It looks like you're using an Ad Blocker.

Please white-list or disable AboveTopSecret.com in your ad-blocking tool.

Thank you.

 

Some features of ATS will be disabled while you continue to use an ad-blocker.

 

Ask any question you want about Physics

page: 257
74
<< 254  255  256    258  259  260 >>

log in

join
share:

posted on Feb, 16 2016 @ 04:21 AM
link   
a reply to: Arbitrageur

Wow you kind of lost track of that explanation got away from you huh?? 😀Simpliest way to explain galaxy formations is to look at densities. In spiral galaxies densities decrease as we move further away. Spiral galaxies for example is created by a spiral density wave. Meaning when new stars are formed blue stars for example they tend to be effected by the spin of the galaxy older stars are evenly distributed throughout the galaxy. The spiral arms are increased density due to new star formation. But there is plenty of stars that travel between the arms including our own solar system.
edit on 2/16/16 by dragonridr because: (no reason given)




posted on Feb, 16 2016 @ 06:56 PM
link   
a reply to: ErosA433

scintillation


Scintillation is a flash of light produced in a transparent material by the passage of a particle (an electron, an alpha particle, an ion, or a high-energy photon)


A photon is just an construct describing some EM wave... so... ripple in EM
And if it is not a photon, it is a charged particle... electron or proton or an ion what in practice IS an charged atom

detection of any charged particles is based on electric interaction, so all the detectors are specifically designed to do so...
I really don't see the argument in your questions that would prove there is something else going on then electric interactions.

sorry to say, but all the gluons, measons or whatever particle from the particle zoo, is just mathematical calculation.
They do not exist except in the theory!!
A theoretical thing that exist ONLY in the theory but has nothing to do with reality


I also have a question for you.
What is the energy needed to move an electron so it radiates ?
beating up to this question... what about the time "it happens"

a hint would be... don't see charges as separated "things" disconnected from the environment, reality is not this way.

...
I postulated ones that any EM wave has the same energy independent of frequency, the only difference is the time those energy is "acting"
Of course the "action" is coupled to real matter, atoms, detectors...

Going down the photoelectric effect...

Charged metals, containing the "free" electrons will discharge if different EM wave will "hit" it.
But not because of different energies !!
The combination of EM wave length and different metals is what makes the electrons to "take off"

EM radiation is just the directional "force" that makes electrons leave the electric atomic bond , IF, they at convenient trajectory.
But there is no "energy" transfer in it.

howsoever...
energy is just a concept to charge money for it.
This is the conspiracy in it all !
edit on 16-2-2016 by KrzYma because: (no reason given)

edit on 16-2-2016 by KrzYma because: (no reason given)



posted on Feb, 17 2016 @ 09:59 AM
link   
a reply to: dragonridr
I'm sure ImaFungi isn't the first person to look at the arms of a spiral galaxy and think those spirals are caused by rotation rather than density, but astronomers figured out why that can't be so. I found a decent animation which shows the increased areas of density in the latter part of the video when it zooms in on one of the spiral arms:

Density Waves and the Stability of Galactic Spiral Arms

It's not perfect but it takes millions of years for that to play out in galaxies so you can't watch that happening in a real live "movie" made within a human lifetime.


originally posted by: KrzYma
howsoever...
energy is just a concept to charge money for it.
This is the conspiracy in it all !
If that's all it is then you're smarter than TPTB running the conspiracy and you've figured out away around that?



posted on Feb, 17 2016 @ 01:54 PM
link   
a reply to: Arbitrageur

What would the conditions have to be for there to be no spiral effect, for the stars 'brightness' to be homogenous in a galaxy?

Could it not have to do with the size or age of stars and the relation of that to their gravitational force, which forces them into such an imagey spiral?



posted on Feb, 17 2016 @ 06:13 PM
link   
a reply to: Arbitrageur

Oh it is suggesting that there are areas with more mass as units of matter per volume, and so over time and therefore space, these greater number of matter per volume equals a greater force of gravity and that which is massive and starts near one another will tend to remain near one another more so then the surrounding lesser density of relative matter per volume.

