\nu} &=& e_L u_L^\mu u_L^\nu - (P_L + \Pi_L) \Delta_L^{\mu \nu} + \pi_L^{\mu \nu}, \\
N^\mu &=& n_L u_L^\mu + V_L^\mu ,\end{aligned}$$ in the Landau frame and $$\begin{aligned}
T^{\mu \nu} &=& e_E u_E^\mu u_E^\nu - (P_E + \Pi_E) \Delta_E^{\mu \nu} \nonumber \\
&+& W_E^\mu u_E^\nu + W_E^\nu u_E^\mu + \pi_E^{\mu \nu}, \\
N^\mu &=& n_E u_E^\mu ,\end{aligned}$$ in the Eckart frame. Here $\Pi$ is the bulk pressure, $\pi^{\mu \nu}$ is the shear stress tensor, $W^\mu$ is the energy dissipation, $V^\mu$ is the baryon diffusion, and $\Delta^{\mu \nu} = g^{\mu \nu} - u^\mu u^\nu$ is the space-like projection. It can be immediately seen that the two frames become identical in the ideal hydrodynamic limit. The dissipative corrections to the energy and the conserved charge densities are neglected for simplicity [@Monnai:2018rgs]. Also I consider a system with a single charge conservation though the extension to general systems should be a straightforward task.

In the following arguments, the vector dissipative currents $W_E^\mu$ and $V_L^\mu$ are considered and the shear and bulk viscous corrections are set aside for simplicity. When the dissipative corrections are much smaller than the equilibrium variables, the difference in the thermodynamic variables of the two frames $\Delta n_{E-L} = n_E 