what if (on colliding galaxies)

On Sep 6, 6:56 am, "Greg Neill" wrote:
Painius wrote:
"Greg Neill" wrote in message
om...
Painius wrote:
I realize that velocity and acceleration are different,
Greg. The velocity component of which i speak is the
velocity along the acceleration vector at any given
point in time.

Okay. For a circular orbit that component is always zero.

How? How can the velocity component always be zero
when the acceleration vector is always a constant value
above zero (circular orbit) in the direction of the center
of the orbit?

Does the orbit radius change in a circular orbit? If you
think so, by how much does it change?

There's no such thing as a perfectly circular orbit, and there is
always a slug or drag coefficient.

~ BG

"Greg Neill" wrote...
in message m...
Painius wrote:
"Odysseus" wrote...
in message
news

Nonsense. From January through June every year the Earth-Moon system
recedes from the Sun, increasing the separation by several million
kilometres. The radial velocity is never great enough to produce
much of a Doppler shift, but to the extent there is any, it will be
to the red during this time. From July through December we approach
the Sun again, making the long-term average radial velocity zero.
And the shape of the Sun has no effect on its gravity (from any
reasonable distance): replace it with the mother of all Borg Cubes
-- having the same mass, of course -- and the planets' orbits would
contnue to curve exactly as they do now.

--
Odysseus

I disagree for reasons already stated again and again.

The radial velocity never goes to zero because the
acceleration (fall) never goes to zero. Ergo, the shift
is always a blue shift, sometimes more blue, sometimes
less blue.

You must see that this is incorrect reasoning. A stone
thrown upwards is always accelerated downwards, yet it
has a positive velocity upwards, comes to a halt (zero
velocity), then a negative velocity as it returns to
the ground. Velocity is not tied in sign to the
acceleration.

Of course! velocity is not necessarily tied in sign to the
acceleration or anything else except...

The sign depends upon the point of reference. One can
also say that as the stone rises higher and higher, its
DOWNWARD VELOCITY is decreasing. Yes, its upward
velocity is decreasing to a zero point. If you assign the
polarity to "upward" as "negative" and to "downward"
"positive", then you have a mental image that is useful
in many ways. However, surely you must see that this
is an arbitrary assignment.

It can also be said that the stone's downward velocity
decreases as it rises, stops, then increases back down
toward the ground. In this case the stone's speed never
changes polarity. In the local example of a stone, such
a point of reference is unwieldy, but it is still a valid way
to picture the motion of the stone.

You might see it better if you think in terms of escape
velocity, and how it decreases as the distance from a
mass increases. For example, the velocity required to
escape the Sun's gravitational field is less at Earth's
orbit than at Mercury's orbit, etc.

A body in a circular orbit is always accelerated downwards,
yet always has a zero radial velocity.

I still have not seen you explain how this can be. It
seems that you're saying that even though a body in
circular orbit always accelerates downward, IOW, the
body always increases or decreases its speed in the
downward direction,

("accelerate" does mean "to increase or decrease
velocity", doesn't it?)

....this somehow allows for the body to still always
have a zero radial velocity. You need to 'splain to me,
Lucy! How exactly does an object maintain zero speed
while "accelerating"?

How in the heck can anything undergo a change in
velocity, and yet maintain a constant velocity, zero or
otherwise?

A body in an
elliptical orbit has sometimes positive, sometimes
negative, and sometimes zero radial velocity. In a
previous post I gave you an equation expressing the
radial velocity with respect to orbit angle. Did you
look at it?

Yes. I'm looking at it now, both here...

*************************************

r(q) = p/(1 + e*cos(q))

whe
r(q) is the radius at orbit angle q measured from
perihelion.
p is a scaling parameter that sets the size of the
orbit (technically it is called the semi latus
rectum).
e is the eccentricity of the orbit.

Perihelion occurs when q is zero (or a multiple of 360 degrees).
Aphelion occurs when q is 180 degrees (or suitable angular
multiples).

If we differentiate the above formula to obtain the radial
"velocity" with respect to changes in angle q, we get:

r'(q) = p*e*sin(q)/(1 + e*cos(q))^2

*************************************

....and here...

http://adsabs.harvard.edu/full/1913PASP...25..208P

(page 210)

Mathematics can be easily manipulated like that. And
sometimes such manipulations, when one is considering
other applications, can lull one into a distorted picture
of reality. Ref.: Ptolemy.

There are three major things to watch out for in math.
In order of most errors made, they are...

