# Why exactly in general relativity are tangent vectors defined as maps from functions to $\mathbb{R}$?

I am basing this on the lectures from the hereaus international winter school on gravity and light.

If $M$ is the manifold of physical spacetime, then at any point $p \in M$, we have a tangent space defined as the set of the velocities of all possible (parametrized) curves through that point. The velocity is defined as $$V_{\gamma(t) , t}(f)=(f \circ \gamma)'(t),$$ where $\gamma: \mathbb R \to M$.

The professor hinted at my question in the video but I still don't fully understand it:

Why is the velocity, and hence the tangent space, defined as a map from a function to the real numbers? Why isn't it defined more naturally as the derivative of $\gamma$ with respect to the parameter of $\gamma$?

Edit based on an answer: If $f$ can be an arbitrary function, then there are uncountably many values of velocity for a given curve over the manifold, whereas in standard newtonian (euclidian) physics, every curve has only one velocity. Why isn't it the same in general relativity (but taking into account curved space)?

• Would Mathematics be a better home for this question? – Qmechanic Feb 5 '17 at 14:41
• I hesitated, because I thought physicists might understand better the reason why it is used this way in general relativity. – user56834 Feb 5 '17 at 14:42
• And also what the physical intuition behind it is. – user56834 Feb 5 '17 at 15:03
• A directional derivative is a map from functions (or better yet, one-forms) to the real numbers. – WillO Feb 5 '17 at 15:20

• I see. But if $f$ can be an arbitrary function, then there are uncountably many values of velocity for a given curve over the manifold, whereas in standard newtonian (euclidian) physics, every curve has only one velocity. Why isn't it the same in general relativity? – user56834 Feb 5 '17 at 15:29
• #Xiaoyi Jing, I followed your advice and started reading the book by Marian Fecko. I've reached the point concerning this question. Indeed the book says that those two definitions are "equivalent". However, I am not quite sure what he means by that. It seems to me that they are not exactly the same. The tangent vector defined as an equivalent class of tangent curves at $P$, is not literally the same mathematical object as the directional derivative of a function $f(M)$ at point $P$. They are "equivalent" in some sense that I find unintuitive, but they're not exactly the same, right? – user56834 Feb 6 '17 at 17:37
In differential geometry vectors are defined as directional derivatives by $V=\frac{d}{d \lambda}=V^{\mu} \partial_{\mu}$, where $\lambda$ is a parameter along $\gamma$, and we used the chain rule of differentiation.