Possibility of star formation around black holes

[dumbest man on earth]

I did? Well my mistake then.

I was stating that I may be wrong - and that you may be right - if you were using the context provided in those 3 Links.

I feel that your motives are simply to disagree and argue.

In that case, I can honestly say that you are wrong about my "motives".

Maybe what? I honestly don't know what you mean.

Explained above.

My motives are to learn more about science from some of the amazingly smart and knowledge people here, share the small amount of knowledge that I have on the subject and expose the pseudoscience that some people try to pass as science.

Sounds good.

F=ma is not applicable here since electomagnetic waves do not have mass.

Jury is still out on that one. Visible light is part of the electromagnetic spectrum, and a photon does appear to have a "mass" of sorts. Maybe momentum?
See this Link : http://www.weburbia.com/physics/photon_mass.html

Obviously the direction changes.

Yes, but does a change in direction demand a corresponding change in velocity/speed?

I don't know what a retro-reflector is so I cannot comment other than to say that F=ma is not applicable to light.

A retro-reflector reflects incoming light back to the source of the light, regardless of the angle of incidence.

Sadly your decision not to learn any science here is your own choice.

I am completely open to and able to "learn any science" from SciForums - it's just that so far I have yet to come across any real science being taught on this Forum. Do not misunderstand me, please! There are a some knowledgeable Posters on this Forum and I have heard that some amazingly smart and knowledgeable people Post herein.

However, the loudest and most prolific Posters on this Forum - do not to me appear to be the amazingly smart and knowledge people - of which you seem to speak.

Any amazingly smart and knowledgeable person knows not to let their Ego corrupt any discussion - let alone any teaching or learning of knowledge.

BTW, I love learning. As I have stated in numerous Posts, the more I learn, the more I realize how so much more there is to learn!
To me learning is not copying/pasting Links to knowledgeable sources though.
The actual learning of any subject is not achieved until full comprehension and understanding of that subject is completed. There are Posters on this Forum who cannot describe - in their own words - the simplest scientific concepts.

Amazing - we agree on something!

Maybe. I have no problem with that.

 

[Russ_Watters]

It really seems that DMOE's mission is just to argue for arguments sake. I thought that maybe there was more to him than that. Oh well.:shrug:

You're right though, that it is pretty telling that he doesn't know and refuses to learn/admit he didn't know - or even beg-out of it via claim of a typo or misread - something that kids probably learn on day 1 of physics.

 

[Russ_Watters]

Yes, but does a change in direction demand a corresponding change in velocity/speed?

That's clumsily worded, so I see two ways to respond:

1. Yes: a change in velocity.
2. No: there is no "corresponding change in velocity", a change in direction IS a change in velocity.

 

[paddoboy]

I feel that your motives are simply to disagree and argue.

That is evident by the vast number of inane questions asked of only those supporting the mainstream section.....
Very few questions if any are put to the pseudo ratbags, and anti brigade that he supports.

 

[Aqueous Id]

Aqueous Id, why do you ignore my questions?

Who are these "home schooled folks" you speak of?

Who are these "school-schooled folks" you speak of?

If you honestly believe that true "knowledge reduces to a matter of preference", then whether you claim to be "home schooled" or "school-schooled" is irrelevant.
The Topic of this Thread is : "Possibility of star formation around black holes".
The Topic of this Thread is not : "Ad Hominem attacks against anyone who disagrees with anyone else so Aqueous Id can jump in and play inane games".

Aqueous Id, which group do you claim membership in : the "home schooled" group : or, the "school-schooled" group?

Honest answers would be appreciated.

If I smoked, I guess this would be the point where I would ask if I could have a cigarette.

Did you ever figure out if acceleration can have a negative sign?

 

[Aqueous Id]

a change in direction IS a change in velocity.

Call it deja vu but I would swear Origin gave that same answer a day or two ago.

Maybe this is like a survey. But with the unusual statistic σ² = 0.

