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In the context of special/general relativity conservation of the stress-energy tensor leads to conservation of momentum/energy density $\nabla_\mu T^{\mu\nu}=0$. This constitutes 4 equations, one for energy and 3 for momentum.

For many systems, however, conservation of energy density follows trivially from conservation of momentum. For instance, for the electromagnetic stress energy tensor, $\nabla_\mu T_{\rm EM}^{\mu\nu}=J_\mu F^{\mu\nu}$. The energy portion is $J_\mu F^{\mu0}=-\vec{E}\cdot\vec{J}$, and the space portion is $J_\mu F^{\mu j}=-(\rho \vec{E} + \vec{J}\times\vec{B})$. Taking the dot product $-\vec{J}\cdot(\rho \vec{E} + \vec{J}\times\vec{B})$ yields $\rho$ times the time component, so the energy lost is already determined by the momentum loss.

For particles, this is also true: the power delivered to a particle is just the velocity times the force.

Is this a general property of $\nabla_\mu T^{\mu\nu}$, that the energy portion follows from the momentum portion? Can anyone point me to a reference explaining this? If not, can someone point me to a counter example?

PS: I have noticed a few papers, which seem to constrain the energy conservation to be related to the momentum conservation to ensure "integrability", although I don't quite see how they connect, or if they are necessarily connected.

EDIT:

The way I worded this was poor. It probably doesn't make sense to say "does energy conservation follow from momentum conservation" since in different frames of reference the two mix in different ways. Rather, I should say Does $\nabla_\mu T^{\mu\nu}$ yield 3 independent equations or 4? Based on experience, I want to say 3, not 4. The two examples above have only 3 independent equations (and without too much work are generalizable to tensor equations with some constraint identity which removes the 4th equation).

For fluid problems in general relativity, it appears that there are only 3 independent equations. Take for instance the stress energy tensor of a perfect fluid:

$T_{\rm pf}^{\mu\nu}=(\epsilon+p)u^\mu u^\nu - p g^{\mu\nu}$

There are 3 degrees of freedom $u^\mu$ because it must satisfy the identity $u_\mu u^\mu=-1$. The density of the particles is another degree of freedom per point in space-time, but there is a conservation of particle number constraint, which gives another constraint. To get well posed equations of motion, there must also be an equation of state, which relates the pressure/energy to the particle number. So to get well-posed equations of motion, it appears there should only be 3 independent equations of motion.

Also, see http://adsabs.harvard.edu/abs/1991A%26A...252..651G where they develop Einstein's equations in spherical symmetric space-time. They explicitly state that one of the dynamical equations of motion is redundant. At the end of section 5 of the paper: "Note that... it is possible to determine the energy density in the fluid frame, $e$, either by means of the equation of state (5), either from the knowledge of $\mathcal{E}$ via Eq. (10). The two values are exactly the same, provided that the pressure $p(s,n_B)$, which appears in the Eq. (25) for $\mathcal{E}$, is thermodynamically consistent with $e(s,n_B)$, i.e. that $p$ obeys Eq. (6)". Eq. (6) in the reference is just the first law of Thermodynamics.

So to reword:

In all the practical circumstances of which I am aware, $\nabla_\mu T^{\mu\nu}$ yields only 3 independent equations, with the 4th being redundant. Is this the case or are there situations where this is not true?

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  • $\begingroup$ Particle moving in a FRW spacetime would have a constant spatial 4-momentum components $p_i$ but its energy $p_0$ would be varying with time. So the answer to a title question is no. $\endgroup$ – A.V.S. May 29 '18 at 8:01
  • $\begingroup$ Thanks @A.V.S... but I'm not asking whether they are related in a simple way, but rather if the conservation of momentum implies conservation of energy. In any case, that was poorly worded. See the edit above for clarification of what I am trying to suss out. $\endgroup$ – juacala May 29 '18 at 20:05
  • $\begingroup$ The divergence of the stress-energy tensor (which is what I am concerned with) is completely local (it's the local interaction of the field with matter at a point). Both examples are of a field (generated by other particles) acting on a test particle. I don't see how that isn't like your example. $\endgroup$ – juacala May 29 '18 at 22:26
  • $\begingroup$ Covariantly for any system, both must be zero, $\nabla_\mu T^{\mu 0}=0$ and $\nabla_\mu T^{\mu i}=0$, in order for energy and momentum to be conserved. I don't think the ordinary derivatives have a meaning in curved spacetimes. $\endgroup$ – juacala May 30 '18 at 16:37
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Noether's theorem links energy conservation to invariance under time and momentum conservation to invariance under space translation of the Lagrangian. The two are therefore unrelated. As an example, for a particle moving in a space dependent but time independent potential energy is conserved while momentum is not.

