Schwarzschild radius

[elas]

Wikipedia gives the value for finding the Schwarzschild radius as approximately equal to 2.95.

A theory under development predicts a value of 2.8615.

I would like a professional opinion on whether the prediction falls within the area of acceptability or probability.

[Klaynos]

Wikipedia gives the value for finding the Schwarzschild radius as approximately equal to 2.95.

A theory under development predicts a value of 2.8615.

I would like a professional opinion on whether the prediction falls within the area of acceptability or probability.

 

What are the errors on the quantities put into the two equations, you would then need to apply error analysis equations to find the bounds of the two predictions....

 

And the Schwarzschild radius for what, and in what units?

 

Are the two equations dimensionally equivalent?

[Spyman]

When the escape velocity is set equal to the speed of light, the radius of a Newtonian Dark Star is exactly the Schwarzschild radius of a Black Hole with General Relativity.

 

Not much of a wiggle room there, if you get a different value you will need a very good explanation of why.

(in my non professional opinion)

[Mr Skeptic]

Don't forget your units!

[swansont]

"The proportionality constant, 2G /c^2, is approximately 1.48×10-27 m/kg, or 2.95 km/solar mass"

 

http://en.wikipedia.org/wiki/Schwarzschild_radius

 

2 is an exact number. c is a defined constant

 

G is known to 4 digits after the decimal

 

http://en.wikipedia.org/wiki/Gravitational_constant

 

Using the above SI value, there is no wiggle room — any discrepancy has to be in he truncated digits. Using the solar mass approximation, it will depend on how well the solar mass is known. I suspect that's known to better than the 3% discrepancy between the two values.

 

http://en.wikipedia.org/wiki/Solar_mass

 

Yes, it is.

 

No wiggle room here.

[Mr Skeptic]

Right, then. The theories make observably different predictions.

[elas]

I calculated the error as 1.3% but that is not material to the debate. The units were deliberately omitted for reasons I do not want to explain until the proposed value is acceptable.The replies make it clear that the proposed value is not acceptable.

 

First let me sincerely thank everyone for the clarity of their replies, not a single vulgar comment; that's a rare acheivement for replies to my submissions. Secondly let me add that all is not lost, I do need to check the units as it would not be the first time I have got them wrong; added to that there is one other possibility that I will calculate with my fingers crossed. Will come back to this forum if I have a positive result,

Many thanks to all

elas

 

(10 minutes later) That was a lot easier than expected. The value 2.8615 was derived using the solar mass but, if the sun were to collapse into a Black Hole it would of course, include the mass of the planets. Redoing the calculation using solar mass plus the mass of the planetary system gives a value of 2.94. I can now make a speculative proposal that there is one equation for finding the nucleus of any force field (strong, electromagnetic or gravitational).

Edited November 6, 2009 by elas

[Martin]

...The value 2.8615 was derived using the solar mass but, if the sun were to collapse into a Black Hole it would of course, include the mass of the planets. ...

 

Why would it necessarily include the mass of the planets?

 

I have another question as well, elas. What value are you using for the mass of the sun? Is it in kilograms? How many kilograms? Different books can differ in the numbers they give (it will depend on the value of Newton's constant which the author of the book is using.)

[Mr Skeptic]

Yay for google calculator!

 

Anyhow, the theories give different predictions. But has anyone actually measured the mass/radius of an actual black hole?

[elas]

Why would it necessarily include the mass of the planets?

 

 

The idea for this equation occurred to me while watching a BBCTV programme On Black Holes in which the expert explained the mathematical relationship between the mass of galaxies and the radii of the Black Holes at the galactic centres. When I read the answers to my original question I realised that the experts used the mass of the system (galaxy) while I had used only the mass of the central body (star); the equation requires the mass value of the system (i.e. the solar system in my example). That is to say that the radius of the field nucleus is determined by the sum of the mass of all bodies with mass, within the field.

