# Is four velocity always given by $U^{\mu} = d x^{\mu}/d\tau$?

I was taught that four-velocity is defined as $${\bf U} = \frac{d \bf x}{d\tau}$$ and that it has the components $$U^{\mu} = \frac{d x^{\mu}}{d\tau}$$ where $$d\bf x$$ is the four displacement and $$\tau$$ is proper time.

My question is simple: is the latter equation (for the components) correct in all coordinate systems?

I tried to figure it out the following way:

$${\bf U} = \frac{d}{d\tau}(x^{\mu}{\bf e_{\mu}})$$ $$= \frac{d x^{\mu}}{d\tau}{\bf e_{\mu}} + x^{\mu}\frac{d{\bf e_{\mu}}}{d\tau}$$ $$= \frac{d x^{\mu}}{d\tau}{\bf e_{\mu}} + x^{\mu}\frac{d x^{\nu}}{d\tau}\frac{\partial{\bf e_{\mu}}}{\partial x^{\nu}}$$ $$= \frac{d x^{\mu}}{d\tau}{\bf e_{\mu}} + x^{\mu}\frac{d x^{\nu}}{d\tau}\Gamma_{\nu \mu}^{\alpha}\bf e_{\alpha}$$ $$= \left(\frac{d x^{\alpha}}{d\tau} + x^{\mu}\frac{d x^{\nu}}{d\tau}\Gamma_{\nu \mu}^{\alpha}\right){\bf e_{\alpha}}$$

and therefore: $$U^{\alpha} = \frac{d x^{\alpha}}{d\tau} + x^{\mu}\frac{d x^{\nu}}{d\tau}\Gamma_{\nu \mu}^{\alpha}.$$

But the only way this component equation agrees with the earlier one is if $$x^{\mu}\frac{d x^{\nu}}{d\tau}\Gamma_{\nu \mu}^{\alpha} = 0$$ for all $$\alpha$$. However, I can't seem to prove/disprove this. Obviously if the Christoffel symbols are zero, then it's trivial. But if there are non-zero Christoffel symbols, then is it still zero?

• In general relativity, the coordinates $x^\mu$ are no longer the components of a legitimate vector --- Minkowski spacetime happens to have a natural vector space structure, but a general manifold does not. So $x^\mu \mathbf{e}_\mu$ is not a well-defined expression. – gj255 Jul 4 '19 at 22:51
• There's a bit of confusion caused by conflating flat Minkowski space of special relativity with the curvilinear space of general relativity. Because Minkowski space has a "linear structure" to it, we can equate coordinates $(x^{0}, x^{1}, x^{2}, x^{3})$ with a vector ${\bf x} = x^{0}{\bf e}_{0} + \cdots + x^{3}{\bf e}_{3}$. Note, however, that in this special case the vectors ${\bf e}_{i}$ do not depend on any parameter like $\tau$. They are static, so $d{\bf e}_{i}/d\tau = 0$. This gives you $\Gamma_{\mu\nu}^{\lambda} = 0$, but this is what you expect in a linear space (special relativity). – Maximal Ideal Jul 4 '19 at 23:28
• The equations $\textbf{U}=d\textbf{x}/d\tau$ and $U^\mu=dx^\mu/d\tau$ and are the same thing written in two different notations, as a matter of definition. If we relaced the $\mu$ with a Latin letter and interpreted this using abstract index notation, it would also mean the same thing. Note that something is clearly not making sense as soon as you get to the equation with the term $x^\mu d\textbf{e}_\mu/d\tau$ in it. This term depends on the coordinates rather than differences of coordinates, whereas everything else in the equation only deals with differences in coordinates. – user4552 Jul 4 '19 at 23:32
• It's not that $x^\mu \mathbf{e}_\mu$ is not well defined - it's the derivative of $\mathbf{e}_\mu$ that makes it more complicated in general because of the changes in direction of the basis vectors $\mathbf{e}_\mu$ on curved surfaces. For instance, the covariant derivative of basis vectors would be $\nabla\mathbf{e_{u}}=\Gamma_{\mu}^{\alpha}\mathbf{e_{\alpha}}$ - and the corresponding covariant derivative of the dual forms would be $\nabla\mathbf{\theta^{u}}=-\Gamma_{\alpha}^{\mu}\mathbf{\theta^{\alpha}}.$ – Cinaed Simson Jul 5 '19 at 0:26

The four-velocity is the tangent four-vector of a timelike world line, that is $$U^\mu = dx^\mu / d\tau$$, and this definition applies to any coordinate system. The difference of the coordinates $$dx^\mu$$ is a vector, i.e. an invariant, and together with the proper time $$\tau$$, an invariant by definition, produces the four-velocity $$U^\mu$$, which is an invariant as well.
However in your figuring out you refer to an object $$x^\mu e_\mu$$ whose difference in general is not a vector, as in an arbitrary coordinate system it does not transform as the coordinates. Your comparison between the two expressions is not justified.
• I think I understand my folly now: $d\bf x$ itself is defined as the four displacement vector, it is not the change in some "four position vector" $\bf x$ (though I see now that my choice of notation is suggestive). So my mistake was thinking of $\bf x$ as a four vector on its own? – J-J Jul 5 '19 at 22:25
• Out of curiosity, is the reason why a four vector like $\bf U$ is well-defined while $\bf x$ is not because $\bf U$ is defined locally and so unaffected by curvy spacetime, whereas $\bf x$ is a vector which stretches from the origin to a point in another region of spacetime, which is a problem in curvy spacetime? – J-J Jul 5 '19 at 22:34
• @J-J. Yes, in a curved manifold you define a vector locally. That means that a vector is ideally attached to a point in spacetime and the set of all the vectors at a point is called the tangent space $T_p$ at that specific point. To compare vectors at different points, you need the covariant derivative $\nabla_\mu$ as the partial derivative $\partial_\mu$ accounts only for a part of the change. – Michele Grosso Jul 6 '19 at 6:40