# Mass-energy equivalence and Newton's Second Law of motion

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

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.