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My understanding of the stress-energy tensor in special relativity (or in general relativity), is that it gives you the flux density of 4-momentum flowing through an oriented 3D hypersurface. So at some point (event) $P$ in spacetime in an inertial frame in SR (or using geodesic coordinates in GR), such that the metric is simply $\eta_{\mu\nu}$ at $P$, and working in relativistic units ($c=1$), I'd expect the 4-momentum $dp^{\alpha}$ flowing through a hypersurface with 3-volume $dV$, oriented along the normal 1-form $n_{\beta}$ to be:

$$dp^{\alpha}=T^{\alpha\beta} n_{\beta} dV$$

Now, I'd like to consider a perfect fluid with $T^{\alpha\beta}=(\rho_0+p_s)u^{\alpha}u^{\beta}+\eta^{\alpha\beta}p_s$ (with rest mass-energy density $\rho_0$, static pressure $p_s$ and 4-velocity flow field $u^{\alpha}$). In particular, I want to consider the situation in a locally inertial co-moving rest frame at $P$, i.e. where $u^{\alpha}=\delta^{\alpha}_0$. If I now consider an infinitesimal hypersurface at $P$, oriented along the $x^1$ coordinate axis, i.e. $n_{\beta}=\delta^1_{\beta}$ and $dV=dx^0 dx^2 dx^3$, the total 4-momentum flowing through the 2D spatial surface with dimensions $dx^2 dx^3$, during time $dx^0$ should be: $$dp^{\alpha}=[(\rho_0+p_s)\delta^{\alpha}_0\delta^{\beta}_0+\eta^{\alpha\beta}p_s]\delta_{\beta}^1dx^0 dx^2 dx^3$$ $$=[(\rho_0+p_s)\delta^{\alpha}_0\delta^1_0+\eta^{\alpha 1}p_s]dx^0 dx^2 dx^3$$ $$=\eta^{\alpha 1}p_s dx^0 dx^2 dx^3$$

Since the metric is diagonal, this implies that the total 4-momentum, flowing through $dx^0 dx^2 dx^3$ has only one component, $dp^1=p_s dx^0 dx^2 dx^3$, i.e. it is entirely spacelike. How can this be? I would have expected any 4-momentum to be timelike, for a material fluid such as the one I'm considering here. Am I fundamentally misunderstanding the nature of the stress-energy tensor? I'd be grateful for any inisghts into my dilemma :)

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2 Answers 2

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Short answer: Pressure is flow of momentum.

For your energy-momentum tensor, spacelike flows are $\sim p$, timelike flow is $\sim\rho$. This makes intuitive sense: If you take a small 3-volume $[t,t+\epsilon]\times [x^2,x^2+\epsilon]\times[x^3,x^3+\epsilon]$ in your hypersurface, your pressure is transmitting a momentum (call momentum $P$ to distinguish from pressure $p$, and somewhat heuristically) $$\int\limits _t^{t+\epsilon}\int\limits _{x^2}^{x^2+\epsilon}\int\limits _{x^3}^{x^3+\epsilon} p \;\text{d}x^2 \text{d}x^3 \text{d}t = \int\limits _{t}^{t+\epsilon} F \;\text{d}t = \int\limits _{t}^{t+\epsilon} \dot P \;\text{d}t = \Delta P\,.$$ Here I've used that $\int p\, \text{d}A$ is a force $F$, and force is change in momentum, $F=\dot P$.

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  • $\begingroup$ Well yes, that's what my calculation also suggests: the momentum flow due to pressure is "purely spacelike momentum". But I don't understand this. How can there be such a thing is "purely spacelike momentum", however it is induced? Microscopically, the momentum associated with the pressure is the random momentum of the fluid particles. But surely those are all timelike, so why does all the timelike momentum not ALSO flow through the spacelike hypersurface? $\endgroup$ Aug 27, 2020 at 11:53
  • $\begingroup$ The momentum isn't purely spacelike, you're just projecting onto a spacelike component. Also, you have chosen a comoving frame, so $T^{\mu\nu}$ is diagonal. For teh energy-momentum tensor of a single particle, $T^{01}\neq0$ if it moves in $x^1$ direction, and it tells you how much energy moves,i.e. the "flow of the timelike poart of the four-momentum". $\endgroup$
    – Toffomat
    Aug 27, 2020 at 12:57
  • $\begingroup$ Actually, working this out in the case of a fluid moving in the negative $x^1$ direction, lead me to a 4-momentum flow with a negative $x^0$ component. That too threw me for a moment. I think I should be thinking in terms of "net" momentum flow, as in outflowing minus inflowing, as I outlined in my answer. Perhaps that's what you implicitly meant by momentum "flow"? $\endgroup$ Aug 27, 2020 at 15:21
  • $\begingroup$ @JamesBates Sure! $\endgroup$
    – Toffomat
    Aug 27, 2020 at 15:44
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I guess I should modify my understanding of the momentum-flow represented by the stress-energy tensor: I should take $T^{\alpha\beta} n_{\beta} dV$ to be the net 4-momentum flowing out through the hypersurface normal to $n_{\beta}$. That is to say, all the 4-momentum flowing out of the hypersurface minus all the 4-momentum flowing in from the other side:

$$dp^{\alpha}_{net}=dp^{\alpha}_{out}-dp^{\alpha}_{in}=T^{\alpha\beta} n_{\beta} dV$$

If $n_{\beta}$ is timelike, then $dp^{\alpha}_{in}=0$ since no 4-momentum associated with matter flows backwards in time. Likewise, for dust without any static pressure ($T^{\alpha\beta}=\rho_o u^\alpha u^\beta$) in a frame where the fluid moves in the positive $x^1$ direction ($u^1 > 0$), momentum flows out of a hypersurface perpendicular to the $x^1$ direction (both $dp^0_{out}$ and $dp^1_{out} \ne 0$), but none flows in ($dp^{0,1}_{in}=0$). If the fluid moves in the negative $x^1$ direction, then no momentum flows out through a hypersurface perpendicular to positive $x_1$ ($dp^\alpha_{out}=0$), but momentum does flow in. In this case, $dp^0_{in}>0$ and $dp^1_{in}<0$, so I should expect $dp^0_{net}=-dp^0_{in}<0$ and $dp^1_{net}=-dp^1_{in}>0$. This is indeed born out by $dp^{\alpha}_{net}=T^{\alpha\beta} n_{\beta} dV=\rho_o u^{\alpha}u^{\beta} n_{\beta} dV$ for $u^0>0$, $u^1<1$ and $n_1=+1$.

Only in the case of a non-zero static pressure accross a spacelike hypersurface, is there a situation with both outflowing momentum and inflowing momentum. On average, for each fluid particle flowing out of the surface with positive $x^1$ momentum (and positive $x^0$ momentum, i.e. energy), there'll be a particle flowing into the surface with equal positive $x^0$ momentum (i.e. energy), but opposite, negative $x^1$ momentum. So in the net momentum, the $x^0$ momenta cancel out (no net gain in energy), but the $x^1$ momenta add up, leading to a purely spatial net momentum transfer.

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