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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 :)

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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.

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