On Fri, 23 Feb 2007 16:50:39 -0000, "George Dishman"
wrote:


"The Ghost In The Machine" wrote in message
...
In sci.physics.relativity, George Dishman

wrote
on Wed, 21 Feb 2007 17:37:16 -0000
:

"The Ghost In The Machine" wrote in
message
...
...
OK, so one gets the following series. Recall that
a+ar+ar^2+...+ar^n = a(1-r^(n+1))/(1-r)

Therefore, given d = distance, v = initial offset,
X = offset multiplier, and t = time:

d = (c+v)*K + (c+vX)*K + (c+vX^2)*K + ... + (c+vX^(n-1))*K
= ncK + v*(1 - X^(n))/(1-X)
= ct + v*(1 - X^(t/K))/(1-X)

where n = t/K for some K.

As K tends to zero,

d = ct + v*(1 - X)/(1-X)
= ct + v

Units? You seem to have lost a "t".

Oops, you're right.

Mind you, I think I see where I goofed; I dropped the K by accident.
A reformulation:

d = (c+v)*K + (c+vX^(1/K))*K + (c+vX^(2/K))*K + ... + (c+vX^((n-1)/K))*K
= ncK + v*K*(1-X^(n/K))/(1-X)
= ct + v*K*(1-X^t)/(1-X)

and now the second term vanishes entirely. That can't be right either.

Of course H. Wilson's formulation is a bit silly if
taken literally; particles don't magically change speed
at midnight.

No but taking the limit as dt tends to zero is the usual
formulation for calculus and discrete steps is the method
for numerical approximation. He thought the time would be
light years so appriximating that by hundreds of one-day
steps is not unreasonable. Given he has found the answer
is of the order of 6 hours, I think he would now use light
minutes if he were to redo the problem.

That's right. I might hjave to use two optional units. One 'lightminutes' and
the other 'lightdays' ..and maybe even LYs.


However, I'm now at a bit of a loss.

Standard assumptions yield

v(t) = v(0) * exp(-kt)

for some constant k. With H. Wilson's modifications this should
yield

v(t) = v(0) * exp(-kt) + c

Actually the decrease of speed depends on the distance rather
than the time we have two equations, the usual integral:

t
s(t) = I v(t) dt
0

and speed as a function of distance:

v(t) = v(0) * exp(-k*s(t)) + c

which is what caused Henry to think it would be a problem.

However since v/c ~ 10^-4 using your simpler version gives
an error in about the fourth decimal. Now Henry was saying
a year ago that R was of the order of "tens to hundreds of
light years" but now is happy with a figure of 6 light hours,
so I wasn't going to worry about four digit accuracy.

I have not been looking at pulsars at because they don't ever have published
'brightness curves'.

The shortest extinction distance I have found with a contact binary period 0.4
days is about 0.3 LYs....100 Ldays.

which can then be integrated without too much difficulty, yielding

d(t) = ct + v*(1 + exp(-kt)/k)

if I've gotten the signs right. In this context k = -log(.99993) =
0.00007000245..., if one uses days as the timeunit.

Given that no star is closer than 4 light years and k is
equivalent to less than a light day, the definite integral
is just the value at infinity.

For practical purposes the ct part also gets ignored. We
aren't worried that the curves are being seen years after
they were emitted, it is only the time relative to the
mean arrival time that is of interest.

Just use my series solution George. It's quite simple.

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.

"Henri Wilson" HW@.... wrote in message
...

The shortest extinction distance I have found with a contact binary period
0.4
days is about 0.3 LYs....100 Ldays.

I'd like to check some software I've done but it is only
accurate for circular orbits. For the binary with the
lowest eccentricity you have looked at, can you tell me
the orbital speed, period and max-min brightness change
and the extinction distance you got. It doesn't matter
if it isn't circular, I would just like a rough check.

Thanks
George

On Mon, 26 Feb 2007 22:29:54 -0000, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .

The shortest extinction distance I have found with a contact binary period
0.4
days is about 0.3 LYs....100 Ldays.

I'd like to check some software I've done but it is only
accurate for circular orbits. For the binary with the
lowest eccentricity you have looked at, can you tell me
the orbital speed, period and max-min brightness change
and the extinction distance you got. It doesn't matter
if it isn't circular, I would just like a rough check.