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  • $\begingroup$ I'm not sure that you're comment is true, because time and position coordinates are not independent in relativity. In a different coordinate frame, the transformation will be mixed in a different way. In any case, the way I asked the question was poor. In any coordinate frame, the question is whether $\nabla_\mu T^{\mu\nu}=0$ yields 3 or 4 independent equations. $\endgroup$ – juacala May 29 '18 at 20:08
  • $\begingroup$ In a different coordinate system energy and momentum are defined differently. My answer is noncovariance, as is your question. The conservation law gives four independent conditions in this particular frame, so in all Lorentz frames. $\endgroup$ – my2cts May 29 '18 at 20:22
  • $\begingroup$ I'm still not convinced. Just because you get 4 separate equations from Noether's theorem doesn't guarantee they are all independent. I get 4 separate equations from $\nabla_\mu T^{\mu\nu}=0$, but the situations I showed only have 3 independent equations. $\endgroup$ – juacala May 29 '18 at 21:53
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Upon reading OP's responses to my comments I seem to understand the source of mutual confusion.

OP implicitly assumes general covariance for tensor equation, or Poincaré covariance if we are talking about flat spacetime. (For simplicity let us restrict further discussion to a flat spacetime).

If we assume Poincaré covariance of the energy momentum tensor, then yes, from equations for spatial components $\partial_\mu T^{\mu i} = 0 $, we can derive $\partial_\mu T^{\mu 0} = 0 $. The reason is simple: if a 4-vector has zero spatial components in all reference frames, then its time component also must be zero. Or in terms of Poincaré group: Lorentz invariant field action plus invariance with respect to spatial translations means invariance w.r.t. time shifts (since Poincaré algebra has $[M_{0i},P_i]=i P_0$).

However, if there is no Poincaré covariance then conservation of energy may no longer be a consequence of momentum conservation. For example, we could write Lagrangian density explicitly depending on time variable: $\mathcal{L}(\varphi,\dot{\varphi},\partial_i \varphi, t)$, then there would be conservation of momentum but not conservation of energy. Alternatively as my2cts suggested, if there is a background dependent on spatial coordinates (so that Lagrangian density has the form $\mathcal{L}(\varphi,\dot{\varphi},\partial_i \varphi, \mathbf{x})$) there would be energy conservation but not momentum conservation. Even when there is both energy and momentum conservation, in general the corresponding equations are independent since $ \frac {\partial {\mathcal {L}}}{\partial \varphi _{,i }}$ and $\frac {\partial {\mathcal {L}}}{\partial \dot\varphi }$ could have completely different structures.

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  • $\begingroup$ Thanks for the clarification. But the lagrangian density cannot depend explicitly on the time (space) variable, since then it would not manifestly be a Lorentz scalar density. If the lagrangian depends on $x^\mu$, your argument is less compelling. So I still don’t think your argument quite shows they are independent. $\endgroup$ – juacala May 30 '18 at 23:33
  • $\begingroup$ I see now. You are saying that for covariant theories there are only 3 equations and in noncovariant theories there are 4. I am only concerned with covariant theories. I think your argument about if 3 components of a 4-vector are zero in all frames then the 4th must also be zero is sufficient to show there are only 3 equations. Thanks. $\endgroup$ – juacala May 31 '18 at 0:07
  • $\begingroup$ I had marked this as the correct answer, but now I don't believe it is sufficient. @AVS makes the argument that if 3 components of a 4-vector are 0 in all reference frames, then the fourth must be. However, one could make the same argument with just 1 component: if 1 component of a 4-vector equation is zero in all reference frames, then all must be. But of course one gets at least 3 independent equations of motion (else all general relativistic fluid models would be unsolvable). $\endgroup$ – juacala Jun 1 '18 at 22:54

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