Units of Velocity Components and Metric Tensor Components I was watching a GR lecture on youtube, and the speaker explains that the units of the components of velocity are $[v^{\alpha}]=\frac{1}{T}$, the metric tensor has units $[g_{\alpha\beta}]=L^2$, and the speed of a curve (in some manifold $M$) has units $[\sqrt{g_{\alpha\beta}v^{\alpha}v^{\beta}}]=\frac{L}{T}$. He also claims that this is something that initially gave Einstein a hard time during the development of GR. Given the first two, I understand the third, but I don't really understand why the first two have those units. He said velocity components and metric tensor components have these units because coordinate distance in some chart has nothing to do with real distance. That explanation doesn't really make sense to me, and I was hoping maybe someone can explain it in a bit more detail.
 A: The units of these quantities vary with the coordinate system.

Consider Minkowski space with the usual Cartesian spatial coordinates. We have
$$ ds^2 = -c^2 \mathrm{d}t^2 + \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2. $$
The nonzero metric coefficients are
\begin{align}
g_{tt} = -c^2 & \sim \frac{L^2}{T^2} \\
g_{xx} = g_{yy} = g_{zz} = 1 & \sim 1.
\end{align}
$4$-velocities come from parameterizing your coordinates with $\tau \sim T$ and differentiating with respect to this parameter. Thus
\begin{align}
v^t & \sim \frac{T}{T} = 1 \\
v^x,\ vy,\ vz & \sim \frac{L}{T}.
\end{align}

Instead we could consider cylindrical coordinates with
$$ ds^2 = -c^2 \mathrm{d}t^2 + \mathrm{d}r^2 + r^2 \mathrm{d}\theta^2 + \mathrm{d}z^2. $$
Now our nonzero metric coefficients are
\begin{align}
g_{tt} = -c^2 & \sim \frac{L^2}{T^2} \\
g_{rr} = g_{zz} = 1 & \sim 1 \\
g_{\theta\theta} = r^2 & \sim L^2.
\end{align}
and our velocity components have units
\begin{align}
v^t & \sim \frac{T}{T} = 1 \\
v^r,\ v^z & \sim \frac{L}{T} \\
v^\theta & \sim \frac{1}{T}.
\end{align}

You could only have $v^\alpha \sim 1/T$ always if you always interpreted your coordinates to be dimensionless but still insisted that proper time had units of time. I've never seen anyone do this, and it's certainly a weird way of going about things. Beyond that, by insisting $g_{\alpha\beta} \sim L^2$, even for the time-time and time-space components, you get that proper volume elements are pure "lengths":
$$ \mathrm{d}V = \sqrt{-g} \, \mathrm{d}x^0 \, \mathrm{d}x^1 \, \mathrm{d}x^2 \, \mathrm{d}x^3 \sim L^4 \not\sim L^3 T. $$
That is, the timelike coordinate itself is associated with a "length" rather than a "time."
