# Physical Motivation for Four-Velocity definition

I'm bothered with the motivation behind defining a four-velocity. In Schutz's A First Course in General Relativity, he uses the concept of a tangent vector at each point of a worldline of a particle given by $$x ^\mu = (ct,x,y,z)$$. And later he states that

$$$$U^\mu = \frac{d x^\mu}{d \tau}$$$$

The mathematical explanation I found for using the proper time as the parameter wich all observer agree, but I can't realize what problems we obtain with instead this definition we use the relation

$$$$U^\mu = \frac{d x^\mu}{dt}$$$$

where $$t$$ is the measure of time in some inertial frame S.

• I don't think you'd be asking this question in Euclidean space. Consider a curve $\vec{r}(\lambda)=(x(\lambda),y(\lambda),z(\lambda))$. Then one can write the tangent vectors as $\vec{T}(\lambda)=d\vec{r}/d\lambda$. OR we could follow your latter suggestion and use $\vec{T}(\lambda)=d\vec{r}/dx$. The tangent vector will still point the right way but no longer is nicely defined and the definition no longer allows you to rotate in a way that mixes the coordinates up since it singles out $x$. Commented Feb 6, 2020 at 0:42
• Doesn’t the book explain somewhere that the four-velocity is defined that way so that it is a Lorentz four-vector? Commented Feb 6, 2020 at 5:19
• @jacob1729 can u give me some example? I'm pretty confused with this topic Commented Feb 6, 2020 at 15:04

The first definition transforms as a four-vector: $$\dfrac{dx^{'\mu}}{d \tau} = \Lambda^{\mu}{}_{\nu} \dfrac{dx^{\nu}}{d \tau}$$.

The second definition transforms not quite as a four-vector: $$\dfrac{dx^{'\mu}}{d t'} = \dfrac{dt}{dt'} \Lambda^{\mu}{}_{\nu} \dfrac{dx^{\nu}}{d t}$$.

This makes sense, since in the first definition you divide the differentials of a four-vector (which themselves also transform as a four-vector) by a scalar (invariant under the Lorentz group).