# Fundamental Principle of Dynamics and equations of geodesics with proper time

I just wanted to have a little precision. In the expression below translating the PFD (Fundamental Principle of Dynamics) in tensor calculus (or more precisely the inertial principle) :

$$a^{\nu}=\dfrac{ \text{D} \, v^{\nu}}{\text{D} \tau}=\dfrac{\text{d}^2 x^{\nu}}{\text{d}\tau^2}+ \Gamma_{\alpha\beta}^{\nu} \,\dfrac{\text{d}\,x^{\alpha}} {\text{d}\tau}\,\dfrac{\text{d}\,x^{\beta}}{\text{d}\tau} = 0$$

why is this relation apparently valid only if we derive from the proper time of the particle $$\tau$$ ?

Could we not express it in relation to the reference time in which the particle is observed?

If so, what is the relationship to move from one to the other? a simple $$\text{d}t=\gamma\text{d}\tau$$ is not enough, is it?

Thank you for your clarification.

## 1 Answer

In special relativity and in general relativity the tensor formalism is used. Reason is that the physical laws written tensorially show the same form whatever the coordinate system. That is what one would expect as the coordinates are just artifacts of the human mind. To guarantee invariance you use the proper time $$\tau$$ to define the four-velocity starting from the coordinate difference $$dx^\mu$$, which is a vector, and then the four-acceleration.

The coordinate time $$t$$ is not an invariant, so can not be used to define a vector.

The relation between the coordinate time $$t$$ and the proper time $$\tau$$ depends on the metric of the reference frame in which you describe the physical events. In general relativity you have $$ds^2 = g_{\mu \nu} dx^\mu dx^\nu$$, where for a timelike trajectory $$ds^2 = - d\tau^2$$. The relation with the time coordinate is not as simple as it is in special relativity, as in your post.