On Dec 3, 2007 8:51 PM, Viktor T. Toth <vttoth at vttoth.com> wrote:
> OK, my understanding is that what Maxima is doing is correct, though
> "correct", when it comes to sign conventions, is often in the eye of the
> beholder.
>
> The Ricci-tensor is formed by contracting the Riemann tensor's contravariant
> index with one of its antisymmetric covariant indices. Which one? This is a
> matter of convention, since either way, we get the same physics, but MTW
> suggest that the convention should be such that, for a sphere, the scalar
> curvature should be positive.
>
Yes, it's a sign convention problem. In books on general relativity,
there are different conventions about Riemann tensor, which
differentiate each other by a sign. For example, the book written by
Steven Weinberg and the book written by Bernard F. Schutz use
different conventions. However when contracting the Riemann tensor to
the Ricci tensor, they use the same convention, which is
corresponding to contracting the second index with the fourth index in
ctensor.
So if you want to get a positive scalar curvature for a sphere, I
think it would be better to change the convention of the Riemann
tensor instead of the contracting convention in order to be consistent
with books on general relativity.
> In ctensor, the antisymmetric indices of the Riemann tensor are the second
> and third indices. Maxima forms the Ricci tensor by contracting the third
> index with the (contravariant) fourth index. Let's see if it makes good
> sense as per MTW. Do we get a positive scalar curvature for the
> 2-dimensional sphere?
>
> load(ctensor);
> ct_coords:[u,v];
> dim:2;
> lg:ident(2);
> lg[1,1]:R^2*cos(v)^2;
> lg[2,2]:R^2;
> cmetric();
> christof(all);
> riemann(true);
> ricci(true);
> scurvature();
> ratsimp(sum(sum(ric[i,j]*ug[j,i],j,1,dim),i,1,dim));
>
> Either calculated by hand (last line), or using the built-in scurvature
> function, I get 2/R^2, which is positive definite. This suggests that we
> have the "correct" choice of sign for the Ricci tensor.
>
> Looking at FLRW, to determine if we get the "correct" sign, let me just
> calculate the tt component of the Einstein tensor using a metric signature
> of [+,-,-,-]. First, however, let's check if the built-in ricci function
> gives the same result as I would get by hand:
>
> load(ctensor);
> ct_coords:[t,x,y,z];
> depends(a,t);
> derivabbrev:true;
> lg:-ident(4)*a^2;
> lg[1,1]:1;
> cmetric();
> christof(all);
> riemann(true);
> ricci(true);
> ratsimp(sum(riem[1,1,i,i],i,1,dim)-ric[1,1]);
>
> The last line gives 0; in other words, both ric[1,1] and explicit
> calculation of the tt component of the covariant Ricci tensor give
> -3\ddot{a}/a.
>
> The tt component of the Einstein tensor should be positive as it coincides
> with the energy density:
>
> einstein(true);
>
> I get 3\dot{a}^2/a^2, which is again positive definite. I can recalculate
> this by hand, too, forming the mixed index Ricci tensor first:
>
> for i thru dim do for j thru dim do mric[i,j]:sum(ric[i,k]*ug[k,j],k,1,dim);
> ratsimp(mric[1,1]-1/2*scurvature());
>
>
> So, I believe we are using the "correct" sign conventions here, insofar that
> quantities that we normally view as positive are, in fact, positive, but I
> admit that it has been (and remains) a source of great confusion for me as I
> tried to sort things out, so please feel free to point out if I missed
> something.
>
>
> Viktor
>
>
>
>
>
>
> -----Original Message-----
> From: Evan [mailto:evanxxx at gmail.com]
> Sent: Sunday, December 02, 2007 11:22 PM
> To: Viktor T. Toth
> Cc: maxima at math.utexas.edu
> Subject: Re: [Maxima] question about ctensor
>
> On Dec 2, 2007 11:28 PM, Viktor T. Toth <vttoth at vttoth.com> wrote:
> > I don't think there is a bug, but please feel free to check and let me
> know
> > if you think otherwise!
>
> For example, for the Robertson-Walker metric, it's easy to calculate
> the tt component of the covariant Ricci tensor from the Riemann
> curvature tensor given by ctensor. The result is 3*diff(a,t,2)/a,
> where a is the scale factor. However the result given by ctensor
> directly is -3*diff(a,t,2)/a. Have I made some mistakes?
>
>