 

PS It should be noted that the radii of nuclei is independent of spin (wikipedia article) therefore nuclear spin must decrease with increase in the speed of the field (because no point on the nuclei surface can exceed the speed of light) this causes the angular momentum to remain stable (i.e.unchanging). This sugests that Angular Momentum originates from the distribution of mass (in the form of two opposing waves [just like ocean waves on planet Earth]) within the nuclei.

 

I have another question as well, elas. What value are you using for the mass of the sun? Is it in kilograms? How many kilograms? Different books can differ in the numbers they give (it will depend on the value of Newton's constant which the author of the book is using.)

 

The Black Hole equation uses the solar mass as 1 unit of mass (see Wikipedia on Black Holes).

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Yay for google calculator!

 

Anyhow, the theories give different predictions. But has anyone actually measured the mass/radius of an actual black hole?

 

The measurements made by observers are the mass values of galaxies (solar mass = 1) and the Black Hole radii in km.

Edited November 7, 2009 by elas
Consecutive posts merged.

[Mr Skeptic]

But how can they possibly have measured the radius of a black hole? Are you sure those aren't calculated from measured mass?

[elas]

But how can they possibly have measured the radius of a black hole? Are you sure those aren't calculated from measured mass?

 

You are correct and I apologise if I have given the wrong impression. My interest lies in comparing a proposed equation for particle radii with the equation used to find the radii of Black Holes (in the belief that nature uses the same structural equatation for all forces). Using the solar mass did not produce a usable value, hence my opening question. Using the Solar system mass does produce a usable value and I can now continue with my speculative theory. Discussion of speculative theories is not allowed, so I shall say no more except to point out that perhaps the difference between the use of the system mass compared with the use of the central body mass might be of some interest to Black Hole theorist.

[swansont]

2.95 ≠ 2.86

 

2.94 ≠ 2.86

 

No reason to include the planet mass in the equation, but even if you did the numbers are still not equal.

 

If the sun were to somehow be replaced by a black hole with equal mass, the planets would not fall into it — the gravity experienced by the planets would be exactly the same. The Schwarzschild radius, per the equation, would be 2.95 km.

[elas]

2.95 ≠ 2.86

 

2.94 ≠ 2.86

 

No reason to include the planet mass in the equation, but even if you did the numbers are still not equal.

 

If the sun were to somehow be replaced by a black hole with equal mass, the planets would not fall into it — the gravity experienced by the planets would be exactly the same. The Schwarzschild radius, per the equation, would be 2.95 km.

 

You are aware of my work on particle radii. You should also be aware of my explanation of (atomic) electron shell in terms of balanced fields (both published ‘elsewhere’). I wanted to see if there is a similar equation that would give the radius of the nucleus and whether the equation would support the balanced field concept. This has nothing to do with the actual formation of Black Holes but, it has everything to do with the equation used to calculate the radius of Black Holes.

In order to find the minimum radius of the nucleus of the solar gravitation field, it is necessary to use all the mass within the solar gravity field. In reality of course the solar system does not contain enough mass to create a Black Hole but, that does not prevent the use in the Black Hole equation as part of a mathematical exercise.

The equations found have been published ‘elsewhere’.

[Mr Skeptic]

Elas, my point is that as far as I know, we only know that your theory makes different predictions than does GR. This is generally considered a bad thing, yes, but unless we have actually observed a black hole's radius (by some method other than GR) and confirmed that the GR prediction is correct, then this difference shouldn't invalidate your theory. Obviously, you'd still need a reason for someone to follow your theory as opposed to GR.

[elas]

Elas, my point is that as far as I know, we only know that your theory makes different predictions than does GR. This is generally considered a bad thing, yes, but unless we have actually observed a black hole's radius (by some method other than GR) and confirmed that the GR prediction is correct, then this difference shouldn't invalidate your theory. Obviously, you'd still need a reason for someone to follow your theory as opposed to GR.

 

Quite the opposite, it is essential to proving my proposition, that my equation produces the same result as GR, preferably it should also explain how and why.

