particular point or frame is in that frame and can
define that frame.
Other objects can move wrt that frame but are not in it.

And that's ridiculous, Henri.

See? You can't answer a simple question. I asked you where you got
this notion, and all you can do is repeat it.
HW: "2+2=7.43"
PD: "Why Henri, wherever did you get that idea?"
HW: "Don't be an idiot, PD. 2+2=7.43"

Draper, most people here - excluding the ratpack - show a few signs of
intelligence. Why can't you?



Now once again: Where did you get the idea that a frame is everything
that is at rest with respect to a defined point?

You really can't answer a simple question, Henri? This isn't about me
going to school, it's about where YOU got this idea.

stop raving Draper. You are a waste of time...

Every time I ask you a question you don't want to answer (or you can't
remember what the answer is), you say it's raving.

Every time you say something stupid and I point it out to you, you say
that any "real physicist" understands what you are talking about and
agrees with it, and then when I ask you what "real physicist" that is,
you suddenly say I am raving. It's like a little pattern with you. I
can ring a little bell and you'll salivate.

I know you're just trying to waste my time Draper.

'If you can't beat 'em, at least slow them down"

A reference frame is simply something that can be used to compare velocities.

Really? How can a reference frame be something that can be used to
compare velocities, if the only things that are in the frame are
things that are mutually at rest?

Or are you saying that the only way to compare the velocity of a stop
sign and a truck is to find the two reference frames in which each of
these things is at rest and then compare the two reference frames?
Sort of a "Have your people talk to my people" kind of thing?

Gord, you really are hopeless Draper.


Are you really this dense?

And what "real physicist" says that a reference frame is defined as
everything that is at rest with respect to a defined point? And which
"real physicist" says that a reference frame is simply something that
can be used to compare velocities? You still haven't answered either
of these questions.

Why don't you ask captain Roberts.

IT'S A BLOODY REFERENCE FOR VELOCITIES.
EVERYTHING THAT IS MUTUALLY AT REST IS IN THE SAME FRAME.

As well as everything that is not mutually at rest -- all in the same
frame. Do you disagree with that?

You are a moron Draper. Are you thinking of an ant walking across a framed
picture?

Objects that are moving wrt a frame ARE NOT in that frame. The ant is NOT in
the frame of the picture.


Of course the ant is not in the frame of the picture. The frame of a
picture forms a boundary around a picture. What does this have to do
with a reference frame, Henri? Were you thinking that a reference
frame is a box around a collection of objects?

No but you apparently are.

PD


"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."
--Jonathan Swift.

On 27 Feb 2007 15:05:45 -0800, "PD" wrote:

On Feb 26, 5:51 pm, HW@....(Henri Wilson) wrote:
On Mon, 26 Feb 2007 19:44:54 -0000, "George Dishman"
wrote:


For those that are interested, here is a basic description of the
mathematical foundation of Henri's simulation of an orbiting binary.

One of the interesting remarks about halfway down is the assumption
that no light pulse ever catches up with a preceding pulse, no matter
how far away the observer is. This somehow gets accomplished by
something resembling extinction, but apparently all he knows is that
the extinction has to happen on a distance much smaller than the
observer distance, to preserve that assumption.

I'm sure Henri is willing to entertain questions about his algorithm.

....Poor fellow....
....really wuld love to be a scientist...

PD

This is how I will have to approach it.

My program assumes that identical pulses are emitted at regular intervals from
some 30000 points around the orbit. (the pulsar does much the same) The true
radial velocities towards the observer are known and are stored in an array.
The emission 'lag' between consecutive pulses is period/30000. ...that provides
the value of t.

The program calculates the maximum and minimum travel times over the observer
distance of the pulses emitted in one cycle . (I actually use three period
cycles to avoid end effects but we will only talk about 1 period here.)
The max and min are divided into 167 equal time divisions, into which each
pulse is appropriately placed as it arrives....so some divisions might end up
with say 27 others with 116, depending on the bunching. I then plot these 167
numbers, equally spaced, across the screen. This gives the bunching pattern of
the pulses over one period, in 'space' at the observer distance. A spatial scan
of that pattern give brightness vs time according to the basic equation, dn/dt
= dn/dx.dx/dt. (where dx/dt = ~c)

Assuming no nett radial velocity of the barycentre, the period of the
brightness curve is the same as that of the source star...so 30000/167 or 180
pulses are emitted per time division at the source end. The time interval
spanned by an interval is period/167 (seconds) at both source and receiver.

[ Because the divisions are sufficiently small, I can assume an average source
velocity for each....which I can calculate and graph (blue curve). I can also
track and average the source velocities of all the pulses that arrive in each
time division. Note: assume no pulse actually overtakes another or even goes
close.]

