Boundary condition on stress-energy tensor

Question

Let's consider a boundary between a continuous substance and vacuum. In the absence of external forces the continuity equation for the stress energy tensor is $$ \partial_\mu T^{\mu\nu} = 0 \tag{*}\label{continuity-eq} $$ (greek letters run from 0 to 3, latin letters from 1 to 3). Let's consider only stationary case $\partial_0 T^{\mu\nu} \equiv 0$ so that this equation simplifies to $$ \partial_k T^{k\nu} = 0 $$

Then it follows that on the boundary with vacuum we have $$ T^{k\nu} n_k = 0 \tag{**}\label{boundary-condition} $$ where $n_k$ is normal vector. For a substance like fluid or gas, stress energy tensor is usually given in this form: $$ T^{\mu\nu} = (\varepsilon+p)u^\mu u^\nu - pg^{\mu\nu} \tag{***}\label{standard-form} $$ where $\varepsilon$ is proper energy density, $p$ is pressure, $u$ is relativistic 4-velocity and $g$ is metric tensor. This expression doesn't satisfy boundary condition $\eqref{boundary-condition}$, because the pressure term $pg^{\mu\nu}$ has non-vanishing contribution.

I'm assuming that $\eqref{standard-form}$ is only valid in the bulk of the material and neglects surface effects, is this correct? How does one deal with the boundary in this case - are you supposed to use some modified expression for stress energy tensor or maybe introduce surface tension as an "external" force in $\eqref{continuity-eq}$?

Answer 1

Yes, you need to take into account external forces acting on the system at the boundary and confining the particles inside. The stress tensor can be defined through the total force $F_\mu=\int f_\mu^\mathrm{int}\:dV$ exerted on the particles in a given volume by the surrounding particles of the body ($f_\mu^\mathrm{int}$ is the force volume density). Assuming the interparticle forces are short-range, the force can be rewritten as the surface integral $F_\mu=\oint\sigma_{\mu\nu}\:ds_\nu$, where $\sigma_{\mu\nu}$ is the stress tensor. This gives us the definition of $\sigma_{\mu\nu}$: the particles located directly above the small surface element $d\mathbf{s}$ act on those located directly below it with the force $\mathbf{e}_\mu\sigma_{\mu\nu}ds_\nu$. It is true when we consider the particles and volumes in the bulk of the system.

On the boundary the situation is slightly different: the particles located exactly on the system boundary are pulled by those located beneath the boundary with the force $-\mathbf{e}_\mu\sigma_{\mu\nu}ds_\nu$ (according to the definition of $\sigma_{\mu\nu}$), which should be compensated with the external force $\mathbf{e}_\mu p_\mu ds$, where $\mathbf{p}=\mathbf{e}_\mu p_\mu$ is the vector of local pressure exerted on the system at given boundary point. Thus we have $\sigma_{\mu\nu}n_\nu=p_\mu$ on the boundary. If the pressure is uniform and normal to the surface then $\sigma_{\mu\nu}n_\nu=-pn_\mu$ (the minus sign is because the pressure force is directed inside the system).

These definitions are from the classical theory of elasticity. If our system contains moving particles then the stress tensor can be decomposed on kinetic and interparticle-interaction parts, $\sigma_{\mu\nu}=\sigma_{\mu\nu}^\mathrm{int}-T_{\mu\nu}$, and in the absence of volume forces we have the continuity equation for momentum density $\nabla_\mu T_{\mu\nu}=f_\mu^\mathrm{int}$, or $\nabla_\mu\sigma_{\mu\nu}^\mathrm{int}=f_\mu^\mathrm{int}$.

Perhaps for a fluid we can neglect $\sigma_{\mu\nu}^\mathrm{int}$ and deal only with the kinetic part $T_{\mu\nu}$, and for a surface tension we can take $\mathbf{p}$ directed inside and equal to the Laplace pressure by absolute value. So, from the preceding formulas, the boundary condition is: $$ T_{\mu\nu}n_\nu=pn_\mu, $$ where $p$ is the Laplace pressure.