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Here's something that's confusing me:

In the fluid approximation in General relativity we only have kinematic terms. We do not have the potential energy of each term. This is not a realistic description of a fluid because there are no collisions between the molecules (a potential energy term would ensure collisions). Why is it legal to throw out this potential energy? (If someone could supply an order of magnitude argument that would be great)

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  • $\begingroup$ when you say potential energy of each term, do you mean curvature associated to each particle? What kind of potential energy are you considering here? $\endgroup$
    – KP99
    Commented Oct 16 at 14:29
  • $\begingroup$ @KP99 I am not considering the curvature of each individual particle. What I simply mean is we have only considered the kinetic energy terms. The particles will go through each other. One would need a relativistic generalisation of the hardsphere model as an example of what I'm talking about. $\endgroup$ Commented Oct 16 at 16:35
  • $\begingroup$ Oh, I would say that the trajectories of fluid particles are integral curves of some time-like/null vector field and so you cannot have two trajectories crossing one another (otherwise the vector would have two directions at the point of collision). This is also true for ordinary fluids / any fluids , even for EM field lines etc $\endgroup$
    – KP99
    Commented Oct 16 at 16:56
  • $\begingroup$ I get you can construct this. What bothers me is you can make some approximations and it still corresponds to reality. Also collisions are discontinuties in the trajectory of motion so I wouldn't worry about the 2 direction argument. $\endgroup$ Commented Oct 16 at 17:21
  • $\begingroup$ A discontinuous trajectory won't be very ideal to work with. But you can extend Boltzmann transport equation on curved spacetime and study collision of particles in a distribution sense ( see sciencedirect.com/science/article/abs/pii/… ) $\endgroup$
    – KP99
    Commented Oct 16 at 17:32

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