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

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.

$$\omega$$ 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 $$\textit{relative velocity}$$ $$v$$

• I think I mostly understand your answer now. You are able to talk about relative velocity between the two timelike vectors because in this case they exist in the same tangent plane. What is unclear is this: if they are both purely timelike, how can they have a relative velocity? Wouldn't they each have a spacelike component relative to the other? Commented Apr 26, 2023 at 12:41
• what is purely timelike ? All massive objects move in timelike trajectories, but we can define relative velocity b/w them..at least thats what we usually do in SR. Yes, when u take the difference, they will have space-like component which is encoded in $v$. This definition of relative velocity becomes ambiguous when we compare velocities b/w different tangent spaces in curved space-time
– KP99
Commented Apr 26, 2023 at 12:56
• I used the term timelike imprecisely. I see what you are saying now. Commented Apr 26, 2023 at 13:11
• No problem, The way u can arrive at the above expressions is that first u write $T_{ab}$ in terms of $\hat{u}$ vector, then write $\hat{u}$ in terms of $u$ and $v$ and simply read the coeeficients. For instance, the coefficients of $u_au_b$ will be $\mu$ etc. So actually, it becomes a local comparison of these quantities
– KP99
Commented Apr 26, 2023 at 13:15
• ω is observer dependent Wrong. $ω$ is calculated w.r.t fluid's rest frame. In other words it is ratio of eigenvalues and not of components of SE tensors. Commented Apr 26, 2023 at 17:39

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

• That was very clear, thanks. Are there any invariant quantities in an FRW universe that do not change with the volume? Commented Apr 26, 2023 at 12:20
• I'm not sure why you are using Lorentz transformation in FRW universe, though it doesn't change the answer qualitatively. Observer dependence is not strictly limited to only the tangent space (where you can apply Lorentz transformation)
– KP99
Commented Apr 26, 2023 at 12:26
• No, but there is no hope to have something invariant globally if it is not even invariant in a local setting. Commented Apr 26, 2023 at 13:32
• What i mean is: you cannot hope to have a scalar quantity invariant along a geodesic if that quantity is not even invariant in a single point of that geodesic, it's meaningless Commented Apr 26, 2023 at 13:36

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.

(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.

• "$\rho$ and $p$ are $\textit{observer independent}$", so how do you define observers in this scenario? The decomposition you have wrote is true if $u$ is divergent free, shear free etc. But if we just represent it in terms of tilted observer, we will get additional terms representing inhomogeneities...and certainly the parameters will change in that case. For reference, I'm using section 1.1.2 from "Dynamical Systems in Cosmology" by J.Wainwright, G.F.R. Ellis
– KP99
Commented Apr 26, 2023 at 18:58
• @KP99: so how do you define observers in this scenario? Observer independent means that we do not need to define them, the quantities can be calculated by any observer at a given point (divergence and shear are characteristics of system of multiple observers so it is not even relevant here). And I outlined calculation of ρ and p in more detail. Commented Apr 26, 2023 at 19:51
• Can you check the reference I mentioned in my comment above? I didn't get why divergence and shear are characteristics of system with multiple observers. The divergence and shear are terms which appear in the cov. derivative of $u$. So in this sense, its about one observer only...unless we are talking about different definitions
– KP99
Commented Apr 26, 2023 at 20:38
• @KP99: I didn't get why divergence and shear are characteristics of system with multiple observers. The divergence and shear are terms which appear in the cov. derivative of u. Precisely. Derivatives means observers at different (nearby) points, i.e. different observers. (OK, one observer evolves along its worldline, but to calculate div and shear we would need all directions not just one). Whereas decomposition of SE tensor depends only on one specific point. Commented Apr 27, 2023 at 3:13
• I see, so we are considering different definitions of observer: by an observer I don't mean just a single time-like trajectory, but a family of them which are integral curves of same vector field $u$. So when we take the derivative, we decompose it into components along $u$ and perpendicular to $u$ (which connects to adjacent trajectory and it is part of single observer definition). There can be many such vector fields $u$, if SET is ideal fluid wrt some $u$, then it need not be ideal for other choice of $u$ (tilted observer) , which I have showed in my answer, i.e. decomposition is not unique
– KP99
Commented Apr 27, 2023 at 3:31