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Derivatives $\nabla_i T^{ik}=0$ of a stress-energy tensor of physical system express conservation laws. Whether contains a stress-energy tensor also the information on the equations of motion of system?

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5 Answers 5

up vote 4 down vote accepted

1) Let us assume that the system is classical and has a Lagrangian formulation

$$ S[\phi]~=~ \int \mathbb{L}, \qquad\qquad \mathbb{L} ~=~{\cal L} ~dx^0 \wedge \ldots\wedge dx^{d-1}, $$

in terms of a Lagrangian density ${\cal L}={\cal L}(\phi,\partial\phi)$ that doesn't depend explicitly on the $d$-dimensional spacetime coordinates $x^{\mu}$.

2) Noether's Theorem for the global translation symmetry $x^{\mu}\to x^{\mu}+\delta x^{\mu} $ yields the following off-shell relation

$$ d_{\mu} T^{\mu}{}_{\nu} ~=~ - \phi^{\alpha}_{,\nu} ~\frac{\delta S}{\delta\phi^{\alpha} }, \qquad\qquad \phi^{\alpha}_{,\nu}~:=~\frac{\partial\phi^{\alpha}}{\partial x^{\nu}} . $$

Thus one may deduce the conservation law

$$d_{\mu} T^{\mu}{}_{\nu}~\approx~ 0$$

from the equations of motion (eom)

$$\frac{\delta S}{\delta\phi^{\alpha} } ~\approx~ 0. $$

However, there are many examples where one can not deduce the other way, in particular, if there are more eoms (labeled by $\alpha$) than conservation laws (labeled by $\nu$). [Here the $\approx$ sign means equality modulo eom.]

3) OP raises an interesting question whether one can deduce the equations of motion from knowledge of the stress-energy tensor $T^{\mu}{}_{\nu}$ itself? This is almost the case, but there are certain terms that the stress-energy tensor $T^{\mu}{}_{\nu}$ cannot see, see Section 4, 5 and 6 below. The question is easiest to solve if one has direct access$^1$ to the canonical stress-energy tensor

$$T^{\mu}_{({\rm can})\nu} ~:=~\phi^{\alpha}_{,\nu} ~\Pi^{\mu}_{\alpha}-\delta^{\mu}_{\nu}~{\cal L},\qquad\qquad \Pi^{\mu}_{\alpha}~:=~\frac{\partial{\cal L}}{\partial \phi^{\alpha}_{,\mu} }. $$

It becomes harder if there are also improvement terms allowed

$$ T^{\mu}{}_{\nu}~=~T^{\mu}_{({\rm can})\nu}+d_{\lambda}f^{\lambda\mu}{}_{\nu},\qquad\qquad f^{\lambda\mu}{}_{\nu} ~=~ - f^{\mu\lambda}{}_{\nu}, $$

and/or if one only knows $T^{\mu}{}_{\nu}$ modulo terms proportional to the eoms (with the additional complication that one doesn't know the eoms to begin with).

4) As a topological-sigma-model-like counterexample, consider a $d$-form

$$ \vartheta ~=~ \frac{1}{d!}\vartheta_{\alpha_1\ldots\alpha_d } ~d\phi^{\alpha_1} \wedge \ldots \wedge d\phi^{\alpha_d}, \qquad\qquad \vartheta_{\alpha_1\ldots\alpha_d }~=~\vartheta_{\alpha_1\ldots\alpha_d}(\phi), $$

in the $\phi$ target space. Let $\omega:=d\vartheta$ be a corresponding $d+1$ form. Now define the Lagrangian $d$-form

$$ \mathbb{L}~:=~\phi^*\vartheta $$

by the pull-back to the $x$-world volume. The corresponding canonical stress-energy tensor vanishes identically $T^{\mu}{}_{\nu} ~\equiv~0$, while the eoms become

$$\phi^*i_{\alpha}\omega ~\approx~ 0,$$

where $i_{\alpha}$ denotes the contraction wrt. the vector field $\frac{\partial}{\partial\phi^{\alpha}}$. The main point is that the vanishing stress-energy tensor $T^{\mu}{}_{\nu} ~\equiv~0$ carries no information about the $d$-form $ \vartheta$. The equations of motion are only trivial, i.e., $0=0$, if $\omega=0$, i.e., if $\vartheta$ is closed.

Still worse, for a generic model, the Lagrangian $d$-form $\mathbb{L}$ could contain such a $\phi^*\vartheta$ term, which the stress-energy tensor $T^{\mu}{}_{\nu}$ cannot see.