 

[Aqueous Id]

That is evident by the vast number of inane questions asked of only those supporting the mainstream section.....
Very few questions if any are put to the pseudo ratbags, and anti brigade that he supports.

Underneath may be a voice crying out. I would hate to think it might be intentional.

 

[origin]

Jury is still out on that one. Visible light is part of the electromagnetic spectrum, and a photon does appear to have a "mass" of sorts. Maybe momentum?
See this Link : http://www.weburbia.com/physics/photon_mass.html

Photons do not have rest mass but they do have momentum.

Yes, but does a change in direction [of a photon from reflection] demand a corresponding change in velocity/speed?

A photon does not bounce off a reflective surface like a tennis ball. A photon can only move at the speed of c. The speed of light in water is less than the speed of light in a vacuum because the photon is absorbed and then reemitted by the electrons (primarily) in the the water molecules. This delay of the absorption/emission is the the decrease in the overall speed of light. The speed of the photons between the interactions with the electrons is c. Likewise, the photon moves towards the reflective surface at c and it is absorbed by the electrons in the reflective material, almost immediately a photon is reemitted back the way it came. So there really is no change in the velocity of 'a' photon during reflection. There is an incident photon that hits the mirror and another photon that is emitted by the mirror. Both photons are moving at c and are moving in straight lines through space.

 

[dumbest man on earth]

Photons do not have rest mass but they do have momentum.



A photon does not bounce off a reflective surface like a tennis ball. A photon can only move at the speed of c. The speed of light in water is less than the speed of light in a vacuum because the photon is absorbed and then reemitted by the electrons (primarily) in the the water molecules. This delay of the absorption/emission is the the decrease in the overall speed of light. The speed of the photons between the interactions with the electrons is c. Likewise, the photon moves towards the reflective surface at c and it is absorbed by the electrons in the reflective material, almost immediately a photon is reemitted back the way it came. So there really is no change in the velocity of 'a' photon during reflection. There is an incident photon that hits the mirror and another photon that is emitted by the mirror. Both photons are moving at c and are moving in straight lines through space.

origin, I have one tiny little quibble concerning your ^^above quoted^^ : "...the photon moves towards the reflective surface at c and it is absorbed by the electrons in the reflective material, almost immediately a photon is reemitted back the way it came."

A normal reflective surface would reflect the photon(s) relative to the angle of incidence - it is a retro-reflector that would reflect it "back the way it came".

An example of a retro-reflector, and an explanation of its purpose can be found here : http://search.newport.com/?q=*&x2=sku&q2=BGR-12.7

Other than that minor quibble - we seem to be in agreement - again.

I do not have to explain that the scalar (absolute value) magnitude of the velocity vector is the speed of the objects motion, do I?

Like what we agreed on previously, there is quite a bit going on in this Thread that has nothing to do with learning science.

origin, I thank you for responding honestly and openly to my queries.

 

[paddoboy]

Photons do not have rest mass but they do have momentum.



A photon does not bounce off a reflective surface like a tennis ball. A photon can only move at the speed of c. The speed of light in water is less than the speed of light in a vacuum because the photon is absorbed and then reemitted by the electrons (primarily) in the the water molecules. This delay of the absorption/emission is the the decrease in the overall speed of light. The speed of the photons between the interactions with the electrons is c. Likewise, the photon moves towards the reflective surface at c and it is absorbed by the electrons in the reflective material, almost immediately a photon is reemitted back the way it came. So there really is no change in the velocity of 'a' photon during reflection. There is an incident photon that hits the mirror and another photon that is emitted by the mirror. Both photons are moving at c and are moving in straight lines through space.

Exactly in all respects origin.....
It is actually a misnomer that a photon slows down in anything other than a vacuum.....taking into account, reflection, refraction etc, it just has a longer path to traverse.

 

[Russ_Watters]

Call it deja vu but I would swear Origin gave that same answer a day or two ago.

