[Widdekind]

As observed by AJB in a recent blog entry,

R^a_a = g^abR_ba = R

T^a_a = g^abT_ba = T

R = -T

Now, this site seems to say, that diag(g^ab) = (1, -1, -1, -1); and, that T = diag(T^a_a) = p ( c² - v_x² - v_y² - v_z²) = p (c² - v²). So, T > 0 ? And so, R < 0 ??

In trying to calculate R_ab, from the definition:

I thought I found, that many terms cancelled, when one of the dummy indexes was set equal, to one of the formal indexes; that led me to an "einstein strike notation" that simplified the expressions, e.g. with a piece like:

g^ba,b(g_ab,a - g_ba,a) - g^ba,b(g_ab,a-g_ba,a)

wherein a dummy index a, b runs over every index, except a, b.

[elfmotat]

Widdekind, on 1 February 2012 - 09:09 PM, said:

R = -T

Yes, this is true with appropriately chosen units.

Widdekind, on 1 February 2012 - 09:09 PM, said:

Now, this site seems to say, that diag(g^ab) = (1, -1, -1, -1);

First off all, the (inverse) metric is not simply diag(1,-1,-1,-1) in general. The metric is related to local matter/energy distribution by the field equations. The Minkowski metric, η_μν, which represents flat spacetime is given by diag(1,-1,-1,-1). The Minkowski metric is a special case of the general metric in a vacuum (i.e. T^μν=0).


Now, on to a bigger point: the signs actually depend on your choice of metric signature. You can use the signature (+,-,-,-) as the above author did, or you can use the signature (-,+,+,+). If you use (-,+,+,+), the Minkowski metric becomes: η_μν=diag(-1,1,1,1). The choice is completely arbitrary. As long as you pick one and stick with it, all of the consequences work out the same. Most people working with General Relativity tend to use the (-,+,+,+) signature simply because it's easier to deal with one minus than than three. People doing more advanced work, such as work involving spinors, tend to use (+,-,-,-) because it's easier to work with in those fields.

Widdekind, on 1 February 2012 - 09:09 PM, said:

and, that T = diag(T^a_a) = p ( c² - v_x² - v_y² - v_z²) = p (c² - v²).

T isn't diag(T^μ_μ), T=T^μ_μ. The "diag()" would imply that the trace of the SET is still a tensor, which it is not. T is a scalar.

Widdekind, on 1 February 2012 - 09:09 PM, said:

So, T > 0 ? And so, R < 0 ??

The sign of T and R depend on your metric signature. Using (+,-,-,-), T>0 and R<0. Using (-,+,+,+), T<0 and R>0. There is no "correct" choice.

Widdekind, on 1 February 2012 - 09:09 PM, said:

In trying to calculate R_ab, from the definition:

I thought I found, that many terms cancelled, when one of the dummy indexes was set equal, to one of the formal indexes; that led me to an "einstein strike notation" that simplified the expressions, e.g. with a piece like:

g^ba,b(g_ab,a - g_ba,a) - g^ba,b(g_ab,a-g_ba,a)

wherein a dummy index a, b runs over every index, except a, b.

I'm not sure what your question is here.

[elfmotat]

elfmotat, on 12 February 2012 - 10:50 PM, said:

T isn't diag(T^μ_μ), T=T^μ_μ. The "diag()" would imply that the trace of the SET is still a tensor, which it is not. T is a scalar.

To avoid confusion, I would like to amend the above to the following: The "diag()" would imply that the trace of the SET is still a rank 2 tensor, which it is not. T is a scalar - a rank 0 tensor.