Paul Andersen gave me this collection of contact binaries:

http://www.astro.utoronto.ca/DDO/res...ries_prog.html

The velocity curves are all incomplete but suggest fairly circular orbits.
You need brightness variations for them. I managed to find a few but hte
problem is that each member has a brightness curve similar to the other but 180
out of phase...so both tend to cancel.

AP Leo
True distance 400 LYs
Mag Variation. 9.43 - 9.98
Period 0.43 days
Velocities 220,-280_____50,-100

I get an extinction distance of about 30 Ldays...but this is difficult because
the mag change of each star is considerably higher than 0.55.



Thanks
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.

"Henri Wilson" HW@.... wrote in message
...
On Mon, 26 Feb 2007 22:29:54 -0000, "George Dishman"

wrote:


"Henri Wilson" HW@.... wrote in message
. ..

The shortest extinction distance I have found with a contact binary
period
0.4
days is about 0.3 LYs....100 Ldays.

I'd like to check some software I've done but it is only
accurate for circular orbits. For the binary with the
lowest eccentricity you have looked at, can you tell me
the orbital speed, period and max-min brightness change
and the extinction distance you got. It doesn't matter
if it isn't circular, I would just like a rough check.

Paul Andersen gave me this collection of contact binaries:

http://www.astro.utoronto.ca/DDO/res...ries_prog.html

The velocity curves are all incomplete but suggest fairly circular orbits.
You need brightness variations for them. I managed to find a few but hte
problem is that each member has a brightness curve similar to the other
but 180
out of phase...so both tend to cancel.

That seems unlikely but I haven't found any curves for
the individual components. In particular SX Crv should
give very different results because of the high mass
ratio of 15:1.

AP Leo
True distance 400 LYs
Mag Variation. 9.43 - 9.98
Period 0.43 days
Velocities 220,-280_____50,-100

I get an extinction distance of about 30 Ldays...but this is difficult
because
the mag change of each star is considerably higher than 0.55.

Ok, I think that's close enough for my purposes,
thanks.

First, the reason I asked is I have done a simple
calculator you might find useful

http://www.georgedishman.f2s.com/Henri/Extinction.html

Enter the brightness variation in magnitudes and hit
return and it will tell you the variation as a ratio.

Enter the orbital speed and period and it then tells
you the extinction distance.

It assumes a circular orbit edge on but given I wrote
it in less than an hour I think that's reasonable.

Second point regarding AP Leo, because of the different
masses, one star is moving at 75 km/s while the other
moves at 250 km/s. If you assume say 0.55 mag for the
faster star, the extinction is 20.35 days. For the
same extinction (the light passes through the same space)
the slower star variation is roughly 0.165 mag.

I think that's valid for you but I have to be careful
regarding the difference in our interpretations.

George

On Tue, 27 Feb 2007 12:48:00 -0000, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Mon, 26 Feb 2007 22:29:54 -0000, "George Dishman"

wrote:


"Henri Wilson" HW@.... wrote in message
...

The shortest extinction distance I have found with a contact binary
period
0.4
days is about 0.3 LYs....100 Ldays.

I'd like to check some software I've done but it is only
accurate for circular orbits. For the binary with the
lowest eccentricity you have looked at, can you tell me
the orbital speed, period and max-min brightness change
and the extinction distance you got. It doesn't matter
if it isn't circular, I would just like a rough check.

Paul Andersen gave me this collection of contact binaries:

http://www.astro.utoronto.ca/DDO/res...ries_prog.html

The velocity curves are all incomplete but suggest fairly circular orbits.
You need brightness variations for them. I managed to find a few but hte
problem is that each member has a brightness curve similar to the other
but 180
out of phase...so both tend to cancel.

That seems unlikely but I haven't found any curves for
the individual components. In particular SX Crv should
give very different results because of the high mass
ratio of 15:1.

it isn't unlikely at all.
It happens.
What is observed is the combined brightness of the pair. Two sine waves 180 out
of phase can just about cancel each other.
My program allows you to se the brightness variation of either member of the
binary or the combined effect.

In the case of SX Crv, the heavy star is probably also about six times larger
in area so will radiate more energy. Even though it is moving a lot more
slowly, its contribution to the combined brightness can still be significant.


AP Leo
True distance 400 LYs
Mag Variation. 9.43 - 9.98
Period 0.43 days
Velocities 220,-280_____50,-100

I get an extinction distance of about 30 Ldays...but this is difficult
because
the mag change of each star is considerably higher than 0.55.

