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### Author Topic: Solving for the gravitational constant.............  (Read 13564 times)

#### Ethos

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##### Solving for the gravitational constant.............
« on: 10/03/2009 23:46:58 »
I believe the following equation has some merit, what do you folks think?

[(hbar*c/e^2)^3]*[(hbar*c/G^3)^3] = (pi^3 * 10^20)

#### DoctorBeaver

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##### Solving for the gravitational constant.............
« Reply #1 on: 10/03/2009 23:56:50 »
erm...

#### JP

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##### Solving for the gravitational constant.............
« Reply #2 on: 11/03/2009 00:44:40 »
Uh.... I think your units are off to say the least.

#### Ethos

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##### Solving for the gravitational constant.............
« Reply #3 on: 11/03/2009 00:52:47 »
Uh.... I think your units are off to say the least.
Please point out which units sir. This equation is balanced and dimensionless and is satisfied both in cgs and SI units. If you have a mind, please solve for G and tell me what your results are...........Ethos

#### JP

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##### Solving for the gravitational constant.............
« Reply #4 on: 11/03/2009 01:00:30 »
I'm assuming e is the electron charge.  The constants hbar, c and G don't have units of charge to cancel that out.

#### Ethos

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##### Solving for the gravitational constant.............
« Reply #5 on: 11/03/2009 03:34:08 »
Yes, e stands for the electron charge. However, hbar times c divided by e squared is the fine structure constant to the minus 1.

(hbar * c/e^2)= approx. 137
(hbar * c/e^2)raised to the third power equals: 2.57338006 E6

[(hbar * c/e^2)]^3 * [(hbar * c/G^3)]^3 = (pi^3 * 10^20)

moving (pi^3 * 10^20) to the left denominator and G^9 to the right numerator,

we have: [(hbar * c/e^2)]^3 times [(hbar * c * pi-1)^3]/10^20] = G^9

[2.57338006 E+6]  X  [1.019159191E-71] = 2.6226839407 E-65

taking 2.6226839407E-65 to the ninth root we get: 6.6727538776 E-8 in cgs units

Because (hbar * c/e^2) is dimensionless and,
because (hbar * c/G^3) is dimensionless,

this equation works just as well when done in SI units.

6.6727538776 E-8 in cgs units and 6.6727538776 E-11 in SI units agrees very closely with NIST figures.

« Last Edit: 11/03/2009 03:36:26 by Ethos »

#### Ethos

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##### Solving for the gravitational constant.............
« Reply #6 on: 11/03/2009 03:50:47 »
I'm assuming e is the electron charge.  The constants hbar, c and G don't have units of charge to cancel that out.
Understanding so little about gravity, how can we be sure that the graviton has no significant charge. Charged particles bend space time because they posess magnetic fields. Even though gravity does not bend space time to the same extent per supposed particle, gravity does bend it. That being said, are we being premature in judging gravity to have no charge?

Understanding that there is no space empty of field, and the bending of space time is accomplished by both the electromagnetic and gravitational forces, I suspect that to assume gravity to have no charge is short sighted. We must honestly ask ourselves these question:

What is charge anyway?
And if magnetism and gravity both bend space-time, why should they be considered so different? Maybe it's only a question of magnitude and local stimuli....................Ethos
« Last Edit: 11/03/2009 04:24:51 by Ethos »

#### JP

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##### Solving for the gravitational constant.............
« Reply #7 on: 11/03/2009 04:45:08 »
Ok.  If you're working in the proper units, you can write the fine structure constant in that way.  If that's the case, the second term needs to also be dimensionless.  I'm confused as to how it gets so.  Let's just check it in SI:

hbar ~ J-s = kg m2s-1
c ~ m s-1
G-3 ~ m-9kg3s6,

so:
hbar c G-3 = (kg m2s-1)( m s-1)(m-9kg3s6)=(kg s)4m-6,

which isn't dimensionless.

#### Ethos

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##### Solving for the gravitational constant.............
« Reply #8 on: 11/03/2009 05:11:40 »
Wouldn't hbar ~ J-s = Kg m2s-2
« Last Edit: 11/03/2009 05:47:09 by Ethos »

#### JP

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##### Solving for the gravitational constant.............
« Reply #9 on: 11/03/2009 05:18:57 »
J-s means J*s, which is the units for hbar, and I'm using * to denote multiplication.  1 Joule is 1 kg m2s-2, so 1 J*s= kg m2s-1

#### syhprum

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##### Solving for the gravitational constant.............
« Reply #10 on: 11/03/2009 07:10:56 »

erm

#### Ethos

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##### Solving for the gravitational constant.............
« Reply #11 on: 11/03/2009 08:20:49 »
Yes, e stands for the electron charge. However, hbar times c divided by e squared is the fine structure constant to the minus 1.

