Question about the Euler tensor for timelike and spacelike metricsignatures

[Jay R. Yablon]

Sent this a few days ago, has not posted yet:

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 p>0, that means T<0 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.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com

[John Baez]

In article <iZBVf.30395$jf2.11905@twister.nyroc.rr.com>,
Jay R. Yablon <jyablon@nycap.rr.com> wrote:

>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 -+++.

I'm a bit too lazy to work through this in detail, so all I can say is this.

If you work out a manifestly coordinate-independent formula like

T^uv=(u+p)U^uU^v + g_uv p

using some signature, and then you whimsically decide to change your
conventions regarding the signature, the formula will still be true
without any changes.

Proof: if this weren't true, particle physicists would have *killed off*
people working on general relativity long ago, because they would never
agree about the basic formulas of physics. The only way people can
tolerate the coexistence of two conventions, +--- and -+++, is that the
basic formulas are right either way.

So, I think your error was changing your equation to

T^uv=(u+p)U^uU^v - g_uv p

when you switched from +--- to -+++. The change in g_uv should do
precisely the right thing, with no extra fiddling.

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[Daryl McCullough]

John Baez says...

>If you work out a manifestly coordinate-independent formula like
>
>T^uv=(u+p)U^uU^v + g_uv p
>
>using some signature, and then you whimsically decide to change your
>conventions regarding the signature, the formula will still be true
>without any changes.

Yes, but something that's interesting is that when we deal
with Clifford algebras, the algebra for (+---) spacetime is
not the same as the algebra for (-+++) spacetime. I don't
think that that difference would allow us to say that we are
*really* one signature instead of another, though.

--
Daryl McCullough
Ithaca, NY

[Aaron Bergman]

In article <e4opag0219a@drn.newsguy.com>,
stevendaryl3016@yahoo.com (Daryl McCullough) wrote:

> John Baez says...
>
> >If you work out a manifestly coordinate-independent formula like
> >
> >T^uv=(u+p)U^uU^v + g_uv p
> >
> >using some signature, and then you whimsically decide to change your
> >conventions regarding the signature, the formula will still be true
> >without any changes.
>
> Yes, but something that's interesting is that when we deal
> with Clifford algebras, the algebra for (+---) spacetime is
> not the same as the algebra for (-+++) spacetime. I don't
> think that that difference would allow us to say that we are
> *really* one signature instead of another, though.

This just ends up being the fact that there are many (8, I think)
different choices for the extending the usual Spin(1,3) -> SO(1,3) cover
to a cover of all of O(1,3). Two of these choices come from the two
Clifford algebras Cl(1,3) and Cl(3,1), but thought of as just a choice
of cover, it has nothing to do with the signature.

Aaron

[Jay R. Yablon]

So how does one account for the fact that this tensor is written as
T^uv=(u+p)U^uU^v + g_uv p for -+++, see, e.g., Stehpani, "Exact
Solutions . . ." equation (5.9), and as T^uv=(u+p)U^uU^v - g_uv p for
+---, see, e.g., Einstein, ". . . The General Theory" (1916), equation
(58) (where u is used in place of u+p)?

What I find confusing are the trace relations: U^uU_u=-1 for -+++ while
U^uU_u=+1 for +---. Yet, g^us g_vs = lambda^u_v no matter what the
signature. Therefore, the trace of T^uv=(u+p)U^uU^v + g_uv p for -+++
is T = -u-p+4p = -u+3p while the trace of the same T^uv=(u+p)U^uU^v +
g_uv p for +--- is +u+p+4p=u+5p. But, the trace of T^uv=(u+p)U^uU^v -
g_uv for +--- is u+p-4p = u-3p which seems to make more sense because
u-3p = -(-u+3p). The u+p+4p=u+5p versus the u+p-4p = u-3p in the trace
equation is what I am trying to sort out.

Jay.

[Jay R. Yablon]

So how does one account for the fact that this tensor is written as
T^uv=(u+p)U^uU^v + g_uv p for -+++, see, e.g., Stehpani, "Exact
Solutions . . ." equation (5.9), and as T^uv=(u+p)U^uU^v - g_uv p for
+---, see, e.g., Einstein, ". . . The General Theory" (1916), equation
(58) (where u is used in place of u+p)?

What I find confusing are the trace relations: U^uU_u=-1 for -+++ while
U^uU_u=+1 for +---. Yet, g^us g_vs = lambda^u_v no matter what the
signature. Therefore, the trace of T^uv=(u+p)U^uU^v + g_uv p for -+++
is T = -u-p+4p = -u+3p while the trace of the same T^uv=(u+p)U^uU^v +
g_uv p for +--- is +u+p+4p=u+5p. But, the trace of T^uv=(u+p)U^uU^v -
g_uv for +--- is u+p-4p = u-3p which seems to make more sense because
u-3p = -(-u+3p). The u+p+4p=u+5p versus the u+p-4p = u-3p in the trace
equation is what I am trying to sort out.

Jay.