I suppose that still does not answer however, why it results in the spiral image;

I could start by asking what would the appears of such a galaxy, if the totality of matter was evenly distributed around the central black hole, how would that progress over time; what is causing the spiral image in that case;

You say, dying stars, and gravitational relations of distance and mass; but the reason I ask why I do not know what is causing the spiral, is because I do not conceive of why over time there would not instead result in any other possible pattern of blotches. densities of areas of matter; So that leads me to believe there is some other source for the spiral image as well, some other factor, like the nature of the central black hole, its potential mere existence or its potential rotation;

I guess a question to ask would be; what percentage of stars in the milky way are traveling in the same direction? And are there two ultimate directions? Considering potential absolute space intergalactic and/or beyond; if you imagine the central black hole like the center of a wheel, or a point on paper, and you drew a bunch of dots around the point to act as stars, and then you had to draw arrows on the dots to indicate which way they were traveling, how would the arrows line up, percentage wise in relation to one another, and what are the factors in causing each masses direction of travel? And what is the black holes direction of travel, and how different would the milky way be, our understanding of it, and potentially of gravity, if the central black hole, rotates, or does not?



posted on Feb, 17 2016 @ 06:47 PM
link   
I have two simple questions, both from my Astrophysics class.

First a definition:
Absolute Magnitude - the brightness of a star (or the Sun, or a planet) as seen from Earth from a (reference) distance of 10 parsecs (pc).
Apparent Magnitude - the brightness of a star (or the Sun, or a planet) as seen from Earth.

1. Is the apparent magnitude of our star, the Sun, not a fixed quantity? No matter where you are in the galaxy, solar system, if someone says 'what is the apparent magnitude of the Sun?' the answer is ALWAYS -26.7, right? Apparent magnitude is always based on the view from Earth. Period.

Our text book is trying to say in the 'answer key' section: 'if you were viewing the Sun from 10 pc away, the apparent and the absolute magnitude is the same '4.8'. That's "absolute horse hockey" IMO. The position of any observer, on a spaceship, on another planet in a 10pc distant solar system, the apparent magnitude does NOT change. It's still -26.7.

The answer key says more specifically 'if a star is at 10pc, then its absolute and apparent magnitude is the same. What the dimwits in the book don't realize is that the Earth is a REFERENCE value and from Earth the Sun is always an apparent magnitude of -26.7. Yes the Sun is A star, but not just any star in this definition. In fact the absolute magnitude of ANY star is still that value as if they were viewed from 10pc from Earth. Vega? Abs Magnitude is always 0. Venus - abs mag is always about -3, (dark sky, no clouds). The ONLY time apparent and absolute magnitude is the same is if the object is ACTUALLY 10pc from Earth.

In fact BOTH the apparent and absolute magnitude systems are based on Earth. One is the brightness at 10pc from -Earth- and the other is the apparent brightness in the sky from Earth (actually the 4.8 value was a 'given' based on an estimate back when those systems were created).

The ONLY time it might change is if/when the Sun becomes a red giant or something in 8 B years. What's wrong with these people.
======

2. Which type of star in the Main Sequence has a LOWER absolute magnitude, a younger type O star, or an older type M star. I argued that almost ALL of the stars on the Main sequence, type O, which is the upper LEFT end, have higher brightness, size and luminosity. The brighter ones have an absolute magnitude of -4.8. That would be brighter than Venus (-3.8 abs mag)

The brightest type M stars on the main sequence (middle part, just near the turn off) have absolute magnitudes of +4.3. That's VERY dim, about 2.5% as bright as the star Vega (0 on the apparent magnitude). The dimmest stars you can see in a small city are around +3.

The Professor says that the type O stars have a lower absolute magnitude. I think he's got it backwards.

Help me out Physics guys.

edit on 17-2-2016 by Maverick7 because: (no reason given)



posted on Feb, 17 2016 @ 08:09 PM
link   
1) Yes apparent magnitude is always as observed from the Earth in what ever configuration it is in, thus the apparent magnitute of the sun is very high (or low depending on how you take the scale) high magnitude = bright = more negative number... this is often cause of some confusion.

