Rigorous computation of four-velocity

Question

In Special Relativity we can consider spacetime to be the Minkowsky space $\mathbb{R}^{1,3}$ which is just $\mathbb{R}^4$ together with the non degenerate symmetric bilinear form $g$ that in some basis has the matrix $(g_{\mu\nu})=\operatorname{diag}(1,-1,-1,-1)$. In particular we can consider the path of a particle through this spacetime as simply a curve $\alpha : I\subset \mathbb{R}\to \mathbb{R}^{1,3}$.

Such a curve can be written as

$$\alpha(\tau) = (\alpha^0(\tau),\alpha^1(\tau),\alpha^2(\tau),\alpha^3(\tau))$$

With $\tau$ being the proper time (i.e. the time measured on the reference frame of the particle itself). In the classical language, one would refer to $\alpha$ as a position four-vector $\mathbf{R}=(ct,x,y,z)$. Following that classical line of thought we could simply in a non-rigorous way consider infinitesimal changes on the coordinates $dt,dx,dy,dz$ which induces a $d\mathbf{R}$ on the position. Dividing by $dt$ we would get

$$\dfrac{d\mathbf{R}}{dt}=\left(c,\dfrac{dx}{dt},\dfrac{dy}{dt},\dfrac{dz}{dt}\right).$$

If one wanted them how $\mathbf{R}$ changes as $\tau$ changes then we would do

$$\dfrac{d\mathbf{R}}{d\tau} = \dfrac{d\mathbf{R}}{dt}\dfrac{dt}{d\tau} = \gamma (c,\mathbf{v})$$

using time dilation $dt = \gamma d\tau$ and $\mathbf{v}$ as the usual velocity. Although this result is true I would like to know how to do it in the rigorous version. The problem in the rigorous version is that for $i=1,2,3$ the derivative of $\alpha^i$ with respect to $t$ is zero, because $\alpha^i$ is a function not defined on $\mathbb{R}^{1,3}$ where $t$ is a coordinate we can differentiate with respect to, it is defined in $I$ where we can only differentiate with respect to proper-time.

Indeed in the rigorous version this $\mathbf{v}$ doesn't seem to make sense, because it seems to be the derivatives of the $\alpha^i$ with respect to $t$.

So in the rigorous version, what is really this $\mathbf{v}$ and how does one show that $\alpha'(\tau) = \gamma(c,\mathbf{v})$?

Answer 1

Mathematically, since the coordinates $\alpha$ are functions of $\tau$, you can, assuming certain assumptions on the curve, reexpress the curve as a function from the time coordinate to the spacetime coordinates by writing $\alpha^\mu(t)=\alpha^\mu((\alpha^0)^{-1}(t))$, which is just a reparametrization of the curve from $\tau$ to $t$. The assumptions are in essence that the function $\alpha^0(\tau)$ be invertible, see last comment. They will always be satisfied, if you don't consider limit cases.

Now to prove that $\alpha'(\tau)=\gamma(c,\vec{v})$, you decompose the total derivative with respect to $\tau$ by expressing it via a function of $t$ as you have done and it is rigorous since we have defined the functions $\alpha^\mu(t)$.

Of course it wouldn't work when $\gamma \longrightarrow \infty$, but then you would be going at the speed of light.

Answer 2

It is incorrect to say that we can only differentiate $\alpha^i$ with respect to proper time. We know that proper time is the parameter along the curve, i.e. a point $x$ on the curve is a function of the proper time $\tau$. This relation is given by the vector $\alpha=\alpha(\tau)$. However, we also have the relation $$\tau(x)=\int^x \mathrm{d}\tau$$ which is just the arc length formula for $\mathbb{R}^{3,1}$. Thus we can write $\alpha(x)=\alpha(\tau(x))$ and take derivatives with respect to coordinates. Note further the relation $$\eta_{\mu\nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}\tau}\frac{\mathrm{d}x^\nu}{\mathrm{d}\tau}=1$$ which is to be solved for $\alpha^i$ in terms of $t$. Thus we can find the components of the curve all in terms of $t$, making the derivatives in the OP possible.