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I haven't been able to find a simple self-contained definition of the stress-energy tensor as used in the Einstein field equations.

Suppose I have $N$ classical (not quantum) point particles with position $x_i$, velocity $v_i$ and mass $m_i$, with forces $F_i(x_1, v_1,...,x_N,v_N)$ in some appropriate reference frame.

How would a physicist then compute the components of $T$? Please state it in the simplest possible self-contained way (i.e. no analogies, no references to elsewhere-defined concepts like flux).

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  • $\begingroup$ Related: physics.stackexchange.com/q/303773/2451 $\endgroup$
    – Qmechanic ♦
    Commented May 24 at 10:17
  • $\begingroup$ @Qmechanic, that question contains a lot of extra terms and formalisms so that I would say it's not a self-contained definition of the stress-energy tensor. As a non-physicist I find it hard to extract a simple definition from it. Ideally I'd like it to be so self-contained that I could just directly turn it into a python program that takes the states of the particles and forces/potential-energy, and outputs a stress-energy tensor (except the python program would have to choose a discretization but ignore that). $\endgroup$
    – user56834
    Commented May 24 at 11:16
  • $\begingroup$ Out of curiosity, what is your interest in the stress-energy tensor as a non-physicist? $\endgroup$
    – my2cts
    Commented May 25 at 9:11
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    $\begingroup$ @my2cts, just trying to understand the Einstein field equations to some extent :) $\endgroup$
    – user56834
    Commented May 26 at 4:42

2 Answers 2

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The energy-momentum tensor of a set of point massive particles (in special relativity), not including energy and momentum in the field due to their interaction, is, by definition,

$$ T_{matter}^{\mu\nu}(\mathbf x,t) = \sum_a m_a \frac{1}{\gamma_a}\delta\big(\mathbf x - \mathbf r_a(t)\big) u_a{}^\mu u_a{}^\nu $$ where $\gamma_a$ is the Lorentz factor of $a$-th particle $$ \gamma_a = \frac{1}{\sqrt{1-\frac{v_a^2}{c^2}}} $$ and $u_a$ is four-velocity of $a$-th particle: $$ u_a{}^0 = \gamma_a c $$ $$ u_a{}^k = \gamma_a v^k $$ or, abusing the index notation, $$ u_a{}^\mu = (\gamma_a c,\gamma_a \mathbf v_a). $$

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  • $\begingroup$ I assume $\delta(x-r_a)$ is the dirac-delta? Also, is the last equation supposed to be $u_a=..$? I thought $u_a^\mu$ is a real scalar. Also, why don't we just cancel out the $\gamma_a$? we divide by $\gamma_a$ in the main equation and then multiply in the four-velocity. $\endgroup$
    – user56834
    Commented May 24 at 13:04
  • $\begingroup$ Yes, it's 3D delta function centered at the particle. $u_a$ is a four-vector,$u_a{}^\mu$ is its $\mu$-th component. $\endgroup$
    – Ján Lalinský
    Commented May 24 at 13:06
  • $\begingroup$ How hard would it be to also define the energy momentum tensor due to their interactions? In the classical case it would be the energy-momentum tensor that results from the potential function $V(x_1,v_1,...,x_N,v_N), if we can still write it that way in GR? $\endgroup$
    – user56834
    Commented May 24 at 13:07
  • $\begingroup$ If $u_a^\mu$ is just a component then why do you define it as a tuple? $\endgroup$
    – user56834
    Commented May 24 at 13:07
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    $\begingroup$ @my2cts I don't have a great reference, after a while I found this paper where they demonstrate this kind of formulae doi.org/10.1007/s10773-017-3489-1 . Treatment of EM field energy and momentum in sections 5,6 is however wrong for point particles. Consistent theory of point particles should provide well-behaved energy-momentum tensor for EM interaction (with terms that integrate to finite numbers). In Frenkelian theories, there is one, see the formula (1) in Stabler's paper doi.org/10.1016/S0031-9163(64)91989-4 $\endgroup$
    – Ján Lalinský
    Commented May 25 at 23:01
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The expression for SET which Ján provides, describes a system of free N particles. And because they are free, they will satisfy the usual conservation rule for energy, momentum and angular momentum. As Ján mentions in the comment, if we now add an interaction force $F_i(\{x_j\}, \{v_j\})$, then the conservation equations need to be modified. For eg, the momentum equation will look like: $$\frac{d}{dt}\sum_i \rho v^a_i = f^a_i$$ where $f_i$ is the interaction force density on i-th particle. The LHS of the above equation can be thought of as $\partial_0T^{a0}_{free}$, where $T^{ab}_{free}$ is the SET of free particles. In general, if you have some interaction terms, then one can generalize the continuity equation for SET as $$\partial_bT^{ab}=F^a$$ As an example, look at divergence of SET for EM in presence of some 4-current. If one can write $F^a = \partial_bV^{ab}$, then the new SET defined as $T^{ab}-V^{ab}$ will satisfy the continuity equation

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  • $\begingroup$ Could you add an actual final definition of the stress-energy tensor? Note I'm not sure what the "continuity equation" is supposed to do. $\endgroup$
    – user56834
    Commented May 26 at 4:47
  • $\begingroup$ You can find few definitions of SET from the wiki link (en.m.wikipedia.org/wiki/Stress%E2%80%93energy_tensor) : like the canonical definition, or Einstein-Hilbert definition as well as Belinfante-Rosenfeld modifications. They require you to start from a Lagrangian theory. Also, recently I found another definition using some phase space distribution function : $T^{ab}=\int dP p^ap^bF(x, p) $ where $F(x, p)$ is the measure of number of particles in a small spatial volume dV centered about x and small momentum volume dP centered about p $\endgroup$
    – KP99
    Commented May 26 at 8:24

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