The standard candle in this is Vega and the scale is set such that its magnitude is 0 (apparent)

The scales really are just to allow comparison between stars of different spectral types set at different distances. How do we know a star is a giant M class for example, because the absolute magnitude is very high but the spectral type says that the surface temperature should be fairly cool... THUS it should be a large radius star.

2) Yes and no. So this might have been your Prof trolling a bit.

since the scale is negative going, it is correct to say because O class stars are lower magnitude (more negative) and such are very bright... and Class M stars are typically higher magnitude (less negative) and such are dimmer.

I suspect it to be this... you are of course correct in your head scratching because even when I studied physics and astronomy we always refered to high magnitude in the standard way of high magnitude (more negative on the scale) is bright.



posted on Feb, 17 2016 @ 10:08 PM
link   

originally posted by: Maverick7
I have two simple questions, both from my Astrophysics class.
...
Our text book is trying to say in the 'answer key' section: 'if you were viewing the Sun from 10 pc away, the apparent and the absolute magnitude is the same '4.8'. That's "absolute horse hockey" IMO. The position of any observer, on a spaceship, on another planet in a 10pc distant solar system, the apparent magnitude does NOT change. It's still -26.7.
What your textbook is trying to say is that if you stayed on Earth and moved the sun 10 pc away the apparent magnitude will change. The sun will appear dimmer, in fact the sun is far less bright even from Pluto. So sorry to say the textbook is right in this case. Maybe you're objecting to the impossibility of moving the sun 10pc away? I agree there's no way to do that, it's just a "what if the sun was that far away from Earth?" scenario. If that's not your objection I don't understand your objection or your confusion.


The ONLY time apparent and absolute magnitude is the same is if the object is ACTUALLY 10pc from Earth.
The way I interpret your text book, that's what it's trying to suggest, hypothetically move the sun to be actually 10pc from earth and what it says will be true, apparent and absolute magnitude is the same, which is what it's saying.


The ONLY time it might change is if/when the Sun becomes a red giant or something in 8 B years. What's wrong with these people.
I think sooner than 8B years, but the sun isn't as constant as you might think. It does change, not a lot, but measurably, maybe a tenth of a percent over the 11 year cycle:

The Inconstant Sun

Maybe it changes more over longer periods of time. We're still investigating the possible connection between the sun and the "Little Ice Age"


The Professor says that the type O stars have a lower absolute magnitude. I think he's got it backwards.
I think eros is saying there's some confusion over what "lower" means. I think he's right about that, though I don't see any trolling here. It's a little confusing. So the professor is right and you're right, the only reason you don't see that you agree is that you're not speaking the same language. His lower is the same as your higher, because lower to him means -4.8 is below -3.8 and you interpret that same difference to mean the -4.8 has a higher brightness than the -3.8.


originally posted by: ImaFungi
I suppose that still does not answer however, why it results in the spiral image;

I could start by asking what would the appears of such a galaxy, if the totality of matter was evenly distributed around the central black hole, how would that progress over time; what is causing the spiral image in that case;
Do you or do you not see the spirals in the diagram I posted here? If you don't, then I can't help you further. If you do, then that gives you good insight to the cause. Those are elliptical orbits that bunch up in places as illustrated.



posted on Feb, 17 2016 @ 11:53 PM
link   

originally posted by: KrzYma
a reply to: ErosA433



A photon is just an construct describing some EM wave... so... ripple in EM
And if it is not a photon, it is a charged particle... electron or proton or an ion what in practice IS an charged atom

detection of any charged particles is based on electric interaction, so all the detectors are specifically designed to do so...
I really don't see the argument in your questions that would prove there is something else going on then electric interactions.

sorry to say, but all the gluons, measons or whatever particle from the particle zoo, is just mathematical calculation.
They do not exist except in the theory!!
A theoretical thing that exist ONLY in the theory but has nothing to do with reality




What do you mean "detection of any charged particles is based on electric interaction"?
In an accelerator some particles are measured via the electromagnetic interaction but there are others that are not.
There is a hadron and muon chamber as well.

