Proving the weyl tensor is zero problem

  • Thread starter Thread starter Airsteve0
  • Start date Start date
  • Tags Tags
    Tensor Weyl Zero
Airsteve0
Messages
80
Reaction score
0

Homework Statement


Show that all Robertson - Walker models are conformally flat.


Homework Equations


Robertson Walker Metric: ds^{2}=a^{2}(t)\left(\frac{dr^{2}}{1-Kr^{2}}+r^{2}(d\theta^{2}+(sin\theta)^{2}d\phi^{2} )\right)-dt^{2}

Ricci Tensor: R_{\alpha\beta}=2Kg_{\alpha\beta}

Ricci Scalar: R=8K

Weyl Tensor: C_{\alpha\beta\gamma\delta}=R_{\alpha\beta\gammaδ}-\frac{1}{2}(g_{\alpha\gamma}R_{\beta\delta}-g_{\alpha\delta}R_{\beta\gamma}-g_{\beta\gamma}R_{\alpha\delta}+g_{\beta\delta}R_{\alpha\gamma}) + \frac{R}{6}(g_{\alpha\gamma}g_{\beta\delta}-g_{\alpha\delta}g_{\beta\gamma})

The Attempt at a Solution


In order for the models to be conformally flat the weyl tensor must vanish, therefore that is what I have tried to show. By subbing in the values for the Ricci tensor and the Ricci scalar (both of which were given in a lecture by my professor) I arrived at the following expression:

C_{\alpha\beta\gammaδ}=R_{\alpha\beta\gamma\delta}-\frac{2}{3}K(g_{\alpha\gamma}g_{\beta\delta}-g_{\alpha\delta}g_{\beta\gamma})

However, as you can see I am left with the Riemann tensor undefined and I cannot show the weyl tensor to be zero. Any help is greatly appreciated, thanks!
 
Physics news on Phys.org
I'm pretty sure your textbook contains the definition of the Riemann tensor, and of the other tensors. Look it up. (You will only need to know the metric to calculate it.) Then, the Ricci tensor is related to the Riemann one through
R_{ij} = R^k_{\, ikj},
and, finally, the Ricci scalar is simply the trace of the Ricci tensor.

So well, I'd recommend you start working from the definitions. It is probably more instructive to derive those forms of the Ricci tensor and scalar yourself, and see if you get the same forms for this specific metric.
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top