Magnetic Stars: Rebuttal to Max Planck Institute (Advanced Draft Release: The Grant Chronicles)

A list of "things Grant doesn't know" would fill a VERY LARGE book.

Number one on the list of "things Grant doesn't know" would be "Grant doesn't know what he's talking about".

indeed, DB. Sadly, item 2 on the list would be "Grant doesn't know that he doesn't know what he's talking about"...


Regards
Krill

Then you attempt to also calculate the pressure, and this is where the wheels come off. Firstly, you can't make the assumption that the pressure is constant throughout the sphere at all times. Although it might be at first, it certainly won't be later on - that's sort of the whole point of the hydrostatic equilibrium equations.

Wrong again Krill the variation of pressure in a hydrostatic equation is based upon the change in radius inside a nebula cloud you can not assume wide variances in pressure within a cloud what would cause this, as gas would be moving from one sector to another trying to equalize.

Secondly, you can't figure out the pressure by dividing the mass by the volume. That would give you the density, not the pressure.

Kilograms are not a unit of force - they're a unit of mass. For cosmological purposes, the difference is sort of important.

In order to determine the pressure, you'd need to use something like the ideal gas law: I went thru this with you already Gravity

P = nRT/V

Grant: So now you are playing stupid the number of moles can be equated to mass. Definition The mass of 1 mole of a compound is called its molar weight or molar mass. The units for molar weight or molar mass are grams per mole. Here is the formula to determing the number of moles of a sample:
mol = weight of sample (g) / molar weight (g/mol)
R is the universal gas constant. So it comes down to p=M/V times the constant R, temperature is negated. But that was not the point the pressure was not set it is a variable and What I show the Volume is directly proportionate to the pressure no matter what nR is. So if you want you figure out the number moles in 50 Suns and apply the gas constant. What ever the number for P when you shrink the radius to a 1/10 the pressure it will increase a 1000 times that constant.

Which is sort of problematic for you: since you're assuming the temperature to be infinitessimally close to absolute zero (another variable you don't want to deal with, but which is very important), you'd have a pressure of, well, zero - which of course is the opposite of what you want.

Grant: Wrong Krill if we add temperature into the equation it will add to the increase of pressure

So now we don't have any sensible numbers at all here - but let's be generous and assume that, in your calculations, we can just call 1kg = 10N to make everything square again (this is of course complete baloney, but hey, who's counting? It's extremely generous in your favor, also).

Grant: What, are in denial Krill the gravitational acceleration numbers are correct you can’t spin saying that there are pressure concentration supported by gravitational focal points? How so if gravity is so weak? The numbers do not lie.
Time to go back to school Krill you have been served.

Really Guys you need to quit while you are behind.

First try and concentrate on that very very low gravitational acceleration and tell us how it collapses a cloud???

"P = nRT/V

Grant: So now you are playing stupid the number of moles can be equated to mass. Definition The mass of 1 mole of a compound is called its molar weight or molar mass. The units for molar weight or molar mass are grams per mole. Here is the formula to determing the number of moles of a sample:
mol = weight of sample (g) / molar weight (g/mol)
R is the universal gas constant. So it comes down to p=M/V times the constant R, temperature is negated."

oh no, not this one again. We did this what, two, three years ago?

temperature does not get "negated" and the number of moles cannot be "equated" to mass (although one is proportional to the other).

I thought we established way back when that your thinking that

"p=M/V"

is an alternate formulation of the ideal gas law arose because you couldn't tell the difference between the letter "p" and the Greek letter "rho"? What you're quoting here is a way of calculating density, not pressure.

You can't, I'm afraid, just toss out temperature.

Regards
Krill

PS will you please, pretty please, stop with saying that mass and weight are the same thing as you do repeatedly here? It sort of makes you look like an idiot

Using that with your own numbers yields a value for "P", what you claim is the initial pressure, of:

2.3 x 10^-15 Pa

and your claimed pressure when the cloud has collapsed to a radius equal to the orbit of Mercury of:

2.3 Pa

Just to be clear, that last figure is:

.000023 Atmospheres

What is this nonsense that you have posted? Oh I get it making up numbers. Don’t you know how to read a table?

