How to find the energy-momentum tensor of a free relativistic particle from its lagrangian?

Question

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)?

Answer 1

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: $$\tag{10} T^{ab}(x) = m \int \frac{d z^a}{d \tau} \, \frac{d z^b}{d \tau} \, \delta^4 (x - z) \, d\tau. $$

Answer 2

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

  1. 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.

  2. 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.