Why is the first component of the energy-momentum tensor $-p_a v^a$? A single particle of mass $m$ travels along a geodesic curve in flat spacetime. A tangent vector $u^a$ to this geodesic, parameterised by the parameter $\tau = \int \left( \eta_{\mu \nu}u^\mu u^\nu \right)^{1/2}$, is referred to as the 4-velocity of the particle. Furthermore, due to this choice of parameter, it can be shown that:
$$
u_a u^a \equiv \eta_{\mu \nu} u^\mu u^\nu = -1
$$
For clarity, abstract index notation is used, along with setting $c = 1$.
Furthermore, the energy—momentum vector $p^a$ for said particle is defined as:
$$
p^a = mu^a
$$
The book I’m reading (General Relativity by Wald) says that the energy $E$ of the particle, measured by an observer with its own 4-velocity $v^a$, is defined as:
$$
E = -p_a v^a
$$
This equation is followed with the statement:

Thus, in special relativity, energy is recognized to be the “time component” of the 4-vector $p^a$.

Here is where I get lost. The statement makes it sound like this is obvious, but I’m struggling to prove this from the equations given.
Basically, in a given Cartesian coordinate system, we have two geodesic curves along with one energy-momentum vector for each curve - $u^a$ for the particle and $v^a$ for the observer. The energy of the particle as measured by the observer is thus given by:
$$
E = -p_a v^a = -\eta_{\mu \nu}p^\mu v^\nu = mu^0v^0 - mu^1v^1 - mu^2v^2 - mu^3v^3
$$
However, the first component of $p$ is simply:
$$
p^0 = mu^0
$$
I don’t see how this can equate with the energy under general circumstances without further restrictions on $v$. I haven’t yet used the fact that $u_a u^a = v_a v^a = -1$, but any attempt in doing so only seems to complicate things - for example, substituting $u^0 = \sqrt{1 + (u^1)^2 + (u^2)^2 + (u^3)^2}$.
So my question is: How does the above definition of the energy $E$ lead to the first component of $p$ being $E$? I’ve seen proofs of this for other definitions but am interested in how this can be derived from the above definition, where energy is the contraction of the particle energy-momentum tensor with the 4-velocity of the observer.
 A: The point is that this number $E$ is the time-component of the vector $p$, as calculated by an observer with velocity $v$. So, the very notion of 'the time-component of a vector' is a-priori meaningless; it only acquires meaning once you specify an observer, hence it is an observer-dependent notion. The observer tells us what they consider to be pure time components, and what they deem to be pure spatial components. Here is a simple linear algebraic fact which will hopefully clarify things:

Let $(V,g)$ be an arbitrary $(n+1)$-dimensional Lorentzian inner product space, and let $v\in V$ be a normalized (for convenience) timelike vector i.e $g(v,v)=-1$. Let $T=\text{span}\{v\}$, and let $S:=T^{\perp}=\{w\in V\,:\, g(w,v)=0\}$ be the orthogonal complement with respect to $g$. Then, $S$ is a spacelike subspace and we have a $g$-orthogonal direct sum decomposition $V=T\oplus S$.

It is a good exercise for you to prove this statement. Here, the timelike vector $v\in V$ plays the role of the instantaneous velocity vector of an observer. Then, this observer's velocity allows us to decompose the vector space $V$ (in general it will be the tangent space at a specific spacetime point) into two mutually orthogonal pieces: $V=T\oplus T^{\perp}=T\oplus S$. The first summand $T$ is a timelike subspace, and it is spanned by the observer's velocity (so is $1$-dimensional), while the second summand $S$ is an $n$-dimensional spacelike subspace of $V$. The elements of the subspace $T$ is what the observer would consider to be 'purely temporal', while vectors in $S$ are what this particular observer would call 'purely spatial'.
So, given any $\xi\in V$, there exist unique $\alpha\in\Bbb{R}$, and $s\in S$ such that
\begin{align}
\xi&=\alpha v+s.
\end{align}
In fact, you can show that $\alpha=-g(\xi,v)$ and $s=\xi-[-g(\xi,v)v]=\xi+g(\xi,v)v$. So, given an arbitrary vector $\xi$, we decomposed it (with respect to the observer!) into a purely temporal portion $av\in T$ and a purely spatial portion $s\in S$.
In your case, you're starting with a vector $p$ (interpreted as 4-momentum), and you're decomposing it relative to an observer's instantaneous velocity vector $v$:
\begin{align}
p&=Ev+\vec{p},
\end{align}
for some unique number $E\in\Bbb{R}$ and some unique vector $\vec{p}\in S$. This is literally the same statement as above, with different notation. So, the number $E=-g(p,v)=-g_{ab}p^av^b=-p_av^a$, which a-priori is just obtained via a linear-algebraic direct sum, is now given the physical interpretation as the energy as measured by an observer.

Edit:
If you don't like direct sums and prefer to think in terms of bases, then here's the explanation. You can't just take an arbitrary basis $\{e_0,e_1,e_2,e_3\}$ such that $g(e_a,e_b)=\eta_{ab}$ for all $a,b$ (i.e an 'orthonormal basis'), and then expect that relative to this basis, the number $E$ will equal $p^0$. You have to involve the observer by decomposing vectors with respect to a basis where the first (timelike) vector is the observer's velocity: $\{v,s_1,s_2,s_3\}$, such that $g(v,v)=-1, g(v,s_i)=0, g(s_i,s_j)=\delta_{ij}$.
So, in your post when you talk about a Cartesian coordinate system blablabla, what you missed out is that the coordinate system has to be 'adapted' to the observer in the above sense.