The gravitational acceleration at the surface of the cloud with a radius of 23 million km is 2.96 meters/sec^2 and the pressure is 10^15 times the initial pressure of the cloud.
Your sinking to a new low Krill.

temperature does not get "negated" and the number of moles cannot be "equated" to mass (although one is proportional to the other).


What is the definition of a mole Krill and please try to leave the mass, grams, and kilograms.

If you are going to play the game don't get caught lying.

If you want to add temperature to the equation be my guest, but you already have a short fall of gravitational acceleration to counter the rise in pressure this will further increase it to the point of containment breach in some conditions

"The gravitational acceleration at the surface of the cloud with a radius of 23 million km is 2.96 meters/sec^2 and the pressure is 10^15 times the initial pressure of the cloud.
Your sinking to a new low Krill."

please explain how this is different from what I posted. Like I said, your numbers and method are nonsense, but...

Your own numbers (once blanks have been filled in, and I explained how) say that the initial pressure is 2.3x10^-15 Pascals.

That would mean that the final pressure (according to you) would be

10^15 x 2.3x10^-15

which is, err, 2.3 Pascals. Just like I said. What's the problem?

Regards
Krill

"temperature does not get "negated" and the number of moles cannot be "equated" to mass (although one is proportional to the other).

What is the definition of a mole Krill and please try to leave the mass, grams, and kilograms.

If you are going to play the game don't get caught lying."


jeez, Grant, what's with all the throwing around constant accusations of "lying" and "sinking to a new low". Do you think that will help you to mask the facts?

oh well...

Mole definition from Wikipedia:

"A mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 12 grams of carbon 12, where the carbon 12 atoms are unbound, at rest and in their ground state. [1] The number of atoms in 0.012 kilogram of carbon 12 is known as Avogadro's number. It is approximately 6.0221415×1023 (2002 CODATA value).

A mole is a dimensionless name for an integer, much like dozen or googol. Although the exact value of the mole is not known at present, it is equal to Avogadro's number, which is known to 1 part in 10 million."

the last part is important. Number of moles is dimensionless, ie does not have units of mass. Thus, you are not entitled to "equate" them - although, as I said, they are proportional under the assumptions you are making.

Regards
Krill

"If you want to add temperature to the equation be my guest, but you already have a short fall of gravitational acceleration to counter the rise in pressure this will further increase it to the point of containment breach in some conditions"

you're not going to be able to add the temperature into the mix without doing a numerical simulation to find the answer. Hell, you need that just to solve for pressure and density. Just like I keep saying. I don't think you understand what "numerical simulation" means... but I posted a link to such a set of results already, as you acknowledged.


But without that, your numbers don't say what you claim. Not even close. As I already showed you.

Regards
Krill

PS what's a "containment breach"? Do you get your material from Star Trek?

"A mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 12 grams
Stop right there, now do you see the word grams Krill, which constitutes mass

you're not going to be able to add the temperature into the mix without doing a numerical simulation to find the answer. Hell, you need that just to solve for pressure and density. Just like I keep saying. I don't think you understand what "numerical simulation" means... but I posted a link to such a set of results already, as you acknowledged.
Get this thru your head add temperature and the numbers drift further away from you in my favor. You will need more gravitational acceleration to contain the same amount mass, but you can't see that.

"Stop right there, now do you see the word grams Krill, which constitutes mass"

so just because the word "grams" appears in the middle of a definition which makes it clear that moles is *mot* a unit of mass, you conclude that it is, in fact, saying that it is a unit of mass. Nice.

Regards
Krill

Grant, you stopped too soon.

A mole is not just 12 grams, it's 12 grams of Carbon 12.

12 grams of hydrogen do not constitute a mole, and a mole of hydrogen does not weigh 12 grams.

12 grams of uranium do not constitute a mole, and a mole of uranium atoms do not weigh 12 grams.

A mole is a fixed number. You can have a mole of pennies, cars, etc. The common factor is the number of items.

This is basic chemistry.

Krill tell how a star forms if the cloud contracts to .000005 of its originalraduis and gravitational acceleration at the surface of the cloud is less than a third of gravitational acceleration of the surface of the Earth?

