Momentum density from stress energy tensor in field theory

Question

Question is similar to one in this link 1.

Let us consider very simple Lagrangian that contains only kinetic energy. Its interpretation follows from saying that the field we are varying $\mathbf{u}$ is a displacement field. In that case, we have the following Lagrangian field density.

$$ \mathcal{L} = \frac{1}{2} \rho \left( \frac{\partial \mathbf{u}}{\partial t} \right)^2 $$

If one evaluates canonical momentum then it is equal to:

$$ \pi = \rho \frac{\partial \mathbf{u}}{\partial t} $$

We have more conservation laws that follow from stress-energy tensor defined in the following way.

$$ T^\mu{}_\nu =\frac{ \partial \mathcal{L}}{\partial (\partial_\mu \varphi_a)} \partial_\nu \varphi_a - \delta^\mu_\nu \mathcal{L} $$

One of the equations is energy conservation and other three are different kind of momentum conservations. The fact that they are different can be shown by computing that, for example, for component $1$ of the momentum ($T^{0}_1$) is equal to:

$$ \mathbf{\Pi}_1 =T^0_1 = \rho \frac{\partial \mathbf{u}}{\partial t} \cdot \frac{\partial \mathbf{u}}{\partial x_1} $$

Questions:

If we consider any hydrodynamics books then what happens is people consider three conservation laws: 1) conservation of mass, 2) conservation of energy (related to $T^0_0$) and 3) conservation of canonical momentum.

Why are three additional conservation equations of stress-energy momentum not considered? It seems to be incredibly powerful tool to have three more equations to deal with.

Only explanation that I have for them not being used is that one can somehow show that canonical momentum equation and energy equation could be combined to express conservation of $\mathbf{\Pi}$. But I am unable to find any references on that. Also it is not that clear to me that it should be possible because the underlying symmetries are different (symmetry of coordinates vs symmetry of fields).

Is it possible that conservation of $\mathbf{\Pi}$ is in some sense redundant taking into account other equations?

Answer 1

The stress-enegy conservation laws are just the usual hydrodynamic equations.

For example, starting from the usual Lagrangian density for a barotropic fluid $$ {\mathcal L} = \rho \dot \phi +\rho \frac 12 (\nabla \phi)^2 +u(\rho) $$ we can compute the canonical energy momentum tensor $$ {T^\mu}_\nu = \frac{\partial {\mathcal L}}{\partial \phi_\mu}\phi_\nu -\delta^\mu_\nu {\mathcal L}. $$

In components we have $$ {T^0}_0 = \rho\,\partial_t \phi - {\mathcal L}, \\ {T^0}_i = \rho \,\partial_i \phi ,\\ {T^i}_0 = \rho \,\partial_i \phi \partial_t\phi,\\ {T^i}_j = \rho \,\partial_i\phi\partial_j\phi - \delta^i_j {\mathcal L}, $$

Using the physical interpretation of the variables in the action as ${\bf v}=\nabla \phi$, and $$ {\mathcal L}=-P = \rho\dot \phi +{\mathcal E} $$ where $P$ is the pressure and ${\mathcal E}$ the total energy density (kinetic plus internal) we can express the components in terms of physical quantities:
$$ {T^0}_0 = -{\mathcal E}, \\ {T^0}_i = \rho v_i,\\ {T^i}_0 = -v_i(P+{\mathcal E}),\\ {T^i}_j = \rho v_i v_j + \delta_{ij}P. $$ Note that energy-momentum tensor is not symmetric. The momentum density is however equal to the mass flux.

The conservation equation $$ \partial_0 {T^0}_0 +\partial_j {T^j}_0=0 $$ becomes (minus) $$ \partial_t {\mathcal E}+ \partial_j\{v_j({\mathcal E}+P)\}=0. $$ This asserts that a decrease in energy in a region arises from two mechanisms: i) the convective efflux of energy $\int {\mathcal E} {\bf v}\cdot {\bf n} \,d|S|$, and ii) the rate of working, $\int P\,{\bf v}\cdot {\bf n} \,d|S|$, by the forces exerted on the neighbouring fluid. Here ${\bf n}$ is the outward normal to the surface element $d|S|$ of the region.

The remaining three conservation equations $$ \partial_0 {T^0}_i +\partial_j {T^j}_i=0,\quad i=1,2,3 $$ become the momentum conservation law $$ \partial_t \rho v_i +\partial_j (\rho v_iv_j +\delta_{ij} P)=0. $$ Here the change in momentum density is due to the $\rho v_iv_j$ momentum advection current together with the momentum flux due one bit of fluid pushing on the other via the force $-\nabla P$. This conservation law is a simple combination of the Euler equation with the mass-conservation equation. Mass conservation follows from the symmetry $\phi\to \phi+\rm const.$ rather than from time-space translation invariance.

Answer 2

Your canonical momentum conservation is the one that is usually not considered in hydrodynamics. It corresponds to symmetry under shifts of $\mathbf{u}$. If you had a "mass term" proportional to $\mathbf{u}\cdot\mathbf{u}$ (or really anything that depends on $\mathbf{u}$ itself and not its derivatives) then this is no longer a symmetry.

In any case, for your Lagrangian, the conservation of the canonical momentum is equivalent to the field equations, which are equivalent to the conservation of the energy-momentum tensor (the last point mike stone shows in his answer).