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I've run into a great deal of confusion on what I expected to be a very simple issue of covariance. I have an equation $$T^{\mu\nu}_{~~~~;\mu} = -G^{\nu}.$$ This is manifestly covariant; so far so good.

Now, let's apply a forward Euler update in time, which I would expect to maintain covariance to first order in $\Delta t$. I write this down in two frames (let's work in Minkowski space, with the two frames related by a Lorentz transformation; one lab frame has greek indices and time interval $\Delta t$, the other, moving with four-velocity $u^{\mu} = (\gamma, \gamma \beta, 0, 0)$ relative to the lab frame has (greek) indices and time interval $\Delta \tau = \Delta t/u^0$): $$\frac{\left(T^{0\nu}\right)^{n+1} - \left(T^{0\nu}\right)^{n}}{\Delta t} = -\left(G^{\nu}\right)^n - \left(T^{i\nu}_{~~~~,i}\right)^{n}$$ $$\frac{\left(T^{(0)(\nu)}\right)^{n+1} - \left(T^{(0)(\nu)}_{~~~~}\right)^{n}}{\Delta \tau} = -\left(G^{(\nu)}\right)^n - \left(T^{(i)(\nu)}_{~~~~,i}\right)^n$$ Now let's assume that I've already applied my advective source terms $T^{i\nu}_{~~~~,i}$ to get $(T^{i\nu}_{~~~~,i})^{n'}$ in a separate operator-split step (i.e. first order), so I just need to compute

$$\frac{\left(T^{0\nu}\right)^{n+1} - \left(T^{0\nu}\right)^{n'}}{\Delta t} = -\left(G^{\nu}\right)^{n'}$$ $$\frac{\left(T^{(0)(\nu)}\right)^{n+1} - \left(T^{(0)(\nu)}\right)^{n'}}{\Delta \tau} = -\left(G^{(\nu)}\right)^{n'}$$ Now, to illustrate the issue, let's specialize to a simple $T^{\mu\nu}$ for a relativistic gas moving with four-velocity $u^{\mu}$ i.e. at rest in the parentheses/comoving frame:

$$T^{(\mu)(\nu)} = \left( \begin{array}{cccc} u & 0 & 0 & 0\\ 0 & u/3 & 0 & 0\\ 0 & 0 & u/3 & 0 \\ 0 & 0 & 0 & u/3\end{array} \right) $$

$$T^{\mu\nu} = \left( \begin{array}{cccc} (\gamma^2 + \beta^2 \gamma^2/3) u & 4 \beta \gamma^2 u/3 & 0 & 0\\ 4 \beta \gamma^2 u/3 & (\beta^2 \gamma^2 + \gamma^2/3)u & 0 & 0\\ 0 & 0 & u/3 & 0 \\ 0 & 0 & 0 & u/3\end{array} \right) $$ and a simple $G^{\nu}$ (chosen to avoid affecting $T^{(i)(j)}$): $$G^{(\nu)} = (0,1,0,0)$$ $$G^{\nu} = (\beta \gamma, \gamma, 0, 0)$$ Now, let's apply this four-force in two ways. First, I'll apply $G^{\mu}$ in the lab frame, to get

$$\left(T^{\mu\nu}\right)^{n+1} = \left( \begin{array}{cccc} (\gamma^2 + \beta^2 \gamma^2/3) u - \beta \gamma \Delta t & 4 \beta \gamma^2 u/3 - \gamma \Delta t & 0 & 0\\ 4 \beta \gamma^2 u/3 - \gamma \Delta t & (\beta^2 \gamma^2 + \gamma^2/3)u & 0 & 0\\ 0 & 0 & u/3 & 0 \\ 0 & 0 & 0 & u/3\end{array} \right). $$

Second, I'll apply $G^{(\nu)}$ in the fluid frame and then transform the updated stress-energy tensor to the lab frame, $(T^{\mu\nu})^{n+1} = \Lambda^{\mu}_{(\nu)}\Lambda^{\nu}_{(\nu)}( T^{(\mu)(\nu)})^{n+1}$, to get

$$ \left( T^{\mu\nu}\right)^{n+1} = \left( \begin{array}{cccc} (\gamma^2 + \beta^2 \gamma^2/3) u - 2 \Delta \tau \beta \gamma^2& 4 \beta \gamma^2 u/3 - (1 + \beta^2)\gamma^2 \Delta \tau & 0 & 0\\ 4 \beta \gamma^2 u/3- (1 + \beta^2)\gamma^2 \Delta \tau & (\beta^2 \gamma^2 + \gamma^2/3)u - 2\beta \gamma^2 \Delta \tau & 0 & 0\\ 0 & 0 & u/3 & 0 \\ 0 & 0 & 0 & u/3\end{array} \right) $$

Now see that, even accounting for $\Delta \tau = \Delta t / \gamma$, these two expressions differ to first order in $\Delta t$, whereas I would expect to lose covariance only to $\mathcal{O}(\Delta t^2)$! Surely I must be missing something very obvious here?

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