Yeah, it sounded familiar.

So do you get it now DMOE?

 

[Beaconator]

Does it matter if the letters are capitalized or not? DMOE is worried that negative signs can't be used. Are people have trouble with their shift keys or something? Is that what this is really about? That would be just freaking awesome. Class dismissed!

It is negative because spin throws off a trajectory. It may make the moving object more precise considering regular momentum, but the initial point it was heading to is impossible for it to reach after angular momentum is applied.

For instance you could have a cat in orbit destined for (0,0) slap its tail in any direction and it would land in an area before (0,0).

Therefore under a coordinate system angular momentum of a photon is created by gravity only. Under equivalence it is an additional momentum and an additional velocity.

 

[origin]

It is negative because spin throws off a trajectory. It may make the moving object more precise considering regular momentum, but the initial point it was heading to is impossible for it to reach after angular momentum is applied.

For instance you could have a cat in orbit destined for (0,0) slap its tail in any direction and it would land in an area before (0,0).

Therefore under a coordinate system angular momentum of a photon is created by gravity only. Under equivalence it is an additional momentum and an additional velocity.

This makes no sense at all.

 

[Trippy]

An increase in speed is called acceleration.
A decrease in speed is called deceleration.

A change in direction is called...a change in direction?...adjusting the vector?...?
At any rate, a change in direction does not demand any change in speed.

A change in direction of a body in motion, when applied to an object with mass, requires a force to be applied to the object to change its direction of motion. IE: It needs to be accelerated.

Think of it this way:

I have an object with mass, moving in a straight line with a velocity of 10 m/s in a northerly direction.
I then change it's direction so that it is moving in a straight line with a velocity of 10 m/s in a westerly direction.

1. Its speed - the magnitude of the velocity - has not changed.
2. Its northerly velocity has decreased by 10 m/s - it has deccelerated in the northerly direction.
3. Its westerly velocity has increased by 10 m/s - it has accelerated in the westerly direction.

Simply by changing the direction the object is travelling in, I have caused it to simultaneously accelerate and deccelerate without ever actually changing its speed.

 

[Trippy]

A change in direction of a body in motion, when applied to an object with mass, requires a force to be applied to the object to change its direction of motion. IE: It needs to be accelerated.

Think of it this way:

I have an object with mass, moving in a straight line with a velocity of 10 m/s in a northerly direction.
I then change it's direction so that it is moving in a straight line with a velocity of 10 m/s in a westerly direction.

1. Its speed - the magnitude of the velocity - has not changed.
2. Its northerly velocity has decreased by 10 m/s - it has deccelerated in the northerly direction.
3. Its westerly velocity has increased by 10 m/s - it has accelerated in the westerly direction.

Simply by changing the direction the object is travelling in, I have caused it to simultaneously accelerate and deccelerate without ever actually changing its speed.

To put some numbers on it, if this was done over a 3 second period then the over all acceleration applied was (approximately) 4.7 m/s/s in a south westerly direction.

 

[Motor Daddy]

Trippy, How far does an object travel in 1 second if its initial velocity was 0 m/s, and it has an acceleration of 10 m/s^2?

Here's a hint, I made a cheat sheet for you.

 

[dumbest man on earth]

An increase in speed is called acceleration.
A decrease in speed is called deceleration.

A change in direction is called...a change in direction?...adjusting the vector?...?
At any rate, a change in direction does not demand any change in speed.

A change in direction of a body in motion, when applied to an object with mass, requires a force to be applied to the object to change its direction of motion. IE: It needs to be accelerated.

Think of it this way:

I have an object with mass, moving in a straight line with a velocity of 10 m/s in a northerly direction.
I then change it's direction so that it is moving in a straight line with a velocity of 10 m/s in a westerly direction.

1. Its speed - the magnitude of the velocity - has not changed.
2. Its northerly velocity has decreased by 10 m/s - it has deccelerated in the northerly direction.
3. Its westerly velocity has increased by 10 m/s - it has accelerated in the westerly direction.