So you say the strong force/gluon is just EM?



posted on Feb, 18 2016 @ 08:07 AM
link   
a reply to: ErosA433

Right, the 'scale' is a backwards scale but I think when astronomers say 'higher magnitude' they mean a 'negative magnitude NUMBER', (or a lower numerical indicator).

I think astronomy should add the word 'number' when speaking of magnitude and standardize on that. A high magnitude NUMBER would be like +4. A lower magnitude NUMBER would be -4. The first dimmer, the second brighter.

To me, in physics, you should not be putting 'tricky' questions on an exam because physics is already 'tricky'. Don't try to trick the student, try to assess their actual knowledge and understanding. I clearly understand the H-R diagram but couldn't understand his poorly worded question.
----

On the other thing here is the actual question in the text:
OK, here is the actual question:

OK, here is the actual question: “If we were observing our Sun from a distance of 10pc, what would be its apparent and absolute magnitude”.

Now, you could imagine that as YOU being out in space on a spaceship observing the Sun, which is REASONABLE. Or you could be imagining that the Sun is moved by MAGIC to a distance of 10 pc. Which is a more reasonable assumption?

It think it's utterly STUPID for an astronomy book to ask such a question thinking it's pedigogically helpful. It is not teaching you anything. It totally omits the fact, as they do often, that absolute magnitude and apparent magnitude are constructs which are based on viewing from EARTH.

A better way to put that would be 'if the Sun was magically moved to 10pc away from EARTH, what would the abs and app magnitude be'. But putting it that way sounds like the person writing the book had a screw loose. Why would you ever need to cogitate that? In fact, there might be a handful of stars in a sphere around Earth at a real distance of 10pc such that they do share the same relative and absolute magnitude but WHO CARES? It's not an important concept. Abs and app mag are already confusing thanks to Hipparchus.


To me, that question at least needs re-written or even omitted. It's stupid, useless, and doesn't reinforce the main aspects of those two concepts, and it's equivocal. Who has moved? The observer, who is observing, not the freaking SUN, fercryinoutloud.

Thanks for chipping in.



posted on Feb, 18 2016 @ 08:24 AM
link   
Here's a pull quote from the book:

Apparent magnitude measures a star's brightness; absolute magnitude measures its luminosity.

This is an incomplete quote IMO. It should say 'by looking at it from Earth, you see how bright it is visually but you don't know if it is a big bright star, just further away. So we use the term App Mag to denote the view from Earth.

But to get the ABSOLUTE luminosity or energy given off we need to know how far away it is because to an observer the star field appears 'flat' or 2D. So we use absolute magnitude which means 'moving the star to 10pc and eliminating the inverse square law'.

Yes the pull quote is right but does it teach you anything? No it tends toward rote memorization, which I think is a very wrong way to teach physics.



posted on Feb, 18 2016 @ 10:07 AM
link   

originally posted by: Maverick7
Now, you could imagine that as YOU being out in space on a spaceship observing the Sun, which is REASONABLE. Or you could be imagining that the Sun is moved by MAGIC to a distance of 10 pc. Which is a more reasonable assumption?
To me it makes absolutely no difference which assumption I use, because I get the same answer with either assumption. You however seem to be stuck on the definition that apparent magnitude can only be defined by what is apparent from earth. When I read a question like that, I would throw the "as viewed from Earth" assumption out the window if I assume I'm viewing the sun from a distant space ship, I would assume that they want me to change the definition of apparent magnitude to suit the question to mean what's apparent to the observer on the space ship.


It think it's utterly STUPID for an astronomy book to ask such a question thinking it's pedigogically helpful. It is not teaching you anything. It totally omits the fact, as they do often, that absolute magnitude and apparent magnitude are constructs which are based on viewing from EARTH.
absolute magnitude is NOT a construct based on viewing from Earth. Apparent magnitude I suppose you could say is, but that is only because everyone we know is on Earth and actually apparent magnitude isn't what people would see from Earth's surface, it's what a satellite in orbit above the Earth would observe without atmospheric interference or losses. If a satellite or probe moves away from the Earth what's apparent to the satellite will change.


