- #1

George Keeling

Gold Member

- 156

- 39

- TL;DR Summary
- What is energy density in the SR equation for energy-momentum tensor for a perfect fluid?

In Special Relativity I'm given the energy-momentum tensor for a perfect fluid:$$

T^{\mu\nu}=\left(\rho+p\right)U^\mu U^\nu+p\eta^{\mu\nu}

$$where ##\rho## is the energy density, ##p## is the pressure, ##U^\mu=\partial x^\mu/\partial\tau## is the four-velocity of the fluid. In the non-relativistic limit with $$

U^\mu=\left(1,v^i\right)\ \ ,\ \ v^i\ll1\ \ ,\ \ p\ll\rho

$$it is easy to get to a continuity equation for energy flow which is fine and dandy.

But I understand the first two non-relativistic limits but not really the third, ##p\ll\rho##. It is 'because pressure comes from random motions of the individual particles, and in this limit these are taken to be small.' I know what pressure is and it's about ##{10}^5\ kg\ m^{-1}s^{-2}## at sea level. But what is the energy density? Should I take the density of the kinetic energy of all the molecules of air? I calculate that at about ##{10}^3\ kg\ m^{-1}\ s^{-2}##. So ##p>\rho##. Must I include energy according to ##E=mc^2##? Then I get ##\rho={10}^{17}\ kg\ m^{-1}\ s^{-2}## which fits the bill nicely but it's not very non-relativistic . I realize that I am not really using the correct units (because ##c=1## here), nevertheless the comparison should work in any units.

T^{\mu\nu}=\left(\rho+p\right)U^\mu U^\nu+p\eta^{\mu\nu}

$$where ##\rho## is the energy density, ##p## is the pressure, ##U^\mu=\partial x^\mu/\partial\tau## is the four-velocity of the fluid. In the non-relativistic limit with $$

U^\mu=\left(1,v^i\right)\ \ ,\ \ v^i\ll1\ \ ,\ \ p\ll\rho

$$it is easy to get to a continuity equation for energy flow which is fine and dandy.

But I understand the first two non-relativistic limits but not really the third, ##p\ll\rho##. It is 'because pressure comes from random motions of the individual particles, and in this limit these are taken to be small.' I know what pressure is and it's about ##{10}^5\ kg\ m^{-1}s^{-2}## at sea level. But what is the energy density? Should I take the density of the kinetic energy of all the molecules of air? I calculate that at about ##{10}^3\ kg\ m^{-1}\ s^{-2}##. So ##p>\rho##. Must I include energy according to ##E=mc^2##? Then I get ##\rho={10}^{17}\ kg\ m^{-1}\ s^{-2}## which fits the bill nicely but it's not very non-relativistic . I realize that I am not really using the correct units (because ##c=1## here), nevertheless the comparison should work in any units.