# How does E=mc^2 relate to KE=0.5mv^2?

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#### chris

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##### How does E=mc^2 relate to KE=0.5mv^2?
« on: 24/08/2012 09:22:09 »
Einstein's formula is E=mc^2; but the formula for kinetic energy is 1/2 mv^2; why the difference?
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#### newway

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##### Re: How does E=mc^2 relate to KE=0.5mv^2?
« Reply #1 on: 24/08/2012 10:35:36 »
I think they are measuring different things.

KE=0.5mv^2      shows the relation between an object's kinetic energy to its speed and mass. Here v is a variable (can change)

E=mc^2 shows the relation between energy and mass, e.g. if something loss certain amount of mass, that amount will be released as such amount of energy. Here "c" (speed of light) is a constant, at least under current physics.

#### lightarrow

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##### Re: How does E=mc^2 relate to KE=0.5mv^2?
« Reply #2 on: 24/08/2012 10:54:38 »
Einstein's formula is E=mc^2; but the formula for kinetic energy is 1/2 mv^2; why the difference?
You have to use both of them []
The Total enenergy E of a free particle with mass m and speed v is, *at non relativistic speeds*:

E = mc2 + 1/2 mv2

because it's the sum of "rest energy" (the first term) and kinetic energy (the second).

But that equation is not much used because it's "half relativistic and half not" since it uses E = mc2 which is usually used only in SR, since in newtonian mechanics is a constant term and it's omitted, while the term 1/2 mv2 is the kinetic energy only at low speeds.

If you use this one, you are always correct, in newtonian mechanics as well as in SR, even with zero-mass particles:

E2 = (mc2)2 + (cp)2

p is the particle's momentum.

for zero-mass particles: p = h$$\nu$$/c ;    $$\nu$$ = particle's frequency
for non zero-mass particles: p = mv/sqrt[1 - (v/c)2] ;    v = particle's velocity.
« Last Edit: 24/08/2012 11:06:16 by lightarrow »

#### evan_au

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##### Re: How does E=mc^2 relate to KE=0.5mv^2?
« Reply #3 on: 24/08/2012 11:04:27 »
Kinetic Energy E=1/2 mv2 follows the pre-Einstein assumption of conservation of matter: "Matter is neither created nor destroyed".

If you are willing to destroy matter (as permitted by Einstein), E = mc2 releases so much energy that you can often ignore the initial kinetic energy.

#### damocles

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##### Re: How does E=mc^2 relate to KE=0.5mv^2?
« Reply #4 on: 24/08/2012 22:10:55 »

In special relativity, E = mc2

but m = m0/√(1 – v2/c2)

using Taylor expansion

= m0 * (1 + 1/2 v2/c2 + 3/8 v4/c4 + higher order terms)

and so E = m0c2 + 1/2 m0v2 + 3/8 m0v4/c2 + higher order terms

energy = energy equiv of rest mass + newtonian kinetic energy + 1st order SR correction ... (etc.)
1 4 6 4 1
4 4 9 4 4
a perfect perfect square square
6 9 6 9 6
4 4 9 4 4
1 4 6 4 1