If we consider a perfect relativistic fluid it has energy momentum tensor
$$T^{\mu \nu} = (\rho + p) U^\mu U^\nu + p\eta^{\mu \nu} $$
where ##U^\mu## is the four-velocity field of the fluid. ##\partial_\mu T^{\mu \nu} = 0## then
implies the relativistic continuity equation
$$\partial_\mu(\rho U^\mu) + p \partial_\mu U^\mu = 0$$
which reduces to the ordinary continuity equation for matter
$$\partial_t \rho + \nabla \cdot (\rho \vec v) = 0$$
in the non-relativistic limit ##v << c## and ##p << \rho##. Charge obeys an identical equation in the same limit, but does it has a relativistic analogue? If not, why?
$$T^{\mu \nu} = (\rho + p) U^\mu U^\nu + p\eta^{\mu \nu} $$
where ##U^\mu## is the four-velocity field of the fluid. ##\partial_\mu T^{\mu \nu} = 0## then
implies the relativistic continuity equation
$$\partial_\mu(\rho U^\mu) + p \partial_\mu U^\mu = 0$$
which reduces to the ordinary continuity equation for matter
$$\partial_t \rho + \nabla \cdot (\rho \vec v) = 0$$
in the non-relativistic limit ##v << c## and ##p << \rho##. Charge obeys an identical equation in the same limit, but does it has a relativistic analogue? If not, why?
0 commentaires:
Enregistrer un commentaire