New look at dimensions:

[Guest loarevalo]

I'm not as smart as you would think.Tell me, what is your conclusion concerning these calculations? I am asking you to show me. Bear in mind that not many people here would ask you that - they would simply prove you wrong. I too am guilty of that - of participating in these threads just to re-prove my theories while proving wrong everyone else's - while the purpose should be to participate in the spirit of enquiry, not advocacy.

 

So, show me.

[infamous]

I'm not as smart as you would think.Tell me, what is your conclusion concerning these calculations? I am asking you to show me. Bear in mind that not many people here would ask you that - they would simply prove you wrong. I too am guilty of that - of participating in these threads just to re-prove my theories while proving wrong everyone else's - while the purpose should be to participate in the spirit of enquiry, not advocacy.

 

So, show me.

These calculations do not yet constitute a theory loarevalo, what they do show however, is a relationship between these constants of nature that have to date gone un-noticed. If you have read any of Buckminster Fuller's work, you will also notice his consideration of the tetrahedron and the significant number 6. I have not presented this work as proof of anything, I only wish for others to investigate the peculiarities involved in these figures. I believe there is information hidden within these anomolies and I'm just asking others to help me discover what those secrets might be. If you can help with this investigation, I would like to hear your thoughts.

[Erasmus00]

It seems to me like its just a math trick. You could start with any two constants, and produce any third by playing around like that. It seems more mathematical coincidence then fundamental property.

-Will

[infamous]

It seems to me like its just a math trick. You could start with any two constants, and produce any third by playing around like that. It seems more mathematical coincidence then fundamental property.

-Will

There's certainly no trick to: e^2/(hbar*c), this is the fine structure constant represented by (a). If you'll investigate the math, this equation is also balanced and dimensionless. No, I don't think it could be called a trick.

[Erasmus00]

There's certainly no trick to: e^2/(hbar*c), this is the fine structure constant represented by (a). If you'll investigate the math, this equation is also balanced and dimensionless. No, I don't think it could be called a trick.

 

That equation can be taken as the deffinition of alpha, and wasn't at all what I was referring to. The relations you come up with, such as e*hbar = 2A/B^110 (to pick one at random) seem like math games rather then fundamental properties.

-Will

[infamous]

That equation can be taken as the deffinition of alpha, and wasn't at all what I was referring to. The relations you come up with, such as e*hbar = 2A/B^110 (to pick one at random) seem like math games rather then fundamental properties.

-Will

Those math games lead to my calculations for finding the gravitational constant (G), which BTW is the equation that I'm refering to here.

 

G^11.1 = (hbar*c/(a*pi))^2.7 * (hbar*re)^1

 

It will take you some effort but, if you will spend the time you will see that this equation is balanced and dimensionless. Not really a trick at all.

[Erasmus00]

Those math games lead to my calculations for finding the gravitational constant (G), which BTW is the equation that I'm refering to here.

 

G^11.1 = (hbar*c/(a*pi))^2.7 * (hbar*re)^1

 

It will take you some effort but, if you will spend the time you will see that this equation is balanced and dimensionless. Not really a trick at all.

 

Your dimensions don't match. G has dimensions of distance^3/(mass*time^2), and the right side of your equation is, as you say, dimensionless. And again, if you raise to arbitrary powers, you could probably build any number you wanted out of any other numbers you wished to start with. Its just math games.

-Will

[infamous]

Your dimensions don't match. G has dimensions of distance^3/(mass*time^2), and the right side of your equation is, as you say, dimensionless. And again, if you raise to arbitrary powers, you could probably build any number you wanted out of any other numbers you wished to start with. Its just math games.

-Will

I believe you need to see this equation written as such:

 

(hbar*c/e^2)^27 * (hbar*c/G^3)^27 * (hr*re/G^3)^10 = (pi)^27

 

Simplified:

 

(hbar *re)^10 * (hbar*c)^27 /(a *(G)^111) = (pi)^ 27

[Turtle]

And again, if you raise to arbitrary powers, you could probably build any number you wanted out of any other numbers you wished to start with. Its just math games.

-Will

___My field is more in discrete mathematics, so I start to loose the picture as you all discuss the calculus here. Nonetheless, I looked at Infamous' equations & found the algebra is correct to the best of my ability; others seem to as well.

___Now Erasmuss00 wishes to say it's correct algebraically but just a "trick" & I have a discrete view of this assesment. You may say it is a "trick" but this implies an intent to deceive & so is a poor word in this case. On the contrary, Infamous is putting it all up front.

We may find the case with this that it is no more a "trick" than Pascal's Triangle, which is always in the "authorized" texts as a "shortcut", not a trick.

