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According to Einstein's Mass-energy equivalence,

$ E = mc^2$ OR $ m = \frac E{c^2}$..... (1)

and According to Newton's Second Law of motion,

$ F = ma$ OR $m = \frac Fa$ ..... (2)

If we compare eq. (1) and eq. (2), we obtain;

$\frac E{c^2} = \frac Fa$..... (3)

If we multiply both the sides of eq. (3) with $c^2$, we get;

$E = \frac Fac^2$ ..... (4)

Is the above relation valid?

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The relation $E = mc^2$ isn't in general applicable unless you're in an inertial frame where system has zero net momentum. For a single particle, this means the particle is at rest and $F = 0 = a$. So you run into a problem immediately in line (2). – David H Apr 9 '13 at 18:54
up vote 4 down vote accepted

Your symbolic manipulations are correct, but the relations you write down do not properly describe Newton's second law in the context of special relativity.

In the context of special relativity, the relativistic momentum of a particle is defined as $$ \mathbf p = \gamma m \mathbf v, \qquad \gamma = (1-\mathbf v^2/c^2)^{-1/2} $$ Using this definition, Newton's second law is written as $$ \mathbf F = \frac{d\mathbf p}{dt} $$ In particular, note that since $\gamma$ has the velocity in it and therefore depends on time, we cannot move the time derivative past $\gamma$ when we differentiate $\mathbf p$ like we would in non-relativistic mechanics for a point particle. So in the context of special relativity, we in general have $$ \mathbf F \neq m\frac{d\mathbf v}{dt} $$ in direct constrast to non-relativistic mechanics. Also, the equation $E=mc^2$ is actually only true if the symbol $E$ represents the rest energy of the particle, the energy it has when it's speed is zero. Otherwise, the total energy of the particle is $$ E = \gamma mc^2 $$ In particular, the energy of a massive point particle in special relativity depends on its speed and increases with increasing speed.

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