Thread: E=MC^2 Help!!

I would like to ask everyone who may know something about E=MC^2

C is the constant right? The Speed of Light?

If light shone from an object traveling at 1/2 the speed of light is still going just as fast as light emmitted from an object traveling 1/4 the Speed of Light in relation to the other object, the Constant can be given the value of 1, 1 being the rate at which light travels regardles of how fast you are going.

Ok so now the constant is 1

So then we have E=M*1^2

or E=M

why did einstein add the extra C^2 then if it can simplify further?

The constant, C, has units attached as it is a speed.
C = 300 000 000 m/s (a little less than). The rest of what you are saying is flat-out wrong. :|

Originally Posted by KALSTER

The constant, C, has units attached as it is a speed.
C = 300 000 000 m/s (a little less than). The rest of what you are saying is flat-out wrong. :|

But couldnt you just as well say that 300,000,000 m/s = 1 gizmo/s? or better yet 1 m/.0000000001s?

Just because we try to translate it into miles, meteres, inches, feet etc... over seconds, miliseconds, minutes, hours, years dayse etc...
doesn't mean it could not be defined as 1 in another unit of measure, or 1m in another unit of time especially giving that the Speed of Light is the universal constant!

Basically Energy = Mass no matter the Speed!

The formula is independent of the system of units. In natural units, the speed of light is defined to equal 1, and the formula expresses an identity: E = m. In the SI system (expressing the ratio E / m in joules per kilogram using the value of c in meters per second):
E / m = c2 = (299,792,458 m/s)2 = 89,875,517,873,681,764 J/kg (≈9.0 × 1016 joules per kilogram)

So one gram of mass — approximately the mass of a U.S. dollar bill

Okay, Thank You!

I get it now, for SI, to find the amount of joules of energy you can get from an object of a certain mass, measured in SI, You use the SI standard for the speed of light.

the natural method would work but is harder for us to comprehend, so that is why C^2 is needed in the equation but both notations are correct in their own units of measure!

Thank You Again!