Ok, I think that's close enough for my purposes,
thanks.

First, the reason I asked is I have done a simple
calculator you might find useful

http://www.georgedishman.f2s.com/Henri/Extinction.html

I can't run it. I think I acidentally uninstalled my Java programs when I was
setting up my Skype phone..

Enter the brightness variation in magnitudes and hit
return and it will tell you the variation as a ratio.

Enter the orbital speed and period and it then tells
you the extinction distance.

It assumes a circular orbit edge on but given I wrote
it in less than an hour I think that's reasonable.

Second point regarding AP Leo, because of the different
masses, one star is moving at 75 km/s while the other
moves at 250 km/s. If you assume say 0.55 mag for the
faster star, the extinction is 20.35 days. For the
same extinction (the light passes through the same space)
the slower star variation is roughly 0.165 mag.

I gather the published brighness variation is for the pair. It is after all
just a point source.

I think that's valid for you but I have to be careful
regarding the difference in our interpretations.

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 28 Feb, 00:20, HW@....(Henri Wilson) wrote:
On Tue, 27 Feb 2007 12:48:00 -0000, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message ...
On Mon, 26 Feb 2007 22:29:54 -0000, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message ...

The shortest extinction distance I have found with a contact
binary period 0.4 days is about 0.3 LYs....100 Ldays.

I'd like to check some software I've done but it is only
accurate for circular orbits. For the binary with the
lowest eccentricity you have looked at, can you tell me
the orbital speed, period and max-min brightness change
and the extinction distance you got. It doesn't matter
if it isn't circular, I would just like a rough check.

Paul Andersen gave me this collection of contact binaries:

http://www.astro.utoronto.ca/DDO/res...ries_prog.html

The velocity curves are all incomplete but suggest fairly circular orbits.
You need brightness variations for them. I managed to find a few but hte
problem is that each member has a brightness curve similar to the other
but 180 out of phase...so both tend to cancel.

That seems unlikely but I haven't found any curves for
the individual components. In particular SX Crv should
give very different results because of the high mass
ratio of 15:1.

it isn't unlikely at all.
It happens.
What is observed is the combined brightness of the pair. Two sine waves 180 out
of phase can just about cancel each other.

Yes but they have to be similar in luminosity
to get close to cancelation.

My program allows you to se the brightness variation of either member of the
binary or the combined effect.

In the case of SX Crv, the heavy star is probably also about six times larger
in area so will radiate more energy. Even though it is moving a lot more
slowly, its contribution to the combined brightness can still be significant.

15 times more massive means it must be a lot hotter
and more luminous. It might actually be the major
contributor but the point is you need to calculate
that from stellar models if that's what you are going
to claim.

AP Leo
True distance 400 LYs
Mag Variation. 9.43 - 9.98
Period 0.43 days
Velocities 220,-280_____50,-100

I get an extinction distance of about 30 Ldays...but this is difficult
because
the mag change of each star is considerably higher than 0.55.

Ok, I think that's close enough for my purposes,
thanks.

First, the reason I asked is I have done a simple
calculator you might find useful

http://www.georgedishman.f2s.com/Henri/Extinction.html

I can't run it. I think I acidentally uninstalled my Java programs when I was
setting up my Skype phone..

You can get it he

http://www.java.com/en/

Enter the brightness variation in magnitudes and hit
return and it will tell you the variation as a ratio.

Enter the orbital speed and period and it then tells
you the extinction distance.

It assumes a circular orbit edge on but given I wrote
it in less than an hour I think that's reasonable.

Second point regarding AP Leo, because of the different
masses, one star is moving at 75 km/s while the other
moves at 250 km/s. If you assume say 0.55 mag for the
faster star, the extinction is 20.35 days. For the
same extinction (the light passes through the same space)
the slower star variation is roughly 0.165 mag.

I gather the published brighness variation is for the pair. It is after all
just a point source.

Yes, so not really much use unless you can find light
curves for the individual stars. Note there are a
number of complicating aspects such as reflection when
considering such a close pair.

George

On 28 Feb 2007 05:36:30 -0800, "George Dishman"
wrote:

On 28 Feb, 00:20, HW@....(Henri Wilson) wrote:
On Tue, 27 Feb 2007 12:48:00 -0000, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message ...
On Mon, 26 Feb 2007 22:29:54 -0000, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message ...

