Quiz: E = m * c^2

Who amougst you is a brave soul. We need a hero

Here it is,

Note: the equations are annotated as if in a computer program

Et = Energy total

Eo = Energy at rest
Ev = Energy kinetic

m = mass
c = speed of light
v = speed of object

Find Eo = m c^2

Knowing
Et = Eo + Ev [eqn1]

and
Ev = 0.5 m v^2

also, from special relativity (note 9)

Et = Eo / ( 1 - (v/c)^2 )^0.5 [eqn2]

from [eqn1] and [eqn2]
Eo + Ev = Eo / ( 1 - (v/c)^2 )^0.5

devide by Eo

1 + Ev / Eo = 1 / ( 1 - (v/c)^2 )^0.5 [eqn3]

we know that the following equation has a series equivalent
(1 - 4 x)^-0.5 = 1 + 2 x + 6 x^2 + ....

let: 4 x = (v/c)^2 or x = (v/c/2)^2

just keep the lower terms 1 + 2 x (note

(1 - 4 x)^-0.5 = 1 + 2 x = 1 + 2 (v/c/2)^2 [eqn4]

from [eqn3] and [eqn4]
1 + Ev / Eo = 1 + 2 (v/c/2)^2

or

Eo = Ev / 2 / (v/c/2)^2

Eo = 0.5 m v^2 / 2 / v^2 * c^2 * 4

Eo = m c^2

Voila, this deserves a kick in the ass :}

Appendix:
---------

note 8 : extra points, explain what the ignored higher terms are.

note 9 : not needed here but remember from special relativity:

z' = z / ( 1 - (v/c)^2 )^0.5

t' = t / ( 1 - (v/c)^2 )^0.5

m' = m / ( 1 - (v/c)^2 )^0.5

E' = E / ( 1 - (v/c)^2 )^0.5

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from: Checkboard (pprune)

Einstein's original paper is online here:

http://www.fourmilab.ch/etexts/einstein/E_mc2/e_mc2.pdf

AA > > thanks for the slightly different topic of translated paper; too bad it's still a pain to read and understand the old terminology. < < AA

Does the inertia of a body depend upon its energy-content?

Translated from the original German, it's only three pages and fairly simple to understand. It basically says that, if the speed of light is a constant regardless of frame of reference, then a body emitting light radiation has its kinetic energy change by the amount of mass it loses in radiation emitted.
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