# Relativistic Derivatives

My problem ist about accelerated motion in SRT. For example consider a charqed particle with mass m and charge q in an homogeneous Electric Field pointing in x-Direction.

We know that $$\frac{dp^1}{d\tau} = \gamma E q$$ and $$m\frac{dx^0}{d\tau^2} = \frac{dp^0}{d\tau} = \gamma E q v$$. Now i've seen that you can use $$\gamma = \frac{dt}{d\tau}$$ to derive the "relativistic Lorentz Force". My question is: How do we know that $$\frac{dt}{d\tau} = \gamma$$ if we have an accelerated motion and thus $$t(\tau)$$ is not just linear. Is it just a pre-condition we have that $$\frac{dt}{d\tau} = \gamma(v)$$ and if yes how do we know that our solution for $$t(\tau)$$ satisfies this condition?

Thank you for your help :)

The relation $$dt/d\tau = \gamma$$ is automatically true for any massive particle's worldline because of the definition of $$\tau$$. Here's why:
If we consider an infinitesimal spacetime displacement $$dx^\mu = (dt, d\vec{x})$$ along a worldline, then the infinitesimal proper time between these two events satisfies $$d\tau^2 = dt^2 - |d\vec{x}^2|$$ which implies that $$\left(\frac{d\tau}{dt}\right)^2 = 1 - \left( \frac{d\vec{x}}{dt} \right)^2 \quad \Rightarrow \quad \frac{d\tau}{dt} = \sqrt{1 - v^2} = \frac{1}{\gamma}.$$ So the particle's "internal clock" will always tick along at such a rate such that $$dt/d\tau = \gamma$$, even if it accelerates.