If 'n' pulses arrive in a particular time division at distance D, then n/180 is
an indication of the bunching there. We also know the last pulse left the
source nt seconds after the first one.
If we know the source velocities of the pulses at each end of a division, we
can calculate how much compression there has been.
Conversely, knowing the compression, we can calculate the DIFFERENCE in source
speeds of the end two pulses

Over distance D (time D/c), the last pulse has moved from a distance ~= c.n.t
away from the first to ~c.180.t away. (c+v is close enough to c, here). The
speed difference is ct(n-180).D/c = (n-180)/D. My green curve is a scaled plot
of this.

So my green curve is a graph of acceleration vs time. I have to integrate that
wrt time to get a velocity curve.

Would you like to comment before I go on?





"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."
--Jonathan Swift.

On 27 Feb, 22:11, HW@....(Henri Wilson) wrote:
On Tue, 27 Feb 2007 09:22:01 -0000, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
.. .
On Mon, 26 Feb 2007 19:44:54 -0000, "George Dishman"
George, you are completely missing the point again. You are ignoring the
crucial factor of the delay in emission times. You are assuming that the
pulses
are emitted simultaneously from all points around the orbit.

No, I am assuming your program already includes that.

It is the initial delay that largely determines the time taken for one
pulse to
catch the next. The fact that one pulse catches the preceding one does not
indicate an 'infinite velocity' at all, as you seem to believe.

Three things determine the time between arrivals, the
time between emissions of 2.95ms, the reduced radial
distance due to radial speed which gives the classical
Doppler and the delta-v between the emissions which
produces a catch-up effect dependent on observer
distance and the extinction distance. I am assuming
your program already includes at least the first three
and you alter the observer distance to simulate the
combination of actual observer distance and extinction
distance if you haven't included extinction explicitly.

Yes you expressed that well.

OK, thanks. At least we agree what needs to be
considered.

f'/f = c/(c-v)

or using the time between arrivals and turning it round
to find v:

v = c(1 - t'/t)

Distance does not appear in this equation.

Correct, the distance influences t' by affecting how
much catch-up effect occurs.

where t is the time between pulse emissions and t' is
the time between arrivals. That gives v=c for t'=0.

If I guess correctly, your brightness curve is

b = t/t'

so

v = c(1 - 1/b)

gives you a direct conversion from your existing green
curve (base ratio, not the log version) but the former
formulation avoids the infinity in brightness at t'=0.

"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."

So what do you call a genius who needs one of the dunces
to do his maths for him ;-)

...but you keep oversimplifying and solving the wrong problem.

I am making an assumption that your current code is correct
and includes all the effects in the way it calculates the
pulse arrival times.

No, there are many complications.
I'm quite confident my program is correct.

If your program is correct, so will the result
found from it.

This is how I will have to approach it.

My program assumes that identical pulses are emitted at regular intervals from
some 30000 points around the orbit. (the pulsar does much the same) The true
radial velocities towards the observer are known and are stored in an array.
The emission 'lag' between consecutive pulses is period/30000. ...that provides
the value of t.

Yes, so far so good.

The program calculates the maximum and minimum travel times over the observer
distance of the pulses emitted in one cycle . (I actually use three period
cycles to avoid end effects but we will only talk about 1 period here.)
The max and min are divided into 167 equal time divisions, into which each
pulse is appropriately placed as it arrives..

"as it arrives" is the key phrase there. I am assuming you
take into account both the location of the source in its
orbit and the radial speed in order to calculate the time
taken. If you are ignoring the location and just dividing
the distance between the barycentres of the binary system
and the solar system by the c+v speed then you will have
another error that I wasn't previously aware of.

Both Androcles and I soon realised that it is not necessary to include the
Rsin(theta) term because it is always very small compared with relative
movement of pulses over many lightyears.
Ignore it George.

For large magnitude variations that would be true but
Cepheids variability ranges over roughly 0.6 to 2.0
mag which is quite small. At 0.6 mag, ignoring the
velocity part will give a phase error of around 40
degrees (mental arithmetic) so definitely cannot be
ignored.

...although, it might be significant for your pulsar.

It is significant for both stars and pulsars for a
brightness variation of less than about 3 to keep
the phase error below about 10 degrees.

