Are 4-vectors really vectors in general relativity?

Question

After watching this video about geodesics, I am now confused about where 4-vectors fit in general relativity. I already checked this question but it didn't help much. I am still learning the formalities of differential geometry. Let me know if I misunderstood anything.

Let $M$ be a 4D spacetime manifold and $(U,\phi)$ be one of its coordinate charts where $U\subseteq M$ is an open patch of the manifold. A trajectory of an apple in $U$ is parametrized by a smooth map $\lambda:\mathbb R\to U$.

I think $x^\alpha(\tau)=\phi(\lambda(\tau))=(cx^0(\tau),x^1(\tau),x^2(\tau),x^3(\tau))$ is what one calls "position 4-vector" but it is a point, not a vector, despite its name.

A "velocity 4-vector" seems to be defined as $u^\alpha=\frac{dx^\alpha}{d\tau}=(c\frac{dx^0}{d\tau}, \frac{dx^1}{d\tau}, \frac{dx^2}{d\tau}, \frac{dx^3}{d\tau})$. I am not sure if it is a point or vector.

Similarly, a "acceleration 4-vector" is defined as $a^\alpha=\frac{du^\alpha}{d\tau}=(c\frac{d^2x^0}{d\tau^2}, \frac{d^2x^1}{d\tau^2}, \frac{d^2x^2}{d\tau^2}, \frac{d^2x^3}{d\tau^2})$.

Let $\{\vec{e}_0, \vec{e}_1, \vec{e}_2, \vec{e}_3\}$ be coordinate basis vectors for the tangent space $T_{\lambda(\tau)}M$.

In the linked video, he defines the "velocity vector" (I thought it refers to "velocity 4-vector") of the apple as $\vec{u}^\alpha=u^\alpha\vec{e}_\alpha$. He defined the "acceleration vector" to be

$$ \begin{align*} \vec{a}^\alpha &= \frac{d\vec{u}^\alpha}{d\tau} \\ &= \frac{du^\alpha}{d\tau}\vec{e}_\alpha+u^\alpha\frac{d\vec{e}_\alpha}{d\tau} \\ &= \frac{du^\alpha}{d\tau}\vec{e}_\alpha+u^\alpha\left(\left(\frac{d\vec{e}_\alpha}{d\tau}\cdot\vec{e}_\gamma\right)\vec{e}_\gamma\right) \\ &= \frac{du^\alpha}{d\tau}\vec{e}_\alpha+u^\alpha\left(\left(\frac{d\vec{e}_\alpha}{dx^\beta}\frac{dx^\beta}{d\tau}\cdot\vec{e}_\gamma\right)\vec{e}_\gamma\right) \\ &= \frac{du^\alpha}{d\tau}\vec{e}_\alpha+u^\alpha\left(\left(\frac{d\vec{e}_\alpha}{dx^\beta}\cdot\vec{e}_\gamma\right)u^\beta\vec{e}_\gamma\right) \\ &= \frac{du^\alpha}{d\tau}\vec{e}_\alpha+\Gamma^\alpha_{\mu\nu} u^\mu u^\nu\vec{e}_\alpha \\ &= \left(a^\alpha+\Gamma^\alpha_{\mu\nu} u^\mu u^\nu\right)\vec{e}_\alpha \end{align*} $$

It was surprising to me because I thought "acceleration 4-vector" $a^\alpha$ can be written as $\vec{a}^\alpha=a^\alpha\vec{e}_\alpha$ but it seems to be false.

I am struggling to see how 4-vectors $u^\alpha$ and $a^\alpha$ fits in the picture. It seems like they are not vectors, contrary to my expectations. Are they points on the manifold?

Answer 1

Your confusion here probably stems from the lack of distinction between "a vector", and "components of a vector".

• A vector $\vec{a}$ is a geometric object - and a 4 vector is just a 4 dimensional vector together with a transformation law.
• The components of a vector $a^\alpha$ are the numbers which act as "weights" to express our vector as a sum of basis vectors, which are a set of vectors denoted: $ \vec{e}_\alpha$. The values of these components depends on the choice of basis.

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Where you have written objects with the arrow on top, e.g. $\vec{a}^\alpha $, you are referring to the geometric vector. The superscript alpha is misleading here - it implies you are talking about the components when you are really trying to talk about the vector as a whole, so I will drop it going forwards.

Any vector, e.g. $\vec{u} $ can be expressed as a sum of basis vectors, so using summation convention, we could write: $\vec{u} =u^\alpha \vec{e}_\alpha $

The numbers $u^\alpha$ are the vector components and $\vec{u} $ is the vector.

When we take a derivative of such an object, we can apply product rule to it:

$\frac{d}{d\tau} { \vec{u} } = \frac{d}{d\tau} ({ u^\alpha \vec{e}_\alpha }) = \frac{d}{d\tau} { (u^\alpha)} \vec{e}_\alpha + u^\alpha \frac{d}{d\tau} { \vec{e}_\alpha }$

The Christoffel symbols $\Gamma$ enter the picture as derivatives of the basis vectors. This is important in curved spacetime or in curved coordinates, where the basis vectors are non-constant: Each point has a different basis set, so if we are following a particle along its world line, the basis would change and the basis vectors would have non-vanishing derivatives with respect to proper time.

From this point on, I hope the confusion vanishes.

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Another possible source of confusion is in general relativity, these 4-vectors are only defined "at a point in spacetime" - i.e. they are elements of the tangent space to a specific point in spacetime (which is a Lorentzian Manifold).

4-vectors at different points in spacetime are not directly comparable and you have to compare them by finding a path between them integrating quantities along such a path.

4-vectors also cannot be used to describe positions in spacetime since in GR spacetime is a manifold not a vector space. You cannot add points on a meaningful way, analogous to how you cannot add GPS coordinates together to get a meaningful quantity.