 

Currently Black Hole gravitation is calculated froim the orbit of nearby stars and the mass is calculated from the gravitational force. Obviously this would include the mass of any planetary system belonging to the Black Hole (just as electrons contribute to the of the mass of an atom).

 

G is the force between two bodies; therefore the constant required for a single body is G/2. This suggest that G/2 is the true Constant of Force. My aim here is to show that G/2 applies to Black Holes (having already speculated on how it applies to particles). In fact the accepted Black Hole equation acts in the opposite direction to a proposed particle radii equation (Black Hole radius divided by Black Hole mass equals G/2 [mass and radius are expressed in the same terms as used in the Black Hole equation]).

 

The equation for field radius is the same as the equation for field nucleus, but in the opposite direction.

Edited November 8, 2009 by elas

[swansont]

Currently Black Hole gravitation is calculated froim the orbit of nearby stars and the mass is calculated from the gravitational force. Obviously this would include the mass of any planetary system belonging to the Black Hole (just as electrons contribute to the of the mass of an atom).

 

G is the force between two bodies; therefore the constant required for a single body is G/2. This suggest that G/2 is the true Constant of Force. My aim here is to show that G/2 applies to Black Holes (having already speculated on how it applies to particles). In fact the accepted Black Hole equation acts in the opposite direction to a proposed particle radii equation (Black Hole radius divided by Black Hole mass equals G/2 [mass and radius are expressed in the same terms as used in the Black Hole equation]).

 

The equation for field radius is the same as the equation for field nucleus, but in the opposite direction.

 

F = GMm/r^2

 

In the calculation of force, both masses are already used, so you'd get the wrong answer if you used the mass of the whole system and the mass of the planet. It's multiplication, not addition in this equation, so if you are going to propose a contribution from each mass, shouldn't you be using [math]\sqrt{G}[/math] ?

[elas]

F = GMm/r^2

 

In the calculation of force, both masses are already used, so you'd get the wrong answer if you used the mass of the whole system and the mass of the planet. It's multiplication, not addition in this equation, so if you are going to propose a contribution from each mass, shouldn't you be using [math]\sqrt{G}[/math] ?

 

My original question was intended to obtain a reply that would enable me to continue with the developement of an idea. Subsequent questions have resulted in my making quick replies without first rigorously checking the development.

 

Somehow I have given you the impression that I am dealing with two bodies (F = GMm/r^2) when in fact I am dealing with one body (Black Hole mass). The following contradicts some of my previous statements as this time, I have taking the time to do a partial check. As usual for me, the problem of finding the correct units is the last problem to be solved. (Currently the decimal point for G is in the wrong place).

 

As with single particles I find that the Black Hole equation can be replaced with: m/r = constant

where m = mass of Black Hole

r = radius of Black Hole

The constant is very close to G (just as the constant in the particle equation is close to G/2) but I have still to sort out the units and confirm this.

 

When I have finished developement I will submit the result to the appropiate forum.

[Mr Skeptic]

As with single particles I find that the Black Hole equation can be replaced with: m/r = constant

where m = mass of Black Hole

r = radius of Black Hole

The constant is very close to G (just as the constant in the particle equation is close to G/2) but I have still to sort out the units and confirm this.

 

That constant is, m/r = c^2/2G. This is from the equation for the Schwarzschild radius.

[elas]

That constant is, m/r = c^2/2G. This is from the equation for the Schwarzschild radius.

 

Thanks, I note that the equation on the Wikipedia Black Hole page does not use c^ and yet presumably it produces the same r value as the equation on the Schwarzchild page. I will check that bearing in mind that Einstein was unable to answer the question 'why c^' (he asked the question of himself).

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PS (added about 12 hrs later)

At long last I believe I can make I can make a logical deduction from all the comments herein and my earlier work on particles.

Recall that my original proposal was that for elementary particles:

 

mr = linear force constant.........................................1

 

After Swansont’s constructive criticism the units were changed and the equation became:

 

mr = G/2.....................................................................2

 

On this forum Mr Skeptic pointed out that the Black Hole constant I was seeking already existed and it is not G or G2.