A better way to put that would be 'if the Sun was magically moved to 10pc away from EARTH, what would the abs and app magnitude be'. But putting it that way sounds like the person writing the book had a screw loose. Why would you ever need to cogitate that? In fact, there might be a handful of stars in a sphere around Earth at a real distance of 10pc such that they do share the same relative and absolute magnitude but WHO CARES? It's not an important concept. Abs and app mag are already confusing thanks to Hipparchus.
As long as the distance between you and the sun is 10pc, it's going to look the same whether you are moved 10pc away from the sun, or whether the sun is moved 10pc away from you, it wouldn't make any difference if you consider "apparent magnitude" to be what's apparent to you the observer, regardless of where you are.


To me, that question at least needs re-written or even omitted. It's stupid, useless, and doesn't reinforce the main aspects of those two concepts, and it's equivocal. Who has moved? The observer, who is observing, not the freaking SUN, fercryinoutloud.
I don't know, when every human we know of is on Earth it's not unreasonable to say apparent magnitude is related to observations from earth. As soon as you posit a hypothetical observer NOT based on Earth, does it really made sense to stick to the "as observed from Earth" definition? To me that scenario begs us to consider an alternative definition. I didn't write the textbook but I can understand the author's thought process. I'm not sure I understand your thought process because as soon as you talk about a 10 parsec distance between the sun and the observer (regardless of which moved), and you know that the distance between the Earth and the sun is far less than that, isn't it obvious you're no longer talking about the Earth-sun distance of 1AU? It's obvious to me anyway.


originally posted by: Maverick7
This is an incomplete quote IMO. It should say 'by looking at it from Earth...
Any such short quote is rarely if ever complete. We talked earlier in the thread about short quotes we see defining what "energy" is, and none are ever complete. Thoroughly understanding a concept almost always (or always?) involves more than can be expressed in any such short concept summary.

Some concepts taught in school are difficult, but this shouldn't be one of the difficult concepts. If a student of mine objected to losing credit for answering a question based on apparent magnitude being observed from Earth, I'd probably give them credit, however I'd probably ask them if the question about a 10 parsec separation between the observer and the sun makes any sense in that context. I don't think it does when there are not 10 parsecs between the Earth and the sun and you know it, so it seems to me like you could try a little harder to make sense out of the question, by asking which assumptions in a hypothetical scenario you're sticking to and why you're sticking to those, if other assumptions would make much more sense.

edit on 2016218 by Arbitrageur because: clarification



posted on Feb, 18 2016 @ 01:05 PM
link   
a reply to: Arbitrageur

Thanks for the detailed reply but you are wrong. Absolute magnitude IS based on the view of a star 10 pc from Earth.

Apparent magnitude IS based on the view from Earth.



The apparent magnitude of an object only tells us how bright an object appears from Earth. It does not tell us how bright the object is compared to other objects in the universe. For example, from Earth the planet Venus appears brighter than any star in the sky. However, Venus is really much less bright than stars, it is just very close to us. Conversely, an object that appears very faint from Earth, may actually be very bright, but very far away.

Absolute magnitude is defined to be the apparent magnitude an object would have if it were located at a distance of 10 parsecs. So for example, the apparent magnitude of the Sun is -26.7 and is the brightest celestial object we can see from Earth. However, if the Sun were 10 parsecs away, its apparent magnitude would be +4.7, only about as bright as Ganymede appears to us on Earth.


In fairness to you many astronomy books leave out 'from the Earth' assuming the person reading about 'an observer' realizes that we have our telescopes on the planet (mostly). But I assure you both abs and app mags are FROM THE EARTH.

Wiki:


The apparent magnitude (m) of a celestial object is a number that is a measure of its brightness as seen by an observer on Earth. The smaller the number, the brighter a star appears. The sun, at apparent magnitude of -27, is the brightest object in the sky.


Now you might quibble that the MODERN definition is broader, but when those terms were created and in every astronomy book I have they say 'FROM EARTH'.

