How to find the energy-momentum tensor of a free relativistic particle from its lagrangian? Consider a free relativistic particle in Minkowski spacetime.  Its standard action is the following, where $\sigma$ is an arbitrary parametrization ($\tau$ is the particle's proper time.  I'm using units so that $c \equiv 1$ and metric signature $\eta = (1, -1, -1, -1)$):
$$\tag{1}
S = -\, m \int \sqrt{\eta_{ab} \, \frac{d z^a}{d\sigma} \, \frac{d z^b}{d\sigma}} \, d\sigma.
$$
The particle's energy-momentum is defined with the help of a spacetime Dirac delta, summed on the particle's world history (the integral limits are implicit, from $\tau_1 = -\, \infty$ to $\tau_2 = +\, \infty$):
$$\tag{2}
T^{ab}(x) = m \int \frac{d z^a}{d\tau} \, \frac{d z^b}{d\tau} \, \delta^4(x - z) \, d\tau,
$$
where $z \equiv z^a(\tau)$ is the particle's cartesian coordinates in spacetime.  The four-velocity is normalized (using the proper time $\tau$):
$$\tag{3}
\eta_{ab} \, \frac{d z^a}{d\tau} \, \frac{d z^b}{d\tau} = 1.
$$
For a simple free particle in Minkowski spacetime, I cannot use the general expression of the canonical field energy-momentum to find (2) (I don't know how to use it here, for a simple particle):
$$\tag{4}
T^a_{\; \, b} = \sum_k \frac{\partial \mathscr{L}}{\partial (\partial_a \phi_k)} \, \partial_b \, \phi_k - \delta^a_{\; b} \, \mathscr{L}.
$$
The action (1) could be translated into an integral defined on the whole of spacetime:
$$\tag{5}
S = -\, m \iint \sqrt{\eta_{ab} \, \frac{d z^a}{d\sigma} \, \frac{d z^b}{d\sigma}} \, \delta^4 (x - z) \, d\sigma \, d^4 x,
$$
so that
$$\tag{6}
\mathscr{L}(x) = -\, m \int \sqrt{\eta_{ab} \, \frac{d z^a}{d\tau} \, \frac{d z^b}{d\tau}} \, \delta^4 (x - z) \, d\tau = -\, m \int \delta^4 (x - z) \, d\tau.
$$
I'm not sure this makes any sense.  Notice that $\mathscr{L}(x) = -\, \eta_{ab} \, T^{ab} \equiv -\, T(x)$.
So how can I find (2) from (1)?
 A: I've found a nice way using a variation of the metric to define $T_{\mu \nu}$ as in general relativity. Here it is (I never saw this before, for the free particle, but it's probably known by some people here).
Since
$$\tag{1}
\int \delta^4 (x - z) \, d^4 x \equiv 1,
$$
the action (1) from the question gives the Lagrangian density $\mathscr{L}$ for an arbitrary coordinates system and metric $g_{\mu \nu}$:
$$\tag{2}
S = -\, m \iint \sqrt{g_{\mu \nu} \, \frac{d z^{\mu}}{d\sigma} \, \frac{d z^{\nu}}{d\sigma}} \, \delta^4 (x - z) \, d\sigma \, d^4 x = \int \mathscr{L} \, \sqrt{-g} \, d^4 x,
$$
so
$$\tag{3}
\mathscr{L}(x) = -\, \frac{m}{\sqrt{-g}} \int \sqrt{g_{\mu \nu} \, \frac{d z^{\mu}}{d\sigma} \, \frac{d z^{\nu}}{d\sigma}} \, \delta^4 (x - z) \, d \sigma.
$$
The parametrization $\sigma$ is arbitrary (the proper time is $\tau$).  Because of the deltas, all functions $f(x)$ could enter the "sigma" integral and be evaluated on the particle's worldline $z \equiv z^{\mu}(\sigma)$.
In general relativity, the energy-momentum is defined from an arbitrary variation of the metric, $\delta g^{\mu \nu}$ (the following could be expressed in several ways, depending on the authors) :
$$\tag{4}
T_{\mu \nu} \, \delta g^{\mu \nu} \equiv 2 \, \delta\mathscr{L} - g_{\mu \nu} \, \mathscr{L} \, \delta g^{\mu \nu}.
$$
I'll use two metric identities:
\begin{align}\tag{5}
\delta g_{\lambda \kappa} &= -\, g_{\lambda \mu} \, g_{\kappa \nu} \, \delta g^{\mu \nu},
& \delta \sqrt{-g} &= -\, \frac{1}{2} \, g_{\mu \nu} \, \sqrt{-g} \, \delta g^{\mu \nu}.
\end{align}
The variation of the Lagrangian density (3) give the following expression:
$$\tag{6}
2 \, \delta\mathscr{L} - g_{\mu \nu} \, \mathscr{L} \, \delta g^{\mu \nu} = \frac{m}{\sqrt{-g}} \int \frac{g_{\mu \lambda} \, g_{\nu \kappa}}{\sqrt{g_{\rho \sigma} \, \frac{d z^{\rho}}{d\sigma} \, \frac{d z^{\sigma}}{d \sigma}}} \, \frac{d z^{\lambda}}{d \sigma} \, \frac{d z^{\kappa}}{d \sigma} \, \delta^4 (x - z) \, d\sigma \, \delta g^{\mu \nu}.
$$
So the energy-momentum tensor is this:
$$\tag{7}
T_{\mu \nu}(x) = \frac{m}{\sqrt{-g}} \int \frac{g_{\mu \lambda} \, g_{\nu \kappa}}{\sqrt{g_{\rho \sigma} \, \frac{d z^{\rho}}{d\sigma} \, \frac{d z^{\sigma}}{d \sigma}}} \, \frac{d z^{\lambda}}{d \sigma} \, \frac{d z^{\kappa}}{d \sigma} \, \delta^4 (x - z) \, d\sigma.
$$
This is parametrization invariant.  To simplify things, I then change $\sigma \rightarrow \tau$ and use
$$\tag{8}
\sqrt{g_{\rho \sigma} \, \frac{d z^{\rho}}{d\tau} \, \frac{d z^{\sigma}}{d \tau}} = 1.
$$
So we get
$$\tag{9}
T^{\mu \nu}(x) = \frac{m}{\sqrt{-g}} \int \frac{d z^{\mu}}{d \tau} \, \frac{d z^{\nu}}{d \tau} \, \delta^4 (x - z) \, d\tau.
$$
Going back to Minkowski spacetime and cartesian coordinates ($\sqrt{-g} = 1$) give the end result:
$$\tag{10}
T^{ab}(x) = m \int \frac{d z^a}{d \tau} \, \frac{d z^b}{d \tau} \, \delta^4 (x - z) \, d\tau.
$$
A: *

*The canonical stress-energy-momentum (SEM) tensor is not appropriate for 2 reasons:

*

*The field theory for the free relativistic particle (1) with dynamical variables $(z^0,z^1,z^2,z^3): [\sigma_i,\sigma_f]\to \mathbb{R}^4$ is a 0+1D worldline theory, i.e. the canonical SEM tensor is a $1\times 1$ tensor, consisting of energy only, not a $4\times 4$ tensor.


*The energy vanishes for reparametrization-invariant theories such as the free relativistic particle (1), cf. e.g. this Phys.SE post.




*The main point is that one instead should use the Hilbert SEM tensor, where one varies wrt. the metric $g_{\mu\nu}$. This works perfectly, as demonstrated in OP's self-answer.
