# Israel Wilson Perjes metric

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1. Nov 22, 2014

### PhyAmateur

Is there any book or reference perhaps on string theory or superstring theory or even advanced general relativity that treats the Israel Wilson Perjes metric using the tetrad formalism in details, i.e, 1-forms and so? (Not spinors methos) I have ran across many papers that just place the spin connection in a very complicated way where factors come out of the blue. I have been trying to understand the twister method and I kind of grasped it but now I want to see how does it work using vielbeins. Please any suggestion would be great!

2. Nov 27, 2014

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Nov 28, 2014

### Ben Niehoff

The IWP metric tensor is very simple, it shouldn't be hard for you to compute the spin connection yourself.

Are you unsure in general how to compute the connection 1-forms from the orthonormal basis 1-forms?

4. Nov 29, 2014

### PhyAmateur

I did, but calculations get really messy when you want to find the Ricci tensors, that is why I needed a guide to know if I am on the correct track.

5. Dec 1, 2014

### Ben Niehoff

Oh, I thought you just wanted the spin connections. Calculations always get messy when you want Ricci tensors. :D

Have you done other Ricci tensors by hand? It would definitely help to do a simpler example first. For example, the Gibbons-Hawking metrics are

$$ds^2 = \frac{1}{V} (d\psi + \vec A \cdot d \vec x)^2 + V (dx^2+dy^2+dz^2),$$
where $V$ and $\vec A$ are functions of $(x,y,z)$ only, and

$$\nabla^2 V = 0, \qquad \vec \nabla \times \vec A = \vec \nabla V.$$
Under these conditions, the Ricci tensor should vanish.

If you can do that, then it is a simple matter to generalize to (Euclidean signature) IWP:

$$ds^2 = \frac{1}{V_1V_2} (d\psi + \vec A \cdot d \vec x)^2 + V_1 V_2 (dx^2+dy^2+dz^2),$$
where

$$\nabla^2 V_1 = 0, \qquad \nabla^2 V_2 = 0,\qquad \vec \nabla \times \vec A = V_2 \, \vec \nabla V_1 - V_1 \, \vec \nabla V_2.$$
Under these conditions, the Ricci tensor should again vanish.

6. Dec 1, 2014

### Ben Niehoff

By the way, you can find the Ricci tensors of both of these metrics very easily if you first break them up in a few steps. First find how the Ricci tensor is changed by a conformal rescaling:

$$ds^2 = \Omega^2 d\hat{s}^2$$
where $\Omega$ is some function, and you should be able to write $R_{\mu\nu}$ in terms of $\hat{R}_{\mu\nu}$. This calculation can be done in about a page.

Next do another general calculation:

$$ds_n^2 = (d\psi + A)^2 + ds_{n-1}^2$$
where $A$ is a 1-form that is not a function of $\psi$. Again you should be able to write $R_{\mu\nu}^{(n)}$ in terms of $R_{\mu\nu}^{(n-1)}$. This calculation is even easier than the previous one.

Once you have these two formulas, you can combine them in various ways to obtain the Ricci tensors of the previously-mentioned metrics.

7. Dec 2, 2014

### PhyAmateur

Thank you a lot for the nice idea!