1) decimal point placement

2) units of measurement

3) polarity (sign)

Your reality check must focus upon 3) above. In the
determination of the spectral shift of the Sun with
reference to the Earth, it is always "blue".

The equation you offered, while very useful in the
application to derive the radial velocity equation
representing the component velocity of binary stars
in the direction of the Sun, and in other applications,
shows once again that it is the "reference point" that
counts.

Acceleration *does not* determine redshift, only radial
velocity does.

It seems that acceleration plays a part in the redshift
of faraway galaxies billions of light years away. Why
not nearer to us? Accepting what you say as correct,
how about the other way around? Might there be a
way to determine acceleration when redshift is known?
Or is this too complicated a derivation?

happy days and...
starry starry nights!

--
Indelibly yours,
Paine Ellsworth

P.S.: Thank *YOU* for reading!

P.P.S.: http://yummycake.secretsgolden.com

"Odysseus" wrote...
in message news
In article ,
"Painius" wrote:

snip

2) a no-mass barycenter about 1,000,000 LY away
in the general direction of Andromeda and Triangulum,
around which all the galaxies in the Local Group orbit,

[...] In 2), this would mean unstable, chaotic orbits that are highly
unlikely given the age of the Local Group (at least 10 billion years,
which is the approx. age of the Milky Way galaxy).

Why unlikely? Have you calculated the period of a stable, orderly orbit
around your "hypermassive barycenter"? I doubt there's been time for
even one cycle! . . .

In one of several scenarios, that of the Andromeda
galaxy orbiting a hypermassive barycenter in an orbit
that has a very low eccentricity, Andromeda might
then have a transverse velocity of, say, 1500 km/sec.

At that rate it would take 1.885 billion years to make
one complete orbit. That's a little over 7 cycles in the
last 13.5 billion years, or 5.3 cycles in 10 billion years.

happy days and...
starry starry nights!

--
Indelibly yours,
Paine Ellsworth

P.S.: Thank *YOU* for reading!

P.P.S.: http://yummycake.secretsgolden.com

On Sep 6, 10:17 am, "Painius" wrote:
"Odysseus" wrote...

in messagenews


In article ,
"Painius" wrote:

snip

2) a no-mass barycenter about 1,000,000 LY away
in the general direction of Andromeda and Triangulum,
around which all the galaxies in the Local Group orbit,

[...] In 2), this would mean unstable, chaotic orbits that are highly
unlikely given the age of the Local Group (at least 10 billion years,
which is the approx. age of the Milky Way galaxy).

Why unlikely? Have you calculated the period of a stable, orderly orbit
around your "hypermassive barycenter"? I doubt there's been time for
even one cycle! . . .

In one of several scenarios, that of the Andromeda
galaxy orbiting a hypermassive barycenter in an orbit
that has a very low eccentricity, Andromeda might
then have a transverse velocity of, say, 1500 km/sec.

At that rate it would take 1.885 billion years to make
one complete orbit. That's a little over 7 cycles in the
last 13.5 billion years, or 5.3 cycles in 10 billion years.

happy days and...
starry starry nights!

--
Indelibly yours,
Paine Ellsworth

P.S.: Thank *YOU* for reading!

P.P.S.: http://yummycake.secretsgolden.com

Such cosmic cycles do exist, as with the 225 million year galactic
cycle or galactic year(GY), as well as the more localized 110,000 year
cycle of what our solar system seems to have associated with the
impressive Sirius star/solar system that used to be worth 7+ solar
masses.

Due to the ongoing loss of galactic core mass (similar to the ongoing
loss of mass from our sun) and subsequent rate of orbital expansion
taking place, whereas in the past our cycles or encounters with Sirius
had been more frequent. In other words, like most galaxies, it seems
our Milky Way has been growing in size, but not in mass, and there’s
no reason to think that Andromeda hasn’t been expanding at a somewhat
greater rate since there’s less core mass to bind what Andromeda has
to offer.

At 7+ solar masses, Sirius used to have a much tighter tidal radius
grasp onto our wussy little and relatively passive solar system. For
all we know, Earth may have once upon a time belonged to the Sirius B
solar system, if not at least the likes of Venus and of our Selene/
moon that’s so entirely unusual.

