The Lagrangian for a (relativistic) point particle with rest mass $m$ and velocity $v$ is: $$L=-\frac{m}{\gamma (v)}$$ (using $c=1$). Over on Wikipedia we can find the Stress-energy tensor for said particle; if the four-velocity is $u^\mu=(\gamma (v),\gamma (v)v)$ then the stress-energy is: $$T^{\mu\nu}=\frac{mu^\mu u^\nu}{\gamma (v)} \delta^3({\bf x}-{\bf x}_p(t)).$$ Interestingly, the negative of the trace of the stress-energy in this case is: $$-\eta_{\mu\nu}T^{\mu\nu}=\frac{m}{\gamma (v)} \delta^3({\bf x}-{\bf x}_p(t))$$ Which, when integrated over all of space, gives the appropriate Lagrangian for the particle, meaning it serves as a Lagrangian density. Is this just some kind of coincidence for point particles, or are there other cases where the Lagrangian density is equal to the negative trace of the stress-energy tensor?


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The delta function is a density, so the correct relation should be with a tensor density, namely: $$\mathfrak{T}^{\mu \nu} \equiv T^{\mu \nu} \sqrt{|g|} = m \frac{\mathrm{d}x^\mu}{\mathrm{d}s} \frac{\mathrm{d} x^ \nu}{\mathrm{d} s} \delta(x - x(s))$$ for a point-like particle of mass $m$ on worldline $x(s)$ parametrized by its arc-length $s$.

The Einstein field equations imply a continuity equation for the stress tensor which, when stated in terms of the tensor density, is $\partial \mathfrak{T}^{\mu \nu}/\partial x^ \nu + \Gamma^\mu_{ \nu \rho} \mathfrak{T}^{ \nu \rho} = 0$.

Upon substitution of the expression for the tensor density, you get:

\begin{align} \frac{\partial \mathfrak{T}^{\mu \nu}}{\partial x^ \nu} &= m \frac{\mathrm{d} x^\mu}{\mathrm{d} s} \frac{\mathrm{d} x^ \nu}{\mathrm{d} s} \frac{\partial}{\partial x^ \nu}(\delta(x - x(s)))\\ &= -m \frac{\mathrm{d} x^\mu}{\mathrm{d}s} \frac{\mathrm{d}}{\mathrm{d} s} (\delta(x - x(s)))\\& = \frac{\mathrm{d}}{\mathrm{d} s} \left(m \frac{\mathrm{d}x^\mu}{\mathrm{d} s}\right) \delta(x - x(s))\\ \Gamma^\mu_{ \nu \rho} \mathfrak{T}^{ \nu \rho} &= \Gamma^\mu_{ \nu \rho} m \frac{\mathrm{d} x^ \nu}{\mathrm{d}s} \frac{\mathrm{d} x^ \rho}{\mathrm{d}s} \delta(x - x(s)) \end{align}

or, collecting terms:

$$\left(\frac{\mathrm{d}}{\mathrm{d}s} \left(m \frac{\mathrm{d}x^\mu}{\mathrm{d}s}\right) + \Gamma^\mu_{ \nu \rho} m \frac{\mathrm{d}x^ \nu}{\mathrm{d}s} \frac{\mathrm{d}x^\rho}{\mathrm{d}s}\right) \delta(x - x(s)) = 0.$$

Taking an integral with respect to x removes the delta function which, assuming the mass (m) is non-zero and constant, results in the geodesic equation:

$$\frac{\mathrm{d}^2 x^\mu}{\mathrm{d}s^2} + \Gamma^\mu_{ \nu \rho} \frac{\mathrm{d}x^\nu}{\mathrm{d}s} \frac{\mathrm{d}x^\rho}{\mathrm{d}s} = 0.$$


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