Perhaps we need a revamping of those terms. I appreciate your fair comment about 'giving me credit' because as a student you want to be sure they know the 'facts' and are not tripped up by 'confusing verbiage'.



posted on Feb, 18 2016 @ 01:13 PM
link   
I'm reluctant to go on and on about these but I thought about them some more:

The thing I think is wrong in the book




OK, here is the actual question: “If we were observing our Sun from a distance of 10pc, what would be its apparent and absolute magnitude”.


Now if you read closely it does not say "IMAGINE" you are observing from a distance of 10 pc. It says 'if WE were observing our Sun'.

Who in their right might would apply MAGIC to answer that question, i.e. moving the Sun 10 pc away from Earth?

I could go out on the Web and find dozens of sites and even in the book itself, it defines is as such:

Page G-1 Glossary, Universe by WH Freedman 10th edition:



Apparent Magnitude: A measure of the brightness of light from a star or other object as measured from Earth (Chapter 17)


I don't know how I can be more clear or how it can be more clear that this question is at best ridiculously vague and since it doesn't use the term "IMAGINE" you are dealing with reality. Reality demands that the SUN is not being moved and the observation is by an astronaut (in the future?) on a ship that is 10 pc away.



posted on Feb, 18 2016 @ 01:24 PM
link   
Further the definition of 'Absolute magnitude' in WH Freedman is vague in the GLOSSARY (but not in the text, see below)

Absolute magnitude - The apparent magnitude that a star would have if it were at a distance of 10 parsecs.

In the text of 'Universe', by Freedman it specifies (from Earth) in both definitions:
i.imgur.com...

Transcribed:

Absolute Magnitudes Apparent magnitude is a measure of a stars apparent brightness as seen from Earth. A related quantity that measures a stars true energy output - that is its luminosity - is called absolute magnitude. This is the apparent magnitude a star would have if it were located exactly 10 parsecs from Earth.

(bold is mine)
Here's a picture:

astronomy.swin.edu.au...

Guess what? It's got a picture of the Earth. Wonder why? It's standard practice to make that analogy, NOT to make an analogy 'from anywhere' even though that might strictly be true.

The absolute magnitude scale was designed to give astronomers ON EARTH a way to 'de-flatten' the visual star field and determine what was really brighter when they viewed the night sky. Yes, it's a given.
edit on 18-2-2016 by Maverick7 because: (no reason given)



posted on Feb, 18 2016 @ 02:14 PM
link   

originally posted by: Maverick7
In fairness to you many astronomy books leave out 'from the Earth' assuming the person reading about 'an observer' realizes that we have our telescopes on the planet (mostly). But I assure you both abs and app mags are FROM THE EARTH.
You cite a quote that says I'm right because it doesn't mention the Earth for absolute magnitude, and you even point out it doesn't mention Earth, yet you then insist you think it does depend on Earth, but why couldn't you view the star from 10 parsecs in any direction to determine absolute magnitude? Only because we have current limitations in our technology that have us sort of confined to the Earth, but those technological limitations don't really affect the definition of absolute magnitude.

Since you're apparently more OK with spaceships moving than with stars moving, pick a star 10 parsecs away, then travel toward that star in your spaceship, then keep going another 10 parsecs, and stop.

Then you'll be 20 parsecs from Earth, but you can look at that star from a distance of 10 parsecs to determine the absolute magnitude. So that would meet the definition you cited of "Absolute magnitude is defined to be the apparent magnitude an object would have if it were located at a distance of 10 parsecs." But you'd be nowhere near the Earth.

Here's the definition from the glossary of a university with a good astronomy department:

www.astro.ucla.edu...

absolute magnitude: the magnitude an object would have at the standard distance of 10 parsecs.
Nothing in that statement about Earth, in fact isn't it true practically all of the stars in the universe are NOT at a distance of 10 parsecs from Earth? Yes, that's true, so that's the main reason I say it's not related to Earth like apparent magnitude is.