___Anytime in working equations with numerous variables, it is only an advantage when one cancels out irrationals or brings the whole to an integer form. The Universe is corpuscular (quantized) & therefore the smooth curves implied by irrational terms represent artifacts of the math system employed. :hihi:

[Erasmus00]

.

___Now Erasmuss00 wishes to say it's correct algebraically but just a "trick" & I have a discrete view of this assesment. You may say it is a "trick" but this implies an intent to deceive & so is a poor word in this case. On the contrary, Infamous is putting it all up front.

We may find the case with this that it is no more a "trick" than Pascal's Triangle, which is always in the "authorized" texts as a "shortcut", not a trick.

 

What I mean when I say Infamous is playing a math game, is that it is just coincidence that things work out the way his formulas show. The fact that the equations aren't particularly elegant (requiring all sorts of arbitrary powers) seems to me to indicate a lack of physical significance. With a bit of effort, you could probably express all of the values of the physical constants in terms of pi and e (being the natural exponent). However, such things would have no real physical meaning.

-Will

[infamous]

What I mean when I say Infamous is playing a math game, is that it is just coincidence that things work out the way his formulas show.

It may be only coincidence Erasmus00 and you have every right to consider it so. Taking this position as you do, could you explain to me why this coincidence includes all these constants. If it is only coincidence, it is one of massive proportion.

[Erasmus00]

It may be only coincidence Erasmus00 and you have every right to consider it so. Taking this position as you do, could you explain to me why this coincidence includes all these constants. If it is only coincidence, it is one of massive proportion.

 

It works because you are allowing yourself to raise any of the constants to any arbitrary power. I should also point out that your relations don't have the right unit. Consider the first: C^2*re = (A)(B^25). Your A is 6/pi your B is 10^1/3, and are unitless. C^2*re has units of distance^3/time^2. Your units don't match up. Your number game only works because you've chosen cgs, if you measure mks, your relation no longer holds. This is true of many of your relations. Its can hardly be considered a massive coincidence when you can raise numbers to any power.

-Will

[infamous]

It works because you are allowing yourself to raise any of the constants to any arbitrary power. I should also point out that your relations don't have the right unit. Consider the first: C^2*re = (A)(B^25). Your A is 6/pi your B is 10^1/3, and are unitless. C^2*re has units of distance^3/time^2. Your units don't match up. Your number game only works because you've chosen cgs, if you measure mks, your relation no longer holds. This is true of many of your relations. Its can hardly be considered a massive coincidence when you can raise numbers to any power.

-Will

Using (6/pi) as a dimenless value times c in what even unit you choose will work quite nicely. Try it. The value of c in cgs or SI units rationalizes the result.

[Erasmus00]

Using (6/pi) as a dimenless value times c in what even unit you choose will work quite nicely. Try it. The value of c in cgs or SI units rationalizes the result.

 

(6/pi)^1/4 *10^(25/3) = 2.53*10^8 (no units).

 

c^2*re = 2.53*10^8 centimeters^3/seconds^2=253 meters^3/seconds^2.

 

Your relationship is off by several orders of magnitude if I measure distance in meters. All of your other relationships work the same way. They only hold in one set of units.

-Will

[infamous]

(6/pi)^1/4 *10^(25/3) = 2.53*10^8 (no units).

 

c^2*re = 2.53*10^8 centimeters^3/seconds^2=253 meters^3/seconds^2.

 

Your relationship is off by several orders of magnitude if I measure distance in meters. All of your other relationships work the same way. They only hold in one set of units.

-Will

Erasmus00 please go to my post # 31, and use these figures. You'll find that I have included the value of c in each of these examples. By using c in what ever unit one chooses, the equation will work weather in cgs or Si units. If your using the word documents that I posted here first, I can see how that would be misleading. In any case, I appreciate your interest.

[Erasmus00]

Erasmus00 please go to my post # 31, and use these figures. You'll find that I have included the value of c in each of these examples. By using c in what ever unit one chooses, the equation will work weather in cgs or Si units. If your using the word documents that I posted here first, I can see how that would be misleading. In any case, I appreciate your interest.

 

(radius of the electron) = (A) * (B^25) * (C^-2).

 

In centimeters, this works, as I've said. In meters, this comes out:

 

2.818*10^ -15 meters = 2.818*10^ -9 seconds^2/meters^2.

 

So the numbers and the units do not agree.

-Will

[infamous]

(radius of the electron) = (A) * (B^25) * (C^-2).

 

In centimeters, this works, as I've said. In meters, this comes out:

 

2.818*10^ -15 meters = 2.818*10^ -9 seconds^2/meters^2.

 

So the numbers and the units do not agree.

-Will

 

 

Your quite right Erasmus00, I see now where I'm going wrong here. Looks like I need to start all over doesn't it. Thanks for straighting me out on this. One question, even though it doesn't work when I change units, why does everything work in cgs units.