0
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Feb 5 '17 at 14:41
  • $\begingroup$ I hesitated, because I thought physicists might understand better the reason why it is used this way in general relativity. $\endgroup$ – user56834 Feb 5 '17 at 14:42
  • $\begingroup$ And also what the physical intuition behind it is. $\endgroup$ – user56834 Feb 5 '17 at 15:03
  • $\begingroup$ A directional derivative is a map from functions (or better yet, one-forms) to the real numbers. $\endgroup$ – WillO Feb 5 '17 at 15:20
1
$\begingroup$

In curved spacetime, you cannot define vector as an interval or arrow because the segment out of the spacetime manifold is not defined. Imagine if you were a 2D ant living on a 2D sphere, then the void "inside" and "outside" the sphere is not defined. To define a tangent vector on a manifold is, in some sense, attaching a direction with a certain magnitude to a point. Imagine the manifold is a 2D surface, with a certain distribution of heat, then calculus tells you how to find the directional derivatives of the heat distribution. This definition does not require you to embed the 2D surface into some higher dimensional space, and therefore is good. More rigorously, you should define a tangent vector at a point as an equivalence class of curves tangent to each other at that point. This definition can be proven to be equivalent with the definition as you posted.

In rigorous mathematical oriented books of differential geometry, people do not explain much about the intuition behind the definition. If you want to know more about the physical motivations, you should read the book "Differential Geometry and Lie Groups for Physicists" by Marian Fecko. This book is a good one for getting started learning geometry.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – user56834 Feb 5 '17 at 15:29
  • 1
    $\begingroup$ Programmer2134, My post wasn't precise. It may contain mistakes. But the book I mentioned explains all the details you want to understand. $\endgroup$ – Xiaoyi Jing Feb 5 '17 at 15:35
  • $\begingroup$ Programmer2134, I should say that the equivalence class consists of a bunch of curves tangent with each other and sharing the same velocity at that point. $\endgroup$ – Xiaoyi Jing Feb 5 '17 at 15:38
  • $\begingroup$ #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? $\endgroup$ – user56834 Feb 6 '17 at 17:37
  • 1
    $\begingroup$ @Progammer2134 As far as I could remember, the proof that the two definitions are equivalent is left as an exercise in Fecko's book. What you need to do is to start from definition A and show that it implies B, and start from B to show that it implies A. If you have great difficulties in this mathematical proof, just send an email to Marian Fecko. He always answers questions regarding his book. $\endgroup$ – Xiaoyi Jing Feb 7 '17 at 1:38
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.