Why is Newton's Law valid in Relativistic mechanics? 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.
 A: $F = \frac{d\vec{p}}{dt}$, 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.
A: The expression
$$\mathbf{F} := \frac{d\mathbf{p}}{dt}$$
is the definition of the measure of force. What you have discovered is that the statement
$$\mathbf{F} = m\mathbf{a}$$
which holds in Newtonian mechanics, does not hold in relativistic (Einsteinian) mechanics. These two are equivalent in the Newtonian context, but not equivalent in the relativistic context. Instead, the latter is indeed a "Law" in that it does not define force, but is a statement thereabout, while the former is an actual definition, in a theoretical context. And this law only holds at low speeds where Einsteinian and Newtonian mechanics approximate each other, and where we use the Lorentz reference frames.
A: In relativity, the letter $\mathbf{F}$ is called the three-force and defined to be $d \mathbf{p}/dt$. There is no way this can be "incorrect", because it's merely a definition.
Now, you make the good point that the quantity $dp^\mu / d\tau$, which is commonly called the relativistic four-force and denoted by the letter $f^\mu$, is "more natural" in certain contexts. But that doesn't make the previous definition wrong! There are advantages and disadvantages to each.

*

*$\mathbf{F}$ is easier to use if you want to track an object's evolution in time in a fixed reference frame, because it directly deals with $t$

*$\mathbf{F}$ directly corresponds to the familiar Newtonian force in the nonrelativistic limit

*$f^\mu$ is a four-vector, so is easy to Lorentz transform, while $\mathbf{F}$ has a complicated transformation

*$f^\mu$ tends to appear automatically in equations defined from other relativistic theories. For example, when you learn electromagnetism in relativistic form, you'll find $f^\mu = q u^\nu F_{\mu\nu}$.

*$\mathbf{F}$ still commonly appears when you're working in nonrelativistic scenarios. For example, if you take the previous equation and decompose it in terms of the electric and magnetic fields, you get $\mathbf{F} = q (\mathbf{E} + \mathbf{v} \times \mathbf{B})$.

*$\mathbf{F}$ also obeys other useful identifies, such as $dE = \mathbf{F} \cdot d \mathbf{x}$.

Both definitions are commonly used and neither is really more "natural". Sometimes, when solving a dynamics problem, I'll use both definitions at different times, if that's the most efficient route.
A: Relativistic equation of motion
\begin{align*}
&\text{NEWTON space}\\\\
&\mathbf F=\frac{d}{dt}\,(m\,\mathbf v)\tag{1}\\\\
&\text{MINKOWSKI space}\\\\
&\frac{d}{d\tau}\left(m\,u^\mu\right)=K^\mu=
\begin{bmatrix}
  K^0 \\
  K^i \\
\end{bmatrix}\tag{2}
\end{align*}
The EOM's must fulfilled two requirements

*

*the force $~\mathbf K~$ must be Lorentz scalar $~(K'^\mu\,K'_\mu=K^\mu\,K_\mu)~$ and

*for $~v\ll c$ you obtain the NEWTON EOM's
with:
\begin{align*}
&\frac{d}{d\tau}=\frac{d}{dt}\,\gamma\\
 &u^\mu=\begin{bmatrix}
          c \\
          \mathbf{v} \\
        \end{bmatrix}\\
&K^0=\frac{\gamma\,\mathbf{v}^T\,\mathbf{F}}{c}\quad,
K^i=   \gamma\,\mathbf{F}\\     
&\text{and}\\
&\gamma=\frac{1}{\sqrt{1-\frac{\mathbf{v}\cdot \mathbf{v}}{c^2}}}
\end{align*}
you obtain the relativistic equation of motion:
with $\mathbf{v}=[v,0,0]^T~$ and $\mathbf{F}=[F,0,0]^T$
$$\dfrac{dv}{dt}=\dfrac{F}{m}\,\left( 1-\dfrac{v^{2}}{c^{2}}\right)$$
thus   $~v\ll c$ you obtain the NEWTON equation of motion  and you can check that $~(K'^\mu\,K'_\mu=K^\mu\,K_\mu)=-F^2~$ Lorenz scalar

\begin{align*}
&\frac{d}{d\tau}(m\,u^\mu)=\frac{d}{dt}(\frac{dt}{d\tau}\,m\,u^\mu)=\frac{d}{dt}(\gamma\,m\,u^\mu)=\frac{d}{dt}\left(\begin{bmatrix}
   \gamma\,m\,c\, \\
   \gamma\,m\,\mathbf v\\
\end{bmatrix}\right)=\begin{bmatrix}
   K^0\\
   K^i\\
\end{bmatrix}
\end{align*}
A: This is kind of like asking "why is a meterstick still one meter in SR?"
Well it's a meter because we define it that way. And it's always going to be exactly one meter in its own reference frame - by definition.
Likewise, force is the transfer of momentum between two objects. The force 3-vector is not a lorentz invariant and does not appear the same in all reference frames.
It can be misleading to see $F = \frac{d\vec{p}}{dt}$ still applying and thinking "oh that's the same as non-SR Force, so it's the same." But that is not true - $\vec{p}$ changes with SR so it's only the same equation as Netwonian mechanics in its own reference frame - and we know Newton's laws still apply when in an its own frame. To expect this expression to also change by adding a factor of $\gamma$ or something would be having your cake and eating it too.
A: Force is by definition change of momentum, just as power is change of energy. The energy-momentum conservation statement $\partial _\mu T^{\mu\nu}=0$ is the SR field theoretical statement of vanishing force and power.