..so some divisions might end up
with say 27 others with 116, depending on the bunching. I then plot these
167
numbers, equally spaced, across the screen. This gives the bunching
pattern of
the pulses over one period, in 'space' at the observer distance. A spatial
scan
of that pattern give brightness vs time according to the basic equation,
dn/dt
= dn/dx.dx/dt. (where dx/dt = ~c)

Assuming no nett radial velocity of the barycentre, the period of the
brightness curve is the same as that of the source star...so 30000/167 or
180
pulses are emitted per time division at the source end. The time interval
spanned by an interval is period/167 (seconds) at both source and
receiver.

[ Because the divisions are sufficiently small, I can assume an average
source
velocity for each....which I can calculate and graph (blue curve). I can
also
track and average the source velocities of all the pulses that arrive in
each
time division. Note: assume no pulse actually overtakes another or even
goes
close.]

If 'n' pulses arrive in a particular time division at distance D, then
n/180 is
an indication of the bunching there.

Right, or say b = n/180 and you plot b as the green curve.

Yes.....well, I actually plot n with a scaling factor.

If you calculate v = c(1 - 1/b) then you get the "willusory"
radial speed so v = c(1 - 180/n). Just plot v as the red
curve and the job is done.

But I don't have to.

Yes you do, that's what is published and what you
have to compare against.

The program has already recorded the true radial velocities of all the pulses
that arrive in a certain time division. It takes the average and plots that.

I concede that when the number of pulses that arrive in a division becomes
quite high, for instance for magnitude changes above about 2, the 'average
speed' will not be a very accurate indicator of what is happening. So at
present my red curve is not to be taken seriously for large star magnitude
variations. It is quite accurate for mag changes below about 1, which covers
most cases we want to investigate.

It is phase shifted for magnitudes below about 3.

If I want greater accuracy, I can increase the number of sample points around
the orbit, shorten the arrival time divisions and lengthen the graph on the
screen. My method is still 100% sound.

Not if it ignores the velocity component of the Doppler.

It MUST give the same answer as your
equation.

My suggestion was an equation you could apply to your
existing output to cater for pulsars. It will be quite
diffferent, but if you have ignored the classical
Doppler then it will be wrong too.

We also know the last pulse left the
source nt seconds after the first one.
If we know the source velocities of the pulses at each end of a division,
we
can calculate how much compression there has been.
Conversely, knowing the compression, we can calculate the DIFFERENCE in
source
speeds of the end two pulses

Over distance D (time D/c), the last pulse has moved from a distance ~= c.n.t
away from the first to ~c.180.t away. (c+v is close enough to c, here).
The speed difference is ct(n-180).D/c = (n-180)/D. My green curve is a scaled
plot of this.

So my green curve is a graph of acceleration vs time. I have to integrate that
wrt time to get a velocity curve.

Would you like to comment before I go on?

The last part seems to be adding an unnecessary layer of
complexity.

My approach if I was to do it your way would be slightly
simpler. You are tracking N=30000 pulses emitted equal
times T apart to take account of the Keplerian orbit. For
each point n you find the distance to the observer and
the speed at emission. From those and the extinction
distance you calculate the time of arrival t(n). Then the
compression ratio is:

[You can assume the distance is the same for all points.
Travel times across the orbit are generally negligible.]

No you can't, the travel time across the orbit is
small (3.8s compared to 1.5 days) but the Doppler
due to the change of distance is significant. You
cannot ignore the 27km/s velocity component.

r = ( t(n) - t(n-1) ) / T

That is related to what I do....but you need lots of sample points for
elliptical orbits.

or better if you want to remove a small phase error:

r = ( t(n+1) - t(n-1) ) / (2T)

yes....better..

The more points the better but you only use two
either side, you don't average lots.

The brightness is

b(n) = 1 / r(n)

and you can trap the infinity at r(n)=0

The "willusory" velocity is

v(n) = c(1 - r(n))

which is always finite.

Then just plot b(n) as the green curve and v(n) as the
red curve versus t(n).

I'm not convinced the method works....I don't think it takes into account the
emission time lag around the orbit.

You just told me to ignore it!

All you are really doing is calculating the compression that adjacent pulses
emitted at a certain point on the orbit will experience at distance D. This
doesn't tell you the overall picture..

It is the whole picture, it includes all the factors
I listed at the top which you said I expressed well.

.but it might work for small magnitude
changes.

The reason I used the current method was that I wanted to be able to
investigate what happens when the critical distance is exceeded. I don't need
to now because it obviously never IS, due to extinction.

That's harder. For your view of stars you get multiple
overlapping spectra so the star looks like several and
there is no single answer for velocity, you would get a
velocity for each image. For pulsars a large number of
copies would look like a source with random pulse times.
A few copies would be recognisable as separate pulsars
with pulse streams that could be separated. The identical
mean rates would then be identifiable as two copies.
Neither of these situations ever arises of course.

George