(hbar * c/e^2)= approx. 137
(hbar * c/e^2)raised to the third power equals: 2.57338006 E6

[(hbar * c/e^2)]^3 * [(hbar * c/G^3)]^3 = (pi^3 * 10^20)

moving (pi^3 * 10^20) to the left denominator and G^9 to the right numerator,

we have: [(hbar * c/e^2)]^3 times [(hbar * c * pi-1)^3]/10^20] = G^9

[2.57338006 E+6]  X  [1.019159191E-71] = 2.6226839407 E-65

taking 2.6226839407E-65 to the ninth root we get: 6.6727538776 E-8 in cgs units

Because (hbar * c/e^2) is dimensionless and,
because (hbar * c/G^3) is dimensionless,

this equation works just as well when done in SI units.

6.6727538776 E-8 in cgs units and 6.6727538776 E-11 in SI units agrees very closely with NIST figures.

Remembering that (hbar * c/e^2) is dimensionless, the first part of the equation will remain the same whether figured in cgs or SI units. Now plug in the values in the second part of the equation with SI values and you'll find the answer for G will also be registered in SI values. This should prove that (hbar * c/G^3) is also dimensionless. Be my guest and do the math yourself.

If this interests you, I have something even more interesting to share........Ethos
« Last Edit: 11/03/2009 08:22:49 by Ethos »

#### Ethos

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##### Solving for the gravitational constant.............
« Reply #12 on: 11/03/2009 09:24:38 »
An even easier exersize would be to figure (hbar * c/G^3) using both cgs units and SI units. If both computations are equal, then this equation is dimensionless.

Let's try in cgs: (1.054572669 E-27 * 2.99792458 E10/(6.67275E-8)^3 = 1.064098E5

Now in SI: (1.054572669E-34 * 2.99792458E8/(6.67275E-11)^3 = 1.064098E5

Looks like a dimensionless relationship to me....................Ethos

#### Vern

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##### Solving for the gravitational constant.............
« Reply #13 on: 11/03/2009 12:01:54 »
Very interesting Ethos; I'm assuming your maths correct; what is the implication? Have you shown a relationship that was not previously known? Are you showing a new relationship between gravity and the electromagnetic field?

#### lightarrow

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##### Solving for the gravitational constant.............
« Reply #14 on: 11/03/2009 13:14:35 »
Remembering that (hbar * c/e^2) is dimensionless, the first part of the equation will remain the same whether figured in cgs or SI units.
No, hc/e2 is dimensionless in c.g.s. only. In M.K.S. the dimensionless expression is 4πε0hc/e2.

Quote
Now plug in the values in the second part of the equation with SI values and you'll find the answer for G will also be registered in SI values. This should prove that (hbar * c/G^3) is also dimensionless.
No, that expression is not dimensionless neither in c.g.s. nor in M.K.S.

#### Ethos

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##### Solving for the gravitational constant.............
« Reply #15 on: 11/03/2009 14:26:10 »
Remembering that (hbar * c/e^2) is dimensionless, the first part of the equation will remain the same whether figured in cgs or SI units.
No, hc/e2 is dimensionless in c.g.s. only. In M.K.S. the dimensionless expression is 4πε0hc/e2.

Quote
Now plug in the values in the second part of the equation with SI values and you'll find the answer for G will also be registered in SI values. This should prove that (hbar * c/G^3) is also dimensionless.
No, that expression is not dimensionless neither in c.g.s. nor in M.K.S.
Dimensionless is a term equal to a ratio, and when changing from cgs to SI, if the ratio does not change, then the equation is dimensionless. If you'll do the math, you'll find that the results are approiate for both cgs and SI units. Whether you agree that these initial terms are dimensionless or not, math does not lie and the ratio stays the same.

#### lightarrow

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##### Solving for the gravitational constant.............
« Reply #16 on: 11/03/2009 14:38:45 »
Dimensionless is a term equal to a ratio, and when changing from cgs to SI, if the ratio does not change
Who told you that it doesn't change? hc/e2 ≈ 137 in c.g.s. while it's ≈ 1.24*1012 J*m*C-2 = 1.24*1012 kg*m3*A-2*s-4 in M.K.S..
« Last Edit: 13/03/2009 13:08:29 by lightarrow »

#### Ethos

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##### Solving for the gravitational constant.............
« Reply #17 on: 11/03/2009 15:21:58 »
Dimensionless is a term equal to a ratio, and when changing from cgs to SI, if the ratio does not change
Who told you that it doesn't change? hc/e2 ≈ 137 in c.g.s. while it's ≈ 1.24*1012 J*m*C-2 = 1.24*1012 kg*m3*A-2*s-4 in M.K.S..
I'll grant you that, but I'm referring to (hbar c/G^3) not changing. All three values have their cgs and SI counterparts. Now for those that don't like me using (hbar c/e^2) in this equation, let's use this equation instead: (hbar c/(re me c^2)) All values in this equation have their cgs and SI counterparts.

