# e=mc2

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#### 151reward

##### e=mc2
« on: March 28, 2010, 08:58:11 PM »
I hope that a critic of Einstien's relativity theories will answer this question for me.  Let me first preface my question with the disclaimer that I am a layman and not a trained physicist.  I understand that the second postulate of SR states that the velocity of light is independent of its source.  Also, the theory states that nothing in the universe can move faster than the speed of light.  My question is this - if these statements regarding the speed of light are incorrect as many critics believe, then why is C, the speed of light, such an integral part of the equation e=mc2?  The equation states that the energy contained in a given quantity of mass at rest is equal to that quantity times the speed of light squared.  If the speed of light is not a constant or is not as essential to understanding the universe as SR would imply, then how did it become part of this equation?  I hope my questions(s) makes some sense.

By the way, I am on the fence concerning this subject.  Both the critics and supporters of relativity theory seem to offer plausible arguments.

#### Parsifal

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##### Re: e=mc2
« Reply #1 on: March 29, 2010, 01:16:30 AM »
I am not a critic of relativity, at least not moreso than it is healthy to be critical of any scientific theory, but E=mc2 is a product of the Theory of General Relativity. Anyone who dismisses relativity should also dismiss this equation.
I'm going to side with the white supremacists.

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#### bowler

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##### Re: e=mc2
« Reply #2 on: March 29, 2010, 03:06:44 AM »
Firstly E=mc^2 has nothing whatsoeevr to do with general relativity. It comes from special relativity. Which makes your point even stronger as of course it doesn't just affect cosmology. Solid state physics, electronics, optics pretty much the entire modern world are based on physics with special relativity as an integral part. Based on the wording of your post I suspect you know this anyway, if you don't then I suggest you take Physics because you have some aptitude.

#### Parsifal

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##### Re: e=mc2
« Reply #3 on: March 29, 2010, 03:09:50 AM »
Firstly E=mc^2 has nothing whatsoeevr to do with general relativity. It comes from special relativity.

Ah, I thought I remembered learning it came from GR. Which part of SR produces that equation?
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#### bowler

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##### Re: e=mc2
« Reply #4 on: March 29, 2010, 03:19:20 AM »
I'm a bit busy to write it out now, as much as I enjoy relativity the mundane day to day task of stabilising fits pays better. So I have just checked wiki which gives a very good introduction on the special relativity page, the links to relativistic mass and mass-momentum relation are also well worth a visit. If you have any questions ill answer when I get home.

#### parsec

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##### Re: e=mc2
« Reply #5 on: March 29, 2010, 05:56:47 PM »
... E=mc2 is a product of the Theory of General Relativity.
This statement is not true.

EDIT:
This is an English translation of the original paper in which Einstein derived the relation in 1905, long before the Principle of Equivalence, and not to mention, General Theory of Relativity were developed.

« Last Edit: March 29, 2010, 06:23:30 PM by parsec »

#### parsec

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##### Re: e=mc2
« Reply #6 on: March 30, 2010, 09:51:00 PM »
Can someone show me the proof that shows it isn't E = .5mc^2
you're dumb and you post dumb things.

#### Parsifal

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##### Re: e=mc2
« Reply #7 on: April 02, 2010, 07:38:47 AM »
you're dumb and you post dumb things.

+1

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#### SupahLovah

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##### Re: e=mc2
« Reply #8 on: April 02, 2010, 09:54:32 AM »

That was a serious question.
The proof that e=mc2 show that e =/= .5mc2
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#### Parsifal

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##### Re: e=mc2
« Reply #9 on: April 02, 2010, 10:24:11 AM »
Why are the formula's different?

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#### parsec

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##### Re: e=mc2
« Reply #10 on: April 02, 2010, 12:11:53 PM »
Kinetic energy of a particle of mass m moving at speed v comparable to the speed of light in vacuum c is:
$K&space;=&space;mc^{2}\left(\frac{1}{\sqrt{1-v^{2}/c^{2}}}&space;-&space;1&space;\right&space;)$

When the speed of the particle is much lower than the speed of light, i.e. when:
$x&space;=&space;\frac{v^{2}}{c^{2}}&space;\ll&space;1$
we can use the generalized binomial formula to expand the square root:
$(1+(-x))^{-\frac{1}{2}}&space;=&space;1&space;+(-\frac{1}{2})(-x)&space;+&space;O(x^{2})&space;=&space;1&space;+&space;\frac{x}{2}&space;+&space;O(x^{2})$
to get the non-relativistic expression for the kinetic energy:
$K&space;=&space;\frac{1}{2}m&space;v^{2}&space;+&space;O\left[\left(\frac{v}{c}\right)^{2}\right]$
As you can see, the non-relativistic expression for the kinetic energy is only the lowest order term in the expansion of the relativistic expression in powers of the small quantity v/c. This approximation breaks down for speed comparable to the speed of light. Indeed, when $\inline&space;v/c&space;\rightarrow&space;1&space;-&space;0$ you can see that the kinetic energy $\inline&space;K&space;\rightarrow&space;\infty$. So, the speed of light in relativity plays effectively a role of an infinite speed that cannot be reached.
« Last Edit: April 02, 2010, 09:59:38 PM by parsec »