$$\mathbf F = \dfrac {m_0 \mathbf a} {\left({1 - \dfrac{v^2}{c^2}}\right)^{\tfrac 3 2}}$$

# Physical Law

The force and acceleration on a body of constant rest mass are related by the equation:
$$\mathbf F = \dfrac {m_0 \mathbf a} {\left({1 - \dfrac{v^2}{c^2}}\right)^{\tfrac 3 2}}$$
where:

• 𝐅 is the force on the body
• 𝐚 is the acceleration induced on the body
• 𝑣 is the magnitude of the velocity of the body
• 𝑐 is the speed of light
• 𝑚0 is the rest mass of the body.

# Proof

Into Newton’s Second Law of Motion:
$$\mathbf F = \dfrac {\mathrm d}{\mathrm d t} \left({m \mathbf v}\right)$$

we substitute Einstein’s Mass-Velocity Equation:
$$m = \dfrac {m_0} {\sqrt {1 - \dfrac {v^2} {c^2}}}$$
where:

• 𝑣 is the magnitude of the velocity of the body
• 𝑐 is the speed of light in vacuum
• 𝑚0 is the rest mass of the body.

The value 𝑚 is known as the relativistic mass of the body.
The factor $\dfrac 1 {\sqrt{1 - \dfrac {v^2} {c^2} } }$ is known as the Lorentz Factor.

to obtain:
$$\mathbf F = \dfrac {\mathrm d} {\mathrm d t} \left({\dfrac {m_0 \mathbf v}{\sqrt{1 - \dfrac {v^2}{c^2}}}}\right)$$

Then we perform the differentiation with respect to time:

$$\frac{\mathrm d}{\mathrm d t} \left({\frac {\mathbf v}{\sqrt{1 - \dfrac {v^2}{c^2} } } }\right)$$ $$= \frac{\mathrm d}{\mathrm d v} \left({\frac {\mathbf v}{\sqrt{1 - \dfrac {v^2}{c^2} } } }\right) \frac{\mathrm d v}{\mathrm d t}$$
$$= \mathbf a \left({\frac {\sqrt{1 - \dfrac {v^2}{c^2} } - \dfrac v 2 \dfrac 1 {\sqrt{1 - \dfrac {v^2}{c^2} } } \dfrac{-2 v}{c^2} } {1 - \dfrac {v^2}{c^2} } }\right)$$
$$= \mathbf a \left({\frac {c^2 \left({1 - \dfrac {v^2}{c^2} }\right) + v^2} {c^2 \left({1 - \dfrac {v^2}{c^2} }\right)^{3/2} } }\right)$$
$$= \mathbf a \left({\frac 1 {\left({1 - \dfrac {v^2}{c^2} }\right)^{3/2} } }\right)$$

Thus we arrive at the form:
$$\mathbf F = \dfrac {m_0 \mathbf a} {\left({1 - \dfrac{v^2}{c^2}}\right)^{\tfrac 3 2}}$$