Darby
Epochal Historian
Re: Yet another time traveling claim
Designer,
Go the simple route that is routinely used in relativistic physics.
Let C =1
"C" appears in the relativistic equation because the underlying assumption (which is proved in the full statement of special relativity) that the speed of light is a constant for all inertial observers. By letting C = 1 you are able to cure the problem of making units match.
Don't worry that you're "cheating". Letting c = 1 is no more arbitrary that letting c = 300,000 km*sec^-1. There's nothing fundamental about the length that we attribute to a kilometer or the duration that we use to define a second. We just need to be consistent in our equation and let the readers know how we have defined the terms.
If you do that then you have two equations (1) and (2)
(1) 1 = mE^2
(2) E = m
(1) is true if and only if E and M = 1. But that's just a math "truth" and is trivial. It isn't true for any other value of E or M. Plug in some numbers, chug them and see the output. It won't equal 1.
(2) is true for any arbitrary value of M
Another problem with (1), because in his saga it is derived from (2) is that C arbitrarily finds itself on the left side and E^2 arbitrarily finds itself on the right side when C was originally squared and E was not. That's simply silly math. How did they migrate?
Einstein didn't just make up E=mc^2. It is a derived equation with a page of proof, and 100 years of physics from the 19th century, that goes before it. E and C don't jump around arbitrarily.
And the equation E = mc^2 is only valid in the rest frame. It gives the rest energy for the mass. The relativistic equation is:
E = (mc^2)/sqrt(1-(v^2/c^2))
And this is where letting C = 1 comes in.
If you let C = 1 then you state velocity as a unitless number between 0 and 1 that represents the portion of the speed of light at which the object under observation is moving WRT the rest/laboratory frame.
So you end up with this equation:
E = m/1-v^2
because you leave out C^2 (1^2 = 1) in the numerator and divisor.
If you want to express the energy in joules then you have to put the real values of C back in. You can also play with the units so that you express mass, energy and length in the same units - kilograms, joules, meters or some other unit.
1 sec = 300,000 km (converted time to length)
1 meter = 10^-8 sec (converted length to time)
You have to go to general relativity to convert mass to length but that equation is:
L (in meters) = G/c^2 * M (in kilograms)
where G is the gravitational constant.
Last...
don't confuse math for physics. Math is the language used to express physical theories. Just because one might "mess around" with an equation doesn't mean that the result expresses a valid physical model. There are a lot of crank "physicists" on the Internet. Most of them are laughing up their sleeves at the people to whom they post. I think that this is the case here.
Designer,
This equation makes no sense from Titorian?
(1) C = M*(E)^2 Titorian
(2) E = MC^2 and from the great man.
So let eliminate the constants C the speed of light.
(1) -> (2) Substituting C
(3) E = M (M*(E)^2)^2
(4) E = M*((M)^2*(E)^4)
(5) E = M^3*E^4
(6) 1/E^3 = M^3
(7) 1/E = M
Go the simple route that is routinely used in relativistic physics.
Let C =1
"C" appears in the relativistic equation because the underlying assumption (which is proved in the full statement of special relativity) that the speed of light is a constant for all inertial observers. By letting C = 1 you are able to cure the problem of making units match.
Don't worry that you're "cheating". Letting c = 1 is no more arbitrary that letting c = 300,000 km*sec^-1. There's nothing fundamental about the length that we attribute to a kilometer or the duration that we use to define a second. We just need to be consistent in our equation and let the readers know how we have defined the terms.
If you do that then you have two equations (1) and (2)
(1) 1 = mE^2
(2) E = m
(1) is true if and only if E and M = 1. But that's just a math "truth" and is trivial. It isn't true for any other value of E or M. Plug in some numbers, chug them and see the output. It won't equal 1.
(2) is true for any arbitrary value of M
Another problem with (1), because in his saga it is derived from (2) is that C arbitrarily finds itself on the left side and E^2 arbitrarily finds itself on the right side when C was originally squared and E was not. That's simply silly math. How did they migrate?
Einstein didn't just make up E=mc^2. It is a derived equation with a page of proof, and 100 years of physics from the 19th century, that goes before it. E and C don't jump around arbitrarily.
And the equation E = mc^2 is only valid in the rest frame. It gives the rest energy for the mass. The relativistic equation is:
E = (mc^2)/sqrt(1-(v^2/c^2))
And this is where letting C = 1 comes in.
If you let C = 1 then you state velocity as a unitless number between 0 and 1 that represents the portion of the speed of light at which the object under observation is moving WRT the rest/laboratory frame.
So you end up with this equation:
E = m/1-v^2
because you leave out C^2 (1^2 = 1) in the numerator and divisor.
If you want to express the energy in joules then you have to put the real values of C back in. You can also play with the units so that you express mass, energy and length in the same units - kilograms, joules, meters or some other unit.
1 sec = 300,000 km (converted time to length)
1 meter = 10^-8 sec (converted length to time)
You have to go to general relativity to convert mass to length but that equation is:
L (in meters) = G/c^2 * M (in kilograms)
where G is the gravitational constant.
Last...
don't confuse math for physics. Math is the language used to express physical theories. Just because one might "mess around" with an equation doesn't mean that the result expresses a valid physical model. There are a lot of crank "physicists" on the Internet. Most of them are laughing up their sleeves at the people to whom they post. I think that this is the case here.