However the fact that a constant exist is proof that:

 

mr = constant...........................................................3

 

is also true of Black Holes.

 

When working in terms of linear force it was shown that the linear force of a composite is the sum of the linear force of all the elementary particles that make up the composite; leading to the possibility that this might also be true of black holes ( as Gravity is a Linear force the terms are interchangeable). In conclusion I would say that it is not so much a question of finding something new, as it is a question of finding the simplest equation that explains the cause. The cause being the nature of elementary particles; the equations of physics are only true if all the so-called elementary particles have the same fundamental content of matter and force; that is to say that there is only one elementary particle.

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Correction

 

Equation 3 should read:

 

m/r = constant

 

that is of course what I meant when I wrote that the equations for nucleus radius and field radius act in opposite directions. This also implies that Black Holes and other nuclei have a structure that is the reverse of shell structure which would in turn explain why the strong force increases with distance while magnetic and gravitational forces decrease with distance. It also suggests that Black Holes contain quarks like any other natural composite particle nucleus.

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Clearly I have lost the interest of the experts which is a pity because after further examination of the values it is clear that the two questionable values are:

 

1) not greater than 2.95

From the Black Hole equation

and

2) the value of G [see Wikipedia]

then note that:

 

1/(m/r) = 1.48E+00 [approximately 1/2 of 2.95 constant]

 

which (to a greater degree of accuracy) in the same units as used in the particle equation is equal to G/2.

 

Implying that the Schwarzchild radius equation can be used to calculate an accurate value for G

 

It also confirms my original speculation that:

 

r(field) = (G/2)/m

r(nucleus) = (G/2)m

 

Gravity, the force of infinity; determines the nature of all with infinity.

Edited November 11, 2009 by elas
Consecutive posts merged.

[Klaynos]

There is no space for approximately or similar, or close to in physics, there is only consistent with or not consistent with, this depends on your errors, your results are not consistent with the accepted, simple to derive equation.

[elas]

There is no space for approximately or similar, or close to in physics, there is only consistent with or not consistent with, this depends on your errors, your results are not consistent with the accepted, simple to derive equation.

 

This leaves me with a problem:

The simple to derive equations are derived using constants of not greater than 2.95 (yet Paul Davies gives the value of this constant as 3) and the G constant which is probably somewhere between the gravitational constant giving by Codata as 6.67428 x 10^-11 and the Fixler et al (2007) value of, G = 6.693 × 10^−11. With what are these values consistent?

Edited November 11, 2009 by elas

[Klaynos]

This leaves me with a problem:

The simple to derive equations are derived using constants of not greater than 2.95 (yet Paul Davies gives the value of this constant as 3) and the G constant which is probably somewhere between the gravitational constant giving by Codata as 6.67428 x 10^-11 and the Fixler et al (2007) value of, G = 6.693 × 10^−11. With what are these values consistent?

 

Measured values are where you start your error analysis. You haven't stated the values correctly as it should be value +/- error... in ALL cases. Have a look at swansonts post where he discusses the errors of some of the constants above...

[elas]

Measured values are where you start your error analysis. You haven't stated the values correctly as it should be value +/- error... in ALL cases. Have a look at swansonts post where he discusses the errors of some of the constants above...

 

Thank-you for clarifying this point. As I mentioned previously, when asking the opening question I did not expect to become involved in an explanation of the work that I was only just beginning, but I will take all the advice received into account as I proceed. The result will appear in the appropiate forum in about one month.

[swansont]

Measured values are where you start your error analysis. You haven't stated the values correctly as it should be value +/- error... in ALL cases. Have a look at swansonts post where he discusses the errors of some of the constants above...

 

Indeed. The Fixler/Kasevich number is 6.693, with a standard error of 0.027 and a systematic error of 0.021, which means it's consistent with the other value, but not measured as precisely.

 

3 is the acceptable way of expressing 2.95, if you are only using one significant digit. Did Davies use this in a paper or a popular book or talk? I expect it's the latter