The shortest extinction distance I have found with a contact
binary period 0.4 days is about 0.3 LYs....100 Ldays.

I'd like to check some software I've done but it is only
accurate for circular orbits. For the binary with the
lowest eccentricity you have looked at, can you tell me
the orbital speed, period and max-min brightness change
and the extinction distance you got. It doesn't matter
if it isn't circular, I would just like a rough check.

Paul Andersen gave me this collection of contact binaries:

http://www.astro.utoronto.ca/DDO/res...ries_prog.html

The velocity curves are all incomplete but suggest fairly circular orbits.
You need brightness variations for them. I managed to find a few but hte
problem is that each member has a brightness curve similar to the other
but 180 out of phase...so both tend to cancel.

That seems unlikely but I haven't found any curves for
the individual components. In particular SX Crv should
give very different results because of the high mass
ratio of 15:1.

it isn't unlikely at all.
It happens.
What is observed is the combined brightness of the pair. Two sine waves 180 out
of phase can just about cancel each other.

Yes but they have to be similar in luminosity
to get close to cancelation.

Yes. There are plenty somewhat like that.
The 'cancelling effect' is quite profound. The individual variations might be 1
or 2 or mags but the combined one only 0.2.

My program allows you to se the brightness variation of either member of the
binary or the combined effect.

In the case of SX Crv, the heavy star is probably also about six times larger
in area so will radiate more energy. Even though it is moving a lot more
slowly, its contribution to the combined brightness can still be significant.

15 times more massive means it must be a lot hotter
and more luminous. It might actually be the major
contributor but the point is you need to calculate
that from stellar models if that's what you are going
to claim.

Ah...but the point of using the BaTh is to produce BETTER models.

AP Leo
True distance 400 LYs
Mag Variation. 9.43 - 9.98
Period 0.43 days
Velocities 220,-280_____50,-100

I get an extinction distance of about 30 Ldays...but this is difficult
because
the mag change of each star is considerably higher than 0.55.

Ok, I think that's close enough for my purposes,
thanks.

First, the reason I asked is I have done a simple
calculator you might find useful

http://www.georgedishman.f2s.com/Henri/Extinction.html

I can't run it. I think I acidentally uninstalled my Java programs when I was
setting up my Skype phone..

You can get it he

http://www.java.com/en/

Yes I'm getting it now.

Anyway I've tested your method and it works.

Enter the brightness variation in magnitudes and hit
return and it will tell you the variation as a ratio.

Enter the orbital speed and period and it then tells
you the extinction distance.

It assumes a circular orbit edge on but given I wrote
it in less than an hour I think that's reasonable.

Second point regarding AP Leo, because of the different
masses, one star is moving at 75 km/s while the other
moves at 250 km/s. If you assume say 0.55 mag for the
faster star, the extinction is 20.35 days. For the
same extinction (the light passes through the same space)
the slower star variation is roughly 0.165 mag.

I gather the published brighness variation is for the pair. It is after all
just a point source.

Yes, so not really much use unless you can find light
curves for the individual stars. Note there are a
number of complicating aspects such as reflection when
considering such a close pair.

That is another problem. There is also the matter of tidal lock, in which case
the facing surfaces will be considerably hotter than the shaded ones. That's a
big separate issue in my opinion.

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 28 Feb, 20:16, HW@....(Henri Wilson) wrote:
On 28 Feb 2007 05:36:30 -0800, "George Dishman" wrote:
On 28 Feb, 00:20, HW@....(Henri Wilson) wrote:
On Tue, 27 Feb 2007 12:48:00 -0000, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in messagenews:9vs6u256siscg7s2023f3gslr815vvt0rb@4ax .com...
On Mon, 26 Feb 2007 22:29:54 -0000, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in messagenews:8qmut21hrngdfqvstuoiul0upjn6h7gq8u@4ax .com...

What is observed is the combined brightness of the pair. Two sine waves 180 out
of phase can just about cancel each other.

Yes but they have to be similar in luminosity
to get close to cancelation.

Yes. There are plenty somewhat like that.
The 'cancelling effect' is quite profound. The individual variations might be 1
or 2 or mags but the combined one only 0.2.

You would need to demonstrate that from the
mass-luminosity relationship but it's not
unreasonable.

