# Defining four-vectors in General Relativity?

Nearly every general relativity textbook I look in has a section on four-vectors. However, this section is usually titled something like 'Four vectors in Special Relativity', 'Special Relativity (Again)' etc. I would therefore like to know the formal definition (without ambiguity) of the following four vectors in general relativity: displacement, 4-velocity, 4-wave vector, 4-momentum (of a massive particle) and 4-momentum (of a photon).

I am pretty sure I know some of these1 since (correct me if I am wrong) at the origin of any 'locally orthogonal' coordinate system these must take the special relativistic values.

1which I will not here reproduce since I want to compare what I think with any answers given without undue influence.

• Why do you think they are different? It's the operators not the operands that differs, so to speak. (That said I qualify myself as curious and with limited relativity understanding.) Mar 18, 2017 at 19:59
• @JMLCarter It is not that I think they are different - more that I don't trust my self to assume they are the same. Any derivation or description I have seen has been in terms of a flat metric not one which is curved. Mar 18, 2017 at 20:07

In General Relativity spacetime is a four dimensional smooth manifold $M$ with a Lorentzian metric tensor $g$, which means that its signature is $(-,+,+,+)$. The underlying intuition behind this, is that spacetime is a four dimensional space that locally resembles the flat spacetime, however globaly might be curved.

Given a point $x\in M$, that is, an event, the main feature of a vector at this point is that it indicates a direction. We want though to talk about the directions "tangent to $M$", in other words, directions such that when moving instanteneously along them, we are moving on top of $M$.

This leads us to define the tangent space $T_xM$ of all tangent vectors to $M$ at $x$.

There are two main ways to do it:

1. We understand $T_x M$ as equivalence classes of curves to first order. For this definition consider two curves $\gamma_1(\tau),\gamma_2(\tau)$ such that $\gamma_1(0)=\gamma_2(0)$. We say that they are equivalent to first order if given any function $f$ on $M$ we have $(f\circ\gamma_1)'(0)=(f\circ\gamma_2)'(0)$. The intuition here is: if this holds, this means that if we move just a little from $x$ along $\gamma_1$ and $\gamma_2$ we don't feel the diference, for any function we pick, the change it feels is the same in both cases. This means we are picking exactly the direction of the curve. If they are equivalent to first order we write $\gamma_1\sim \gamma_2$. We then call the tangent vector to the curve $\gamma'(0)=[\gamma]$ the equivalence class by this relation. The tangent space $T_x M$ is the space of all such equivalence classes.

2. The second way to do it is the following: we define $T_x M$ to be the space of all directional derivative operators. Remember that in flat space a vector $v$ could produce the directional derivative of a function $f$ by $v\cdot \nabla f$? This also encodes the idea of direction. Thus a vector is defined as a directional derivative operator. To make this precise, we define a tangent vector as a mapping of functions $v(f)$ such that (i) it is linear and (ii) it obeys Liebnitz rule, $v(fg)=v(f)g(x)+f(x)v(g)$ (recall we are defining a vector at $x$). The space of these so-called derivations is also the tangent space $T_x M$.

Each definition has its advantage, but they give rise to the same space. To show this, consider $[\gamma]$ an equivalence class of curves, this defines a derivation by $[\gamma](f)=(f\circ\gamma)'(0)$. This mapping of an equivalence class of curves to a derivation is a vector space isomorphism between the two spaces.

The point is: once you have a coordinate chart $(U,\phi)$ in $M$ it is obvious that the partial derivative operators

$$\dfrac{\partial}{\partial\phi^\mu}\bigg|_{x}$$

are derivations at $x$. They are corresponding to the equivalence class of the coordinate lines drawn on spacetime by the coordinate system. Interestingly you can show that they form a basis of $T_x M$. Thus any vector $v\in T_x M$ can be decomposed as

$$v = v^\mu \dfrac{\partial}{\partial \phi^\mu}\bigg|_{x}$$

and this gives rise to the well-known components of vectors $v^\mu$.

In that setting, a worldline is a curve $\gamma : \mathbb{R}\to M$, its four velocity is its derivative $\gamma'(\tau)$. If this is the worldline of a massive particle of mass $m$, its four momentum is $p = m\gamma'$ and so forth.

Special Relativity is just the specific case of flat spacetime $M = \mathbb{R}^4$ and $g_{\mu\nu}=\eta_{\mu\nu}$. The only difference is that then you have a global cartesian coordinate system $(t,x,y,z)$ and you can identify globaly a vector with its components $v^\mu$ with respect to the associated basis.

This is just an overview to allow you to get the idea. To really understand this and learn how to work with it I recommend:

1. The book Gravitation by Misner, Thorne and Wheeler
2. The book Spacetime and Geometry by Sean M. Carroll
3. The YouTube lectures on General Relativity from the International Winter School on Gravity and Light by Frederic Schuller.