~ Brad Guth Brad_Guth Brad.Guth BradGuth

Painius wrote:
"Greg Neill" wrote...
in message m...
Painius wrote:
"Odysseus" wrote...
in message
news
Nonsense. From January through June every year the Earth-Moon
system recedes from the Sun, increasing the separation by several
million kilometres. The radial velocity is never great enough to
produce much of a Doppler shift, but to the extent there is any,
it will be to the red during this time. From July through December
we approach the Sun again, making the long-term average radial
velocity zero. And the shape of the Sun has no effect on its
gravity (from any reasonable distance): replace it with the mother
of all Borg Cubes -- having the same mass, of course -- and the
planets' orbits would contnue to curve exactly as they do now.

--
Odysseus

I disagree for reasons already stated again and again.

The radial velocity never goes to zero because the
acceleration (fall) never goes to zero. Ergo, the shift
is always a blue shift, sometimes more blue, sometimes
less blue.

You must see that this is incorrect reasoning. A stone
thrown upwards is always accelerated downwards, yet it
has a positive velocity upwards, comes to a halt (zero
velocity), then a negative velocity as it returns to
the ground. Velocity is not tied in sign to the
acceleration.

Of course! velocity is not necessarily tied in sign to the
acceleration or anything else except...

The sign depends upon the point of reference. One can
also say that as the stone rises higher and higher, its
DOWNWARD VELOCITY is decreasing. Yes, its upward
velocity is decreasing to a zero point. If you assign the
polarity to "upward" as "negative" and to "downward"
"positive", then you have a mental image that is useful
in many ways. However, surely you must see that this
is an arbitrary assignment.

You would have to choose a reference frame that is moving
at a velocity higher that of the maximum velocity of
the stone in the Earth reference frame in order for this
to occur. The only frame of reference of interest for
the redshift problem is one firmly attached to the Earth.


It can also be said that the stone's downward velocity
decreases as it rises, stops, then increases back down
toward the ground. In this case the stone's speed never
changes polarity.

Of course it does. You just said it yourself; it slows,
stops, then moves again inthe opposite direction.

In the local example of a stone, such
a point of reference is unwieldy, but it is still a valid way
to picture the motion of the stone.

You might see it better if you think in terms of escape
velocity, and how it decreases as the distance from a
mass increases. For example, the velocity required to
escape the Sun's gravitational field is less at Earth's
orbit than at Mercury's orbit, etc.

A body in a circular orbit is always accelerated downwards,
yet always has a zero radial velocity.

I still have not seen you explain how this can be. It
seems that you're saying that even though a body in
circular orbit always accelerates downward, IOW, the
body always increases or decreases its speed in the
downward direction,

("accelerate" does mean "to increase or decrease
velocity", doesn't it?)

No. Acceleration is merely a change in velocity, which
can affect either or both the speed or direction. In
a circular orbit, only the direction changes.

Draw a circle. Can you point to a place on the circumference
that is further or closer to the center than any other point
on the circumference? No, the radius is constant. So what
do you make of circular orbits?


...this somehow allows for the body to still always
have a zero radial velocity. You need to 'splain to me,
Lucy! How exactly does an object maintain zero speed
while "accelerating"?

As explained previously, if an accelerating force is applied
to a body always at right angles to its diection of motion,
a circular trajectory results. The speed remains constant
while the direction changes continuously.


How in the heck can anything undergo a change in
velocity, and yet maintain a constant velocity, zero or
otherwise?

See above.


A body in an
elliptical orbit has sometimes positive, sometimes
negative, and sometimes zero radial velocity. In a
previous post I gave you an equation expressing the
radial velocity with respect to orbit angle. Did you
look at it?

Yes. I'm looking at it now, both here...

*************************************

r(q) = p/(1 + e*cos(q))

whe
r(q) is the radius at orbit angle q measured from
perihelion.
p is a scaling parameter that sets the size of the
orbit (technically it is called the semi latus
rectum).
e is the eccentricity of the orbit.

Perihelion occurs when q is zero (or a multiple of 360 degrees).
Aphelion occurs when q is 180 degrees (or suitable angular
multiples).

If we differentiate the above formula to obtain the radial
"velocity" with respect to changes in angle q, we get:

r'(q) = p*e*sin(q)/(1 + e*cos(q))^2

*************************************

...and here...

http://adsabs.harvard.edu/full/1913PASP...25..208P

(page 210)

Mathematics can be easily manipulated like that. And
sometimes such manipulations, when one is considering
other applications, can lull one into a distorted picture
of reality. Ref.: Ptolemy.

There are three major things to watch out for in math.
In order of most errors made, they are...

1) decimal point placement

2) units of measurement

3) polarity (sign)

Your reality check must focus upon 3) above. In the
determination of the spectral shift of the Sun with
reference to the Earth, it is always "blue".

