Consider the Einstein Field equations $$R_{\mu\nu}-\frac12Rg_{\mu\nu}=\frac{8\pi G}{c^4}T_{\mu\nu}.$$ Typically, the stress energy tensor $T_{\mu\nu}$ is assumed to be a perfect fluid, which implies that $T_{\mu\nu}$ takes the form $$T_{\mu\nu}=(\rho+p)U_{\mu}U_{\nu}+pg_{\mu\nu},$$ which implies that $T_{00}=\rho$ and $T_{ii}=p$. Furthermore the Einstein field equations can be rearranged to solve for the stress energy tensor if you first specify a metric. Why is it that in nearly all cases we assume the stress energy tensor takes on the form of a perfect fluid and or dust, and what happens when we are trying to model something that is not a perfect fluid, i.e How do we know what stress energy tensor to choose given a certain situation. For example a wormhole, wormholes are modeled by anisotropic fluids.
1 Answers
"Why is it that in nearly all cases we assume the stress energy tensor takes on the form of a perfect fluid and or dust?"
Because in that case the non-linear Einstein's field equations reduce to ordinary linear differential equations: a homogeneous second-order on $\sqrt{g_{tt}}$ and a non-homogeneous first order on $g_{rr}^{-1}$. See for example https://physics.stackexchange.com/a/679431/281096.
"How do we know what stress energy tensor to choose given a certain situation?"
A good orientation is to choose the stress-energy tensor which is compatible with Special Relativity Theory (SRT). Otherwise, with some restrictions (energy conditions), you are free to define it. You may like to read the answer to similar question on this platform, too, see https://physics.stackexchange.com/a/90323/281096 .
- 1,831
-
How do we come about these energy conditions? – aygx Feb 10 '22 at 16:14
-
See for example papers of Prado Martin-Moruno and Matt Visser "Classical and semi-classical energy conditions", https://arxiv.org/abs/1702.05915, or better, Carlos Barcelo and Matt Visser "Twilight for the energy conditons?", https://arxiv.org/abs/gr-qc/0205066 , if you mean sources on this topic. – JanG Feb 10 '22 at 19:02
-
If these papers above are not quite what you look for, I would advice to read Dennis Lehmkuhl article entitled "General Relativity as a Hybrid theory: The Genesis of Einstein’s work on the problem of motion", https://arxiv.org/abs/1803.09872v1 . where Einstein's view on stress-energy tensor is presented. – JanG Feb 10 '22 at 19:18
-
Thank you for the sources, for the energy conditions, I just expand the summations and do the necessary calculus involved and that will be my energy momentum tensor correct? – aygx Feb 11 '22 at 13:13
-
Yes, it should be. Show later the result here. – JanG Feb 11 '22 at 13:16
-
Will do. When expanding our summations, how would we get the summations out of component form? For instance: $$T_{\mu\nu}t^{\mu}T^{nu}=T_{00}t^0t^0+T_{01}t^0t^1+...+T_{33}t^3t^3$$. Like how do we Identify the values of the vector fields and the components of the Stress energy tensor if we dont have a general expression to use? – aygx Feb 11 '22 at 13:49
-
Sorry, but I do not understand what you try to accomplish. Try to explain. – JanG Feb 11 '22 at 16:27
-
My apologies for the unclear question. The definition for a perfect fluid follows as $$T_{\mu\nu}=(\rho + p)U_{\mu}U_{\nu}-pg_{\mu\nu}$$ . The energy conditions follow as $T_{\mu\nu}V^{\mu}V^{\nu} \ge 0$ for any time vector and so on for null vectors. When we expand those sums we get $$T_{\mu\nu}V^{\mu}V^{\nu}=T_{00}V^0V^0+T_{01}V^0V^1+...+T_{33}V^3V^3.$$ If we can construct the tensor from the energy conditions, that implies we dont know what form the tensor takes. Each summation term has a component of $T$ in it. How can I find the values of those components and the values of V? – aygx Feb 11 '22 at 18:19
-
No, from energy conditions one cannot derive the energy-stress tensor. You can postulate some $T_{\mu \nu}$ or you can postulate one of the metric components and calculate the second (see my first link) getting whole metric tensor. Using it you can calculate pressure and energy density and thus the tensor $T_{\mu \nu}$. First then you can check what energy condition it satisfies, if any. – JanG Feb 11 '22 at 19:04