Until now, we have avoided using forces in Relativity. We did so for a good reason. The definition of force even in classical mechanics is somewhat problematic, let alone in relativity. Nonetheless, let us try and see whether we can generalize classical relations to get a consistent definition of forces also in relativistic mechanics.
In classical mechanics, we could in principle define the force as one of the following. Through the momentum (Newton's second law):
First, the force as defined using the momentum is given by
On the other hand, defining the force using the acceleration (while assuming a fixed mass) yields a result which is inconsistent with the previous definitions. The force is given by
In classical mechanics, we could in principle define the force as one of the following. Through the momentum (Newton's second law):
$$ {\bf F} = { d{\bf p} \over dt}. $$
We could do so through the work:
$$ W= \int_1^2 {\bf F} \cdot d {\bf x}. $$
Or, if we differentiate with time, through the power:
$$ {\bf F}\cdot {\bf v} = { dE \over dt}. $$
We will see that these definitions are consistent with the relativitic energy defined in previous sections. On the other hand, the definition:
$$ {\bf F}_\mathrm{bad} = m { d^2{\bf x} \over dt^2}, $$
is bad. The underlying reason has to do with the fact that the relativistic mass is not constant, and we have to take its change into consideration as well. First, the force as defined using the momentum is given by
$$ {\bf F} = {d {\bf p} \over dt} = {d \over dt} {m_0 {\bf v} \over \sqrt{1-v^2/c^2}} = {m_0 {\bf \dot v} \over \sqrt{1-v^2/c^2}} + {m_0 {\bf v} ({\bf v} \cdot {\bf v}) / c^2 \over ({1-v^2/c^2})^{3/2}}. $$
Thus, the power (=work per unit time) associated with this force, is given by:
$$ {d E \over dt} = {\bf F} \cdot {\bf v} = {m_0 ({\bf \dot v \cdot {\bf v}}) \over \sqrt{1-v^2/c^2}} + {m_0 v^2 ({\bf v} \cdot {\bf v}) / c^2 \over ({1-v^2/c^2})^{3/2}}, $$
and after factorizing the terms:
$$ {d E \over dt} = {m_0 ({\bf \dot v \cdot {\bf v}}) \over \sqrt{1-v^2/c^2}} \underbrace{\left[1 + {v^2/c^2 \over 1-v^2/c^2} \right]}_{1 / (1-v^2/c^2)} = \gamma^3 m_0 ({\bf \dot v \cdot {\bf v}}) $$
On the other hand, we can estimate the power obtained if we use the relativisitic energy to define the force:
$$ {d E \over dt}
= {d \over dt}
\left( \gamma(v) m_0 c^2 \right)
= m_0 c^2 {d \over dt}
\left( 1\over \sqrt{1-{v^2\over c^2}} \right)
= m_0 c^2
{ {\bf v}\cdot{\bf \dot v}/c^2
\over \left( {{1-{v^2\over c^2}}}\right)^{3/2}
} = \gamma^3 m_0 ({\bf \dot v \cdot {\bf v}}), $$
Thus, the definition through energy (work) and the definition through momenta are equivalent. On the other hand, defining the force using the acceleration (while assuming a fixed mass) yields a result which is inconsistent with the previous definitions. The force is given by
$$ {\bf F}_\mathrm{bad} = m {d^2 {\bf x} \over d t} = m_0 \gamma {\bf \dot v},
$$
such that the power is then
$$ {\bf F}_\mathrm{bad}\cdot {\bf v} = \gamma m_0 ({\bf \dot v}\cdot {\bf v})
\neq \gamma^3m_0 ({\bf \dot v}\cdot {\bf v}) ,
$$
clearly different from the momentum and energy related definitions.