You seem to be contradicting yourself from my perspective. First you say it's not ok to move stars around to talk about apparent magnitude. But then you seem to have no problem with saying that's what you need to do to determine a star's absolute magnitude, since according to you you'd have to move it to 10 parsecs away from Earth to determine that. So which is it? Can you or can't you move stars around? If you were consistent in your argument you'd say you can't move stars around in either case and you'd have to travel to a place 10 parsecs from the star to determine absolute magnitude, meaning you won't ever be on earth to do that (with maybe a few exceptions though I don't know if any star is EXACTLY 10 parsecs from Earth, some are probably close to that).


originally posted by: Maverick7
Guess what? It's got a picture of the Earth. Wonder why?
Could it be because all the people we know of live on Earth?

edit on 2016218 by Arbitrageur because: clarification



posted on Feb, 18 2016 @ 03:46 PM
link   

originally posted by: Arbitrageur
*Those are elliptical orbits* that bunch up in places as illustrated.



Ok, that is the information I was after. I was wondering what causes the elliptical orbits in the first place, what percentage is caused by the central black hole (percentage of the stars elliptical orbits) and then where, spatially, like from distance 0 being central black hole to distance 100 being edge of galaxy; I am basically wondering, if the black hole causes the elliptical orbits of stars nearest the central black hole and then those stars cause the elliptical orbit of the stars +x distance further from the black hole, which cause the elliptical orbit +x+1 distance further from the black hole etc. and if that is the case is it not as if the black hole is causing all those stars to elliptically orbit the black hole?

And it is thought, before or after or during early energy and material settling, before super massive black hole maybe at milky way center, whatever matter and energy and gravity field existed prior to becoming the milky way, always had some aspect of orbit to it, it was always as somewhat of a gravitational object, rotating, and so then another question is; after wondering how much of the milky way the black hole itself is responsible for causing stars to orbit, by process of utilizing stars within its gravitational reach to attract stars outside its gravitational reach,, though the question turns into, is that technically within its gravitational reach then.. because stars are orbiting, and it is thought the stars outside the edge should reach escape velocity;

So the question is, why, what caused, the original rotation of the pre milky way material which formed milky way, and how long in this process until the central black hole was formed, and did the surrounding stars themselves rotation form the central black hole.



posted on Feb, 18 2016 @ 04:33 PM
link   
Beating a dead horse on magnitude of stars:

On this page, a respected amateur astronomer (just below) says:

www.tristateastronomers.org...

Apparent magnitude is how bright the object appears to us on Earth. A certain object can have an extremely high Absolute Magnitude if is is far away it will appear to be quite dim to us. Proxima Centauri is over 4 light years away. That star, incidentally has a very low absolute magnitude so it is not a very bright star.


This is the reverse of the way SOME astronomers say it, but I wonder if 'conversational Astronomy' is different than 'textbook Astronomy'?

======
Wikipedia gives the best verbiage:


The brighter an object appears, the lower the value of its magnitude, with the brightest objects reaching negative values


Even better would be ‘the lower its value on the magnitude scale’.

Why? Because ‘absolute magnitude NUMBERS’ (or values) look like arithmetic numbers but they are not. They are VALUES.

Where astronomers get mixed up is by relating a value with a minus sign in front of it with an arithmetic negative number. By doing so they actually show their ignorance of proper mathematical notation.

Putting a symbolic value on something does not == greater or lesser like an arithmetical negative number. Now they do use that in their equations when trying to relate abs mag numbers, but the scale itself is a value.

The word 'magnitude' has a simple meaning: 'size; great size'. Talking about 'lower absolute magnitude' as great brightness is an oxymoron. 'Lowered - Greatness'? Really?

I think to eliminate this all astronomers should be saying 'Magnitude Number''. Then they could say:

"The Sun is very bright and has a Low Magnitude Number". Nobody would be confused. But being bright, it has great 'magnitude', absolutely speaking.


OK, I'm done.



posted on Feb, 19 2016 @ 01:16 AM
link   

originally posted by: ImaFungi
Ok, that is the information I was after.


finally!

I was wondering what causes the elliptical orbits in the first place, what percentage is caused by the central black hole (percentage of the stars elliptical orbits) and then where, spatially, like from distance 0 being central black hole to distance 100 being edge of galaxy; I am basically wondering, if the black hole causes the elliptical orbits of stars nearest the central black hole and then those stars cause the elliptical orbit of the stars +x distance further from the black hole, which cause the elliptical orbit +x+1 distance further from the black hole etc. and if that is the case is it not as if the black hole is causing all those stars to elliptically orbit the black hole?
You asked a lot of great questions there and in the following paragraphs, and I wish we had good answers, but I think we only have suggestive, partial and incomplete answers or ideas about these things.

