General Tensor contraction: Trace of Energy-Momentum Tensor (Einstein metric)

tetris11
Messages
22
Reaction score
0
Okay so I have:

Eqn1) Tij=\rhouiuj-phij = \rhouiuj-p(gij-uiuj)

Where Tij is the energy-momentum tensor, being approximated as a fluid with \rho as the energy density and p as the pressure in the medium.


My problem:
Eqn2) Trace(T) = Tii = gijTij = \rho-3p

My attempt:

Tr(T) = Tii = gij[\rhouiuj-p(gij-uiuj)]
= [\rhogijuiuj-pgijgij+pgijuiuj)]
= \rhou - p + pu

which doesn't equal rho-3p (eqn2) as required, so I've done something wrong.
I think I've contracted incorrectly but I don't know why... please help?
 
Physics news on Phys.org
What are the following?

g^{i j} g_{i j} = ?

g^{i j} u_i u_j = ?
 
Well,
g^{i j} u_i u_j = 1

g^{i j} g_{i j} = ??
uh... g? or 0?

Might need to help me out here, maths isn't my first language...
 
Cheers man, that actually makes complete sense - but just for the record:

gij gij = δii = n, where n is number of dimensions?

I'm just wondering how you knew it was four without knowing how dimensions it was.
Tensors aren't all 4-d, right?
 
tetris11 said:
Cheers man, that actually makes complete sense - but just for the record:

gij gij = δii = n, where n is number of dimensions?

I'm just wondering how you knew it was four without knowing how dimensions it was.
Tensors aren't all 4-d, right?

Well this is the 'Special & General Relativity' board so it was probably just a good guess?
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
Back
Top