Apologies for this dumb question. But why can we assume the trace of the stress energy tensor is 0?

theoretical-physics.net/dev/fluid-dynamics/general.html

@MoreAnonymous the trace of the stress-energy tensor is not 0 in general, although it is for special cases like CFTs

are you actually talking about the matrix $\mathbb T$ in section 7.1.2 of your link?

In that case you're looking for the "volumetric-deviatoric decomposition" of the stress (not stress-energy) tensor

The volumetric component ($-p\mathbb I$) is the trace part, because it contains the normal stress in each direction on an infinitesimal fluid packet at rest (it should be something like $\frac13\delta_{ij}\sum_k \sigma_{kk}$), and the trace is coordinate-invariant

Since the deviatoric component ($\mathbb T$) is the remaining part, it's clearly traceless

@NiharKarve Cheers ...

@NiharKarve This helps. Serves me right for skimming through

@NiharKarve any idea what the equation of state for our universe is?

Most famously the trace of the EM field SET is zero

But obviously for a perfect fluid, $\mathrm{Tr}(T) = \rho + 3 p$

@Slereah which should be a constant

right since the divergence of the stress energy tensor is 0

whats the value of that constant?

I think I've gotten something wrong

Why isnt the trace of the stress energy tensor constant?

Why would it be

If you consider a null dust for instance the trace is just $T = \rho$

Just given by the energy density over the spacetime

There's no reason it should be constant

but the divergence of the stress energy tensor is 0

and the divergence of the metric is 0

so by product rule?

Somewhere Im obviously doing something silly ... sigh

@Slereah Also can u not post stuff life this? Its disheartening to see this if anything

The divergence of the SET being zero is just implying the local convervation of currents in this case

it's just the classic $$\dot{\rho} = \nabla p$$

was thinking of this

$$ \nabla_\mu( T^{\mu \nu} g_{\mu \nu}) = $$

and realised my mistake

Am I the only one who when sometimes thinking of classical mechanics confuses himself?

And kind of embarrassed as well :3

yes, you are the only person in the world that ever has any confusion about physics :p

yes, speaking of

are cases when a classical model is "complete"

does it mean there is no quantum phenomena (even if a quantum model can accurately describe the scenario?)

or to put another way, is classical model incomplete even if it gives correct results for a given case

basically, is the world quantum only when it matters (ie. when classical fails)