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How is the stress energy tensor obtained? In most textbooks, it's simply stated as

$$T^\mu{}_\nu=(\rho+P)U^\mu U_\nu-P\delta^\mu{}_\nu$$

I can see why this makes sense for a comoving observer at rest wrt. the perfect fluid. But I don't understand how the general case is arrived at. Would someone kindly explain?

Qmechanic
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Alex
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  • Answers here: https://physics.stackexchange.com/q/411934/226902 and https://physics.stackexchange.com/q/365940/226902. Lagrangian for perfect fluid; https://physics.stackexchange.com/q/308613/226902 – Quillo Jun 27 '23 at 15:11

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The answer to this depends on what you're starting from. If you know the Einstein tensor, then you can find the stress-energy tensor from the Einstein field equations. If you know the Lagrangian density, then you can find the stress-energy tensor by variation with respect to the metric. If you know the rate at which energy-momentum is being transported along four orthogonal axes, then that corresponds to the stress-energy tensor.

  • Thank you, Ben, would you mind telling me what the Lagrangian density is? And/or notes on how I might do the variation wrt the metric? Also, are there any links about the transport along 4 orthogonal axes -- I am not at all familiar with this approach? Thanks again! – Alex Oct 20 '13 at 10:02
  • @Alex: For transport of momentum, see the WP article, http://en.wikipedia.org/wiki/Stress-energy_tensor . I don't have a good reference handy for the variation with respect to the metric. –  Oct 20 '13 at 14:29