One key is of course gravity, because it can pull something like a cloud of gas together.
The other key I think is watching what happens to an ice skater who is spinning with her arms extended, then pulls her arms in, what happens? She starts spinning faster, because of conservation of angular momentum. So what I am fairly certain of is whatever the right answer to your question is, those two factors figure in prominently in the evolution of solar systems and disc-shaped galaxies like ours.

Watch this simulation of solar system formation showing how particles with circular orbits coalesce into objects with elliptical orbits, so this shows how elliptical orbits can evolve and while this shows elliptical planetary orbits, the same thing can happen with stars too, right?

Planetary System Formation Simulation (200 AU View)


I know it's a solar system simulation and you're asking about galaxies, and there's at least one more key factor affecting galaxy evolution, and that's mergers, which are thought to be common. We see our galaxy currently in the process of merging with a smaller one and think a larger merger with Andromeda might happen in the future. Such mergers have been simulated and unsurprisingly, rotations result and a disk forms as seen in this simulation by Steinmetz:

arxiv.org...


Fig. 2. The formation of a bulge and the rebirth of a disk. The sequence shows the formation of a bulge by the merger of two almost equal mass ‘pure disk’ systems at z∼3. After z=3 smooth accretion of gas regenerates the disk component around the bulge.


a reply to: Maverick7
I agree it would be less confusing to have a brightness or magnitude scale where the higher numbers meant brighter. However, the way that scale originated is essentially what amounts to an "apparent distance" scale where 2000 years ago we didn't know the true distance, but we could say it looks like the brighter stars are closer and the dimmer stars are further away. It was a 6 point scale from 1-6 with 1 being the smallest apparent distance (brightest) and 6 being the largest apparent distance (dimmest), so there was some logic to the higher numbers inferring larger apparent distance in the origin of that idea.

So I guess the mistake was made in the 19th century to make that old "apparent distance" scale into a magnitude scale which is why it seems inverted now, but I guess it's hard to toss out 1900 years of history and that's why the old scale wasn't abandoned.


"The Sun is very bright and has a Low Magnitude Number". Nobody would be confused.
That's an optimistic viewpoint. I think people will still find it confusing that the scale is inverted whether you add the word "number" as you suggest, or not. The main problem is that the scale is inverted, not whether the word "number" appears or not.

edit on 2016219 by Arbitrageur because: clarification



posted on Feb, 19 2016 @ 05:24 AM
link   
a reply to: joelr



What do you mean "detection of any charged particles is based on electric interaction"? In an accelerator some particles are measured via the electromagnetic interaction but there are others that are not. There is a hadron and muon chamber as well.


You have no clue, do you ?




The muon is an elementary particle similar to the electron, with electric charge of −1 e and a spin of 1⁄2, but with a much greater mass (105.7 MeV/c2)


this the theory tells !

I call it a "heavy electron"




Of the hadrons, protons are stable, and neutrons bound within atomic nuclei are stable. Other hadrons are unstable under ordinary conditions;


let me translate it for you !

Hydrons are protons +1 charge, neutron is proton (+1) -- electron (-1) conjunction (0) and unstable outside atoms.






So you say the strong force/gluon is just EM?


NO, I say it does not exist in the form it is theorized in particle physics.
Nobody can explain why atoms nuclei holds together in the picture we have right now, so strong force has been invented to hold the theory.
Nobody ever observed or measured the so called "strong force", it is based on calculations only and the idea, coulombic force changes sign at atomic levels.
So even if this theory would be right, it is en electric force, not electro-magnetic "force" like radiation


edit on 19-2-2016 by KrzYma because: (no reason given)

edit on 19-2-2016 by KrzYma because: (no reason given)



new topics

top topics



 
74
<< 254  255  256    258  259  260 >>

log in

join