Simply by changing the direction the object is travelling in, I have caused it to simultaneously accelerate and deccelerate without ever actually changing its speed.

Hi. Trippy, it only seems proper to welcome you to this Thread.

I feel kind of special, seeing as how you joined this Thread just to present your position on a 6-day old Post I made. Especially since origin and myself seemed to have cleared up that misunderstanding nearly 2-days ago.

Trippy, I have no problems with my thinking, so I prefer making my own decisions on which way to "think". Thank you anyway.

Speaking of "Think of it this way:" - I have decided to follow the instructions contained in your "avatar" icon or picture. From now on I will most likely : "KEEP CALM AND DON'T FEED THE TROLL"

So...if by chance you get the impression that I am ignoring any Posts by anyone - I may just be following the instructions that I mentioned previously.

BTW, are you counting this as the same "one other Thread"?
I only ask because it is sometimes hard to be "on the same page", so to speak, when the "same book" is not even being utilized by all parties in the "interaction"!

 

[Trippy]

Trippy, How far does an object travel in 1 second if its initial velocity was 0 m/s, and it has an acceleration of 10 m/s^2?

Here's a hint, I made a cheat sheet for you.

Don't need the cheat sheet - the answer is 5m. Distance is the area under a velocity-time graph - when I studied kinematics in highschool, I couldn't remember the formulae themselves, so instead of memorizing them, I learned how to derive them.

To put it another way.

a = 10 m/s/s
u = 0 m/s
t = 1 s.

a=(v-u)/t
so:
v = at + u
which means that v = 10 m/s

Using those values for the variables, all of the formulae you cited give the same answer - 5m.

Using the above assertion, that the distance in a velocity-time graph is the area under the graph, and assuming constant acceleration.

From a standing start, the graph represents a triangle, and so the area under it is half the base times height, which is just a special case of the first equation on the list where u=0. The rest of those formulae are essentially derivations from that starting part.

 

[Beaconator]

This makes no sense at all.

Either play or watch tennis. Note the difference in trajectories. A ball hit with topspin (every ball you see a pro hit), travels faster and lands before a flat ball hit with the same force. Yet a ball hit with backspin will travel further. With backspin it is also possible for the ball to be hit a little higher, land and spin back toward its starting point. It happens in golf too when a wedge is used. Some points in tennis have ended with a ball that hits the opponents side and spins back over the net out of the opponents reach.

 

[Motor Daddy]

Don't need the cheat sheet - the answer is 5m. Distance is the area under a velocity-time graph - when I studied kinematics in highschool, I couldn't remember the formulae themselves, so instead of memorizing them, I learned how to derive them.

To put it another way.

a = 10 m/s/s
u = 0 m/s
t = 1 s.

a=(v-u)/t
so:
v = at + u
which means that v = 10 m/s

Using those values for the variables, all of the formulae you cited give the same answer - 5m.

Using the above assertion, that the distance in a velocity-time graph is the area under the graph, and assuming constant acceleration.

From a standing start, the graph represents a triangle, and so the area under it is half the base times height, which is just a special case of the first equation on the list where u=0. The rest of those formulae are essentially derivations from that starting part.

So what you're saying is that it takes 1 second to travel the 5 meters and get to a velocity of 10 m/s. If at that exact t=1 second point you started decelerating to 0 m/s at the acceleration rate of 10 m/s^2, then presumably it would decrease velocity from 10 m/s to 0 m/s in 1 second, hence, it took you 2 seconds to travel 10 meters. Do the same in reverse direction and it takes you the same two seconds. So if at the exact point in time (t=2 seconds) you reach the turnaround point and you return to the start point in an additional 2 seconds, then you've completed a 20 meter round trip journey in 4 seconds. 20 meters in 4 seconds is a speed of 5 m/s, not 10 m/s.