No! It is sometimes redward, sometimes blueward,
sometimes nil. It must be so because the Earth
repeatedly approaches and withdraws from the Sun,
thus yielding radial velocities alternately
inward directed and outward directed.


The equation you offered, while very useful in the
application to derive the radial velocity equation
representing the component velocity of binary stars
in the direction of the Sun, and in other applications,
shows once again that it is the "reference point" that
counts.

No, the equations are generally applicable to any elliptical
orbit where the point of view is from the primary focus;
they yield the length of the radius vector and the radial
velocity as measured from the orbit focus.


Acceleration *does not* determine redshift, only radial
velocity does.

It seems that acceleration plays a part in the redshift
of faraway galaxies billions of light years away. Why
not nearer to us? Accepting what you say as correct,
how about the other way around? Might there be a
way to determine acceleration when redshift is known?
Or is this too complicated a derivation?

The acceleration is determined by noting the change in
velocity with distance over time. It requires a method
for determining distance independant of redshift, hence
the use of cerftain types of supernova as standard
candles.

On Sep 6, 9:51*am, "Painius" wrote:

Mathematics can be easily manipulated like that. *And.... can lull one into a
distorted pictureof reality. *Ref.: Ptolemy.

Heh. Basing on false premise. Like space being a universally-isotropic
'Nothing' all the way back to the BB, devoid of a cosmological density
gradient. :-)

Rogue galaxies, Paine? lmao!

Now you sound too much like BradBoi!

I don't think you can find anything about actual rogue galaxies.

Saul Levy


On Sat, 06 Sep 2008 11:45:42 GMT, "Painius"
wrote:

Maybe, but i submit its unliklihood due to the
presence of small rogue galaxies that appear to
be gravitationally bound to the three big spirals.
I think those rogues would have huge red shifts
and would be zipping away from us if there were
no hypermassive object at the barycenter.

I honestly haven't personally checked their red-
shift values, but i haven't read anything that is
particularly unsettling about them, either.

Taking a simple case for illustration, ignoring the rest of the Local
Group, supposing our Galaxy and M31 to be orbiting each other without
being perturbed by any other objects, and neglecting their approach
speed, what would you expect their average tangential speeds to be? The
circumference of their orbits would be on the order of a megaparsec ...

--
Odysseus

A megaparsec doesn't seem like much, O. Let's see...

The radius vector would have to average about 1 / 2pi or
0.159 megaparsecs = 519,000 light years. At the present
distance of Andromeda this would indicate a highly
eccentric orbit around a barycenter that would be maybe
about 3-400,000 light years from the center of the Milky
Way. If the Andromeda galaxy is at or near apoapsis
(apgalaxion?), then the tangential velocity of Andromeda
with respect to the Milky Way would be relatively slow,
perhaps 500 km/sec or less.

In this special case, i would not expect there to be any
significant mass at the barycenter, and the orbits of the
Milky Way and Andromeda around the barycenter would
be stable.

We should debate the Borg on Venus, Paine! lmao!

Saul Levy


On Sat, 06 Sep 2008 13:44:16 GMT, "Painius"
wrote:

"Saul Levy" wrote in message...
.. .

Paine just doesn't get it, Odysseus. I've given up on him on this
topic.

Saul Levy

"Giving up" never results in a win, but often results
in a draw.

There will always be another battlefield. g

In article ,
"Painius" wrote:

How? How can the velocity component always be zero
when the acceleration vector is always a constant value
above zero (circular orbit) in the direction of the center
of the orbit?

The velocity component toward the primary is zero; the velocity
component along the tangent to the orbit is not zero.

A while ago someone (you?) asked whether it's better to look at the
acceleration components from the standpoint of the orbiting object or
from someone looking overhead. Let's look at it both ways.

First, from the object itself: There's an acceleration toward the
primary; that's because of gravity. There's zero velocity toward the
primary and a constant velocity tangent to the orbit, in the direction
of that motion.

All of those things *must* be true. The acceleration toward the center
*must* be true because that's how gravity works. The masses of the
objects divided by the square of the distance times a constant gives you
the acceleration between them. The velocity toward the center is
constant; that *must* be true because the distance toward the center
never changes. That comes from the definition of a circle: the locus of
points at some constant distance from the center.
Finally, the velocity in the direction of the orbit is a constant, and
that doesn't change either: that's how the object is moving around.