5) Consider for simplicity the special case of point mechanics where $d=1$. Then the stress-energy tensor is just the energy function $h:=T^0{}_0$. Similarly, we change notation $x\to t$ and $\phi^{\alpha}\to z^I$. The one-form $\vartheta$ is a presymplectic potential

$$\vartheta ~=~\vartheta_I~ dz^I, \qquad\qquad \vartheta_I~=~\vartheta_I(z), $$

and the two-form $\omega:=d\vartheta$ is a presymplectic two-form. (If the two-form $\omega$ is non-degenerated, it becomes a symplectic two-form.) The Lagrangian is

$$L~=~\vartheta_I~\dot{z}^I,$$

which can be viewed as a Hamiltonian system with zero Hamiltonian. The corresponding energy function vanishes identically $h~\equiv~0$, while the equations of motions becomes

$$\omega_{IJ}~\dot{z}^J ~\approx~ 0.$$

6) To be explicit, consider $n$ non-relativistic charged particles in a magnetic vector potential ${\bf A}$. Working in units where $c=1$, the Lagrangian reads

$$L~=~\sum_{i=1}^n \frac{m_i}{2} \dot{\bf r}^2_i + \sum_{i=1}^n q_i\dot{\bf r}_i\cdot {\bf A}({\bf r}_i) - V({\bf r}_1, \ldots, {\bf r}_n),$$

where we have added a potential $V=V({\bf r}_1, \ldots, {\bf r}_n)$ to be more general. The corresponding energy function

$$h ~=~\sum_{i=1}^n \frac{m_i}{2} \dot{\bf r}^2_i + V({\bf r}_1, \ldots, {\bf r}_n)$$

does not know about the magnetic potential ${\bf A}$. However the eoms contain the Lorentz force. Here the magnetic potential ${\bf A}$ plays the role of the presymplectic potential from Section 5.

--

$^1$ It seems admittedly rather artificial that one would know in advance whether a given stress-energy tensor $T^{\mu}{}_{\nu}$ is canonical or not, because if one doesn't know the eoms, one wouldn't know the Lagrangian density ${\cal L}$ as well.

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Concerning magnetic example: The energy deals only with the length of velocity. However, momentum conservation deals with its direction and gives the remaining e.o.m's: $\sum \frac{\partial L}{\partial \dot{r_i}}=P$ Differentiating this w.r.t. time gives you the remaining e.o.m's –  Terminus Feb 1 '12 at 13:00
    
In point mechanics ($0+1$ world volume dimensions), the stress-energy tensor is just a $1×1$ matrix $T^0{}_0$ carrying only energy $h$. Momentum conservation along the target space directions would be an additional piece of information not implied by OP's setup. –  Qmechanic Feb 2 '12 at 10:14

The case of electrovacuum.

In General Relativity, from the Ricci tensor, we know the stress-energy tensor and vice-versa (by Einstein equation).

There are some conditions which, if are satisfied by the Ricci tensor (or equivalently, by the stress-energy tensor), are equivalent to the fact that the equations of motions are those from electromagnetism, and we can even determine the electromagnetic tensor (up to an overall "phase factor", or even uniquely in "charge without charge"). They are called the Rainich conditions, and were proposed in 1924-25 by G.Y. Rainich.

These conditions were rediscovered by Misner and Wheeler, as part of the charge-without-charge subprogram of geometrodynamics. For this, see Geometry of gravitation and electromagnetism, by L. Witten, and chapter 9 from Gravitation: An Introduction to Current Research, ed. L. Witten, and section 5.3 from Spinors and Space-time: Spinor and twistor methods in space-time geometry, R Penrose, W Rindler - 1986.

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There certainly are systems that are fully described by their energy-momentum conservation as a simple example let us take one-dimensional particle in a potential. The energy is conserved: $$ \frac{m \dot{x}^2}{2} + V(x) = E $$ Differentiating this expression w.r.t. time one gets

$$ \frac{2 m \dot{x} \ddot{x}}{2} + \frac{\partial V(x)}{\partial x} \dot{x} = 0 $$

$$ (m \ddot{x}+ \frac{\partial V(x)}{\partial x} )\dot{x} = 0 $$

This means that either E.O.M.'s are satisfied or the particle is at rest anywhere. If one can rule out the latter case by physical arguments, one arrives at a derivation.

The same is true for e.g. scalar field. However, as demonstrated in other reply (Sigma-model) -conservation laws might be spoiled in the way that they do not bring any information. This is fortunately not the case for classical fields.

