F.T.L. Morse, the Forgotten Descendant

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21 years 9 months ago #3787 by Patrick
Replied by Patrick on topic Reply from P
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>With all due respect, the term c^2 by itself is not useful as a physical expression. c is the velocity of light. c^2 is only meaningful in the context of a whole physical expression, but by itself has meaning only as a mathematical term. <hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
I'm going to streeeetch my neck out here so bare with me. Call me stupid if you want to. <b>WARNING!!!! In M.H.O.</b>
C^2 is equal to the speed of gravity. All *mass* has gravity, "pure" energy has no gravity as it has no mass. In order to convert mass to energy it needs to overcome its own internal gravitational pull, it does this at c^2.

How could we test this?
I'm not a scientist but here is an idea.
In a contained environment, measure the gravitational pull of a specific piece of mass, convert it to energy within the contained environment, remeasure the gravitational pull. My suspicion is there will be none. I believe this is why c^2 is part of E=mc2, mass past the pull of gravity equals energy and energy divided by gravity equals mass. Therefore, gravity equals energy divided by mass.
Now, you should be able to get around the *clocking* issue by measuring its gravitaional pull instead of its speed.




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21 years 9 months ago #3789 by Quantum_Gravity
we need some real world communication angles and setups that would be used in any situation, like real world angles, first straight across as this A
B, maybe we can test vanderwall forces with having a setup where point A and B are up and down form each other.

The intuitive mind

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21 years 9 months ago #3608 by MarkVitrone
Replied by MarkVitrone on topic Reply from Mark Vitrone
Nice suggestion QG, that is what I am talking about. Lets design a vanderwaal foces test laterally. I will research this and make a few calls and let ya'll know. MV

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21 years 9 months ago #3614 by Quantum_Gravity
are we talking realistic distances here like a 1/4 mile across a lake like anothe experiment. we need to specifie distance as if we consider to have point A and B.

The intuitive mind

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21 years 9 months ago #3972 by Jim
Replied by Jim on topic Reply from
How does anyone know c2 as in e=mc2 is a valid constant and what the true value of it is? It seems to me there is a lot of bogus math done to make c2 equal the speed of light squared. How do you know c2 is exactly equal to the speed of light squared? Why can't it be a little less than that?

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21 years 9 months ago #3647 by Patrick
Replied by Patrick on topic Reply from P
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>It seems to me there is a lot of bogus math done to make c2 equal the speed of light squared. How do you know c2 is exactly equal to the speed of light squared? Why can't it be a little less than that?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
I don't know if this will help any but here you go.
The following is how the E, m, and c2 concept works in a back and forth manner in the approximate conversion of one kilogram (2.205 lbs.) of mass into energy expressed in Joules:

(<b>E</b>)=mc2; (1 kg. times 2.9979 x 10 8 times 2.9979 x 10 8) = <b>8.9874 x 10 16 Joules.</b>

(<b>m</b>)=E/c2; (8.9874 x 10 16 Joules divided by "c2" 2.9979 x 10 8 times 2.9979 x 10 8) = <b>1 kg</b>

(<b>c2</b>)=E/m; (8.9874 x 10 16 Joules divided by 1 kg.) = "c2" <b>2.9979 x 10 8 times 2.997 x 10 8.</b>

"c^2" the speed of light squared. Is c^2 actually the speed of gravity?






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