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4:50 AM
Apologies for this dumb question. But why can we assume the trace of the stress energy tensor is 0?
 
5:42 AM
@MoreAnonymous the trace of the stress-energy tensor is not 0 in general, although it is for special cases like CFTs
 
5:52 AM
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
 
 
3 hours later…
8:46 AM
@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?
 
 
6 hours later…
3:02 PM
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
 
4:13 PM
The divergence of the SET being zero is just implying the local convervation of currents in this case
it's just the classic $$\dot{\rho} = \nabla p$$
Or something like that
 
 
1 hour later…
5:17 PM
was thinking of this
$$ \nabla_\mu( T^{\mu \nu} g_{\mu \nu}) = $$
and realised my mistake
 
5:55 PM
Am I the only one who when sometimes thinking of classical mechanics confuses himself?
And kind of embarrassed as well :3
 
 
5 hours later…
fqq
11:22 PM
yes, you are the only person in the world that ever has any confusion about physics :p
5
 
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)
 

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