Quastions about Zee's book EGR in a nutshell

[sergiokapone]

Quastions about Zee's book "EGR in a nutshell""

Now I'm reading a book Einstein Gravity in a Nutshell by A. Zee, and I had some misunderstanding.
Zee writes (page 89):

To make sure that you follow this discussion, I suggest you try this fun exercise. Suppose
you were given a space described by the metricds ds^2=dr^2+r^2d\theta^2. This is of course a plane as flat as Kansas, but suppose you didn’t know that. Calculate the curvature by first transforming polar coordinates into locally flat coordinates at the point (r,θ)=(r_∗,0) by going through all the steps here. Then extract the combination of the B_{\mu\nu,\lambda\sigma}s giving the intrinsic curvature. By the end of this straightforward exercise, you will probably agree that there ought to be a better way to get at the curvature.

before that, he wrote,

So, look at our space around a point P. First, for writing convenience, shift our coordinates so that the point P is labeled by x=0. Expand the given metric around P out
to second order: g_{\mu\nu}(x)=g_{\mu\nu}(0)+A_{\mu\nu,\lambda}x^{\lambda}+B_{\mu\nu, \lambda \sigma} x^{\lambda}x^{\sigma}+...
(The commas in the subscripts carried by A and B are purely for notational clarity, to separate two sets of indices.)

Trying to solve the problem. I expanded in a series metrics as mentioned g_{\theta\theta}(r) = r^2_*+2r_*(r-r_*)+2(r-r_*)^2
I see that B_{\mu\nu, \lambda \sigma}=B_{\theta\theta,r r}=B_{22,11}=1.

How can I get intrinsic curvature?
Below, he still writes

Show that for D=2, the combination 2B_{12,12}−B_{11,22}−B_{22,11} measures intrinsic curvature.

I turns out that the curvature is not zero, it should be.

 

[Bill_K]

I expanded in a series metrics as mentioned g_{\theta\theta}(r) = r^2_*+2r_*(r-r_*)+2(r-r_*)^2

Should be g_{\theta\theta}(r) = r^2_*+2r_*(r-r_*)+(r-r_*)^2

 

[sergiokapone]

Should be g_{\theta\theta}(r) = r^2_*+2r_*(r-r_*)+(r-r_*)^2

Yes, right, B's=1 as I sad.

 

[PeterDonis]

I turns out that the curvature is not zero, it should be.

You've only expanded $g_{\theta \theta}$. What about the other components?

 

[sergiokapone]

You've only expanded $g_{\theta \theta}$. What about the other components?

But g_{rr}=1.

 

[PeterDonis]

But g_{rr}=1.

It is in the global metric, but is it in the locally flat coordinates centered on the point (r*, 0)?

 

[PeterDonis]

It is in the global metric, but is it in the locally flat coordinates centered on the point (r*, 0)?

To expand on this some more: "locally flat coordinates" means Cartesian coordinates, i.e., you are supposed to set up a local chart centered on the point (r*, 0) in which the metric at that point is $ds^2 = dx^2 + dy^2$. So the metric coefficients you should be expanding to 2nd order are really not $g_{rr}$ and $g_{\theta \theta}$ anyway; they are $g_{xx}$, $g_{yy}$, and $g_{xy} = g_{yx}$. You know that $g_{xx} (0) = g_{yy} (0) = 1$ and $g_{xy} (0) = g_{yx} (0) = 0$ since the (x, y) chart is locally flat at its origin. So in order to expand out the first and second-order terms for the locally flat metric coefficients, you just need to know the coordinate transformation from the global $(r, \theta)$ coordinates to the local $(x, y)$ coordinates, so you can transform the line element in general--i.e., not just at the single point (r*, 0).

 

[sergiokapone]

you just need to know the coordinate transformation from the global $(r, \theta)$ coordinates to the local $(x, y)$ coordinates, so you can transform the line element in general--i.e., not just at the single point (r*, 0).

Ok, assuming that I do not know that the above metric is flat, how do I know this transformation?

 

[Bill_K]

Ok, assuming that I do not know that the above metric is flat, how do I know this transformation?

Zee apparently tells you how he wants you to do this. Your quote from the book says,

Calculate the curvature by first transforming polar coordinates into locally flat coordinates at the point (r,θ)=(r∗,0) by going through all the steps here.

 

[sergiokapone]

Zee apparently tells you how he wants you to do this.

Yes, but how to do it? That is the question.I have this difficulty. I need to give step by step, then I'll do.

 

[Bill_K]

I think you must go on to the next page. Normally the curvature would depend on both the A's and B's. Since the expression you gave involves B's only, Zee must expect you to transform the coordinates further to eliminate the A's.

 

[sergiokapone]

Well, put the problem in another way. Let us forget for a while about Zee. How to find locally-flat coordinates (x,y) on the sphere at a north pole, say? What it means to find? It is clear that they are Cartesian, so I need to find transition formulas.

 

[Bill_K]

From what you've said, I still don't know just what Zee is planning to do. What he's calling "locally flat coordinates" are more usually called quasi-Cartesian coordinates or normal coordinates at a given point P. To find them, one can

a) Choose a set of orthonormal vectors X_i at P.
b) Solve the geodesic equations to construct a set of geodesics starting at P with initial tangent X_i.
c) Use these geodesics as a set of coordinate axes in the neighborhood of P, defining the normal coordinates as distance measured along each axis.

Just a roundabout way of finding coordinates in which the Christoffel symbols vanish at P, which is what Zee's A coefficients are.