Trying to confirm that the trace of the energy-momentum tensor divided by the energy density is NOT invariant

Question

I am analyzing this question in the FRW universe with a perfect fluid. The trace of the energy momentum tensor $$T^{\mu \nu} g_{\mu \nu} = \rho - 3p $$ is of course an invariant quantity. It does, however, in the case of an FRW metric, change with time as the universe expands.

Is it possible to define a quantity such that all co-moving observers will view it as an 'invariant mass squared' but that is not dependent on volume?

What about $$\frac{T^{\mu \nu} g_{\mu \nu}}{\rho}$$ The energy density, by itself, is not an invariant quantity. If we allow $p=\omega \rho$ (perfect fluid), then $$\frac{\rho - 3p}{\rho} = 1 - 3\omega $$ This quantity looks invariant to me, though I am not convinced this is correct. Would the lack of invariance be the result of $\omega$ being a non-constant function? If, for example, the universe were dust for a co-moving observer, then $\omega = 0$. But, if the observer is not co-moving, they will perceive a pressure in the dust the co-moving observer did not, forcing $$0 > \omega < 1/3$$ ^^ I realized that last sentence is wrong. If I were moving relative to the fluid, the fluid will have a bulk motion in the direction of movement. This is distinct from pressure which is a result of random movement in all directions. So no, a moving observer won't see 'pressure', they will perceive a momentum density flux.

Answer 1

$\omega$ is observer dependent. By an observer, I mean an unit time-like velocity vector field $u$ describing his trajectory. Let $\hat{u}$ and $u$ be two unit time like vectors such that $$ \hat{u}^a=\Gamma (u^a+v^a)$$ where $u^av_a=0$, $\Gamma=(1-v^2)^{-1/2}$. Now if energy density and pressure measured by $\hat{u}$ are $\hat{\mu}$ and $\hat{p}$ respectively, then wrt observer $u$, these quantities will be $$\mu=\hat{\mu}+\Gamma^2v^2(\hat{\mu}+\hat{p})$$ $$p=\hat{p}+\frac{1}{3}\Gamma^2v^2(\hat{\mu}+\hat{p})$$ Thus if $\frac{\hat{p}}{\hat{\mu}}=\omega$, then $\frac{p}{\mu}=\frac{\omega+\frac{1}{3}\Gamma^2v^2(1+\omega)}{1+\Gamma^2v^2(1+\omega)}$ where we clearly see the dependence on $\textit{relative velocity}$ $v$

Answer 2

As you said, it's not an invariant quantity. Since

$\rho -3p = T^{\mu \nu}g_{\mu \nu} \longrightarrow T^{\mu' \nu'}g_{\mu' \nu'} = T^{\mu \nu}g_{\mu \nu}\Lambda_{\mu}^{\mu'}\Lambda_{\nu}^{\nu'}\Lambda_{\mu'}^{\mu}\Lambda_{\nu}^{\nu'}=T^{\mu \nu}g_{\mu \nu} = \rho - 3p$

And since that $\rho \longrightarrow \rho^{'} = \Lambda^{0'}_{0}\rho$

the fraction goes to:

$\frac{\rho - 3p}{\rho}\longrightarrow\frac{\rho - 3p}{\Lambda^{0'}_{0}\rho} $

That obviously is not invariant unless the Lorentz transformation is a pure rotation, for which $\Lambda^{0'}_{0}=1$.

Even allowing for $p=\omega \rho$ won"t remove the $\Lambda$ factor in the denominator

Answer 3

Other existing answers assume that $ρ$ is the local energy density as measured by some observer. However, when discussing spacetimes with ideal fluid sources (FRW cosmologies being one class of such spacetimes) the usual interpretation is that $ρ$ is a fluid's energy density in its rest frame. This means that $ρ$ and $p$ are parameters of the stress-energy tensor decomposition: $$T^{αβ} \, = \left(ρ + p \right)u^α u^β + p\, g^{αβ},$$ where $u^α$ is the 4-velocity of the fluid's rest frame.

These quantities, $ρ$ and $p$ are observer independent, since they correspond to eigenvalues of $T^α{}_β$ and not to the tensor components in observer's reference frame. Recall, that eigenvalues could be obtained from the roots of characteristic polynomial, $\mathop{\mathrm{det}}(\lambda\delta^α_β-T^α_β)$, which is a scalar under Lorentz transformations, since it could be written using pair of Levi-Civita tensors.

So, any algebraic function of these scalar quantities would itself be invariant, including ratios $\omega=p/ρ$ or a quantity defined by OP $(ρ-3p)/ρ$. However, that last quantity is definitely not “invariant mass squared” by any interpretation.

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(Addition): Since there seems to be some confusion with the invariant character of $ρ$ and $p$, let us outline in more details how can we find these quantities at a given point in any coordinate system (without additional data).

1. Write the characteristic polynomial: $$ P(\lambda)=\mathop{\mathrm{det}}(\lambda\delta^i_j-T^i_j)=E_{ijkl}E^{qrst}(\lambda\delta^i_q-T^i_q)(\lambda\delta^j_r-T^j_r)(\lambda\delta^k_s-T^k_s)(\lambda\delta^l_t-T^l_t), $$ where $E^{…}$ is the Levi-Civita tensor. Note, that r.h.s. is a tensor expression that does not have any free indices, i.e. it is a scalar, which means that coefficients of this polynomial are the same in any reference frame/coordinate system and so are the roots.

2. Find the roots of characteristic polynomial $\lambda_i$ ($i=1..4$). If the stress–energy tensor belongs to an ideal fluid then those would be real and three of the roots would coincide. A single root (denote it as $λ_1$) would be the energy density in the fluid's rest frame: $\rho=\lambda_1$, while the triple root would be the pressure up to a sign: $p=-λ_2=-λ_3=-λ_4$.

That is all, there are no additional assumptions on reference frames, observes, spacetime etc.