[hbar c/(re, me, c^2)]^3 * [hbar c/G^3]^3 = pi^3 *10^20

All values in the above formula have their cgs and SI counterparts. If you'll do the math, using cgs units wil produces G in cgs. Likewise, using SI units will produces G in SI.
« Last Edit: 11/03/2009 15:28:18 by Ethos »

#### Ethos

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##### Solving for the gravitational constant.............
« Reply #18 on: 11/03/2009 15:25:28 »
Very interesting Ethos; I'm assuming your maths correct; what is the implication? Have you shown a relationship that was not previously known? Are you showing a new relationship between gravity and the electromagnetic field?
Yes, but untill we can reach an understanding about the equation I'm using, it would be of little use to expound upon it. Thanks for the interest..................Ethos
« Last Edit: 11/03/2009 17:35:34 by Ethos »

#### JP

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##### Solving for the gravitational constant.............
« Reply #19 on: 11/03/2009 16:06:38 »
Now for those that don't like me using (hbar c/e^2) in this equation, let's use this equation instead: (hbar c/(re me c^2))

Why don't you just call that first term α-1, the inverse of the fine structure constant, which is dimensionless?  But even if you do that, it doesn't change the fact that the second term has the wrong units and is not dimensionless.  Having the same value in c.g.s. and SI does not mean something is dimensionless.  Any quantity with units of mass2/distance3 will have the same value in c.g.s. and SI, but it has dimensions.  Take for example:
1 kg2/m3=1 g2/cm3=106g2/m3=10-6kg2/cm3

#### LeeE

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##### Solving for the gravitational constant.............
« Reply #20 on: 11/03/2009 16:50:56 »
Charged particles bend space time because they posess magnetic fields....

...and the bending of space time is accomplished by both the electromagnetic and gravitational forces...

Are you really sure about that?  Empirical evidence suggests that electromagnetism does not bend spacetime.

#### Ethos

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##### Solving for the gravitational constant.............
« Reply #21 on: 11/03/2009 17:08:28 »

I appreciate the feedback jpetruccelli, and I also appreciate the manner in which you handle yourself when faced with someone that needs better understanding about the nature of physics. I confess, there is much I need to learn about this field and you have shown patience with me that others have not. I didn't come here to argue with anyone and find myself rather disturbed when others confront my understanding in an aggressive manner. To that, I again thank you for your patience  and ask to further correspond about this issue with you via private message. I wish to no longer stir the pot, as it were, in public forum because I think you can help me with this idea in private. If that would be OK, let me know and I'll get back with you. Thanks for the civil behavior you've demonstrated on my behalf..........Ethos

« Last Edit: 11/03/2009 21:46:46 by Ethos »

#### lightarrow

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##### Solving for the gravitational constant.............
« Reply #22 on: 12/03/2009 11:16:35 »

I appreciate the feedback jpetruccelli, and I also appreciate the manner in which you handle yourself when faced with someone that needs better understanding about the nature of physics. I confess, there is much I need to learn about this field and you have shown patience with me that others have not. I didn't come here to argue with anyone and find myself rather disturbed when others confront my understanding in an aggressive manner. To that, I again thank you for your patience  and ask to further correspond about this issue with you via private message. I wish to no longer stir the pot, as it were, in public forum because I think you can help me with this idea in private. If that would be OK, let me know and I'll get back with you. Thanks for the civil behavior you've demonstrated on my behalf..........Ethos
I really hope you wasn't talking about me when you say that others confronted your understanding in an aggressive manner, because I really hadn't the slight intention to be so, and, infact, I objectively wasn't.

#### Ethos

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##### Solving for the gravitational constant.............
« Reply #23 on: 13/03/2009 10:08:51 »
Now why would you get that idea?

#### lightarrow

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##### Solving for the gravitational constant.............
« Reply #24 on: 13/03/2009 13:06:40 »
Now why would you get that idea?
So it was my having coloured in red some of my post?
If it's so, I can immediately change it; I usually use blue and red just because they are the best seen colours among the available ones.

#### The Naked Scientists Forum

##### Solving for the gravitational constant.............
« Reply #24 on: 13/03/2009 13:06:40 »