15 times more massive means it must be a lot hotter
and more luminous. It might actually be the major
contributor but the point is you need to calculate
that from stellar models if that's what you are going
to claim.

Ah...but the point of using the BaTh is to produce BETTER models.

Well that won't happen since we know ballistic
theory is wrong from Sagnac but it's an
interesting mental what-if excercise.

First, the reason I asked is I have done a simple
calculator you might find useful

http://www.georgedishman.f2s.com/Henri/Extinction.html

I can't run it. I think I acidentally uninstalled my Java programs when I was
setting up my Skype phone..

You can get it he

http://www.java.com/en/

Yes I'm getting it now.

Anyway I've tested your method and it works.

Excellent. I was really intended for high
brightness ratios but it should be close
at lower values too.

Yes, so not really much use unless you can find light
curves for the individual stars. Note there are a
number of complicating aspects such as reflection when
considering such a close pair.

That is another problem. There is also the matter of tidal lock, in which case
the facing surfaces will be considerably hotter than the shaded ones. That's a
big separate issue in my opinion.

Yes, however they can get a separate indication
of rotation rates in some cases. However testing
will be much easier on reasonably widely separated
binaries where stellar interactions are less
important.

George

On 1 Mar 2007 00:25:03 -0800, "George Dishman"
wrote:

On 28 Feb, 20:16, HW@....(Henri Wilson) wrote:
On 28 Feb 2007 05:36:30 -0800, "George Dishman" wrote:
...

What is observed is the combined brightness of the pair. Two sine waves 180 out
of phase can just about cancel each other.

Yes but they have to be similar in luminosity
to get close to cancelation.

Yes. There are plenty somewhat like that.
The 'cancelling effect' is quite profound. The individual variations might be 1
or 2 or mags but the combined one only 0.2.

You would need to demonstrate that from the
mass-luminosity relationship but it's not
unreasonable.

15 times more massive means it must be a lot hotter
and more luminous. It might actually be the major
contributor but the point is you need to calculate
that from stellar models if that's what you are going
to claim.

Ah...but the point of using the BaTh is to produce BETTER models.

Well that won't happen since we know ballistic
theory is wrong from Sagnac but it's an
interesting mental what-if excercise.

No we don't George.
Your use of a rotating frame led to hidden errors.

First, the reason I asked is I have done a simple
calculator you might find useful

http://www.georgedishman.f2s.com/Henri/Extinction.html

I can't run it. I think I acidentally uninstalled my Java programs when I was
setting up my Skype phone..

You can get it he

http://www.java.com/en/

Yes I'm getting it now.

Anyway I've tested your method and it works.

Excellent. I was really intended for high
brightness ratios but it should be close
at lower values too.

Yes, so not really much use unless you can find light
curves for the individual stars. Note there are a
number of complicating aspects such as reflection when
considering such a close pair.

That is another problem. There is also the matter of tidal lock, in which case
the facing surfaces will be considerably hotter than the shaded ones. That's a
big separate issue in my opinion.

Yes, however they can get a separate indication
of rotation rates in some cases. However testing
will be much easier on reasonably widely separated
binaries where stellar interactions are less
important.

Yes...and these are the ones whose curves I can match perfectly.


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 1 Mar, 09:29, HW@....(Henri Wilson) wrote:
On 1 Mar 2007 00:25:03 -0800, "George Dishman"
wrote:
On 28 Feb, 20:16, HW@....(Henri Wilson) wrote:
On 28 Feb 2007 05:36:30 -0800, "George Dishman" wrote:
....
Ah...but the point of using the BaTh is to produce BETTER models.

Well that won't happen since we know ballistic
theory is wrong from Sagnac but it's an
interesting mental what-if excercise.

No we don't George.
Your use of a rotating frame led to hidden errors.

The proof you agreed was actually done in the
non-rotating frame.

That is another problem. There is also the matter of tidal lock, in which case
the facing surfaces will be considerably hotter than the shaded ones. That's a
big separate issue in my opinion.

Yes, however they can get a separate indication
of rotation rates in some cases. However testing
will be much easier on reasonably widely separated
binaries where stellar interactions are less
important.

Yes...and these are the ones whose curves I can match perfectly.

We'll discuss that when we get on to considering
the optical equations. First there is a lot to be
learned from J1909-3744.

George