# Calculating four-momentum from energy-momentum tensor

How does one calculate four-momentum from the energy-momentum tensor $T_{\mu\nu}$? Would it be the eigenvector of the four-momentum tensor? If so, which eigenvector should be considered (in case of multiple eigenvectors)? If the answer is the time-like eigenvector, would this eigenvector be unique or it's possible there might be more than one time-like eigenvector? Also, what is the physical interpretation of the other eigenvectors?

• The four-momentum of what given the energy-momentum tensor? – AccidentalFourierTransform Feb 17 '18 at 23:00
• The four-momentum of the mass corresponding to the energy-momentum tensor. – physics_2015 Feb 17 '18 at 23:02
• Sorry, I still don't understand what you mean. What is the four-momentum of a mass? What is the mass corresponding to an energy-momentum tensor? – AccidentalFourierTransform Feb 17 '18 at 23:02
• Let's say an object has the energy-momentum tensor $T_{\mu\nu}$. How does one calculate the four-momentum of this object from $T_{\mu\nu}$? – physics_2015 Feb 17 '18 at 23:05
• Point particles don't have an energy-momentum tensor; the latter is a field-theoretic concept. – AccidentalFourierTransform Feb 17 '18 at 23:11

You extract an energy-momentum vector from the stress-energy-momentum density by integration over a space-like 3-surface (whereas integration over a time-like surface would yield momentum fluxes). If your space-like surface is defined by the condition $t=\text{const}$, that means integrating the first column $T_{\mu0}$.
For point particles, momentum is a covector $$p_\mu\mathrm dx^\mu$$ For continuous media (fields and fluids), we have to promote this to a momentum density $$\rho_\mu\mathrm dx^\mu\otimes\mathrm dV$$ where $$\mathrm dV = \mathrm dx\wedge\mathrm dy\wedge\mathrm dz$$ is the 3-dimensional volume element. That is no longer an invariant object, as a change of frame will mix-in $$\mathrm dt\wedge\mathrm dx\wedge\mathrm dy\qquad \mathrm dt\wedge\mathrm dx\wedge\mathrm dz\qquad \mathrm dt\wedge\mathrm dy\wedge\mathrm dz$$ By purely geometric reasoning, the energy-momentum vector thus has to be promoted to a tensor $$T_{\mu\nu}\mathrm dx^\mu\otimes\star\mathrm dx^\nu$$ where we have represented the surface elements via their normal vectors by Hodge duality.
• OK. Thanks. I'm still not very clear. Let's say we have a point particle. Does it mean we have to integrate $T_{\mu0}$ of this particle to get its four-momentum? If so, over which volume one has to integrate? – physics_2015 Feb 18 '18 at 0:17