# Momentum density from stress energy tensor in field theory

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?

• There are no "three additional conservation equations of stress-energy momentum". The time derivative of momentum equals the divergence of the stress distribution. Commented Jul 30, 2019 at 20:34
• @my2cts are you saying that momentum conservation equations are satisfied if and only if canonical momentum equations are satisfied? Commented Jul 30, 2019 at 21:28

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.

• Thanks for your answer -- this is a really interesting approach I have never seen before. What about theories which do not limit velocity fields being irrotational or incompressible, or anything else, and where just as in the case of my Lagrangian the field you are varying is displacement field $\mathbf{u}$? Also, I am not sure why mass conservation follows from field symmetry, can you please explain that more? Commented Aug 6, 2019 at 1:56
• @Daniels Krimans. See chapter 1 in my physics 508 lecture notes (or the CUP book with Paul Goldbart based on it: goldbart.gatech.edu/PostScript/MS_PG_book/bookmaster.pdf). Much of the material is in the end of chapter excerises. A link to the notes is at courses.physics.illinois.edu/phys508/fa2018/amaster.pdf Commented Aug 6, 2019 at 12:23
• thank you very much Commented Aug 6, 2019 at 18:25

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).

• Thanks for the answer! I am not sure I understand why you say that canonical momentum conservation is usually not considered in hydrodynamics. Euler equation is an equation for canonical momentum $\rho \mathbf{v}$. And I am not sure what kind of mass term you are talking about. In free fluid there is no term that depends on the field $\mathbf{u}$. Also, I am not sure why you think that conservation of canonical momentum is equivalent to conservation of energy-momentum tensor. Can you please expand on that a little? Commented Aug 6, 2019 at 1:54
• Well to start off, there is a difference between the Euler-Langrange equations gotten by varying the action and the hydrodynamic Euler equation. It is true that if only derivatives of the fields appear in the Lagrangian (no "mass terms"), then conservation of canonical momentum is equivalent to the Euler-Lagrange equations. This is pretty trivial and has nothing to do with hydrodynamics. However the Lagrangian you wrote (you are missing a spatial derivative term by the way) is not a hydrodynamic Lagrangian in the sense that it leads to the Euler equations (try varying it!). Commented Aug 6, 2019 at 5:45
• What you wrote down is basically a theory of three free massless scalar fields. In this case it is true finding the Euler-Lagrange equations is the standard approach. You do not even need to consider the energy conservation equation here separately, it is implied by the Euler-Lagrange equations. But in hydrodynamics Euler-Lagrange equations are not the standard approach. Starting with conservation equations like from the energy-momentum tensor is standard. In fact some people take this as the definition of hydrodynamics, when everything is governed by conservation laws and an equation of state Commented Aug 6, 2019 at 6:22
• You can do an Euler-Lagrange approach in hydrodynamics though. Probably the most well known involves something called Clebsch potentials. The simplest form of this involves a single scalar $\phi$ and it looks a lot like a theory of a single (not three) free massless scalar a lot like you wrote. In all of these variational descriptions of hydrodynamics I've seen, the Euler-Lagrange equations are equivalent to the conservation of the energy-momentum tensor (and potentially internal conservation laws). If you start with one set of equations you can derive the other. Commented Aug 6, 2019 at 6:28
• i really appreciate your insights, thanks! Commented Aug 6, 2019 at 18:24