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Comment to the answer(v1): If you in the last sentence mean classical systems as opposed to quantum systems, then the last sentence is not true. There are classical counterexamples. –  Qmechanic Feb 1 '12 at 10:19
    
I would be really glad if you posted any. Bu, either way, my answer could be changed to 'reasonable classical' as i am sure this is the case for Young-Mills (including electromagnetism) and scalar fields. –  Terminus Feb 1 '12 at 12:04
    
E.g., a point charge in a background magnetic field is a classical counterexample, cf. my answer. –  Qmechanic Feb 1 '12 at 12:33
    
The energy deals only with the length of velocity. However, momentum conservation deals with its direction and gives the remaining e.o.m's: $\sum \frac{\partial L}{\partial \dot{r_i}} = P $ Differentiating this w.r.t. time gives you the remaining e.o.m's –  Terminus Feb 1 '12 at 12:58
    
In point mechanics ($0+1$ world volume dimensions), the stress-energy tensor is just a $1\times 1$ matrix $T^0{}_0$ carrying only energy $h$. Momentum conservation along the target space directions would be an additional piece of information not implied by OP's setup. –  Qmechanic Feb 1 '12 at 15:28

Simple answer, your equation has only four components, but there is one equation of motion for every field variable in the system which is more than four, so the answer is no.

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Ok, then it is possible to consider as EOM statement $ \nabla_i T ^ {ik} =G ^ {k} $, where $G ^ {k} $ - 4-force? –  Sergio Jan 31 '12 at 7:11
    
The conservation laws give you some information about the equations of motion but not all of it. If further constraints are applied such as requiring spherically symmetrical solutions and the number of fields is small you may find that the conservation laws are sufficient to solve the problem –  Philip Gibbs Jan 31 '12 at 7:18
    
Although No may be formally correct, this answer(v1) ignores the fact that the stress-energy tensor almost provides full information about the system. –  Qmechanic Feb 1 '12 at 10:26
    
That depends on how many field variables you have in your system. –  Philip Gibbs Feb 1 '12 at 11:19
    
E.g., if you were told that the energy function of a system is $h = \frac{1}{2} \sum_{i=1}^n m_i \dot{x}^2_i + V(x_1, \ldots, x_n)$, wouldn't you have a pretty good idea of what the equations of motion are, even if $n$ is a huge number? –  Qmechanic Feb 2 '12 at 10:08

The situation is similar to conservation of energy, momentum etc. in classical mechanics. Consider for instance motion of the mass $m$ in the potential $V(r)$. In general case the energy is conserved, and the momentum is not. We know the expression for energy:

$E=\frac{mv^2}{2}+V(r) = const$

The equation of motion is

$\frac{d\vec{v}}{dt}=-\nabla V(r)$

and it cannot be derived from expression for energy. However, knowing the integral of motion makes easier finding solution of the equations of motion.

Similarly, Maxwell equations (that are equations of motion) cannot be derived from electromagnetic field stress-energy tensor.

On the other hand, the expression for stress -energy tensor can be used to derive expressions for some other important physical quantities.

If we consider compound system of charge + electromagnetic field, we can derive the Lorentz force acting on charge from the expression of EM field's stress-energy tensor and Maxwell equations:

$\frac{\partial T^k_i}{\partial x^k}=-\frac{1}{c}F_{il}J^l$

What is important here is that we use expression for the EM field stress-energy tensor and Maxwell equations (that are equations of motion for the EM field) to derive the force acting on charge.

But knowing the force acting on charge is not enough to derive equation of motion for the charge. We also need to know the "inertial properties" of the charge for that.

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Strange, but in the literature often there is a statement that the equations $\nabla_i T^{ik}=0$ are the equations of motion. And that the equations of motion can be derived from conservation laws, seemed to me strange. That is the reason of asking this question. For example, in the textbook of Landau and Lifshits "Hydrodynamics" the Euler equations of motion of an ideal liquid are deduced from these equations. –  Sergio Jan 27 '12 at 10:18
    
And I disagree with statement that it is impossible to derive the Maxwell equations from $\nabla_i T^{ik}=0$. At least, from this equation it is possible to derive the second pair of the Maxwell equations $F^{ij}_{,k}+F^{ki}_{,j}+F^{jk}_{,i}=0$. –  Sergio Jan 27 '12 at 10:20
    
As far as I can see from Landau and Lifshits "Hydrodynamics", in derivation of Euler equation they use the Newton's second law (which is the equation of motion itself): force acting on element of fluid is equal to acceleration of this element multiplied by its mass. Regarding the Maxwell equations, see "Do Maxwell's Equations overdefine the electric and magnetic fields?" on physics stack exchange, where it is explained that second pair of Maxwell equations are only consequences of the first pair equations and initial conditions. –  Murod Abdukhakimov Jan 27 '12 at 10:43
    
It is necessary to look head named "relativistic hydrodynamics" first paragraph. –  Sergio Jan 27 '12 at 11:30
    
and second paragraph –  Sergio Jan 27 '12 at 12:10

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