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Question about the Euler tensor for timelike and spacelike metric signatures



 
 
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Old March 25th 06 posted to sci.physics,sci.physics.electromag,sci.physics.particle,sci.physics.relativity
Jay R. Yablon
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Default Question about the Euler tensor for timelike and spacelike metric signatures

I have been working with the Euler (perfect fluid) tensor lately, and had a
question about the formulation of this tensor in relation to the spacetime
metric signature.

For a spacelike signature -+++, this tensor is T^uv=(u+p)U^uU^v + g_uv p,
where u is dust density, p is pressure density, and U^u is the velocity four
vector, and U^uU_u=-1 on account of signature -+++.

The trace T=T^u_u of the above is then T = -u-p+4p = -u+3p. If we have an
equation of state u=p ("stiff matter"), then T=2p=2u, and then the above
tensor becomes T^uv=T U^uU^v + (1/2) g_uv T

For a timelike signature +---, I believe (please help me here) that we must
write T^uv=(u+p)U^uU^v - g_uv p, that is, the g_uv p term now has a minus
sign and U^uU_u=+1 on account of signature -+++.

The Kronecker delta is the same no matter what, so the trace is now T=u+p-4p
= u-3p. For the same equation of state u=p for stiff matter, then T=-2p=-2u
and the tensor then becomes T^uv=-T U^uU^v + (1/2) g_uv T.

Since the dust density p0, that means T0 for the energy tensor where the
metric has a timelike signature +---, and it seems odd to me to have a trace
density 0.

Am I missing something here?

Maybe I should use T^uv=-(u+p)U^uU^v + g_uv p for the timelike signature
+---? Then T=-u-p+4p = -u+3p just as in the spacelike case -+++, then, for
stiff matter u=p, the trace is T=2p=2u.

Then, the final tensor becomes T^uv=-TU^uU^v + (1/2) g_uv T.

What is the right answer / approach here?

Thanks.

Jay.
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