Now let's look at it from above, with accelerations and movements in an
arbitrary coordinate plane. Let's say you've got a star in the middle of
some graph paper and an planet orbiting in a circular orbit that's a few
inches in radius. Pick some point on that circular orbit, say in the
upper right quadrant, put the planet there. It's moving
counter-clockwise.

The planet has an acceleration toward the star. Draw that vector and
then draw its vertical and horizontal components. The planet has a
velocity at right angles to that acceleration. Draw that vector and its
vertical and horizontal components. So right now the planet is above and
to the right of the star; it's accelerating toward the star and moving
to the upper left.

Now let's think only about the vertical components for a moment. There's
a downward acceleration and an upward speed. Eventually the downward
acceleration will shave off that upward speed until it reaches the top
of the orbit. Then it will continue to accelerate downwards and start to
actually move downwards. At the point when it is even horizontally with
the star, the downwards acceleration is zero -- it flips! and the
downward speed is maximum. Thereafter, the acceleration is upwards, so
it slows down until it reaches the bottom farthest away point, stops,
reverses direction, and starts moving upwards again.

The situation horizontally is the same except shifted by 90°. Which
makes sense, of course, because the horizontal components of the vectors
are shifted that way.

At the very top or the very bottom of the vertical movement, there is no
vertical movement; the velocity is zero for an instant. At the very left
or the very right of the horizontal movement, there is no horizontal
movement: for an instant, it stops and changes direction.

So the answer to your question is, "like that, because it *has* to be
that way."

--
Timberwoof me at timberwoof dot com http://www.timberwoof.com
People who can't spell get kicked out of Hogwarts.

"Timberwoof" wrote...
in message
...
In article ,
"Painius" wrote:

How? How can the velocity component always be zero
when the acceleration vector is always a constant value
above zero (circular orbit) in the direction of the center
of the orbit?

The velocity component toward the primary is zero; the velocity
component along the tangent to the orbit is not zero.

A while ago someone (you?) asked whether it's better to look at the
acceleration components from the standpoint of the orbiting object or
from someone looking overhead. Let's look at it both ways.

First, from the object itself: There's an acceleration toward the
primary; that's because of gravity. There's zero velocity toward the
primary and a constant velocity tangent to the orbit, in the direction
of that motion.

All of those things *must* be true. The acceleration toward the center
*must* be true because that's how gravity works. The masses of the
objects divided by the square of the distance times a constant gives you
the acceleration between them. The velocity toward the center is
constant; that *must* be true because the distance toward the center
never changes. That comes from the definition of a circle: the locus of
points at some constant distance from the center.
Finally, the velocity in the direction of the orbit is a constant, and
that doesn't change either: that's how the object is moving around.

Now let's look at it from above, with accelerations and movements in an
arbitrary coordinate plane. Let's say you've got a star in the middle of
some graph paper and an planet orbiting in a circular orbit that's a few
inches in radius. Pick some point on that circular orbit, say in the
upper right quadrant, put the planet there. It's moving
counter-clockwise.

The planet has an acceleration toward the star. Draw that vector and
then draw its vertical and horizontal components. The planet has a
velocity at right angles to that acceleration. Draw that vector and its
vertical and horizontal components. So right now the planet is above and
to the right of the star; it's accelerating toward the star and moving
to the upper left.

Now let's think only about the vertical components for a moment. There's
a downward acceleration and an upward speed. Eventually the downward
acceleration will shave off that upward speed until it reaches the top
of the orbit. Then it will continue to accelerate downwards and start to
actually move downwards. At the point when it is even horizontally with
the star, the downwards acceleration is zero -- it flips! and the
downward speed is maximum. Thereafter, the acceleration is upwards, so
it slows down until it reaches the bottom farthest away point, stops,
reverses direction, and starts moving upwards again.

The situation horizontally is the same except shifted by 90°. Which
makes sense, of course, because the horizontal components of the vectors
are shifted that way.

At the very top or the very bottom of the vertical movement, there is no
vertical movement; the velocity is zero for an instant. At the very left
or the very right of the horizontal movement, there is no horizontal
movement: for an instant, it stops and changes direction.

So the answer to your question is, "like that, because it *has* to be
that way."


Not good enough, TW, but thank you for trying!
Seriously, i know you put a lot of thought into all
this. But it's just too simplified, and it feels as if
you've left out something very important. For one
thing, an answer like your bottom-line conclusion
is not conducive to changing people's minds or
reducing misconceptions.

happy days and...
starry starry nights!

--
Indelibly yours,
Paine Ellsworth

P.S.: Thank *YOU* for reading!

P.P.S.: http://yummycake.secretsgolden.com