4
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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:

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.