In relativistic kinematics, we derive momentum of a body as $$p=\frac{m_0\vec v}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma m_0\vec v$$
Then, $$\vec F=\frac{d\vec p}{dt}\tag{1}$$
$$
\implies\vec F=\frac{d(\gamma m_0\vec v)}{dt}
$$
By doing the differentiation we get,
$$\vec a=\frac{d\vec v}{dt}=\frac{F_0}{m_0\gamma}-\frac{(\vec F.\vec v)\vec v}{m_0c^2\gamma}$$
We find that acceleration need not be in the direction of net external force.
I have a question as to why equation $(1)$ even holds. How can we assume that $(1)$ holds? The expression of every quantity gets changed in relativistic mechanics like momentum, kinetic energy etc. Why any sort of factor is not multiplied to it or added to it?
In Newtonian mechanics $(1)$ is the result of everyday observation. We can also verify the law by doing experiment on linear air track as in that case friction gets reduced very much thus help in analyzing $(1)$ clearly.
In relativistic mechanics also, is Newton's Law, i.e. expression $(1)$, merely the consequence of observations only or is there any other reasoning to it also?
Addendum-1
While deriving the expression of momentum we get $$p=m_ov\gamma=m_o\frac{dx}{dt}\frac{dt}{d\tau}=m_o\frac{dx}{d\tau}$$
where $\tau$ is the proper time interval (time span in the frame in which the particle appear to be at rest).
So, $p=m_o\frac{dx}{d\tau}\neq m_o\frac{dx}{dt}$
In Newtonian mechanics, $t=\tau$ (existence of a universal time which flows independent of any inertial frame).
Why we assume $F=\frac{dp}{dt}$ in relativistic mechanics?
It can be $F=\frac{dp}{d\tau}$ also.
We can also see that $F=\frac{dp}{d\tau}$ is somewhat more fundamental because it gets reduced to $F=\frac{dp}{dt}$ in low velocity moving frames.
That is my doubt actually.
Addendum-2
I have read all the answers, but one point is still not clear to me.
Force is a measure of interactions which is acted on the particle. In newtonian mechanics (meaning working with very low velocity compared to light), the measure of interaction is given by Newton's law $F=\frac{dmv}{dt}$. There is no distinguishment between any sort of time meaning there is a universal time which flows independent of any reference frame, provided velocity of reference frame is very low compared to light.
But while working with objects moving with velocity near the speed of light. Then the measure of net interaction acted on the particle is $\frac{dmv\gamma}{dt}$ or $\frac{dmv\gamma}{d\tau}$ where $\tau$ is the proper time interval (time interval in the frame of moving object)?
In books, they directly say $F=\frac{dmv\gamma}{dt}$ even when particle moves with very high velocity.
But if we consider both the possibilities $F=\frac{dmv\gamma}{dt}$ or $\frac{dmv\gamma}{d\tau}$, then both reduces to the Newton's law in very low velocity as $\tau\to t$ and $\gamma\to 1$ in case of very low velocity.
Best Answer
$F = \frac{d\vec{p}}{dt}$, where $p = \gamma m \vec{u}$, is a defined quantity in SR. The theoretical justification is that $\gamma m \vec{u}$ is conserved in collisions and approaches $m \vec{u}$ in the low-speed limit. (The justification for defining momentum as $\sum m \vec{u}$ in Newtonian physics is that this quantity is conserved if no external force acts on the system. We want to identify a similarly-conversed quantity in SR.) The other justification is that it agrees with experimental results.