"Krill tell how a star forms if the cloud contracts to .000005 of its originalraduis and gravitational acceleration at the surface of the cloud is less than a third of gravitational acceleration of the surface of the Earth?"

well, if the only thing holding it from collapsing further is the pressure of the gas, and that pressure is .002% of the air pressure on the Earth's surface, it's going to keep collapsing, isn't it?

Regards
Krill

OK grasp for straws, don't you need to know the total mass of the gas within the confinds of the nebula cloud? Once you know its composition and you dou you multiply in the constants Krill. But it is mass dependent now enough of this.

"The simulation followed the collapse of an interstellar gas cloud more than 1 light-year in diameter and containing a mass of gas and dust equal to 50 Suns. The end result, after 266,000 years, was the formation of a cluster of stars typical to our galaxy."

actually that is as close to correct as you can get, that 1lyr in diameter is about the size of the galactic center radius if there were a black hole at the center I believe

well, if the only thing holding it from collapsing further is the pressure of the gas, and that pressure is .002% of the air pressure on the Earth's surface, it's going to keep collapsing, isn't it?
Krill read the factor 10 problem over

"well, if the only thing holding it from collapsing further is the pressure of the gas, and that pressure is .002% of the air pressure on the Earth's surface, it's going to keep collapsing, isn't it?
Krill read the factor 10 problem over"

I read what you wrote. The .002% of Earth atmospheric pressure is what you get *after* accounting for your 15 orders of magnitude. All of this, of course, is according to your own ludicrous "mathematics". All I did was fed in the numbers that you hadn't bothered to do.

Regards
Krill

I read what you wrote. The .002% of Earth atmospheric pressure is what you get *after* accounting for your 15 orders of magnitude.

Why are you bother with figuring out variables just get this through your head every time the radius shrinks to .1 of original the original pressures increases 1000 times and gravitational acceleration increases 100 times so there is a factor 10 deficit for every step no matter what the initial pressure or gravitational acceleration is. After a few steps the cloud will not collapse but slow to an equilibrium.

I think I see what you are getting at. No on first thought it will not keep collapsing, not on the whole, you will have part of the system forming a black hole in gravitational collapse. Because the center of the galaxy is fueled hydrodynamically, I would think upon near collapse you would have something similar to a toroid shape eventually a biconcave disk flattened and depressed at the center with a dumbbell shaped cross-section. I guess I am thinking about this in terms of biology since there is no representation of what it actually looks like I can find. There would be interactiong between the hydrophilic and hydrophobic sections, with the hydrophobic sections gravitating back toward itself.

"Why are you bother with figuring out variables just get this through your head every time the radius shrinks to .1 of original the original pressures increases 1000 times and gravitational acceleration increases 100 times so there is a factor 10 deficit for every step no matter what the initial pressure or gravitational acceleration is. After a few steps the cloud will not collapse but slow to and equilibrium."

I understand what you're saying. In fact, it's kind of like what I put to you as a yes/no question derived from your assertions (however wrong they might be):

at some radius, there will be an equilibrium, yes?

Soooo... we have to figure out what that number is.

According to your "mathematics", for something of a mass of 50 suns that radius will be smaller than the orbit of Mercury.

Regards
Krill

KRILL,
IT IS NICE TO FIGURE OUT EVERY DETAIL, BUT IF I DID I WOULD GET BOGGED DOWN WITH NUMBER CRUNCHING. THE MAIN THING HERE IS THAT THE WAY A STAR IS FORMED IS NOT ALONG CURRENT THEORIES. THIS CAN ALSO BE APPLIED TO THE HYDROSTATIC STABILITY OF THE SUN AND PLANETARY FORMATION.

Now I need to finish the web page that you have been waiting forever for.

just make it as simple as you can, don't get too bogged down, it gets confusing lol

Krill if you want to do it, because I am not, Set up a differental equation dependent on a variable radius after initial pressure and gratiational acceleration of the nebula cloud is set on opposite sides of the hydrostatic equation to equal 0.

According to your "mathematics",

Its funny how its my mathematics, the formulas used are standard in any text book and the numbers don't lie. large amount of pressure, weak gravitational acceleration and its still weak if it was collasped to .000005 of its original size. You need a new theory.