Welcome to Cadabra Q&A, where you can ask questions and receive answers from other members of the community.
+1 vote

Hello, I'm doing the calculation of the Tide tensor, and although the expression is correct at the end, the sign of the expression is incorrect. That makes me think that the problem is in some configuration of the simmetries of the Riemann tensor. The problem might be a result of a poor use of the system by me. I'll put the code for Reissner-Nordström spacetimes as an example

{r,t,\phi,\theta}::Coordinate; {\mu,\nu,\rho,\sigma,\lambda,\kappa,\chi,\gamma, \tau, \beta, a, b}::Indices(values={t,r,\phi,\theta}, position=fixed); \partial{#}::PartialDerivative; g_{\mu\nu}::Metric. g^{\mu\nu}::InverseMetric. R^{\rho}_{\sigma\mu\nu}::RiemannTensor.

`ss:= { g{t t} = 1-2*M/r + q2/r2,
{r r} = -1/(1-2*M/r + q2/r2), g{\theta\theta} = -r**2, g{\phi\phi}=-r2 \sin(\theta)2 }.

complete(ss, $g^{\mu\nu}$);`

`ch:= \Gamma^{\mu}{\nu\rho} = 1/2 g^{\mu\sigma} ( \partial{\rho}{g{\nu\sigma}} +\partial{\nu}{g{\rho\sigma}} -\partial{\sigma}{g_{\nu\rho}} ):

evaluate(ch, ss, rhsonly=True);`

rm:= R^{\rho}_{\sigma\mu\nu} = \partial_{\mu}{\Gamma^{\rho}_{\nu\sigma}} -\partial_{\nu}{\Gamma^{\rho}_{\mu\sigma}} +\Gamma^{\rho}_{\mu\lambda} \Gamma^{\lambda}_{\nu\sigma} -\Gamma^{\rho}_{\nu\lambda} \Gamma^{\lambda}_{\mu\sigma}; substitute(rm, ch) evaluate(rm, ss, rhsonly=True);

ex:=K^{a}_{b}=R^{\mu}_{\rho\nu\beta}ei^{a}_{\mu}i^{\rho}_{t}i^{\nu}_{b}i^{\beta}_{t}; substitute(ex, rm) viel:={i^{t}_{t}=Q/f,i^{r}_{t}=-j,i^{\theta}_{t}=0,i^{\phi}_{t}=0,i^{t}_{r}=-j/f,i^{r}_{r}=Q,i^{\theta}_{r}=0,i^{\phi}_{r}=0,i^{t}_{\theta}=0,i^{r}_{\theta}=0,i^{\theta}_{\theta}=1/r,i^{\phi}_{\theta}=0,i^{t}_{\phi}=0,i^{r}_{\phi}=0,i^{\theta}_{\phi}=0,i^{\phi}_{\phi}=1/(r*\sin(\theta)),ei^{t}_{t}=f*Q*/((Q**2-j**2)),ei^{r}_{t}=-f*j/((j**2-Q**2)),ei^{\theta}_{t}=0,ei^{\phi}_{t}=0,ei^{t}_{r}=-j/((j**2-Q**2)),ei^{r}_{r}=Q/((-f**2+Q**2)),ei^{\theta}_{r}=0,ei^{\phi}_{r}=0,ei^{t}_{\theta}=0,ei^{r}_{\theta}=0,ei^{\theta}_{\theta}=r,ei^{\phi}_{\theta}=0,ei^{t}_{\phi}=0,ei^{r}_{\phi}=0,ei^{\theta}_{\phi}=0,ei^{\phi}_{\phi}=r*\sin(\theta)}; substitute(viel, $j=(Q**2-f)**(1/2)$) substitute(viel, $f=1-2*M/r + q**2/r**2$) evaluate(ex, viel, rhsonly=True);

in General questions by (230 points)

Hi Gabriel. I'd suggest that you improve the quality of your post.

First, the code is not well rendered, so if I copy from the post and run it, if fails! I spent about 20 minutes modifying it just to go through the evaluations.

Second, explain your notation!

And finally, it good be great if you share the obtained and expected results. By the way, it might be as simple as a confusion with the signature of a term, I mean the usual (+,-,-,-) vs (-,+,+,+) confrontation.

I'm sorry, I have to mention that I'm using the cadabra cloud, so codes on it don't really work on the desktop version I was told.

To explain the code, I'm using the Reissner-Nordström metric with the (-,+,+,+) signature. To calculate the tides tensor i'm using a vierbein, the first vierbein is chosen to be the velocity of a particle in-going radially.

1 Answer

0 votes

Just to be clear, I got the correct result. The problem was solved when I changed the indices that where summed over by the vierbeins, i.e

K^{a}_{b}=R^{\mu}_{\rho\nu\beta}ei^{a}_{\mu}i^{\rho}_{t}i^{\nu}_{b}i^{\beta}_{t} -> K^{a}_{b}=R^{\mu}_{\rho\nu\beta}ei^{a}_{\mu}i^{\rho}_{t}i^{\beta}_{b}i^{\nu}_{t}

by (230 points)

Ok, that is excellent! However, that interchange is related to my comment on your question, i.e. some times people defines the Ricci scalar with a different sign.